• Save
Linked in slides - Quantum Fields
Upcoming SlideShare
Loading in...5
×
 

Linked in slides - Quantum Fields

on

  • 481 views

PART II - Continuation of PART I - Quantum Mechanics.

PART II - Continuation of PART I - Quantum Mechanics.

Statistics

Views

Total Views
481
Views on SlideShare
481
Embed Views
0

Actions

Likes
0
Downloads
0
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Linked in slides - Quantum Fields Linked in slides - Quantum Fields Presentation Transcript

    • PART II – QUANTUM FIELDS From First Principles August 2014 – R1.7 Maurice R. TREMBLAY
    • “A poet once said, ‘The whole universe is in a glass of wine.’ We will probably never know in what sense he meant that, for poets do not write to be understood. But it is true that if we look at a glass of wine closely enough we see the entire universe. There are the things of physics: the twisting liquid which evaporates depending on the wind and weather, the reflections in the glass, and our imagination adds the atoms. The glass is a distillation of the earth’s rocks, and in its composition we see the secrets of the universe’s age, and the evolution of stars. What strange array of chemicals are in the wine? How did they come to be? There are the ferments, the enzymes, the substrates, and the products. There in wine is found the great generalization: all life is fermentation. Nobody can discover the chemistry of wine without discovering, as did Louis Pasteur, the cause of much disease. How vivid is the claret, pressing its existence into the consciousness that watches it! If our small minds, for some convenience, divide this glass of wine, this universe, into parts – physics, biology, geology, astronomy, psychology, and so on – remember that nature does not know it! So let us put it all back together, not forgetting ultimately what it is for. Let it give us one more final pleasure: drink it and forget it all! ” Richard Feynman TYPICAL WINE FLAVONOIDS Epicatechin Three primary acids are found in wine grapes: TARTARIC ACID (C4H6O6) 2,3-dihydroxybutanedioic acid Tartaric acid is, from a winemaking perspective, the most important in wine due to the prominent role it plays in maintaining the chemical stability of the wine and its color and finally in influencing the taste of the finished wine. [Wikipedia] MALIC ACID (C4H6O5) hydroxybutanedioic acid Malic acid, along with tartaric acid, is one of the principal organic acids found in wine grapes. In the grape vine, malic acid is involved in several processes which are essential for the health and sustainability of the vine. [Wikipedia] CITRIC ACID (C6H8O7) 2-hydroxypropane-1,2,3-tricarboxylic acid The citric acid most commonly found in wine is commercially produced acid supplements derived from fermenting sucrose solutions. [Wikipedia] RESVERATROL DERIVATIVES trans cis R = H; resueratrol R = glucose; p Ice Id Malvidin-3-glucoeide Procyanidin B1 Quercetin Resveratrol (3,5,4'-trihydroxy-trans- stilbene) is a stilbenoid, a type of natural phenol, and a phytoalexin produced naturally by several plants. [Wikipedia] In red wine, up to 90% of the wine's phenolic content falls under the classification of flavonoids. These phenols, mainly derived from the stems, seeds and skins are often leached out of the grape during the maceration period of winemaking. These compounds contribute to the astringency, color and mouthfeel of the wine. [Wikipedia] Prolog
    • Contents “It is more important to have beauty in one’s equations than to have them fit experiment … because the discrepency may be due to minor features that are not properly taken into account and that will get cleared up with further development of the theory….” Paul Dirac, Scientific American, May 1963. PART II – QUANTUM FIELDS Review of Quantum Mechanics Galilean Invariance Lorentz Invariance The Relativity Principle Poincaré Transformations The Poincaré Algebra Lorentz Transformations Lorentz Invariant Scalar Klein-Gordon & Dirac One-Particle States Wigner’s Little Group Normalization Factor Mass Positive-Definite Boosts & Rotations Mass Zero The Klein-Gordon Equation The Dirac Equation Determining the structure of the proton: a Feynman diagram for deep inelastic scattering process. The diagram shows the flow of momentum when a high energy electron e (·) scatters (hense the exchange of a photon g with momentum q) from a quark (·) taken from the wavefunction of the proton p (·). This is a simple case called the Parton Model invented by Richard Feynman. We assume that the parton () has negligeable (i.e., a small fraction x of) transverse momentum with respect to the proton p, so the parton momentum x p is in the same direction as the proton momentum p, that is, the parton has momentum x p m , where 0 £ x £ 1. Finally, momentum conservation forces us to have the equality p¢=x p m + q given vertex couplings of the form ±i eg m where the gamma matrices satisfy g m g n + g n g m = 2 gmn. 2014 MRT
    • “Quantum field theory is the way it is because […] this is the only way to reconcile the principles of quantum mechanics […] with those of special relativity. […] The reason that quantum field theory describes physics at accessible energies is that any relativistic quantum theory will look at sufficiently low energy like a quantum field theory.” Steven Weinberg, Preface to The Quantum Theory of Fields, Vol. IP. ART 0 – PHYSICAL MATHEMATICS PART I – QUANTUM MECHANICS Introduction Angular Momentum Quantum Behavior Postulates Quantum Angular Momentum Spherical Harmonics Spin Angular Momentum Total Angular Momentum Momentum Coupling General Propagator Free Particle Propagator Wave Packets Non-Relativistic Particle 2014 MRT PART III – THE HYDROGEN ATOM What happens at 10-10 m The Hydrogen Atom Spin-Orbit Coupling Other Interactions Magnetic & Electric Fields Hyperfine Interactions Molecules Appendix The Harmonic Oscillator Electromagnetic Interactions Quantization of the Radiation Field Transition Probabilities Einstein’s Coefficients Planck’s Law A Note on Line Broadening The Photoelectric Effect Higher Order Electromagnetic Interactions References
    • Review of Quantum Mechanics We will provide only the briefest of summaries of PART I - Quantum Mechanics, in the generalized version of Dirac. This will also strengthen our mathematical conventions. POSTULATE #1: Physical states are represented by rays in Hilbert space. A Hilbert space is a kind of complex vector space; that is, if | F ñ and | Y ñ are vectors in the space (often called ‘state-vectors’ or ‘kets’) then so is h | F ñ + x | Y ñ, for arbitrary complex numbers h and x . It has a norm; for any pair of vectors there is a complex number á F | Y ñ, such that: F Y = Y F F Y Y F Y F Y x + x = x + x + = + 1 1 2 2 1 1 2 2 F F Y F Y F Y 2 * 1 2 * 1 1 2 2 1 * h h h h where the asterisk ( * ) indicates that the complex conjugate is taken. The norm á Y | Y ñ also satisfies a positivity condition: A ray is a set of normalized vectors: Y Y ³ 0 and vanishes if and only if the state-vector (or ket) is null: | Y ñ = 0. Y Y = 1 with | Y ñ and | Y ñ belonging to the same ray if | Y ñ = x | Y ñ, where x is (as above) an arbitrary complex number with the extra condition that it’s magnitude is unity: | x | = 1. 2014 MRT
    • POSTULATE #2: Observables are represented by Hermitian operators. These are mappings | Y ñ ® A | Y ñ of Hilbert space into itself, linear in the sense that: A x Y +h F =x A Y +h A F The above relation satisfies the reality condition: A† = A where the daggar ( † ) definies the Hermitian Operator and it indicates that the complex conjugate (i.e., replacing i ® -i ) and transpose (i.e., matrix elements are ‘transposed’ on either side of the diagonal, Ai j ® A j i ), A† = A*T, is taken. For any linear operator A the adjoint A† is defined by: F A†Y = AF Y = Y AF * The state represented by a ray R has a definite value a for the observable represented by an operator A if vectors | Y ñ belonging to this ray are eigenvalues of A with eigenvalue a : A Y =a Y ( for Y in R) An elementary theorem tells us that for A Hermitian, a is real, and eigenvalues with different a s are orthogonal. 2014 MRT
    • POSTULATE #3: If a system is in a state represented by a ray R, and an experiment is done to test whether it is in any one of the different states represented by mutually orthogonal rays R1, R2, …, then the probability of finding it in the state represented by Rn is: ( ) 2 P R ® Rn = Y Yn where | Y ñ and | Yn ñ are any vectors belonging to rays R and Rn , respectively. Another elementary theorem gives a total probability unity (i.e., they add up to 100%): n åP ( R ® Rn ) = 1 if the state-vectors | Yn ñ form a complete set. 2014 MRT
    • Let us now review symmetries: A symmetry transformation is a change in our point of view that does not change the result of possible experiments. If an inertial observer O (i.e., he ain’t movin’) sees a system in a state represented by a ray R or R1 or R2, …, then an equivalent observer O (i.e., he’s moving away!) who looks at the same system will observe it in a different state, represented by R or R1 or R2, …, respectively, but the two observers must find the same probabilities: PO ( R ® Rn ) = PO ( R ® Rn ) For any such transformation R ® R, of rays we may define an operator U on Hilbert space H , such that if | Y ñ is in ray R then U | Y ñ is in the ray R, with U either unitary and linear : U U = a + b =a +b Y F Y F or else antiunitary and antilinear : F Y F Y U U U U U F Y F Y * * * = U a + b =a U + b U Y F Y F This is a fundamental theorem from Wigner (1931): Any symmetry transformation can be represented on the Hilbert space of physical states by an operator that is either 2014 linear and unitary or antilinear and antiunitary. MRT
    • As mentionned in the condition á F | A† Y ñ = á A F | Y ñ = á Y | A† F ñ* above, the adjoint of a linear operator L is defined by: F L †Y º LF Y This condition cannot be satisfied for an antilinear operator, because in this case the right-hand side of the above equation would be linear in | F ñ, while the left-hand side is antilinear in | F ñ. Instead, the adjoint of an antilinear operator A is defined by: F A Y º AF Y = Y AF † * With this definition, the conditions for unitarity or antiunitarity both take the form: U†U = 1 Multiplying this unitary operator by it’s inverse U - 1 we get U and there is always a trivial symmetry transformation R ® R, represented by the identity operator U = 1. This operator is, of course, unitary and linear. Continuity then demands that any symmetry (e.g., a rotation or translation or Lorentz transformation) that can be made trivial by a continuous change of some parameters (i.e., like angles or distances or velocities) must be represented by a linear unitary operator U (rather than one that is antilinear and antiunitary). 2014 MRT U† =U-1 † (U U -1) = 1 Ä U -1 = U -1. Thus:
    • A symmetry transformation that is infinitesimally close to being trivial can be re-presented by a linear ‘unitary operator’ that is infinitesimally close to the identity: with e a real infinitesimal (e.g., an infinitesimal change in the coordinates d x m or an angle d j ). For this to be unitary and trivial, T must be Hermitian and linear, so it is a can-didate for an observable. Indeed, most (and perhaps all) of the observables of physics (e.g., angular momentum or momentum) arise in this way from symmetry transformations. The set of symmetry transformations has certain properties that define it as a group. If T1 is a transformation that takes rays Rn into Rn, and T2 is another transformation that takes Rn into Rn, then the result of performing both transformations is another symmetry transformation, which we write T2 T1 (T1 then T2) that takes Rn into Rn. Also, a symmetry transformation T which takes rays Rn into Rn has an inverse, written T -1, which takes Rn into Rn, and there is an identity transformation T = 1, which leaves rays unchanged. The unitary operators U(T ) act on vectors in the Hilbert space, rather than on rays. If T1 takes Rn into Rn, then acting on a vector | Yn ñ in the ray Rn, U(T1) must yield a vector U(T1)| Yn ñ in the ray Rn, and if T2 takes this ray into Rn, then acting on U(T1)| Yn ñ it must yield a vector U(T2) U(T1)| Yn ñ (again U(T1) then U(T2) but this time on | Y ñ) in the ray Rn . But U(T2 T1)| Yn ñ is also in this ray, so these vectors can differ only by a phase f n (T2,T1) : U T U T Y = fn 2 1 U T T Y Þ U T U T = i fn ( T2 , T1 ) U T T For f = 0, U(T ) furnishes a representation of the group of symmetry transformations. 2014 MRT ( ) ( ) e ( 2 1 ) ( 2 ) ( 1 ) e ( 2 1) ( , ) 2 1 n i T T n U = 1+ ie T +O(e 2 )
    • As an example of symmetry, consider a transformation (parametrized by the variable q, e.g., an angle j , a translation a or a Lorentz boost z ) on the space-time coordinates x m : where Lm ö ÷ ÷ ø æ = = º + L = ® å= m m [ ( )]m ( ; ) tanh n is a constant matrix (a function of the velocity v of a ‘moving’ frame). Under an infinitesimal transformation of the variable q , the coordinate differential d x m is given by: ç ç è c x x x a f x v v m m z b n n n 3 0 q dxm d q ¶ q q =0 º ¶ all f m (x;q ) so that the state vector |y ñ will transform according to (i.e., by using Taylor’s expansion): y ® y = y + » y + ¶ y y q q ( ) ( ) ( ) ( ) ( ) ( ; ) ( ) y q y e e y é y q q ù all in which the real infinitesimal is e = dq and the generator for the parameter q is given by: 3 å= = ¶ ¶ ( ; ) ¶ = - ¶ 0 0 m m q m q q x T i f x all q y q m m m m q m m m m m q m m m m m m m m m m ( ) ( ; ) ( ) ( ) ( ) [ ( 2 )] 0 0 x id T x i T O x x id i f x x x x x d f x x x x dx x dx T = + = + + ú ú û ê ê ë ¶ ¶ ¶ = + - ¶ ¶ ¶ ¶ º + ¶ ¶ å å å = = 1 1 1  (Generator) all 2014 MRT
    • A finite set of real continuous parameters q a describe a group of transformations T(q ) with each element of the group connected to the identity by a path (i.e., U U -1 = 1) within the group. The group multiplication law U(T2) U(T1) = U(T2T1) thus takes on the form (i.e., a connected Lie group): T(q )T(q ) = T( f (q ,q )) with f a(q ,q ) a function of the q s and q s. Taking q a = 0 as the coordinate of the identity, we must have: f a (q ,0) = f a (0,q ) =q a As mentionned above, the transformation of such continuous groups must be represen-ted on the physical Hilbert space by unitary operators U[T(q )]. For a Lie group these op-erators can be represented by a power series (e.g., in the neighborhood of the identity): [ ( )] 1 1 U T q i q aT q q T According to f a(q ,0) = f a(0 ,q ) =q a above, the expansion of f a(q ,q ) to second-order must take the form: 2014 MRT = + å + åå + c b bc b c a a 2 where Ta, Tbc = Tcb, &c. are Hermitian operators independent of the q s. Suppose that the U[T(q )] form an ordinary (i.e., f = 0) representation of this group of transformations, i.e.: U[T(q )]U[T(q )] =U[T( f (q ,q ))] ( , ) = + Ååå +... f a q q q a q a f q q c b a b c bc with real coefficients f a bc. The addition of the second-order term was highlighted by Å.
    • Applying the multiplication rule U[T(q )] U[T(q )] =U[T( f (q ,q )] and using the series U[T(q )] =1 + iq a Ta + ½q bq c Tbc + … above with q ® f a (q ,q ) = q a + q a + f a bc q b q c , we get: 2 1 [ 1 i q at 1q b q c t ] [ 1 i q a T 1 q b q c T ] 1 i ( q a q a f a b bc q q c )T ( q b q b )( q c q c )T 2 + a + bc + ´ + a + bc + = + + + + a + + + bc + 2 (The S were also omitted to get space.) The terms of order 1, q , q , and q 2 automatically match on both sides of this equation – from the q q terms we get a non-trivial condition: = - - å a a a Tbc TbTc i f bcT Since we are following Weinberg’s development, he points out that: This shows that if we are given the structure of the group, i.e., the function f a(q ,q ), and hence its quadratic coefficient f a bc , we can calculate the second-order terms (i.e., Tbc) in U[T(q )] from the generators Ta appearing in the first-order terms. (A pretty amazing fact I would say!) However, as he points out: There is a consistency condition: the operator Tbc, must be symmetric in b and c (because it is the second derivative of U[T(q )] with respect to q b and q c) so the equation Tbc = -Tb Tc - i Sa f a bc Ta above requires that the commutation relations be: where C a º - = å [Tb ,Tc ] TbTc TcTb i C bcT a a a bc are a set of real constants known as structure constants: 2014 MRT a cb C a bc= - f a bc + f Such a set of commutation relations is known as a Lie algebra.
    • As a special case of importance, suppose that the function f a(q ,q ) is simply additive: f a (q ,q ) =q a +q a This is the case for instance for ‘translations’ in spacetime, or for ‘rotations’ about any one fixed axis (though not both together). Then the coefficients f a bc in the function f a(q ,q ) =q a + q a + Sbc f a bc q b q c vanish, and so do the structure constants C a bc = - f a bc + f a cb. The generators then all commute: Such a group is called Abelian. In this case, it is easy to calculate U[T(q )] for all q a. Again, from the group multiplication rule U[T(q )]U[T(q )] =U[T( f (q ,q ))] and the function f a(q ,q ) = q a + q a above, and taking e = q /N , we have for any integer N : N ö N çè é U T U T þ ý ü î í ì ù úû êë ÷ø [ (q )] = æ q and hence: [Tb ,Tc ] = 0 2014 MRT Letting N ® ¥, and keeping only the first-order term in U[T (q /N )], we have then: N [ (q )] lim 1 q a a a N T N é U T i ù ú úû ê êë ö ÷ ÷ ø æ ç ç è = + å ®¥ i T a å i T a U T a U q q [ (q )] = e Þ (q ) = e
    • Lorentz Invariance Lorentz invariance is needed to replace the principle of Galilean invariance and the discovery of the non-conservation of parity in weak interactions (1956) has reemphasized that an invariance principle and its consequences must be experimentally verified. One key invariance principle in quantum mechanics and quantum field theory is that: Different equivalent observers make the same predictions as to the outcome of an experiment carried out on a system. This statement means that observer O will attribute the vector |yO ñ to the state of the system, whereas observer O will describe the state of this same system by a vector |yO ñ. y = y O O O O O O |fOñ |yOñ The vectors | fO ñ and | yO ñ seen by observer O and vectors | fO ñ and | yO ñ seen by observer O. A unitary transformation U(L) (a function of the Lorentz transformation) relates both systems. We shall call the vector |yO ñ the translation of the vector |yOñ. Stated mathematically, the postulate above asserts that if |yOñ and |fO ñ are two states and |yO ñ and |fO ñ their translations, then: 2 2 O O O O f y = f y If all rays in Hilbert space are distinguishable, it the follows from the above equation – as a mathematical theorem (Wigner, 1931) – that the correspondence |yO ñ ® |yO ñ is effected by a unitary or anti-unitary operator, U( O , O ), the operator U is completely determined up to a factor of modulus 1 by the transformation L which carries O in O. We write: y O =U(L)y O where U depends on the coordinate systems between which it affects the correspondence and U( L = 1 ) = 1 if L is the identity transformation 1, i.e. if O and O are the same coordinate system. |fOñ |yOñ O O U(L) U L U L f y f y ( ) ( ) = 2014 MRT áfO |U(L)|yOñ
    • For special relativity, as an example, we consider the inhomogeneous Lorentz transformations. A relativity invariance requires the vector space describing the possible states of a quantum mechanical system to be invariant under all relativity transformations, i.e., it must contain together with every | y ñ all transformations U(L) | y ñ where L is any special relativity transformation. The transformed states can always be obtained from the original state by an actual physical operation on the system. Consider for example a Lorentz transformation along the z-axis with velocity v. The transformed state, which arises from the momentum eigenstate | YO ( p,s )ñ, is given by U(v)| YO ( p,s )ñ. This is the state of the system as seen by observer O. It is, however, also a possible state of the system as seen by O and which can be realized by giving the system a velocity -v along the z-axis. | U(v) YO ñ | YO ñ v O O z, z p The state vector | YO ñ seen by observer O and the vector | YO ñ seen by observer O moving away from O at velocity v. A unitary trans-formation U(v) brings state | YO ñ into state | YO ñ. Here are two typical problems*: 1.Suppose that observer O sees a W-boson (spin one and mass m ¹ 0) with momentum p in the y-direction and spin z-component s . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the W state? 2.Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? sees sees 2014 MRT * S. Weinberg, Quantum Theory of Fields, Vol. I – Foundations, 1995 – P. 104-105. y x p s m Solving these two problems is the goal of the following slides and they involve modern concepts starting with particle definitions as unitary representations. x
    • The Relativity Principle Einstein’s principle of relativity states the equivalence of certain ‘inertial’ frames of reference. It is distinguished from the Galilean principle of relativity, obeyed by Newtonian mechanics, by the transformation connecting coordinate systems in different inertial frames. If the contravariant vector x m = (ct , r) are the coordinates in one inertial frame (x 1, x 2, x 3) i.e., r = x i ( i = 1,2,3 ), as Cartesian space coordinates, and x 0 = c t a time coordinate, the speed of light being c ) then in any other inertial frame, the coordinates x 3 3 åå åå = = = = d x d x º d x d x 0 0 Frame Frame ¶ = ¶ m m x x hrs h Here his the diagonal 4´4 matrix, with elements (i.e., the Minkowski metric): h= +1, mn 00 h= h= h= -1 and hº 0 for m ¹ n. These transformations have the special property 11 22 33 mn that the speed of light c is the same in all inertial frames (e.g., a light wave traveling at speed c satisfies | d r /d t | = c or in other words Shd x m dxn = c 2d t 2 - d r 2 = 0, from which it mn mn 2014 MRT follows that Smn hmn d x m dxn = 0, and hence | d r /d t | = c .) m must satisfy: O O 3 0 3 0 m n m n mn m n m n hmn h or equivalently stated as the Principle of covariance: åå ¶ ¶ m n s mn r x x The covariant vector can be given as xm = Sn hmn xn = (ct , -r). The norm of the vector Sm xm x m = (x 0) - Si (x i)2 = c 2 t 2 - | r |2 is a Lorentz invariant term.
    • Poincaré Transformations Any coordinate transformation x m ® x m that satisfies Smn hmn (¶ x m /¶ x r )(¶ x n /¶ x s ) = hrs is linear and allows us to define the Poincaré Transformations: m m a x x x x a x x + L + L + L + L = + L =å= with a m arbitrary constants (e.g., ‘leaps’), and Lm n a constant matrix satisfying the condition: å( mn L å[L )L ] = ( L [L )L ] = ( )L = L = L = [ ]L  and now, when multiplying with the inverse of the matrix hmn Lm r , we get from this: h r h h h h h h h h n m 2014 MRT m m m m m m n n n 3 3 2 2 1 1 0 0 3 0 ååhmn L L =h rs m n n s m r The matrix hhas an inverse, written h mn , which happens to have the same mn components: it is also diagonal matrix, with elements: h 00 = +1,h 11 = h 22 = h 33 = -1 and h mn º 0 for m ¹n. To save on using summation signs ( S ), we introduce the Summation Convention: We sum over any space-time index like m and n (or i, j or k in three dimensions) which appears twice in the same term, if they appear only once ‘up’ and also only once ‘down’. As an example, and while also enforcing this tricky summation convention, we now multiply hLm Ln = habove with h s t Lk and when inserting parentheses for show: m n r s rs t r k k k r r k r k m n r k mn s s st s t st s t mn m n mn st st s t h s th n k = Ln Lk s t ( m = 0,1,2,3)
    • These transformations do form a group. If we first perform a Poincaré transformation x m ® Lm n x n + a m, and then a second Poincaré transformation x m ® x m, with: x x a xn ar ( sum over ρ and ν) 2014 MRT m = L m r r + m = L m r (L r n + ) i.e. then the effect is the same as the Poincaré transformation x m ® x m, with: xm = Lm L x + L a + a ( r ) n m r m r n ( r ) Taking the determinant of hm n Lm r Ln s = hrs gives: so Lm n has an inverse, ( L-1)n s , which we see from hm n Lm r Ln s = hrs and takes the form: L- r = L =h h L rs m  nm s r n n Note ( 1) (Det L)2 = 1 The transformation T = T(L,a) induced on physical states therefore satisfy the group composition rule: T(L,a) T(L,a) = T(LL,La + a) and the inverse of this T (L, a) transformation is also obtained from T (L, a) T (L, a) = T (L L, La +a) above to be T(L-1,-L-1a) such that: T(L,a) T(L-1,-L-1a) = 1 The whole group of transformations T(L,a) is properly known as the inhomogeneous Lorentz group, or Poincaré group. It has a number of important subgroups – notably T(L,0) which we will look at in greater detail in a little while.
    • So, in accordance with the discussion in the previous slides, the transformations T(L,a) induced a unitary linear transformation on a state vector in the Hilbert space: y (xm ) ® y (x) =U(L,a)y (x) and this unitary operator U satisfies the same group composition rule as T(L,a): U(L,a)U(L,a) =U(LL,La + a) For example, we will soon discuss the wave function in its momentum representation: Y( pm ; j,mj ) ® Y( p,mj ) =U(L, p) Y( p,mj ) for which the same composition rule applies: 2014 MRT U(L, p) ºU[LL( p)]=U[L(Lp)] U[L-1(Lp)LL( p)]
    • Now it is time to study three-dimensional rotations and add relativity to the overall description. To this effect we will exploit pretty much all the group symmetry properties! Let us recall a few facts about homogeneous Lorentz transformations: 3-axis. This can be represented in 2014 MRT 0 0 3 0 3 x = x - x = x - x g b z z ( ) cosh sinh 1 1 x = x 2 2 = 3 3 0 3 0 x x x x x g b z z ( ) cosh sinh x x = - = - that is, assuming propagation in the direction of the x or z z cosh 0 0 sinh 0 1 0 0 0 0 1 0 xm = [L(xm - xn ,z )]m x v ù é ù é 0 1 2 x x x z b g b z z ù 0 1 2 sinh = æ = tanh sinh cosh 1 é x x x z g = = ÷ø and v = | v | is the relative velocity of the two frames. b z b ö çè - = = cosh 1 2 and with as well as c ú ú ú ú ú û ê ê ê ê ê ë ú ú ú ú û ê ê ê ê ë - - = ú ú ú ú ú û ê ê ê ê ê ë 3 3 sinh 0 0 cosh x x z z matrix form as: where n n
    • The explicit matrix representation of a restricted homogeneous Lorentz transformation in the x 1-direction (rotation in the x 0 - x 1 plane) is given by: ù z z cosh sinh 0 0 é - Λ( 1 - 0 , ) z z sinh cosh 0 0 0 0 1 0 and the infinitesimal generator M 10 for this rotation is defined as: ù é 0 1 0 0 1 0 0 0 0 0 0 0 z K M d x x Similarly, the infinitesimal generators M 20 and M 30 for rotations in the x 2 - x 0 and x 3 - x 0 planes respectively, are given by: ú ú ú ú û ê ê ê ê ë - = 0 0 0 1 x x z ú ú ú ú û ê ê ê ê ë º = = = 0 0 0 0 Λ( - , ) 0 1 0 10 1 z z d ù ú ú ú ú û é ê ê ê ê ë z K M d x x º = = ù ú ú ú ú K M d x x , û é ê ê ê ê ë º = = 0 0 1 0 0 0 0 0 1 0 0 0 = = 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 Λ( - , ) 0 0 0 0 Λ( - , ) 0 3 0 30 3 0 2 0 20 2 z z z z z d d 2014 MRT
    • The infinitesimal generators in the x i - x j plane, i.e. spatial rotations, are: ù ú ú ú ú û 0 0 0 0 0 0 1 0 0 1 0 0 2014 MRT é ê ê ê ê ë - º = ù ú ú ú ú û é ê ê ê ê ë 0 0 0 0 0 0 0 1 0 0 0 0 - º = ù ú ú ú ú 0 0 0 0 0 0 0 0 0 0 0 1 û é ê ê ê ê ë - º = 0 0 0 0 0 1 0 0 0 0 1 0 12 3 31 2 23 J1 M , J M and J M where we define M mn = - M nm. In matrix form these terms come together as: ù ú ú ú ú û é ê ê ê ê ë K K K 1 2 3 K J J 1 3 2 K - J J 2 3 1 - - = ù ú ú ú ú ú û é ê ê ê ê ê ë 01 02 03 M M M 01 12 13 M M M 02 12 23 M - M M - - = 0 0 0 0 0 0 0 0 3 2 1 03 13 23 K J J M M M Mmn
    • The general result for M mn can now be written alternatively as: 2014 MRT An arbitrary infinitesimal Lorentz transformation, by expanding according to L(w) = exp(-½ i M mn wmn / ) in a power series, can be written as: where we can use hm n Lm r Ln s = hrs to show that w mn (or w m n ) is antisymmetric w mn = - wnm : = + å mn mn mn M i ω 2 Λ(ω) 1  h h d d = + + = + + = + + = + + which implies w r s + ws r = 0 QED. L = ω·S - ζ ·K and A = eω·S-ζ·K where w and z are constant 3-vectors. ( ω )( ω ) n h d d d d ( ω ω ) n m m m m h h h ω ω rs rs s r n m r n s ms rs rn n s m r s r s r mn n s s r r rs mn h ω ω
    • A finite rotation in the m -n plane (in the sense m ton ), is again obtained by exponentiation: 2014 MRT m n mnz Λ(x - x ,z ) = eM One verifies that the infinitesimal generators, M mn , satisfy the following commutation rules: [Mmn ,Mrs ] =h mrMns +hns Mmr -h ms Mnr -hnr Mms
    • For the rotation acting on the space-time coordinates, note that the time coordinate is unaffected (hence only latin indices): x f ( x ; ) [ R ( n ˆ, )] k xi xk (1 cos )[n ˆ ´ ( n ˆ ´ r )] k sin [ n ˆ ´ r ] k ö ÷ ÷ø m º m j = j = + - j + j æ = ¶ ç çè ¶ ˆ[(cosh 1) ˆ sinh ] r r n r n = + - · - z z cosh ˆ sinh 0 = - ¶ ¶ n r ˆ ˆ ( ; ) = - ¶ ¶ ¶ ¶ ¶ · = - ¶ = j k i i j k i j k i j k k k k x i x x i x x i x i f x e e z z j m ( n ˆ r ) ( n ˆ ) ( n ˆ ) n J ´ hence: = e ¶ j k i i j k x J i x ¶ î í ì = - · = z z ct ct ct i in which we made use of the relation x j = hi j x j = -x j, h being the Minkowski metric. It can be shown that the generators for rotation are equivalent to the generators for the SO(3) Special Orthogonal group (which are Hermitian). Thus, the representation for a finite rotation acting on the wavefunction is unitary and it is given by U(R) = exp(-ij nˆ · J / ). 2014 MRT where v = c tanhz nˆ . So, we get:
    • The general Lorentz transformations for a simple spatial rotation, LR, is given by: 1 0 0 0 0 m R n 0 [ R ( ˆ, j )] i ˆ ù ú ú ú ú û é ê ê ê ê 0 ë L = ( ) j n in which the three-dimensional spatial rotations, R(n,j ), are elements in the simple orthogonal group SO(3). However, this is not relavant for evaluating the Wigner coefficients since it is trivial to show that both the Wigner transformation and Lorentz rotation, LR, belong to the same little group H(L,k). The convenional way of characterizing the Lorentz transformation, L, is described by the generators for boosts with Ki = M0i = -Mi0 and rotations J i = ½ e i j k Mj k, that is: mn mn ω 1 2 (ω) e i M  - L = Note that the parameters associated with the Lorentz transformation are given by the anti-symmetric tensor w mn. (In addition, the matrix representations of the Lorentz generators in the four-vector coordinates are given by: ù úû d 0 0 é é = iab êë - ù J = úû êë i ai bi i i i i K d e 0 0 0 and where the indices a, b represent the rows and columns, respectively.) 2014 MRT
    • ù é 0 3 x x z + z Now, consider a Lorentz boost along the x3-axis: ù é ù é = ù é z z cosh sinh cosh 0 0 sinh 0 1 0 0 0 0 1 0 1 2 0 1 2 0 1 2 x x x x x x x x in which we recall that coshz = g and sinhz = bg where g = 1/Ö(1 - b 2) and b = v/c. In four-vector notation we have: ú ú ú ú ú û ê ê ê ê ê ë + = ú ú ú ú ú û ê ê ê ê ê ë ú ú ú ú û ê ê ê ê ë ú ú ú ú ú û ê ê ê ê ê ë z z z z sinh cosh sinh 0 0 cosh 0 3 3 3 x x x x xm = L m x º f x [ (z )] n m n ( ;z ) and the associated generator is given as: since L0 m = - ¶L n m z z z = = + ¶L = ¶ úû é = - ¶L = - é ¶ ¶ z + ¶ 0|z = 0 = coshz |z = 0 = 1, &c. m + ¶L ù úû é êë ¶ n z - ¶ ¶ ¶ ù êë ¶ ¶ ù ú ú û ê ê ë ¶ ¶ + ¶L ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ = - ¶ = = = = 3 3 0 0 3 0 0 3 3 3 0 3 3 3 0 0 0 3 0 3 0 3 0 0 0 0 0 0 0 0 3 ( ; ) x x x i x x x x i x x x x x x x x i x x i x x K i f x z z z z m z z z z 2014 MRT
    • Similarly: = é ¶ - ¶ x i K ¶ - ¶ = úû i i 0 0 i i (xi 0 x0 i ) x x x ù êë ¶ ¶ Consider that the state-vector of the system is given by |y ñ, hence the action of K i on the state is: † = é ¶ - ¶ x i x K i i i i y y y y ¹ úû 0 0 ( ) ( ) (x) [K (x) ] x x x x ù êë ¶ ¶ This implies that the generator for the Lorentz boost, K i, is not Hermitian and hence the exponentiation of the generator (i.e., exp(-i z i K i / )) will not be unitary. The representation of the Lorentz boost acting on the wavefunction is not unitary and hence is not trace-preserving. We can summarize the effects of the rotations and the Lorentz boost into one second-rank covariant tensor: Mmn = i (xm ¶n - xn ¶m ) in which J i = ½ e ijk Mj k and K i = Mi0. 2014 MRT
    • These generators (i.e., P m and M mn ) obey the following commutation relations, which characterize the Lie algebra of the Poincaré group (and adding  and c for reference): [Ji , J j ] = ie i j k Jk ; [Ji ,K j ] = ie i j kKk and [Ki ,K j ] = -ie i j k Jk The first of these is the usual set of commutators for angular momentum, the second says that the boost K transforms as a three-vector under rotations, and the third implies that a series of boosts can be equivalent to a rotation. Next we have: Ji Pj i e i j k Pk Ki P i cPi Ki Pj i d i j  The rotation J i and boost K j generators can be written in covariant notation M mn and the commutation relations are then re-written as: M M = - i h M - h M + h M - h M [ , ]  ( ) rs mn mr ns nr ms ns mr ms nr M P = - i P - P [ , ] ( ) mn l nl m ml n = [ , ] 0 m l h h P P  2014 MRT [ , ] [ , 0 ] [ , ] P0 c =  ; =  and = [Ji , P0 ] = 0 ; [Pi , P0 ] = 0 and [Pi , Pj ] = 0 where P 0c = H, the Hamiltonian, and finally all components of P m should commute with each other: Together, these equations above form the Lie algebra of the Poincaré group.
    • For a simple translation a m of the coordinates x m : we get: xm º f m (x;am ) = xm + am ( ; ) 1 î í ì when = = m n = ¹ = f x a ¶ ¶ m d n = m n n m 0 when m a 0 a and using: m ù é i i f x a 1 1 = - m ¶ = - ¶   i n d = - ¶ ö çè n m n n ¶ ¶ x i x P i i.e., the momentum operator. For a finite translation a m , the state-vector |y ñ of a relativistic system (expressed by the Dirac wavefunction y ) will transform as: i P m a - y m y  m y (x ) ® (x) = e (x) Note that the generators P 0 = -i  ¶0 and Pk = -i ¶k are known as the Hamiltonian and momentum operators and they are Hermitian since their associated eigenvalues are defined to be real. This implies that the representation for a finite translation acting on the wavefunction (e.g., as you will see soon it is given by U(1,a) = exp(-i Pm am / )) is actually unitary. 2014 MRT y y e y e y n m n ÷ø = æ + ú ú û ê ê ë ¶ ¶ ¶ = + - ¶ = i P a x a    0 ( ; ) all we obtain:
    • The contravariant generators for the space-time translations a m are defined by Pm, in which the time translation, P 0 and the spatial translation, P i are the Hamiltonian and momentum operators, respectively, of a free particle. The finite translations acting on the space-time coordinates are well-defined (with no Lorentz transformation): m P a i m m - T I a  ( 4´4 , ) = e with the corresponding covariant generators P m =hmn Pn . The Lorentz transformation L can be described by an antisymmetric second-rank tensor w mn which is defined by the parameters in L: ωi j i ω0i ωi0 jk =e i j k and z = = - Here z i is the Lorentz boost along the i-th axis, and q k are the parameters involved in the rotation along the axial vector. The generators for these Lorentz transformations are given by M mn have been explicitely derived previously. Here the generators for the spatial rotation J i and Lorentz boost K i are given by: 1 J i = e i j kM and K = M j k i i 2 0 The Lorentz transformation can be described by its generators (without translation): mn ω mn 1 2 (ω) e i M  - L = 2014 MRT
    • In general, the elements of the Poincaré group are given as: i Mmn i P a m m 1 - - mn ω T (L, a ) = e 2  e  and it produces the following space-time coordinate transformation: Similarly, the momentum of a free particle also transforms according to: in which P xm ® xm = Lm x + a n m n Pm ® P¢m = Lm P + P n m n m is the momentum of the particle in the new coordinate frame. By definition, a momentum contravariant four-vector is given by Pm P m = (P 0 /c)2 - (P i )2 = (E/c)2 - | p |2 = mo 2c 2 is another Lorentz invariant term as well. In this context, mo is defined as the rest mass of the particle. 2014 MRT
    • For the inhomogeneous Lorentz group, the identity is the transformation Lm n = d m n , a m = 0, so we want to study those transformations with: n Lm =d + ω and a =e m m m n m n both w m n and e m being taken as infinitesimal. The Lorentz condition hm n Lm r Ln s = hrs reads here: where ws r º hm s w m r and w n m h h = L L n m r m s rs mn = + + n h d d ( r ω r )( s ω ) mn = + + +O 2 ω ω (ω ) h sr sr rs s ws r. Keeping only the terms of first order in w in the m r º h m s Lorents condition hm n Lm r Ln s = hrs , we see that this condition now reduces to the antisymmetry of ws r : ωrs = -ωsr An antisymmetric second-rank tensor in four dimensions has (4 ´ 3) /2 = 6 independent components, so including the four components of em, an inhomogeneous Lorentz transformation is described by 6 + 4 = 10 parameters. 2014 MRT
    • Since U(1,0) carries any ray into itself, it must be proportional to the unit operator, and by a choice of phase may be made equal to it. For an infinitesimal Lorentz transformation Lm n = d m n + w m n and a m = e m, U(1 + w,e ) must then equal 1 plus terms linear in wrs and e r . We write this as: e rs e P U i M i (1+ ω, ) = 1+ 1 - r + ω ... 2 r rs   Here P r and M rs are e- and w-independent operators, respectively, and the dots denote terms of higher order in e and/or w. In order for U(1 + w,e ) to be unitary, the operators P r and M rs must be Hermitian: (Pr )† = Pr and (Mrs )† = Mrs Since wrs is antisymmetric, we can take its coefficients M rs to be antisymmetric also: 2014 MRT As we shall see, P 1, P 2, and P 3 are the components of the momentum operator, M 23, M 31, and M 12 are the components of the angular momentum vector, and P 0 is the energy operator, or Hamiltonian. Mrs = -Msr
    • Let us consider the Lorentz transformation properties of P r and M rs. We consider the product: U(L, a)U(1+ w,e )U -1(L, a) According to the composition rule T (L, a) T (L, a) = T (L L, La +a) with T = U(L, a) the product U(L-1,-L-1a) U(L,a) equals U(1,0), so U(L-1,-L-1a) = U -1(L,a), i.e., U(L-1,-L-1a) is the inverse of U(L,a). It follows from U (L, a) U (L, a) = U (L L, La +a) that, in sufficient detail to show these important group operations to that they be understood, we have: - - - 1 1 1 U a U U a U a U U a U a U a L + L = L + L -L = L + L + -L + e e e ( , ) (1 ω, ) ( , ) ( , ) (1 ω, ) ( , ) ( , ) [(1 ω) , (1 ω)( ) ]  - - 1 1 a a a - - 1 1 U a U a = L + L - + L + ( , ) [(1 ω) , (1 ω) ] - - a a a a L=L = L= + L =- + L + ; ; (1 ω) & (1 ω) - - 1 1 U a a = L + L -L + L + L + [ (1 ω) , (1 ω) ] - - - 1 1 1 U a a a = L + L - LL + L + L L [ (1 ω) , ( ) ω ] - - 1 1 [ (1 ω) , ω ] 1 1 1 ω; ; & 1 1 U a - - L= + = L=L =-L º L + L L e - L L e e e e e  2014 MRT where Lm n and a m are here the parameters of a new transformation, unrelated to w and e . In the end the transformation rule is given by: U(L, a)U(1+ w,e )U -1(L,a) ºU[1+ LwL-1,Le - LwL-1a] since L1L-1= 1.
    • ù where we have exploited the antisymmetry of wrs , i.e., wrs = -wsr , and we have used the inverse ( L-1)n 2014 MRT Using U(1 + w,e ) = 1 + ½ (i /) w rs M rs - (i /) e r P r to first order in w and e we have then: U a i ω M i P U - 1 ( , a ) 1 1 i - 1 M i - 1 a P L L - L - L L + = L úû     i M 1 1 = + L L ω ( ) i P a P - L - L L i ω M 1 1 i a P a P a P a P r m ( ) 1 2 + L L é - + + r m s e e s s mn m n n m n  i P r m r - L r m rs m s n m m n m n n m n r rs m s mn n r rs m n m n rs r m m r r m mn n rs r m m m mn mn r r ( , ) 1 1 L é + rs - rs e e e i M a P a P i P 1 1 = + L L - - - L úû êë = + L L ù êë - -       ω ( ) 2 ( ) 2 ω 1 2 [ ω ( ) ] 2 ( ω ) ( ω ) 2 2 1 1 r = hn m h r s Lm r . Equating coefficients of wrs and er on both sides of this equation we find: - 1 r U ( L , a ) P U ( L , a ) = L P L - 1 L = L L - - r m m s mn m n n m n r m rs U a M U a M a P a P ( , ) ( , ) ( ) s = Ln
    • Next, let’s apply these rules rules to a transformation that is itself infinitesimal, i.e., Lm n = d m n + w m n and a m = e m, with infinitesimals w m n and e m unrelated to the previous w and e . By using U(1 + w,e ) = 1 + ½(i /) w rs M rs - (i /) e r P mn M e P P P i ω , ω 2 1 = úû é -  i é ω mn M - e P , M = ω M + ω M - e P + e P úû 2 Equating coefficients of wrs and er on both sides of these equations, we would find these commutation rules: 2014 MRT m r = [ , ] 0 i P M P P m rs m r s ms r = - h h [ , ] i M M M M M M mn rs n r ms m r n s s m rn sn r m h h h h P P = + - + [ , ]   This is the Lie algebra of the Poincaré group. r m m m r m mn ù êë s rn r s s r n r ms m m mn m mn ù êë 1  and r and keeping only terms of first order in w m n and e m, our equations for P r and M rs become:
    • In quantum mechanics a special role is played by those operators that are conserved, i.e., that commute with the energy operator H = P 0. We just saw that [Pm, Pr ] = 0 and (i/) [Pm, Mrs ] = h m r Ps – h m s Pr shows that these are the momentum three-vectors: 2014 MRT and the angular momentum three-vector: These are not conserved, which is why we do not use the eigenvalues of K to label physical states. In a three-dimensional notation, the commutation relations may be written: and J H = P H = H H = K H = iP [ , ] [ , ] [ , ] 0 [ , ] i i i i , and J J i J J K i K K K i J = e = e = - e = = [ , ] [ , ] [ , ] i j i jk k i j i jk k i j i jk k and J P i P K P iH e d [ , ] [ , ] where i, j, k, &c. run over the values 1, 2, and 3, and e i j k is the totally antisymmetric quantity with e 123 = +1. The commutation relation [ Ji , Jj ] = i e i j k Jk is the angular-momentum operator. P = [P1,P2 ,P3] J = [J 23, J 31, J 12 ] = [-J 32 ,-J 13,-J 21] and the energy P 0 itself. The remaining generators form what is called the ‘boost’ three-vector: K = [J 10, J 20, J 30 ] = [-J 01,-J 02 ,-J 03] i j i jk k i j i j
    • Now,* there is one peculiar consequence to one of these commutators – the two boost generators are: [Ki ,K j ] = -ie i jk Jk This commutator means that two boosts Bi and Bj in different directions (i.e., the indices i and j can’t equal each other at the same time) are not equivalent to a single boost B: where B is some boost. The reason things aren’t equal is the factor Wn ´ m(W), the Wigner Rotation where W is the Wigner Angle (i.e. a true space-time rotation although to be realistic, for practical reason it is usually an infinitesimal one.) 2014 MRT Bnˆ Bmˆ =Wnˆ ´mˆ (W)B By using B = W B W -1, the expression Bn Bm = Wn ´ m (W) B above can be re-written as: BnBm = WnW BWn- 1 ˆ ˆ ˆ ´m ˆ ˆ ´m ˆ W Wn´m W = BWn´m W ( ) ( ) ˆ ˆ ( ) ˆ ˆ ( ) * Credit for developing this in the way it is shown here and in the next few slides and example is given pretty much as it is in Entanglement in Relativistic Quantum Mechanics, E. Yakaboylu, arxiv:1005.0846v2, August 2010.
    • For example, let us use Lorentz transformations as boost matrices along the x- and y-direction, respectively, as defined by: ù 0 é ù é - g g b 0 0 ˆ ˆ ˆ y y y 0 1 0 0 0 0 0 0 0 0 g g b ˆ ˆ ˆ x x x g b g B and B 0 0 1 0 The ordered (i.e., reading from right-to-left as the are applied) product gives: ù g g g g b g b ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x y x x y y 0 0 0 2014 MRT é - ù é - ù é 0 0 0 0 g g b ˆ ˆ ˆ x x x g g b y x g g b g g b b g 0 0 1 0 0 0 ˆ ˆ ˆ y y y 0 1 0 0 0 0 ˆ ˆ ˆ x x x Notice that the result is non-symmetric. Now we can write this matrice for By Bx as: Byˆ Bxˆ =Wyˆ´xˆ (W)Bf =W-zˆ (W)Bf where the ‘arbitrary’ Wigner Rotation matrix here is given by: ú ú ú ú û ê ê ê ê ë - - - = ú ú ú ú û ê ê ê ê ë - ú ú ú ú û ê ê ê ê ë - - = 0 0 0 1 0 0 0 1 0 0 0 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x y y x y x y x x x y y y g b g g b g g b g B B ù ú ú û 1 0 é = W ê ê W z ˆ ( ) R z ˆ ë ù ú ú ú ú û é ê ê ê ê ë 1 0 0 0 W W 0 cos sin 0 - W W W = - 0 - ( ) 0 sin cos 0 0 0 0 1 ú ú ú ú û ê ê ê ê ë - - = ú ú ú ú û ê ê ê ê ë - = 0 0 0 1 0 0 0 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y y y y x x x x g b g ˆ ˆ
    • The result of the (group) matrix multiplication is (remember that M -1 = h M T h ): 0 g g g g b g b ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x y x x y y 0 0 g b g ˆ ˆ ˆ x x x g g b g g b b g ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x y y x y x y 0 0 0 1 ù Bf W B B é é - ù g g - g g b - g b é 1 0 0 0 W - W 0 cos sin 0 0 sin cos 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ y x y x x y y z y x - W+ W W- W - W ù 0 g b g g b g g g b b g cos sin cos sin sin 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x x y x y x y x y x y g b g g b g g g b b g sin cos sin cos cos 0 The ‘Wigner Angle’ W can be obtained by demanding that the Bf matrix be symmetric in, say, its M23 and M32 components: and after solving for the ratio sin W /cos W = tan W we get: 2014 MRT ú ú ú ú û ê ê ê ê ë - W- W W+ W W = ú ú ú ú û ê ê ê ê ë - - - ú ú ú ú û ê ê ê ê ë W W = - W = - 0 0 0 1 0 0 0 0 1 ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1 ˆ x x y x y x y x y x y -g yˆ sinW =g xˆ sinW+g yˆg xˆb yˆb xˆ cosW g g b b tan ˆ ˆ ˆ ˆ y x y x ˆ ˆ g + g y x W = - or g g ˆ ˆ y x 1 cos g g b b ˆ ˆ ˆ ˆ y x y x 1 sin ˆ ˆ ˆ ˆ + + W = - + W = - y x y x g g g g and
    • By replacing sin W and cos W in the boost matrix Bf one gets: ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Note that we can read Bn Bm = Wn ´ m (W) Bf (e.g., By Bx = Wy ´ x (W) Bf = W- z (W) Bf in the example above) backward to note that any boost B in the n-m (e.g., the x-y plane in the example above) can be decomposed into two mutually perpendicular boosts (in order) followed by a Wigner rotation (using group algebra – we mean it’s inverse)*: 2014 MRT ù ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ë g g - g b g - g b ˆ ˆ ˆ ˆ ˆ ˆ ˆ x y x x y y y b g b g ˆ ˆ ˆ ˆ 2 ˆ x x y y g b g x x y + + 2 g g 1 1 y x g g + g ˆ ˆ ( ) ˆ ˆ ˆ y x y + 2 ˆ 2 ˆ g g ˆ ˆ x y b g b g ˆ ˆ ˆ ˆ x x y y + - g b g + - = - W = Bf W - B B 0 0 0 1 2 1 1 g g ˆ ˆ g g ˆ ˆ ˆ ˆ ˆ x x y g b ˆ ˆ x y x y y y 0 0 0 1 ( ) ˆ ˆ 1 ˆ z y x Notice now that this Bf matrix is symmetric. So, as a result in this case (i.e., a boost along the x-direction followed by a boost in the y-direction), we obtain a boost along ‘some’ direction given by W = tan-1[- bx gx byg y /(gx + gy)] in the x-y plane. ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ B =W- 1 ˆ ˆ (W)B B n ´ m nˆ mˆ ˆ ˆ ˆ ˆ This result will be used later when we discuss particle representation in quantum field theory using Wigner basis states and especially when we calculate one first hand… * Generic composition of boosts: An elementary derivation of the Wigner rotation, R. Ferraro and M. Thibeault, Eur. J. Phys. 20 (1999) 143-151.
    • Since we will be using this soon let us look at a general example. Suppose that a parti-cle of mass mo is seen from system O with momentum p along the +z-axis. A second ob-server sees the same particle from a system O moving with velocity v along the +x-axis: 1 0 0 0 0 cos 0 sin 0 0 1 0 ù ú ú ú ú û L ˆ ´ ˆ ˆ p ˆ L ˆ p W ˆ p x z y x z y é - é ê ê ê ê ë g b g 0 0 ˆ ˆ ˆ z z z 0 1 0 0 0 0 1 0 b g g ˆ ˆ ˆ z z z ù 0 0 0 0 g b g ˆ ˆ ˆ x x x b g g ˆ ˆ ˆ x x x 0 0 1 0 é é ù g b g b g W W sin 0 cos ˆ ˆ ˆ ˆ ˆ z z z z z W - W ù é ù é - 0 cos 0 sin 0 0 1 0 b g g g sin 0 cos 0 0 0 0 1 g b g ˆ ˆ ˆ x x x b g g ˆ ˆ ˆ x x x 0 0 1 0 ˆ ˆ ˆ ˆ z z z z g g g b g b g g b g b g W- W W+ W sin cos 0 cos sin ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x z x z z x x x z z x x - b g g - b g b g W+ g W - b g b g W- g W sin cos 0 cos sin ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x x z x x z z x x x z z x 0 0 1 0 b g g g W W = ú ú ú ú û ê ê ê ê ë W W ú ú ú ú û ê ê ê ê ë - = ù ú ú ú ú û ê ê ê ê ë W - W W W ú ú ú ú û ê ê ê ê ë + + ú ú ú ú û ê ê ê ê ë - L = L - L = =- - sin 0 cos 0 0 0 1 0 sin 0 cos 0 0 0 0 0 1 ( ) ( ) ( , ) ˆ ˆ ˆ ˆ z z z z Since Lx´z(Lp) is symmetric we can extract the [Lx´z(Lp)]3 2 = [Lx´z(Lp)]2 3 ˆ ˆ ˆ ˆ ˆ ˆ components: ö 2014 MRT ÷ ÷ø - b g b g W- g W = g W Þ - g + g W = b g b g W cos sin sin ( )sin cos ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x x z z x z x z x x z z æ ç çè b g b g ˆ ˆ ˆ ˆ x x z z + Þ W = - b g b g ˆ ˆ ˆ ˆ x x z z tan arctan + W = W = - W x z x z ˆ ˆ ˆ ˆ sin cos g g g g
    • g b g ˆ ˆ ˆ x x x 0 0 0 0 0 0 1 0 0 m = [moc,0,0,0]T g g b g g b g 0 x z x x x z z b g g g b g b g 0 g g x z b g g x x z x x z x x x z z we then have (with k ˆ ˆ 0 px Now, since: x v ­ é - L(v) y W-y(L,p) g g b g g b g 0 é - - ù ˆ ˆ ˆ ˆ ˆ ˆ ˆ x z x x x z z b g g g b g b g 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x x z x x x z z g b g 0 0 ˆ ˆ ˆ z z z 0 0 1 0 0 1 0 0 0 0 1 0 ù m c m c é - = ù é ù é - - o ˆ ˆ o m c ˆ ˆ ˆ o ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0 0 0 1 0 This provides us with the three cartesian values for the boosted momentum: é 2014 MRT y mo Lp pz ˆ ˆ ( ˆ ˆ ) Lp = - pxˆ i + pzˆk = moc -b xˆg xˆg zˆ i +b zˆg zˆk Suppose that observer O sees a particle (mass mo ¹ 0) with momentum pz in the z-direction. ù A second observer O moves relative to the first with velocity v in the x-direction. How does O describe the particle’s motion? ù ú ú ú ú û ê ê ê ê ë - = ú ú ú ú û ê ê ê ê ë ú ú ú ú û ê ê ê ê ë - L = z z z z z z x x x x z ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 0 0 0 0 0 0 0 1 ( ) b g g b g g b g g L p ú ú ú ú û ê ê ê ê ë ú ú ú ú û ê ê ê ê ë ú ú ú ú û ê ê ê ê ë - L = m c L p k ˆ ˆ o ˆ ˆ ˆ 0 0 0 ( ) z z z z z x z b g b g g m a standard rest momentum for massive particles): ù ú ú ú û é- = ê ê ê ë ù ú ú ú û é ê ê ê ë L L p p L m c b g g o ˆ ˆ ˆ x x z o ˆ ˆ z z 1 2 3 0 b g m c p ˆ ˆ ( ˆ ˆ ) Lp = -b xˆg xˆg zˆ moc i + b zˆg zˆ moc k = moc -b xˆg xˆg zˆ i + b zˆg zˆ k x z z Lz(p) which when applied in the Figure will look like:
    • The Poincaré Algebra The inhomogeneous Lorentz transformation (or Poincaré group), L = {L,a}, is defined by: xm = (Lx)m = Lm x + a n m n and is a space-time translation, i.e., as the product operation of a translation by a real vector a m = (t , a) and a homogeneous Lorentz transformation, Lm n (the translation being performed after the homogeneous Lorentz transformation.) It can conveniently be represented by the following matrix equation: where the last coordinate, i.e., 1, has no physical significance and is left invariant by the transformation. The generators for infinitesimal translations are the Hermitian operators Pm, and their commutative relations with the Hermitian generators for ‘rotations’ in the xm - xn plane, M mn = - M 2014 MRT ù ú ú ú ú ú ú û é ê ê ê ê ê ê ë ù ú ú ú ú ú ú û é ê ê ê ê ê ê ë 0 0 3 0 2 0 1 0 0 L L L L a 1 1 3 1 2 1 1 1 0 L L L L a 2 2 3 2 2 2 1 2 0 L L L L a 3 3 3 3 2 3 1 3 0 L L L L = ù ú ú ú ú ú ú û é ê ê ê ê ê ê ë 0 1 2 3 0 1 2 3 x x x x a x x x x 1 0 0 0 0 1 1 nm are as expressed in contravariant form: i[Pm ,Mrs ] =h mr Ps -h ms Pr The commutation rules of these generators with themselves are the Poincaré Algebra: [Pm , Pr ] = 0 and i[Mmn ,Mrs ] =hnrMms -h mrMms -hsmMrn +hsn Mrm
    • Lorentz Transformations Now, those transformations with a These transformations also form a group. known as the homogeneous Lorentz group. If we first perform a Lorentz transformation L: xm = Lm x r = L m L r x (L-1)r = L =h h L rs m We note from (Det L)2 = 1 above that either Det L = +1 or Det L = -1; those transformations with Det L = +1 form a subgroup of either the homogeneous or the inhomogeneous Lorentz group. Furthermore, from the 00-components of hLm Ln = hm n r s rs and Ln s t = h s Lk t h nk, we have: m = 0 form a subgroup with to the Poincaré group: i i 0 0 0 0 (L0 ) = 1+ L L = 1+ L L with i summed over the values 1, 2, and 3. We see that either L0 0 ³ +1 or L0 0 £ -1. Those transformations with L0 0 ³ +1 form a subgroup. Note that if Lm n and Lm n are two such L s, then: i i 2 0 0 L0 Lm º (LL) = L L + L L + L L + L L 3 3 0 0 2 2 0 1 0 0 1 0 0 0 0 0 0 0 m T(L,0)T(L,0) = T(LL,0) 2014 MRT Taking the determinant of hm n Lm r Ln s = hrs gives: so Lm n has an inverse, ( L-1)n s , which we see from hm n Lm r Ln s = hrs and takes the form: s nm r n n n n r r (Det L)2 = 1
    • The problem of classifying all the irreducible unitary representations of the inhomo-geneous Lorentz group (i.e., the Poincaré group) can again be formulated in terms of finding all the representations of the commutation rules above by self-adjoint operators. Believe it or not, the following scalar operators: P 2 = Pm P and W 2 = w w = 1 M M P P -M M P P 2 wm e P M 2 commute with all the infinitesimal generators, M mn and Pm, and constitute the invariants of the group. 2014 MRT n ns m ms s s mn m mn m m where n rs mnrs = 1 They are therefore multiples of the identity for every irreducible representation of the inhomogeneous Lorentz group and their eigenvalues can be used to classify the irreducible representations. The set of all four-dimensional translations is a commutative subgroup of the inhomogeneous Lorentz group. Since it is commutative, the irreducible unitary representations of this subgroup are all one-dimensional and are obtained by exponentiation.
    • 0)2 - 1], and similarly the three-vector [L0 1,L0 2,L0 3] has length Ö[(L0 0)2 - 1], so the and so: L0 Li º L 0 L 1 + L L + L L £ L - L - LL 0 ³ L L - L - L - The subgroup of Lorentz transformations with Det L = +1 and L0 0 ³ +1 is known as the proper orthochronous Lorentz group. ( 0 ) 2 0 1 ( 0 )2 0 1 0 3 3 0 0 2 2 0 0 1 0 i But (L0 0)2 = 1 + Li 0 Li 0 = 1 + L0 i L0 i shows that the three-vector [L1 0,L2 0,L3 0] has length Ö[(L0 scalar product of these two three-vector is bounded by: ( ) 0 0 0 2 0 2 0 0 0 ( 0 ) 1 ( 0 )1 2014 MRT
    • If we look over the Lorentz transformation properties of P r and M rs is the case for homogeneous Lorentz transformations (i.e., a m = 0), we get: 2014 MRT - 1 U a P U a P L L = L L s mn n r m L L = L ( , ) ( , ) rs r m m r - 1 U ( , a ) M U ( , a ) M These transformation rules simply say that M rs is a tensor and P r is a vector. For pure translations (i.e., with Lm n = d m n ), they tell us that P r is translation-invariant, but M rs is not.
    • The group of Lorentz transformations contains a subgroup which is isomorphic to the familiar three-dimensional rotation group. This subgroup consists of all Lm n of the form: 2014 MRT ù ú ú ú ú û é ê ê ê ê ë 1 0 R = 0 R Λ where R is a 3´3 matrix with R RT = RT R = 1. We call such a LR a ‘spatial’ rotation. Every homogeneous Lorentz transformation can be decomposed as follows: Λ = ΛR (R 2)Λ(L1)ΛR (R1) where LR(R1) and LR(R2) are spatial rotations and L(L1) a Lorentz transformation in the x1- direction. If we set m = n = 0 in the equation Llm hlr Lrn = hmn , we then obtain: so that L0 0 ³ L + L + L + = L + = L å= ( ) 1 ( ) 1 ( ) ( ) ( )2 1 0 ³ 1 or L0 0 £ -1. 0 2 3 0 1 2 2 0 3 1 2 0 2 0 i i Remember that a Lorentz transformation for which L0 0 ³ 1 is called an orthochronous Lorentz transformation. A Lorentz transformation is orthochronous if and only if it transforms every positive time-like vector into a positive time-like vector. The set of all orthochronous Lorentz transformations forms a group: the orthochronous Lorentz group.
    • The problem of finding the representation of the ‘restricted’ Lorentz group is equivalent to finding all the representations of the commutation rules above. The finite dimensional irreducible representation of the restricted group can be labeled by two discrete indices which can take on a values the positive integers, the positive half-integers, and zero. To show this, let us define the operators: 2014 MRT and their commutation rules are: ( Momentum ) ( Angular momentum ) (Boost) 1 2 3 P P P [ , , ] 23 13 21 M M M [ , , ] 01 02 03 M M M [ , , ] = = = P J K J J = e J [ , ] i j ijk k K K = - e J [ , ] i j ijk k J K = e K [ , ] i j ijk k From these operators, we can construct these invariants of the group: and mn Mmn M 2 1 J2 -K2 = mn rs e mnrs M M 8 1 - J ·K = which commute with all the Ji and K i. They are therefore the invariants of the group and they are multiples of the identity in any irreducible representation. The representations can thus be labeled by the values of these operators in the given representation.
    • The operator corresponding to the translation by the four-vector a m is given by: i Pm a m - U (1, a ) = e  In an irreducible representation, the operation of translation by a thus corresponds to multiplying each basis vector |Y( p, mj) ñ by exp(-i Pm am /): where E(p)=wp=Ö(p2c2+mo i P m a - - = = ω 2c4) is the one-particle energy. As seen earlier, a rotation Rq by an angle q = | q | about the direction of q = q p is represented on the Hilbert space by: 2014 MRT (1, ) ( , ) e ( , j ) e ( , j ) i P a U a Y p mj m Y p m m Y p m m   hence i J θ θ - · U Y p mj =  Y p m ( ,0) ( , ) e ( , j ) A boost K in the direction of the momentum z: p ζ p = - - 1 sin 1 ˆ sin p p m c i m c ic ζ K ζ ö = ÷ ÷ ø - · =  ö ÷ ÷ ø æ ç ç è æ ç ç è i o o p U Y p mj Y p m ( ,0) ( , ) e ( , j ) and similarly for evolution: e ( , ) e ( , ) j i t j i H t p m p m p  Y =  Y ˆ
    • To make the range of values of the label more transparent, let us introduce the following generators: 1 which satisfy the following commutation rules: J J i J [ , ] e i j ijk k K K i J [ , ] = = i j ijk k = J K | jmj, j'm'j ñ where j, mj, j' and m'j are integers or half-odd integers, -j £ mj £ j, -j' £ m'j £ j' 2014 MRT ( ) ( ) 2 1 2 Ji = i Ji + i Ki and Ki = i Ji - i Ki It follows from the commutation rules that a finite dimensional irreducible representation space, V j j' can be spanned by a set of (2 j +1)(2 j' +1) basis vectors and in terms of which the J and K operators have the following representation: and [ , ] 0 i j e ± ¢ ¢ = ± ¢ ¢ = ± + ± ¢ ¢ , ; , , ; , J j , m ; j , m ( J i J ) j , m ; j , m ( j m )( j m 1) j , m 1; j , m j j j j j j j j ¢ ¢ = ¢ ¢ J j m j m m j m j m j j j j j 3 1 2    ± ¢ ¢ = ± ¢ ¢ = ¢ ¢ ¢± ¢ + ¢ ¢ ± , ; , , ; , K j m j m K i K j m j m j m j m j m j m , ; , ( ) , ; , ( )( 1) , ; , 1 j j j j j j j j ¢ ¢ = ¢ ¢ ¢ K j m j m m j m j m j j j j j 3 1 2   
    • These are thus a denumerable infinity of non equivalent finite-dimensional (in general they are non-unitary) irreducible representations. These can be labeled by two non-ne-gative indices ( j, j' ) where j, j' = 0, 1/2, 1, 3/2,…. The dimension of the representation is (2 j + 1)(2 j' + 1) and D( j, j' ) is single-valued if j + j' is an integer and double-valued. A quantity which transforms under D(0,0) is called a scalar, one which transforms under D(1/2,1/2) a four-component vector, one which transforms under the (1/2,0) representation a two-component spinor. A quantity which transforms under (0,1/2) is called a conjugate spinor. For the D(0,1/2) and the D(1/2,0) representations, an explicit matrix representation of the infinitesimal generators can be given in terms of the Pauli matrices with: situation. For example, under an infinitesimal rotation e about the i-th axis: 2014 MRT (1/ 2,0) 1 J = - i J = - i s s i i i i K (1/ 2,0) 1 s K (0,1/ 2) 1 s i i i i 2 2 2 1 (0,1/ 2) 2 = - = + A two-component spinor, x, transforms under spatial rotation as in the three-dimensional ( ) 3 2 1 ξ® ξ = (1+ ies i )ξ R and under an infinitesimal Lorentz transformation in the x i-direction, this spinor transforms according to: Note also that the quantity x* x is not a scalar. ( ) 4 2 1 ξ® ξ = (1+ s ie ) ξ Λ
    • Lorentz Invariant Scalar The irreducible representations of the inhomogeneous Lorentz group can now be classi-fied according to whether Pm is space-like, time-like, or null vector, or Pm is equal to zero. For the last case, Pm = 0, the complete system of (infinite dimensional) unitary representations coincides with the complete system of (infinite dimensional) unitary representations of the homogeneous Lorentz group which we studied earlier. The representations of principle interest for physical applications are those for which P 2 = m2c 2 = positive constant, and those for which P 2 = 0. o Let us first discuss the case P 2 = mo 2c 2. In that case, P 0 /| P 0|, the sign of the energy, commutes with all the infinitesimal generators and is therefore an invariant of the group. Defining a Lorentz invariant scalar product within the vector space by integrating over a set of p (with P + ψ p m p m ψ x d p 0 = åò =åò 0 f f Y( , ) Y( , ) ( ) f ( , ) ( , ) j = mj p m ψ p m d p j j j m j j P P * 2 1 1 2014 MRT 2 = mo 2c 2, P 0 = +Ö( p 2 + mo 2c 2) = E /c) and summing over the index mj: where j = 0, 1/2, 1, 3/2, 2,…. So, we have hinted that an irreducible representation of the type P 2 > 0, P 0 > 0 is labeled by two indices (mo , j ), where mo is a positive number and j is an integer or half-integer. The index mo characterizes the mass of the elementary system, the index j the angular momentum in its rest frame, i.e., the spin of the elementary system. The fact that the irreducible representation is infinite dimensional is just the expression of the fact that each elementary system is capable of assuming infinitely many linearly independent states.
    • Klein-Gordon & Dirac For each (mo , j ) – and a given sign of the energy – there is one and only one irreducible representation of the inhomogeneous Lorentz group to within unitary equivalence. For j half-integral, the representation is double-valued. For j = 0, the representation space is spanned by the positive energy solutions of the relativistically covariant equation for a spin-0 particle – the Klein-Gordon equation: 2014 MRT j r t = - Ñ + j r ¶ - ¶   ( , ) ( 2 4 ) ( , ) 2 c m c t o 2 2 2 2 2 t For j = 1/2 by the positive energy solution of the Dirac equation (free particle case): i ¶ y r t  = -  α · Ñ + b y r ( , ) ( 2 ) ( , ) i c moc t t ¶ c m m c x r m y r p p p For j = 1 by the positive energy solutions of the Proca equation (not discussed). ö æ ö æ f ( ) 1 ¶ f ¶ f ¶ f = - n - m ¶ f n - m 1 m 2 c 4 x t x x x x We will discuss both the Klein-Gordon and Dirac equations later. ( ) 2 4 o n m m n m n m m f f f i x ïþ ïý ü ïî ïí ì + ÷ ÷ ø ç ç è ¶ ¶ ÷ ÷ ø ç ç è ¶ ¶ ¶ ¶  ( 2 ) 0 - g ¶m + o y = Y( ,t) º ψ(x ) = (x) = ò +¥ Y( p,mj ) d ×Y(t) -¥ or such that in what follows, it applies also to states that get acted on by the Dirac equation:
    • One-Particle States The physical states of particles are described by the Wigner basis states | Yk mo( j, mj) ñ (which are equivalent to the states | Y(k, mj) ñ) for a unitary irreducible representation of the inhomogeneous Lorentz group (Poincaré group) with pm p m = p02 – p2 = (moc)2. Our goal is to find eigenkets of | Yp mo( j, mj) ñ as they appear following an homogeneous Lorentz transformation group U(L,a) product on a state-vector | Yp mo( j, mj) ñ is as follows: Λ 1 Λ v L U U p j m Y Y [ ( )] [ ( )] ( , ) U a j m U a m o = pm j km j E p o o U a m - o 1 1 Λ v L U L p U U L p U p Y j m = L L [ ( )] [ ( )] [ ( )] [ ( )] ( , ) ( ) E p ( ) ( , ) ( , ) ( , ) ( , ) km j o These states form the Hilbert space of the theory and the momentum states | Y( p, mj) ñ can be obtained from the standard state | Yp mo( j, mj)ñ º | Y( p, mj) ñ by a unitary transformation: Y( p,mj ) = L( p) Y(k,mj ) and L ( p) is some standard Lorentz transformation matrix that depends on p = p å+ x º x = E L p e [ ( L , )] ( , ) ( ) ( ) 1 ( ) ( ) ¢ =- ¢ L - · - j m j pE j j m m i Et j j j W p j m ( ) E p ( ) (2π ) p r  D Y 3/ 2  y m y m and the momentum states are normalized over intermediate states and we get the coordinate ket: 2014 MRT The spin j corresponds to the eigenvalues J 2 = j ( j + 1)2 of J 2 and J3 = mj  (mj = j, j – 1, …, -j ). Also, U(1,a)| Yk mo( j, mj)ñ means the same thing as exp (- i k m am ) | Yk mo( j, mj)ñ ( p m =  k m ).
    • The Lorentz transformations associated with the Poincaré group can be constructed in which the quantum state differs by only a mixture of the internal spin indices mj (i.e., run-ning over the discrete values j, j – 1, …, -j ) but possess the same physical observables. The effects of the inhomogeneous Lorentz transformations (e.g., acting on the Dirac fields) can be elucidated by considering the particle states obtained from the irreducible unitary representations of the Poincaré group. The unitary operation representing the Lorentz transformations acting on the Poincaré generators are given by the equations U(L,a) Pr U m Pm and U(L,a) Mrs U -1(L,a) = [L-1]r m [L-1]s n (Mmn -1(L,a) = [L-1]r - am Pn - a ). n Pm Since the momentum four-vector commutes among each other according to [Pm Thus, the one-particle state is an eigenvector of the momentum operator: Pm Y( p,mj ) = pm Y( p,mj ) and the state transforms accordingly for a space-time translation: i p m a - - 1 º = ( , ) ( , ) e ( , j ) e ( , j ) i P a U a Y p mj m Y p m m Y p m m   , Pn ] = 0, the particle states can be characterized by the four-momentum P m together with additional internal degrees of freedom mj . The internal degrees of freedom pertain to the spin vector which can be affected by transformations in the space-time coordinates. 2014 MRT
    • The components of the energy-momentum four-vector operator Pm all commute with each other (i.e., [Pm , Pn] = 0), so it is natural to express physical state-vectors in terms of eigenvectors of the four-momentum. This four-momentum Pm is a trusted observable! We now introduce a label mj to denote all other degrees of freedom (i.e., all other total angular momentum orientations), and thus consider state-vectors | Y ( p, mj) ñ with: Pm Y( p,mj ) = pm Y( p,mj ) Under space-time translations, the states | Y ( p,mj) ñ transforms as: n - m 1 = i p a U a Y p mj Y p m m  ( , ) ( , ) e ( , j ) m m m = - = L- ( Λ ) ( , ) ( Λ )[ 1( Λ ) ( Λ )] ( , ) ( Λ ) ([ 1] ) ( , ) P U Y p m U U P U Y p m U n P Y p m j j j 2014 MRT We must now consider how these states transform under homogeneous Lorentz transformations U(L,0) º U(L) is to produce eigenvectors of the four-momentum with eigenvalues Lp (where we use U -1 (L,a) Pm U (L,a) = [L-1]m n Pn ): n m Λ Λ p U p m p U p m = L = L Y Y ( ) ( , ) ( ) ( ) ( , ) j j m n Hence U(L) | Y ( p,mj) ñ must be a linear combination of the state vectors | Y (Lp, mj) ñ: å¢ = ¢ L L ¢ U(Λ) Y( p,mj ) Cm m ( , p) Y( p,mj ) j j j m It is thus natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible.
    • In other words, if a system is confronted with a homogeneous Lorentz transformation L, the momentum p is changed to L p. According to Pm | Y( p , m) ñ = p m | Y( p, m)ñ, the one-particle jjstate must possess an eigenvalue of L p as well: m m m L = L - L L = L L n ( ,0) ( , ) ( )[ 1( ) ( )] ( , ) ( )[ ] ( , ) P U Y p m U U P U Y p m U P Y p m in which U(L,a) Pm U n j j j m p U Y p m = L L ( ) ( ) ( , ) j m Pn has been used for the Lorentz transformed -1(L,a) = [L-1]n momentum generator. It can be seen then that U(L) | Y( p, mj) ñ is a linear combination of the states | Y(Lp, m¢j) ñ, where: j å+ j U( ) Y( p,m ) D( ) ( , p) Y( p,m ) L = ¢ L L ¢ j m m ¢ =- m j j j j j ( j ) (L, p) is termed the Wigner coefficient and, as said previously, they depend on the irreducible representations of the Poincaré group. As a special example, the Casimir operator Pm P m cannot change the value of the momentum (i.e., its an invariant – as also stated previously): P m P m Y ( p , mj ) = p m p m Y ( p , mj ) = ( moc )2 Y ( p , mj ) in which Dm¢j mj and this, believe it or not, does leave the momenta of all the particle states invariant. 2014 MRT
    • Wigner’s Little Group Hence, to distinguish each state, the standard four-momentum given by k m = [moc,0,0,0] is chosen, from which all momenta can be achieved by means of a pure Lorentz boost: p = L( p) k (N.B., the standard four-momentum is non-unique and it also depends on the charac-teristics of the particle, e.g., whether it is a massive or a massless particle). Notice that a simple three-dimensional rotation, W (which is an element of the Poincaré group), will render the standard four-momentum invariant: m åW m k = k n n n which implies that there exists a subgroup of elements consisting of some arbitrary Wigner rotations, W, and this subgroup is called the little group. ( , ) 2 ( , ) ( , ) ( , ) ( , ) H Y k mj = moc Y k mj & p Y k mj = 0 and J Y k mj = mj Y k mj It is important to distinguish that this little group is not unique in the Lorentz group but it is actually isomorphic to other bubgroups under a similarity transformation. This is because there is no well-defined frame for the standard momentum k m due to the equivalence principle in special relativity. The definition of the little group is dependent on the choice of the standard momentum as well as the Lorentz transformation L. 2014 MRT In this momentum rest frame, the state | Y( k,mj) ñ is specified in terms of the eigen-values of the Hamiltonian H = p0, the momentum p, and the z-component of the total angular momentum operator J as:
    • So, the only functions of p m that are left invariant by all proper orthochronous Lorentz transformations Lm n are the invariant square p 2 = Smn hmn p m pn, and p When L( p) is applied to k we get the momentum p. When L is applied next to p it becomes L p. When L-1(L p) is L-1( L p) L( p) applied to L p we recover k ! 2 £ 0, also the sign of p 0. Hence, for each value of p 2, and (e.g., for p 2 £ 0) each sign of p 0, we can choose a ‘standard’ four-momentum, k m, and express any p m of this class as: where L m =å p m Lm ( p) k n n n n is some standard Lorentz transformation that depends on p m, and also implicitly on our choice of the standard k m. We can define the states | Y ( p, mj) ñ of momentum p by: Wigner’s little group is the Lorentz transformation L-1(L p) L L( p) that takes k to L( p) k = p, and then to L p, and then back to k, so it belongs to the subgroup of the homogeneous Lorentz group. Operating on this equation with an arbitrary homogeneous Lorentz transformation U(L), we find: k m = [moc,0,0,0] pm = Sn Lm n L p = S ( p) kn n L0 n pn L Y( p,mj ) = N( p)U[L( p)] Y(k,mj ) N( p) is a numerical normalization factor to be chosen on the next slide and where U[L( p)] is a unitary operator associated with the pure Lorentz ‘boost’ that takes [mc ,0] into [ p 0, p]. oΛ ΛL U p m N p U p k m ( ) ( , ) ( ) [ ( )] ( , ) = L - L ( ) [ ( )] [ ( ) ( )] ( , ) The transformation W m n (i.e., L-1(Lp) L L(p) = W(L, p)) thus leaves k m invariant: Sn W m n k n = k m. For any W m n satisfying this relationship, we have: 1 1 j j j N p U p U p p Y k m Y Y L L ΛL  = = j å+ U[W( , p)] Y(k,m ) D( ) [W( , p)] Y(k,m ) L = ¢ L ¢ ¢ =- m j j j j m m j j j where the coefficients Dm¢j mj ( j) [W (L, p )] furnish the representation of the little group. 2014 MRT
    • The Wigner coefficient for the standard four-momentum can be evaluated as: å å = ¢ ¢ º ¢ Note that the Wigner coefficients form a representation of the little group, which means that for any little group elements W, W ¢, the group multiplication property holds: ¢ = ¢ ¢ º ¢ U WW k m WW k U W U W k ( ) ( , ) ( ) ( , ) ( ) ( ) ( , ) which implies directly that: U(W)U(W¢) ºU(WW¢) ¢ ¢ + ¢ =- j j j j j j m m m j j m j j j U(W) ( p,mj ) m m (W, k) (k,m ) (W) (k,m ) Y D( ) Y D Y thus: å å ¢ ¢ ¢ ¢ ¢ = ¢ ¢ s s s s s s s s s , ( ) ( ) ( , ) j j j m m m j W W Y k Y Y Y D D D ¢ ¢ =å ¢ ¢ Ds s (WW ) Ds m (W ) Ds m (W) j j j m However, this choice for the normalization condition leads to problems with the subsequent transformation equations relating to the particle states which involve some tedious momentum-dependent constants. 2014 MRT
    • Normalization Factor Using W = L-1 L L and inserting into U(L) | Y( p,m) ñ = N U[L(Lp)] U[L-1(Lp) L L(p)] | Y( p,m) jsñ and using U(W) | Y( k,m)ñ = SD(W) | Y( k,m¢) ñ we get: jm¢j m¢j mjjå¢ U(Λ) Y( p,mj ) = N( p) Dm ¢ m [W( L , p)]U[L( L p)] Y(k,mj ¢ ) j or, recalling the definition | Y ñ = N U[L(p)] | Y ñ, we finally get: U p m N p [ ( , )] ( , ) (Λ) Y( , ) ( ) D W Y j m m p p mj The normalization condition is achieved using the scalar product: ( , ) ( , ) ( ) ( ) 2 ¢ ¢ = d ¢ d k¢- k Y p mj Y p mj N p mj mj The normalization factor N( p) is sometimes chosen to be N( p) = 1 but then we would need to keep track of the p 0/k 0 factor in scalar products. Instead, the convention is that: 2014 MRT for which å¢ ¢ L L ¢ L = j j j m N p ( ) j j m 0 0 N ( p ) = k p p mj p mj - m¢j mj Y( ¢, ¢ ) Y( , ) =d (p¢ p)d N p k = L = L E p p ( ) ( ) ( ) N p ( ) ( ) 0 ( ) ( ) 0 0 0 E p p N p p L Þ L L = and
    • We choose the following normalization condition (c.f. Weinberg QTF I): This implies that the Wigner coefficients cannot sum up to unity, but instead up to a phase factor that depends on the momentum of the particle, p, and the Lorentz transformation, L: † * D D L L ¢ = L L L L ¢ p U U p p p p m p m Y s Y s Y Y ( , ) ( ) ( ) ( , ) ( , ) ( , ) ( , ) ( , ) Y Y which implies that: p mj k mj p m¢j mj Y( , ¢ ) Y( , ) = (2 0 )d (3) (p - p¢)d s s 0 * 0 D D ¢ ¢ ¢ 0 p p p å L * L = 0 s ¢ s ds s ¢ L ( ) ( , ) ( , ) p j j j m Dm Dm j j j j j j j j j j m m m m m m m m m m j j p p p p p p ¢ ¢ ¢ ¢ ¢ å å ¢ = = L L L s s d d s s s s , , ( , ) ( , ) 2 ( , ) ( , ) (2 ) and finally: D p D p p L L = I ( , ) ( , ) ´ L 0 2 2 0 † p ( ) The transformations mentionned above whereby the representations of the Poincaré group is being derived from the little group is termed as the method of induced representations. 2014 MRT
    • The explicit form for W(L, p) is: W(L, p) = L-1(L, p)L L( p) ( k ¾L¾(¾p)® p ¾¾L® Lp ) and it can be shown that this is an element of the little group associated with the standard vector k m. (Note that L(Lp) is the Lorentz boost of the momentum (Lp) ê êë é j cosh p ˆ sinh q q = i i j ˆ sinh (cosh 1) ˆ ˆ p p p L p ( ) q d q where the angle q is defined by tanhq = Ö[(p + - i j 0)2 – mo 2c 4]/p ù ú úû 0 and the components of the particle momentum are given by p i (i = 1, 2, 3). The Lorentz boost L( p) transforms the standard momentum k m to momentum p m. In addition, the general Lorentz transformation LL for ‘pure boost ’ only is written as: ù ú úû é cosh n ˆ sinh x x L = i j ê êë + - i j i j L n ˆ sinh x d (cosh x 1) n ˆ n ˆ m from standard momentum vector k m ). Then the Wigner transformation corresponding to the present choice of k m is a SO(3) rotation. Furthermore, the Lorentz boost L( p) can be computed from the boost generator K= M= i (x¶- x¶) where L( p) = exp(-i cq p ˆ · K /). i i0 i 0 0iThe explicit form of L( p) is given by: ˆ 2014 where n is a unit vector along the direction of the boost. This transforms the momentum p m to a Lorentz-transformed momentum (Lp) m. MRT
    • Considering now a Taylor expansion of the Wigner transformation W(L, p) = L-1(Lp) L L( p) as well as the explicit form of the matrix for L(p) just obtained for an arbitrary Lorentz transformation L: W L p = L L p L L p ( , ) ( ) ( ) ic ˆ 1 mn M ic q Λ mn q in which the Lorentz boosts, L( p) and L(L, p) and the arbitrary Lorentz transformation, L, are parametrized in terms of its generators. In addition, the spatial component of the Lorentz-transformed momentum is given by pL = Lp (Note that the corresponding Lorentz boost for the Lorentz transformed momentum starting from the standard vector, k m, is given by tanhq ¢ = | pL | / (Lp)0, in which L(Lp) k = Lp ). Consider the inifinitesimal Lorentz transformation being parametrized by the anti-symmetric tensor w mn (expressed to first order terms of w ): (ω) 1 1 ωmn M 2 L = - mn + thus: ¢p ·K - - p·K - º ω ˆ 2 1 e e e   p a pa i [ωmn M ] p p [ω] p a b b a b a (L ) @ - º + mn b 2 2014 MRT
    • A Taylor series expansion of the Wigner transformation is considered as fallows: thus: + ω 0   = + W L p = W L p + ¶ W L p ( , ) ( , ) ω ( , ) ω 0 - = 1 mn L p L p = + ¶ L L ω mn 1 ω [ ( ) ( )] ¶ ¶ = ω 0 ω mn mn W p L p (ω) L ( p ) ω L ( p ) (ω) L p ù + L ¶L é ö çè ù é L @ + ¶ L ( , ) 1 ω ( ) é = + ¶ L ù é ù 1 ω ( ) In the ‘second term’ of this last equation the expression can be re-written as:   + úû êë ÷ø + L æ- ú úû ê êë ¶ + ú úû ê êë ¶ ú úû ê êë L ¶ = - = - = - = - ( ) 2 ( ) ω ( ) ω ( ) ω ω ω 0 1 ω 0 1 ω 0 1 ω 0 1 L p L p L p i M L p mn mn mn mn mn mn mn mn é p 0 æ + - ù é ¢ ¢ ö æ = + ¶ úû é ¶ L 0 0 0 ω ( ) p p p p p p p p p in which the substitution Lp = p¢ has been made and, in general, L-1( p¢) =h LT( p¢)h . ù ú ú ú ú ú û ê ê ê ê ê ë ö ÷ ÷ø ç çè ú ú ú ú ú û ê ê ê ê ê ë ¢ ÷ ÷ø ç çè - ¢ + ¢ - ¢ - ¢ ¶ ù êë ¶ = = - p p 2 2 o o o 2 o ω 0 2 2 o o o 2 o ω 0 1 ω 1 1 ( ) 1 ω ω p m c m c m c m c p m c m c m c m c L p L p j k j k j k i j i j j i d d mn mn mn mn 2014 MRT
    • Now, since (Lp)a @ pa - (i /2)[wmn Mmn]a b pb º pa + [w]a b pb, we get (  =c = 1): ù é ¶ mn p i [ω M ] p [ω] p a b b a b @ - º mn b mn a ¶ L ω ( ) mn ú úû ω 2 = ω 0 ê êë and since x i = w0i = -wi0 and q k = ei j k wi j as well as J i = e i j k Mjk and Ki = Mi0: p i i M i M i j M p i i K J p b = - 0 + 0 + b = - x + q b [ω ω ω ] 2 ù ú úû 0 x 0 x e q k j i i i j k p p ( ˆ ) p p ξ 0 i i ( ˆ ) p p ´ θ ù ú úû é ê êë é ê êë ê êë é - - 0 x ( ˆ ) p p ξ i i - · = = º i ù ú úû j ù ú úû · [2 2 ] 2 ù ú úû é ê êë j j i j p x i i j k j 0 p p x - e q 0 = é ê êë k k i k p p p m 0 x a b [ω] in which the axial vector is given by q = [q 1,q 2,q 3]† and mi = (p ´ q)i as well as: ù i p 1 [ p ] pˆ pˆ [ p ] ω ú ú û é ω ( ˆ ) 0 0 ê ê i i i m n n x x mn - - - = úû ë ù é ¶ êë ¶ ¢ å = mn ω 0 p m p m p mn ˆ 2014 At this junction, the sum Sm n p m[p0x n - | p |mn] describes the relation between the components of the momentum and its Lorentz boost. MRT ˆ
    • Assuming that any sum over the cross-terms is zero for m ¹ n, that is: åpˆ m [ p0 x n - p m n ] = p0 (pˆ · ξ) mn in which the boost vector is given by x = [x 1,x 2,x 3]†. With the axial vector, q = [q 1,q 2,q 3]†, q and x become the parameter vectors associated with the infinitesimal Lorentz transformation L. In addition, the vector normal to both the axial and momentum vectors is given by m = p ´ q. 0 p o o 0 0 p p p p j j x b b 1 ˆ [ω] ˆ [ω] ( ˆ ) ˆ ˆ 1 ˆ ˆ x d 0 k k k x ù · - + æ - - - ÷ ÷ø p 0 0 0 0 p p p p b b 1 ( ˆ ) ˆ 1 ( ˆ ´ ) 2( ´ ˆ ) - ÷ ÷ø ö 0 i p 1 ( ˆ ) ˆ 1 ( ˆ ) 2( ˆ p ξ p m J p æ æ - - + · ÷ ÷ ø ù m K p ö æ 0 0 p ξ p ξ p p ξ m p p m p p ξ m p p ξ p ξ p m p m p p ξ p · ú úû é ê êë ö æ ç çè ö æ ö · + - ú úû é é æ é p p i p ê êë - · ÷ ÷ø ç çè 0 0 º - - ù ú ú ú ú ú û ê ê ê ê ê ë ïþ ïý ü ïî ïí ì - ÷ ÷ ø ç ç è ö ç ç è - - + · ÷ ÷ø ç çè - = ù ú ú ú ú ú û é ê ê ê ê ê ë ö ÷ ÷ø æ + - ç çè ù ú ú ú ú ú û ê ê ê ê ê ë ïþ ïý ü ïî ïí ì · ÷ ÷ ø ç ç è ö ç çè - + - = ù ú úû é ω ( ) ê êë ¶ L ¶ = - ´ ´ ) 0 1 ( ˆ ) ˆ ( ˆ ) ( ) ω 0 o o o o 0 o o o o o o o o o o 0 o o o o o o ω 0 1 p m m m m p m m m p m m m p m p p m m m m p p p p p m p p m m m m m m L p L p i i i n n i k n j k j k j k i i i j j i i j x e mn mn ˆ Our equation for w mn [¶L-1(Lp) /¶ wmn ]|w = 0 L( p) above (i.e. the ‘second term’) can be simplified by using the Lorentz boosts from the matrices for L( p) and LL further up, and this can be worked out as: 2014 MRT
    • The third term in our Taylor expansion for W(L,p) above (i.e. the term w mn [L-1(Lp) (-½ i Mmn ) ]|w = 0 L( p) ) can also be simplified in terms of its matrix elements (Note that the expression (-½ i w mn Mmn ) can be re-written as the matrix [ w ] ): 0 p o o p p m o p p x x e q d 0 0 0 p o o p k l k l k 0 ( ˆ ) ˆ ( ) ö p m x p p l ilm m i l i l ( ˆ ) ˆ ( ˆ ˆ ˆ ˆ ) x ù - + ú úû ö æ æ - - ù ö æ - + - ÷ ÷ø d x x x e q 0 p 0 1 ( ˆ ) ˆ 0 p ( ˆ ) ˆ 1 ( ˆ 0 p ´ ´ ) ö æ ö æ ö 0 0 i p p ( ˆ ) 1 ( ˆ p ξ θ p m J é é ù ö æ + - - - · ÷ ÷ ø p ù æ m K p 0 p p m o p m o ö æ - - æ - - æ ö 0 p ξ p ξ p ö æ æ m ξ p p m p p ξ m p p ξ m p p ξ m p p ξ · ú úû é ê êë ÷ ÷ø ç çè · + - - + - ú úû p p 0 0 p p é i p ê êë - · ÷ ÷ø ç çè é é é 0 º - - - ù ú ú ú ú ú ú û ê ê ê ê ê ê ë ïþ ïý ü ïî ïí ì ÷ ÷ø ç çè - + - ÷ ÷ø ç çè - · ÷ ÷ ø ç ç è - + · ÷ ÷ø ç çè - = ù ú ú ú ú ú û ê ê ê ê ê ë - ÷ ÷ ø ç ç è ö ç çè ö ç ç è - · ÷ ÷ ø ç ç è = ú ú ú ú ú û ê ê ê ê ê ë ÷ ÷ø ç çè ù ê êë - ú ú ú ú ú û ê ê ê ê ê ë ¢ ¢ ÷ ÷ø ç çè ¢ - ¢ - ¢ - ¢ = ´ ´ ) 2 1 ( ˆ ) ˆ 1 ˆ ˆ 0 1 ˆ ˆ [ ( )][ω][ ( )] o o o o o o o o 0 o o o o o o 0 o o o o 0 o o o 0 o o o o o 1 m m m m m m m p m m m m p m p m m p m m m m p m m m p m m p p m m m m p p m m m m L p L p m m ilm m i i i k k k l i i i i i l l l k l j j km m i j i j i j x q e in which w mn [¶( p¢)a /¶ wmn]|w = 0 above has been used for the matrix reprentation of the arbitrary infinitesimal Lorentz transformation L(w). 2014 MRT
    • After some careful re-grouping of the terms from all the matrix elements given in w mn [¶L-1(Lp) /¶ wmn]|w = 0 L(p) and [L-1( p¢)][ w ][ L(p)] just obtained, the boost terms in our expansion for W(L, p) has cancelled out completely and leaving behind the rotation terms: Thus, we obtain: where W( p) é 0 0 é W L p = + 0 - ( p -m ) ( ˆ )n -q n e i jn [ ( , )] 1 o W p i p m k = -e i j k [wi j - ( p é é n ´ ( ) ( ) o 1 ω 1 = + · i w0 j - p p Ω J j w0 i ) /( p ù p ξ θ J ù ù ù ù 0 + mo)] is the Wigner angle and J k = ½ e i j k M i j is the rotation generator for the Poincaré group. ú úû ê êë - - = + - ú úû ê êë - - - = - · ú úû ê êë é + - L º - 1 ( ) ( ω ω ) ( ) 1 1 ( ) ( ) ( , ) 1 0 0 o 0 0 o 0 i p p p M p m i p p M p m i i j i j i j j i n i j i j j i i j n x x e q ú ú ú û ê ê ê ë ú úû ê êë 0 p ξ p ´ 2014 MRT
    • The Wigner angle can be re-written as a sum of the contributions from the rotation and the boost: ( ˆ ˆ ) Ω ( p ) θ p ξ o o 0 p m 0 n p ´ + p º - θ ´ x p + m º - Here the angle of rotation are represented by the Euler angles q k = e i j k wi j and the boost parameter x i = x n i = w0 i. The finite Wigner transformation is given by: [ (ω), ] lim ω , ®¥ é Ω·J = ù ú úû ê êë ö ÷ ÷ø æ ç çè ö çè ÷ø L = Læ i N N p N W p W e ˆ 2014 MRT
    • The representation matrix Dms s ( j) [W(L,p)] can be constructed from the angular momentum generators J i explicitly, depending on the angular momentum of the particle. For example, spin-½ particles will be considered to have appropriate generators as given by J = ½ s according to the isomorphism between the proper Lorentz group and the SU(2) Ä SU(2) algebra. The Wigner transformation corresponding to an arbitrary Lorentz transformation L is given by: W(L, p) = eiΩ·J where W ( p)k = -½e i j k [( p i w j 0 - p j w i 0 ) /( p 0 + mo)] and J k = ½ e i j k M j k in the absence of rotation. This can also be written as: o Ω º - p´ ξ x n ´ p 0 o 0 ˆ ( ) p m p m p + = + Here the boost parameters are represented by x i = x n i = wi 0 . For an infinitesimal variation ˆ wmn , the transformation matrix is (e.g., spin-½ case described above): ù ú ú û Ω Ω Ω Ω Ω W p I i I i p é ê ê ë é ö ÷ ÷ø æ + · ù æ ç çè + ö - ö é 1 1 ˆ + ÷ ÷ ø ö æ æ + · n p Ω ç ç è ö ù ú ú û æ é 1 1 ê ê ë ö ÷ ÷ø · + ÷ ÷ø æ ç çè æ + = - ù ú ú û ê ê ë ö ÷ ÷ø æ ç çè ö - ÷ ÷ ø ç ç è ú ú û ê ê ë ÷ ÷ø æ ç çè - @ ÷ ÷ø ç çè ÷ ÷ ø ç ç è ö ç çè L = ´ ´ ´ 3 o ˆ 0 o 0 2 o ˆ 2 2 0 2 3 2 2 2 2 (1/ 2) 2( ) 1 3! 2( ) 2( ) 2! 1 3! 2 2! 2 2 2 sin 2 [ ( , )] cos p m p m i p m I n p n p σ Ω Ω Ω σ Ω σ Ω x ´ x ´ x ´ D 2014 MRT
    • The representation matrix becomes (i.e., spin-½ case): ü ì ù ù ö çè é ö çè 1 1 2 3 ö æ ù é é ö p æ  1 1 1 W p p p by using the well known Pauli spin vector s relation (s · a)(s · b) = (a · b) + i s · (a ´ b) and remembering that W = W /| W |. Thus, with the well known sinx =x - x 3/3! +x 5/5! -… and cosx = 1 – x 2/2! + x 4/4! -… identities: ù é ö æ ö æ 0 (1/ 2) Ω D W p p [ ( , )] 0 For the sake of convenience, I2´2 is assigned as the 2 ´ 2 identity matrix, and s as the ‘vector’ that is comprized of the Pauli 2 ´ 2 matrices [s1,s2,s3]. (Note that the Wigner angle | W | = W is dependent on both the rotation and boost parameters for an arbitrary Lorentz transformation L. In addition, notice that an additional normalization factor of Ö[p 0 /(L p)0] has been appended to the matrix so as to maintain the condition given D†(L, p) D(L, p) = p 0/(L p)0 I2´2). ú úû ê êë ÷ ÷ø ç çè · + ÷ ÷ø ç çè L L = 2 ( ˆ )sin 2 cos ( ) Ω σ Ω 1 i p ïþ ïý ïî ïí ú ú û ê ê ë + ÷ ÷ ø ç ç è + · - ú ú û ê ê ë + ÷ ÷ ø ç ç è - L = ú úû ê êë + ÷ø æ · + ÷ø + æ · L = L L = ´ ´ ·   0 2 2 0 2 0 2 2 0 2 0 0 (1/ 2) 3! 2 2 ( ˆ ) 2! 2 ( ) 2! 2 1! 2 ( ) e ( ) [ ( , )] Ω Ω Ω σ Ω Ω σ Ω σ Ω σ I i p I i i p p i D 2014 MRT ˆ
    • In the absence of rotation, the arbitrary Lorentz boost can be parametrized according to the matrix defined by LL above. We can construct the matrix by the associative property of the representation matrix: D [ W ( L , p )] = p /( L p ) x x ´ x in which: - (1/ 2) (1/ 2) 1 W L p = L L p L L p [ ( , )] [ ( ) ( )] - (1/ 2) 1 (1/ 2) (1/ 2) D D D L p L p [ ( )] [ ] [ ( )] D D º L L The representation matrix D(1/2)[W(L,p)], in the absence of rotation, is this given by: + æ ù úû é êë ö çè ÷ø ö çèæ W é æ W · + ÷ø p L = ù úû êë ö ÷ø - · æ ÷ø çè ö + · æ ÷ø çè ö çè + L + ´ ´ ´ 2 ( ˆ )sin 2 cos ( ) 2 [ ( ˆ )]sinh 2 ( ˆ )sinh 2 ( ) cosh ( )[( ) ] 0 2 2 0 o 2 2 0 2 2 o 0 o 0 0 0 (1/ 2) σ m p n σ p n I i p I p m I i p m p m x q x q x q ´ ö çè æ ÷ø ö çè æ cosh cosh 1 2 ÷ø sinh sinh ( ˆ ˆ ) 2 1 1 2 ( ˆ ˆ ) 2 sinh 2 sinh ˆ sin Ω ö 2 çè ö çè ö çèæ ÷ø æ + ÷ø ö çè cosh cosh 1 2 n p æ ÷ø ö çè · ÷ø sinh sinh ( ˆ ˆ ) 2 1 1 2 ( ˆ ˆ ) 2 sinh 2 sinh 2 cosh 2 cosh cos Ω ö 2 çè n p n p m n p + + · = ÷ø æ æ + + · = ÷ø æ x q x q x q x q & where coshq = p 0/mo and m = n ´ p represents the axis of rotation of the equivalent Wigner transformation (e.g., of the Dirac spinors in the spin-½ case). Here the parameters x and n ˆ are defined in the same manner as the general Lorentz boost as given by the LL matrix. ˆ ˆ 2014 MRT ˆ
    • 0 U Λ Y p m = k L - L Λ Y = L W L Y  U L p U L 1 p L p k m k p 0 j j U L p U p k mj ( ) ( , ) [ ( )] [ ( ) ( )] ( , ) [ ( )] [ ( , )] ( , ) 0 j ( ) 0 U p m k å Λ D = L L ¢ L p W p k m p Y Y ( ) ( , ) ( ) [ ( , )] ( , ) j j m ¢ m j j å ¢ D = L L ¢ W p L p Y k m [ ( , )] ( ) ( , ) j Our transformation thus becomes: s (Lp) Ä Lr n L 0 p j  m j ( j ) m m 0 0 0 m k p ¢ j ¢ j j U( ) ( p,m ) ( p) ( ) [ ( , )] ( , ) ¢ L L ¢ = L å- ¢ = m j j j j m m j j j W p p m p 0 0 Λ Y D Y where W a m (L, p) = (L -1)a n m (p) is the Wigner Rotation and E is the relativistic energy and since U(W ) | Y( k,mj)ñ = Sm¢j Dm¢j mj ( j) [W(L, p)]| Y( k,m¢j )ñ: 2014 MRT since N(L p )U[ L(L p )] | Y( k, m¢j )ñ =| Y(L p , m¢j )ñ Þ U[L(L p )] | Y( k, m¢j )ñ = N -1(L p )| Y(L p , m¢j)ñ where the normalization constant is N(L p ) = Ö[k0/(Lp)0]. Finally we have:
    • Mass Positive-Definite To define the mass positive-definite, we must recall a few definitions such as the unitary quantum transformation for coordinate and angular momenta: where P = {P 1, P 2, P + = - r + 1 rs + U e e r P M i i (1 ω, ) 1 ω ... 2 3}, J = {M   23, M 31, M 12} and K = {M rs 10, M 20, M 30} and their properties are P r † = P r, M rs † = M rs and M rs = - M s r. Now for the definition of invariance under a unitary transformation of a homogeneous Lorentz Transformation L and a uniform translation a: ( , ) (1 ω, ) 1( , ) ( , ) ( , ) U L a U + U L a Y p mj = Y p mj e - which also applies to both momenta: ( , ) 1( , ) ( , ) 1( , ) s ( mn m n n m ) n r m r m mn m U L a PmU- L a = L P & U L a M U- L a = L L M - a P + a P For the momentum, they commute with each other: [Pm ,Pn ] = 0 but for the angular momentum they do not: i [Pm ,Mrs ] =hmr Ps -hms Pr nor do the momenta together commute: i [Mmn ,Mrs ] =hnrMms -hmrMns -hsmMrn +hsn Mrm As comparison, we show the case when L = 1 then a = 0 and a rotation around the 3-axis: i P a i m J pˆ - - · =  =  m q U a and U R (1, ) e ( 3,0) e ( U(L,0) =U(L)) 2014 MRT
    • =åå W Þ ¢ =å ¢ W pˆ D[R( )] ,m ,m zˆ ,m pˆ D( ) ( ,q ,j ) ,m zˆ + - + ¢ - ¢ j m j m j m j m ( )!( )!( )!( )! j k j j j j m m k j m k j m k m m k æ W ö çè ö çè j q j q j [R( )] j ,mj j ,mj [R( )] j ,mj mj [R( )] m D D D( ) W = ¢ W ¢ = W ¢ j å å i ¢ ( ) - · - · ˆ W ¢ W = ¢ = ¢ j m [ R ( )] j , m e j , m j , m e j , m j m j j j i ik j J Ω J p D   ¢ ¢ j j j j m j j m m m j j j j j j j im j m m k m m k im k j j j j e 2 sin 2 e cos !( )!( )!( )! ( , , ) ( 1) 2 2 2 ( ) + - ¢ - ¢ - + ¢ ÷ø ÷ø æ W + - - ¢ - ¢ - + D W =å -               m m m m 2014 MRT Recall also that your mathematical toolbox consists of these fundamental relations: and with Finally: In this instance, we take (i.e., Weinberg’s QFT definition – he also uses Q for our W ): j j m m M i D( ) (1 ) d ( ) ¢ +W = ¢ + W [ ] ¢ j j m j m j ik ik mj mj 2 where Rik = d ik + Wik with W ik = -W k i infinitesimal and for ms running over j, j – 1, …, - j . Also: J M iM J i J j m j m Þ ± = ± = ± + ( ) ( ) ( )( 1) ± ¢ j j ¢ j j ¢ j j ± Þ = = d m m j m m j m m ) (3 j j j j j j j z m m j j m m j j m m j j ( ) 12 J M J m ¢ ¢ ¢ d    ( ) ( ) 1 ) (2 ( ) 1 ) (3 1 ) (2 3
    • Now, let us expand | Y(L p ; j,mj) ñ = |Y( L p)ñ| ,s ; j,mj ñ in the case of an electron (s = ± ½ ): U p j m p 0 ¢ L ¢ L L L = L  , ;  , [ ( , )] , ; , , ;  , ( ) ( ) ( ; , ) ( ) ( ) å å ¢ Y D Y j s m ms m m m W p s j m s m m p 0 ( ) m m m j s j j s j s p  Using J z| ,s ; j,mj ñ = mj| ,s ; j,mj ñ we found out that that | ,s ; j,mj ñ is an eigenfunction of J z and since J z commutes with J 2014 MRT s å å= ± ( 1 2 ) j m m = = = ± , ; , , ; , , ; , , ; , 1 2 , ; ,     C     s j m s m m s m m s j m s m m m = m + m j s s j m m j s s s     s m m s j m m m m s j j s j s j s º = + + - + - + - + , ; , , ; , ( , ) ( )!( )!( )!(2 1) C d s j j s k    j s ( 1)! m m s m s m j m j m - + - + - + - ( 1) (  )!(  )!( )!( )!( )!( )! å + - - + - - - - + + - - + ´ + + + s s j j   k s s j m m k s j k m k s m k j m k j s m k s !( )!( )!( )!( )!( )!            2 the eigenfunction of J z is simultaneously an eigenfunction of J 2 which means that | ,s ;m,ms ñ is not an eigenfunction of J 2. To get around this, we constructed a linear combination of | ,s ;m,ms ñ instead: such that | ,s ; j,mj ñ is simultaneously an eigenfunction of Jz and J 2. The quantities C j m ms = á, s ; m ,ms | , s ; j,mj ñ are numerical coefficients which are known as Clebsch-Gordan (CG) coefficients – and for our benefit j =  + s and mj = m + ms. A general formula for these coefficients is due to Wigner: with  = 0, 1, 2, …, s = ± ½, j = 0, 1/2, 1, 3/2, …, m = ,  - 1, …, -, ms = ± s & mj = j, j – 1, …, -j.
    • Boosts & Rotations Point P at the tip of the distance vector xm ={ct,r} is given in 4-dimensional Rectangular (Cartesian) Coordinates by the intersection of constant x, constant y and constant z planes and time t. The speed of light c is a constant of motion (the same everywhere!) ˆ ˆ ˆ ˆ ˆ ˆ Del Ñ = ¶/¶ x i + ¶/¶ y j +¶/¶ z k = [¶/¶ x, ¶/¶ y, ¶/¶ z ] is now: Ñ ¶ , , , , Ñ m t x y z t The gradient Ñy = ¶y /¶ x i + ¶y /¶ y j +¶y /¶ z k of a scalar function y : é ¶ ù é = ¶ ÷ ÷ø æ - ¶ - ¶ - ¶ Ñ f = ¶f f f f f f m , , , ,Ñ = ¶ úû é - ¶ - ¶ - ¶ The divergence Ñ · A = ¶Ax /¶ x +¶Ay /¶ y +¶Az /¶ z gives: a at x y a z a t m m a a ù Ñ = ¶ Ñ + · a = ¶ ¶ + ¶ ¶ ¶ + ¶ + ¶ ¶ ¶ t z y x t úû êë ù êë ¶ ¶ ¶ ¶ t x y z t x r O The Laplacian Ñ · Ñ º Ñ2 = ¶ 2/¶ x 2 + ¶ 2/¶ y 2 +¶ 2/¶ z 2 leads to the D’Alembertian: v P 2014 MRT A am = [at ,ax ,ay ,az ] = [at ,a] y f z If the scalar product is A · B = A x Bx +Ay By +Az Bz, then: am bm = atbt - axbx - ayby - azbz = atbt - a · b úû êë - ¶ ö ç çè ¶ ¶ ¶ ¶ ˆ ˆ ˆ t = c t Electrodynamics provides the differential equations for the potentials:  2f = r /eo and  2A = j /eo and the continuity equation: ¶f /¶ t + Ñ · A = 0. In this new in four-dimensional notation: Am = {f, A} we get  2Am = jm /eo and Ñm Am = 0. y z x y z x x0 = c t v The transformation laws which give f and A in a moving system in terms of f and A in a stationary system. Since Am = {f, A} is a four-vector, the equation must just look like t = g ( t - | v |z /c2 ) and z = g ( z - | v |t ) with g = (1 - | v |2/c 2)-1/2 except that t is replaced by f , and r is replaced by A. Thus: 2 A c A z A A A A A 2 2 - 1 ( c ) 1 ( c) y y x x z z v v v v - - = = = - = f f f , , , If the vector A = A x i + Ay j +A z k = [ A x , Ay , Az ], then we have the definition of the 4-vector am : 2 2 2 Ñ Ñ = ¶ Ñ 2 2 m m t x y z 2 t So = (1/c 2) ¶ t - Ñ 2 =h¶ mn 2 2 2 2 2 2 - = = ¶ ¶ - ¶ ¶ - ¶ ¶ - ¶ ¶ ¶ m ¶n = ¶ m ¶m with ¶m = ¶/¶m .
    • The Lorentz transformation transforms frame xn (say x, y, z, t ) into frame x m (say x , y , z , t ): é g - g 0 0 v 0 1 0 0 0 0 1 0 n Lm = δ + ω and a = ε n m m n m m m é ù é ù ù é t z ( ) x y t x y t x y = L L = (δ + ω )(δ + ω s ) = sr + (ωsr + ωrs ) +O[Terms in ω2 ] r hrs hmn h h U ( L , a ) = U (1 + ω,ε) = 1 + 1 i ω rs i rs M - ε r P 2   n s r r s mn r m m n n ù ú ú ú ú ú û ê ê ê ê ê ë - - = ú ú ú ú û ê ê ê ê ë ú ú ú ú ú û ê ê ê ê ê ë - + - = = L = ú ú ú ú û ê ê ê ê ë ( ) 0 0 1 ( 1) ˆ ˆ z t c2 z c c x x z v v v vv g g g g n mn m assuming the inertial frame is going in the x 3 (+z) direction. If both w m n and e m are taken to be an infinitesimal Lorentz transformation and an infinitesimal translation, respectively: This allows us to study the transformation: The linear unitary operator U = 1 + i e T was constructed: where the generators G of translation ( P m ) and rotation ( M mn ) are given by P 1, P 2, and P 3 ( the components of the momentum operator ), M 23, M 31, and M 12 ( the components of the angular momentum vector ), and P 0 is the energy operator. 2014 MRT x2 = y x3 = z x1 = x x2 = y x3 = z x1 = x x0 = c t b =| v |/c
    • For a boost in an arbitrary direction with velocity v, it is convenient to decompose the spatial vector r into components perpendicular and parallel to the velocity v: . Then only the component in the direction of is ‘warped’ by the gamma factor: r = r v r = r r = r - r = r - ( r · v ) v r = ( r · v ) v 2 o g -g and or and || || ^ ^ ^ || || 2 o 2 2 2 2 2 2 o 2 2 1 1 1 1 1 1 1 1 1 1 b g g g b g - = ö çè + ÷ø æ = - Þ = - = ÷ ÷ø ö æ - ç çè = - · = m m c c c m c p v v v p and 2 2 2 c t t v v t r · v = - || g g r = r|| + r^ where the gamma factor g (which is a function of the rapidity b = | v | /c ) is: ö ÷ø Now r can also be written as: r r + g r v -g v r g r v and g r v = é - · 2 ^ || 2 ( 1)( ) ( ) = + - = æ - · úû çè ù êë c t t t t v and the relativistic momentum p can be written as: p E = - =g p p v p o 2 o 2 2 2 o 2 c m c ( ) ( ) 0 ( ) m c m c c - = = and 2014 MRT
    • These equations can be expressed in matrix form as: n m n v g g é = = x ct x ù é ù é - ù é = ù 1 m m ù é = 1 x n = L g g x x ct x x x ct c c ct x x ú ú ú ú ú û ê ê ê ê ê ë = L ú ú ú ú û ê ê ê ê ë ú ú ú ú û ê ê ê ê ë - + - = ú ú ú ú û ê ê ê ê ë ú ú ú ú ú û ê ê ê ê ê ë 2 3 0 T T 2 3 0 v 1 ( 1)vˆvˆ r r where 1 is the identity matrix, v is velocity written as a ‘column vector’, vT is its transpose (i.e., the equivalent of a ‘row vector’) and v is its unit vector. ù ú ú ú ú ú ú ú ú ú û More generally for a boost in any arbitrary direction [ b x = vx /c, b y = vy /c, b z = vz /c]: é ê ê ê ê ê ê ê ê ê ë g - b g - b g - b g b b ˆ ˆ 2 ˆ g b b b g g b - + - - - 1 ( 1) ( 1) ( 1) 2 ˆ b b ˆ ˆ b - - + - - ( 1) 1 ( 1) ( 1) b b ˆ ˆ b g g b b ˆ ˆ b b g b - - - + - = ù ú ú ú ú û é ê ê ê ê ë v g - g - + - m L = ˆ ˆ ˆ ˆ 2 ˆ 2 2 2 ˆ 2 2 2 ˆ 2 2 2 ˆ ˆ ˆ ˆ T T ( 1) ( 1) 1 ( 1) ( 1) ˆ ˆ ( ) b b g b b g b g b b g g b b g b g g n z x z y z z y x y y z y x x y x z x x y z v 1 vv v c c 2014 MRT ˆ ˆ ˆ ˆ ˆ ˆ
    • ù p m L p p p m L p p p m L p p p m L p p ( ) ( ) 0 ( ) 0 ( ) 0 ù ú ú ú ú ú ú û é 0 0 1 1 ( ) 0 ( ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ p m L p p = p L p p L p p L p 0 2 2 p p p p p p ( ) 0 ( ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ p m L p p = p L p p L p p L p 0 3 3 1 1 1 p p p p p p n Now, since we are using a momentum p represention we generally have for a boost in any arbitrary direction k é ê ê ê ê ê ê ë ˆ 1 ˆ 1 ˆ 1 p p p g g g g - - - ˆ 1 1 ( 1) ˆ ( 1) ˆ ˆ ( 1) ˆ ˆ p g - + g - p g - p p g - p p ˆ 1 ( 1) ˆ ˆ 1 ( 1) ˆ ( 1) ˆ ˆ p g - g - p p + g - p g - p p - - - + - = ù ú ú ú û é ê ê ê ë 2 T - g g - + - = 2 3 1 3 2 3 2 3 2 3 22 2 1 2 2 1 2 1 3 2 1 2 1 2 3 2 2 2 1 1ˆ p i 2 T ˆ 1 ( 1) ˆ ˆ ( 1) ˆ ˆ 1 ( 1) ˆ 1ˆ ( 1) ˆ ˆ ( ) p p p p p p p p p g g g g g d g n m i i j i j L p m = [ k 0 = moc, ki = 0 ] with p n = [ p0 = E /c , pi ]: where 1 is the unit/identity matrix, p is the momentum written as a column vector, pT is its transpose (another row vector) and p is the momentum direction unit vector and since g =1/Ö(1–b 2) = 1/Ö(1 – | v | 2/c2), we have g b =Ö(g 2 – 1) and p =g mo v with | p | =g mo | v |. With these, we also have a unit vector p parallel to the direction of v : 2014 MRT ˆ ˆ ˆ c m c i o 2 4 o 2 2 p p p ˆ p ˆ p and g m i i i + = Þ º º p p The result is: ú ú ú ú ú ú û ê ê ê ê ê ê ë = = = = = 3 3 3 3 3 2 2 3 3 3 1 1 3 0 3 o 3 3 2 3 2 2 2 2 2 2 2 1 1 2 0 2 o 2 3 1 3 2 1 2 1 1 1 1 1 0 o 1 3 3 0 o 2 0 2 0 o 1 0 1 o 0 0 0 0 o 0 ( ) 0 ( ) ˆ ˆ ( ) ˆ ˆ ( ) ˆ ˆ ( ) p p p p p p p m L p p p L p p L p p L p L m p
    • 2014 MRT Since the maximum momentum can be mo c and the unit time vector labelled ê t (which allows us to set the standard momentum as p 0 = E /c): [ ] [ˆ ] ì ï ï í ï ï î ù ú ú ú û é Ä ê ê ê ë m c = Ä = x y z vp=c t k k ˆ e ˆ e e o e ˆ kx ky kz [ ] m ˆ ˆ and ki Ä êj is the input standard momentum direction p =  k. This gives us the column kets: k ù ú ú ú ú û é = ê ê ê ê ë ù ú ú ú ú ú k 0 k and 0 k > = û é = ê ê ê ê ê ë 0 k m m 0 0 0 0 mo 0 mo
    • Given the geometry involved, a quick review is required. Point P at the tip of the distance vector r is determined in Spherical Coordinates by the intersection of a constant q (i.e., a cone), constant r (i.e., a sphere) and constant j (i.e., a half-plane) surfaces. y Constant j plane x z qˆ r P rˆ jˆ j q P Constant q cone Constant r sphere x z q r · rdqd r dj rsinq dj O y j dq d The gradient of a scalar function f = f (r,q ,j ) The scalar product of a vector function F = F(r,q ,j ) q ¶ f f r θ = ¶ + ¶ f q q ¶j ( sin ) 1 q q j + ¶ ˆ 1 ˆ 1 q q j ¶ + ¶ 1 ( 2 ) 1 2 Ñ F + ¶ r F ¶ · = ¶ F r F r r r r sin sin dV = r2 sinq drdq dj d = dr rˆ + rdq θˆ + r sinq dj jˆ 2 2 2 2 2 2 2 2 2 2 d dr rd r d dr r d r d = + + = + + dr r d d d q q j q q j ( ) ( sin ) sin 2 2 2 2 2 2 q q sin where = + W W = +  · y x z r r = r(r,q ,j ) = rrˆ d is an infinitesimal differential increment of length : Ñ jˆ sin ¶ ¶ r f r r The vector product of a vector function F = F(r,q ,j ) ö - ¶ j q F r q j F rF ( ) ˆ j Ñ´ 1 æ æ + 1 1 ¶ sin ö æ - ¶ rF 1 ( ) ˆ ˆ ( sin ) sin ÷ ÷ø ç çè ¶ ¶ + ¶ ö + ÷ ÷ø ç çè ¶ ¶ - ¶ + ÷ ÷ø ç çè ¶ ¶ ¶ = q q j q q q j r r F r r r r F F r θ Produces another vector perpendicular to the plane formed by Ñ and F with unit vectors given. The Laplacian of a scalar function f = f (r,q ,j ) + ¶ ÷ø r ¶ f Ñ = ¶ 1 1 æ + 1 ¶ Ñ2F = Ñ(Ñ·F) -Ñ´(Ñ´ F) ˆ Oj i kˆ dV is an infinitesimal differential increment of volume : 2014 MRT 2 2 2 2 2 2 2 2 sin sin sin q j q q q q ¶ ö + ÷ø æ çè ¶ ¶ ¶ ö çè ¶ ¶ f r f r r r r f Laplacian of a vector function F = F (r,q ,j ) (identity) We can make the following geometric objects into physical realities if we substitute the speed v for f = f (r,q,j ) and the momentum p for the vector F = F (r,q,j ). Manifold ˆ p
    • To calculate this rotation, we need to choose a ‘standard boost’ L(p) which carries the four-momentum from k m = [moc ,0,0,0] to p m. From above, this is conveniently chosen as: J2 j , m = [ j × ( j + 1)]2 2 j , m Þ J = j × ( j + 1)   j j J j m = m j m Þ J = m J L p p 0 = = = q g p p p ( ) = ( ) = ˆ sinh q = = g - 1 ˆ ( ) ˆ ˆ (cosh 1) ( 1) ˆ ˆ j i j i j i j i j i i i i i i L p m c L p L p m c p p p p ( ) cosh 2 o 0 0 o 0 0 = d + q - = d + g - 0/moˆ c. where q = p/moc. Here pi is the same unit vector pi /| p |, and g = Ö(| p | 2 + mo 2c 2) /mo c = p 2014 MRT , ,   z j j j z j | YO ( p, mj ) ñ mo Jz = mj  p Suppose that observer O sees a particle (spin-mj and mass mo ¹ 0) with momentum p in the y-direction and spin z-component mj . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the relativistic particle’s state? Note that for a massive particle (i.e., of mass mo and four-mo-mentum pm = [p 0c, p]) the standard boost L( p) may be written as: ù ú ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ê ë p p 1 3 p p 2 3 ö ÷ ÷ø æ 0 p p p p - ÷ ÷ø p p p p p 1 1 1 1 0 1 2 2 2 p p - ÷ ÷ø p p p 1 1 1 1 æ ö ö æ æ 0 2 1 - ÷ ÷ø 1 æ + - 0 2 - + ÷ ÷ø 0 3 p p p - + ÷ ÷ø ç çè ö æ æ 0 2 1 p p p - ÷ ÷ø ç çè ö æ æ ç çè 0 - ÷ ÷ø ç çè ö ç çè ö ç çè - ÷ ÷ø ç çè ö ç çè ö ç çè = 2 3 2 2 o 3 2 2 2 o 3 1 2 0 0 p p 2 o 0 p p 1 2 3 p p o 2 2 o 2 2 o 2 2 o o 2 2 o 2 2 o 2 2 o o o o o 2 o 1 1 1 1 ( ) p p m c p m c p m c m c p m c p m c p m c m c p m c p m c p m c m c m c m c m c m c L p
    • Now, suppose we take pˆ to have polar (q ) and azimutal (j ) angles: Then we can take R(p) as a rotation by angle q around the 2-axis, which takes col[0, 0, 1] into col[sinq , 0, cosq ], followed by a rotation by the angle j around the 3-axis: 1 é i J i J - Rp R p = p = p R R R   = Þ úû | Y( p, mj)ñ = | p ñ ´ | mˆ j ñ J pˆ = [ sinq cosj , sinq sinj , cosq ] where 0 £ q £ p and 0 £ j £ 2 p. j q 1 0 j q 3 2 [ ( ˆ )] e e 0 ( ) ( ) ( ˆ ) ( ˆ ) ( ˆ ) 3 2 U R R R ù êë 2014 MRT In this case, for a particle of mass mo = 0 and helicity-s, we have: U B p m B p p ( ) ( , ) ( ) ( ) ( ) ( , ) R j j j p p L p L p ( ) ( ) [ ( ) ( ) ] ( , ) å = B D R = ¢ m j p p W p p ( ) ( , ) ( ) ( ) [ ( , )] ( , ) å U B p m B D = ¢ - B ( ˆ ) since R p = R (p) p and B(| p ˆ |) is a Lorentz boost. We finally get: j U(B) ( p,m j ) = (B ) B ( ) j , m ¢ j [ R ( )] j , m j ( p , m ¢ ) 0 j with the little-group element W(L, p) (the Wigner rotation) given bWy: ( R, p) = R-1( R p) RR( p) = R( Rpˆ )B-1(pˆ ) R-1( Rpˆ ) RR(pˆ ) B( p )R-1(pˆ ) ˆ First Rotation ù é 0 0 q ù é 0 0 | k ñ= col[0,0,1] » U[R( p)]| ˆ Y( p,m)ñ jU[L(R p)R(p)B(| p |)]| Y(k,mjˆ )ñ ù é 0 0 Suppose that observer O sees a massless particle (helicity-s and mass m= 0) with unit o momentum p ˆ in the z-direction with basis ket | k ñ = col[0,0,1]. R(p) rotates | p ñ by angle q around 2the y-axis into col[sinq , 0, cosq ]. Then R(p) 3rotates | p ñ by the angle j around the z-axis. ú ú ú û ê ê ê ë 1 1. B( p) col[sinq , 0, cosq ] y x j z Second Rotation | pˆ ñ ú ú ú û ê ê ê ë 1 2. R2 (q )B( p) 3. ú ú ú û ê ê ê ë = 1 B(R p) R3(j )R2 (q )B( p) pˆ å ¢ - - - ¢ + ¢ =- ¢ + ¢ =- - ¢ = ¢ j j j j j j j j j m j j m m j m j j j j m m j j m m B R B R R B R k m p B B m p B p m p B U p m p ( ) ( ) [ ( ˆ ) ( ) ( ˆ ) ( ˆ ) ( ) 1 ( ˆ )] ( , ) ( ˆ ) ( ) 1 1 0 0 ( ) 0 0 ( ) 1 0 0 0 0 1 Y Y Y Y Y Y Lp  Lp p p p p p R p p p R R R D R R å+ ¢ =- m j j p 0 Y p p q Y ˆ ˆ ˆ ˆ E/cs
    • Then, for a three-dimensional rotation R , we have the following Wigner rotation about the 3-axis with L(p) = R(pˆ) B(| p |) R-1(pˆ) and L( B p) = R( R pˆ) B-1(| p |) R-1( R pˆ): - 2014 MRT ù ú ú ú ú ú û As just shown, it is very important to note that when Lm é ê ê ê ê ê ë g g 0 0 1 0 1 0 0 0 0 1 0 - - = g g 1 0 0 ( ) 2 2 B p n is an arbitrary three-dimensional rotation R(3) , the Wigner rotation W(L,p) is the same as R(3) for all p. To see this, note that the boost L m n ( p ) above may be expressed as L(p) = R(p) B(| p |) R-1(p), where R(p) is a rotation that takes the 3-axis into the direction of p, and: ˆ ˆ ˆ ( ˆ ) g g 0 0 1 0 1 0 0 0 0 1 0 1 0 0 ( ˆ ) 1 - R R R W L L p = p p ( , ) ( ) ( ) - - - 1 1 1 [ ( ˆ ) ( ) ( ˆ )] [ ( ˆ ) ( ) ( ˆ )] R R R p B p p p B p p 1 0 0 0 W W 0 cos sin 0 0 sin cos 0 0 0 0 1 R R R R ( ˆ ) g g 0 0 1 0 1 0 0 R R 0 0 1 0 1 0 0 ( ˆ ) 1 2 2 - 1 2 2 1 ù é ù é ù é p p p p - ú ú ú ú ú û ê ê ê ê ê ë - - ú ú ú ú û ê ê ê ê ë - W W ú ú ú ú ú û ê ê ê ê ê ë - - = = R R R R g g g g
    • Then, for an arbitrary rotation R we see that R-1( R pˆ) R R(pˆ) since it is given by: j j a a é - b b g g ù ú ú ú é = - ê ê ê ù ú ú ú é - ê ê ê ë ù ú ú ú û ê ê ê ù ú ú ú é ê ê ê - Rp R p = - 1 j j j ( ) ( ˆ ) ( ˆ ) 3 R R = R 1 ( ˆ ) p - R L p L  ( )  W p - p - p p p - p ( , ) ( ˆ ) 1( ) 1( ˆ ) ( ˆ ) ( ) ( ˆ ) R = R R R p R B R R B R ˆ ˆ 1 Even in this general case, the rotation R-1( R p) R R(p) takes the 3-axis into the direction p, and then into the direction R p, and then back to the 3-axis, so it must be just a rotation by some azimutal angle j around the 3-axis: 2014 MRT cos sin 0 sin cos 0 0 0 1 cos sin 0 a a sin cos 0 0 0 1 cos 0 sin 0 1 0 b b sin 0 cos cos sin 0 g g sin cos 0 0 0 1 û ë û ë û ë j ˆ ˆ 1 é 1 0 ù i J - Rp R p = p = p R R R Þ =  úû j 3 [ ( ˆ )] e R 0 ( ) ( ˆ ) ( ˆ ) ( ˆ ) 3 U R êë The state of a moving massive particle can be shown to have exactly the same trans-formation under rotation as in non-relativistic quantum mechanics (i.e., W(L= R, p) = R ). and: In conclusion, the Wigner’s formula ( J3 µj ) for the d-matrix elements d m¢j mj ( j ) (j ) still form a Unitary Group U(j ) and most of all, the whole of the dictionary we have learnt already for the Spherical Harmonics, the Clebsch-Gordan coefficients and the Winger-Eckart Theorem apply in the same exact way in relativistic field theory as it does in quantum mechanics – as it does for any problem in which D( j ) [W (L, p)] for all spin- j ! m¢j mj
    • Quantum Fields – Problem 1. Now, for the W-boson problem (see Figure). This first problem requires us to take pˆ as: p0 = E c and p = p Þ pm = E c p ˆ [ 0, ˆ 2 , 0] [ , 0, 2 , 0] and the unit momentum is, by definition for massive particles: We can take [0, 0, 0]T into p2 using L(p) as a boost along p2 followed by a boost L(b3 ) with rapidity b3 along the 3-axis, which takes [0, p2, 0]T into something like [0, p2 , - p3]T: ù ú ú ú é 0 p p = Lp û ê ê ê ë n n = Þ L = L Þ L 0 0 p L ( p)km ( p)m [ L( )k]m m L( ) 2014 MRT For a particle of rest mass MW > 0 and spin s = 1 we have the one-particle state which is mass positive-definite for the W-boson: 0 U p p L = + = L W L ¢ ( )] ( , 1) ( ) ( ) ( , ) å D ¢+ Y Y s s s 0 1 s [ 0 ¢=- + 1,0, 1 (1) p p = L W L - p ( ) ( ) ( , 1) D - + p Y D + W L p Y ( ) ( ,0) ( ) ( , 1) ] (1) 0 1 1 (1) 0 + 1 (1) 1 1 p D + W L + + + p Y with the Wigner Rotation W(L, p), which is given by the following group multiplication: = [0, p2 , - p| Lp ñ ˆ ˆ 3 ]T MW | k ñ = [MW c,0, 0, 0]T L L(pˆ ) ù ú ú ú ú û ˆ y é W ê ê ê ê ë 0 0 0 ( ) M c L p z W p2 ˆ ˆ W(L, p) = L-1(L p)ΛL( p) k [ M c, 0, 0, 0]T = W m Suppose that observer O sees a W boson (spin-1 and mass MW ¹ 0) with momentum p2 in the y-direction and spin z-component Jz = s  = . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the | Y( p,s )ñ state? ˆ ˆ ˆ ˆ
    • W3 2 1 L p = L3 2 Lp L L p - ( , ) ( ) 3 2 ( ) as anticlockwize around the positive 1-axis. We will need this later. Then we have: é é é ù 2 p 0 0 p W W 0 1 0 0 p c 2 2 g b g 0 0 3 3 3 0 1 0 0 0 0 1 0 b g g W W W p p p g g b g 0 ( ) 0 1 W W 0 1 0 0 p c 2 2 W W W p p where the gamma factor is g3 = Ö(1 - b3 2 2 ù ù 2 2 0 p 2) (which is a function of the rapidity b 3 = | v3 | /c along the 3-axis). 2014 MRT First we calculate the Wigner Rotation: ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ë - - + + - = ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ë + + ú ú ú ú ú û ê ê ê ê ê ë - - L = 3 2 2 3 3 0 3 3 2 0 2 3 3 2 2 3 0 3 2 0 2 2 0 3 3 3 3 2 0 0 ( ) 0 1 0 0 0 1 0 0 ( ) b g b g g M c M c M c p M c M c M c M c M c p M c M c M c M c L p W W 1 ´ =- ´
    • ù ú ú ú ú ú ú ú û 2014 MRT p M c é ê ê ê ê ê ê ê ë M c 2 2 2 p + M c 0 × × 2 p g 3 2 p - + g = ù ú ú ú ú ú ú ú ú ú ú û Then we have with the definition of the basis momentum k g g b g 0 p p W W 0 1 0 0 p c ( ) é ê ê ê ê ê ê ê ê ê ê ë 0 0 0 2 2 2 4 2 p c M c p c + M c - + = W W W ù ú ú ú ú ú ú ú ú ú ú û 2 2 é ê ê ê ê ê ê ê ê ê ê ë 2 2 p 2 2 2 4 c M c c + M c 0 1 0 - + = p ù ú ú ú ú ú ú ú ú ú û é é ê ê ê ê ê ê ê ê ê ë - = ù ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ë - × = ù ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ë ù ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ë p - - + + - L = 2 2 2 3 3 2 2 2 2 4 2 0 g 3 3 2 3 2 2 2 4 g 3 3 2 3 0 p 2 g 3 3 0 3 2 0 3 3 2 2 0 3 3 2 2 3 3 0 3 3 2 0 2 3 3 2 2 3 0 3 3 2 0 0 0 0 ( ) p M c c p c c p c p c p c M c M c p M c M c M c M c M c M c p M c M c M c M c L p k W W W W W W W W W W W W W W W b g b g b g b g b g b g b g g m p p m and since p = p2: ù ú ú ú û é ê ê ê ë 2 p - + = ù ú ú ú û é ê ê ê ë L L p p L L = + 2 2 2 3 3 2 1 2 3 2 2 2 0 3 2 0 p M c p p p M c W W b g g and and thus:
    • k 2 2 2 2 p + M c v v Finally we get the eigenvector for the boosted momentum Lp : 2 2 2 + p i j k j k j ˆ 1 ˆ ˆ 1 0ˆ ˆ ˆ ˆ 2 2 2 3 3 2 2 2 2 2 2 3 3 2 ö çè ÷ø - æ = - - L = + - + = - c p c p M c p p M c p W W W b b b g k p M c v v k j 2 2 2 2 p + M c v v ö æ p p j k j ˆ 1 1 ˆ ˆ 1 ˆ ˆ ˆ 2 2 2 2 2 2 3 2 M c ö c çè c p ö c çè p p p c W W W ÷ø - æ + ÷ ÷ø ç çè = - ÷ø - æ L º L = - = - so that: x L3(v) z col[M | k ñ W c,0, 0, 0] = L2( p) W-1(W) 2014 MRT y We see that the effect of a Lorentz boost L2( p) on the initial unit momentum p2 followed by a Lorentz transformation L3(v) which has the effect of creating (as seen from O) a momentum vector p3 in the negative 3-axis direction: Jz =  p2 p3 Suppose that observer O sees a W boson (spin-1 and mass MW ¹ 0) with momentum p2 in the y-direction and spin z-component Jz = s  = . A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the | Y( p,s )ñ state? p M c ö M c æ ö c çè c p W W 2 2 2 3 1 1 ÷ø - æ + ÷ ÷ø ç çè = - v v The effect of the Wigner rotation W (L, p) will produce a rotation (negative in this case) around the 1-axis which results in an angle W between the 2-axis and the boost vector Lp º p2 j – p3 k.* J MW ˆ ˆ ˆ ˆ *It is sad to say but in many texts authors suggest that to find the answer to this problem you only have to calculate the Wigner rotation W (L, p) = L–1(Lp)LL(p), then extract the angle W and calculate D ( j ) m¢j mj = exp(i W J / ). The derivation that follows will be a little more practical. ˆ
    • Using the group properties described earlier, with the vector product of the 3- and 2-axis: ù 3 2 1 W 1 W = L L W - L p L p L- p L p W ´ 1 ´ ´ 3 2 L × = 3 2 L 3 2 L L W - and since L(Lp)L-1(Lp) = 1 we can then go through the matrix multiplication with the order: 1 0 0 0 0 1 0 0 W - W 0 0 cos sin 2014 MRT W3 L p = L3 - 1 2 1 2 Lp L L p ( , ) ( ) 3 2 ( ) ´ =- ´ ù ú ú ú ú ú ú ú ú ú û g b g 0 0 3 3 3 0 1 0 0 0 0 1 0 é ê ê ê ê ê ê ê ê ê ë 2 p 0 0 p W W 0 1 0 0 0 2 2 p c 2 2 ( ) 0 1 p W W W 2 2 p p 0 cos sin p W W W 0 1 0 0 é ù é ù 2 2 2 2 2 p c W W ù W úû êë + + - W úû é p c 2 2 êë + + W W ù ú ú ú ú ú û é ê ê ê ê ê ë - g - b g = ú ú ú ú û ê ê ê ê ë W W ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ë + ù + ú ú ú ú ú û é ê ê ê ê ê ë - - L = L - L = - p 0 0 sin cos sin ( ) cos 1 ( ) 0 1 0 0 3 3 3 0 1 0 0 0 0 1 0 0 0 0 0 sin cos 0 0 0 1 0 0 ( ) ( ) ( , ) 2 0 2 2 0 2 2 0 3 3 3 2 0 2 2 0 3 3 3 1 3 2 3 2 1 M c p M c M c p M c M c M c M c M c M c p M c M c M c M c L p L p W p W W W W W b g g b g g ´ ( ) ( ) ( ) ( ) 1( ) 3 2 1 1 3 2 1 - - - W´ =-W-  L ´ p L p W 1 we get: ( ) [ ( ) ( )] ( ) 1( ) 3 2 1 1 - then:
    • 2014 MRT We finally obtain for the boost matrix L3 ´ 2 (Lp) as a function of the deflection angle W: ù ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ë p p p g g b g b g W+ W W- W ù W úû 0 cos sin sin cos W W W 0 1 0 0 é ù úû cos W - 1 + êë 2 2 2 p c + é p c 2 2 0 1 êë 2 2 + + sin ( ) ( ) W W W W W p p p p - W+ W - W+ W L = 0 cos sin sin cos ( ) 3 2 3 3 3 2 2 3 3 0 3 3 2 0 2 2 0 2 3 3 2 3 3 2 2 3 0 3 3 2 b g b g g b g g M c M c M c M c p M c M c p M c M c M c M c M c L p W W W ´ Since L3 ´ 2 (Lp) is symmetric we can ‘extract’ the [L3 ´ 2 (Lp)]3 2 = [L3 ´ 2 (Lp)]2 3 components: p b g g W W W p g b g W = W ù ïþ ïý ü ïî ïí ì ù úû êë é 2 2 2 p c 2 2 2 p c + é - + + W + W = W úû êë + - + sin cos ( ) 1 sin cos sin ( ) 1 2 2 0 2 3 3 3 3 2 2 0 2 3 3 M c M c p M c M c M c p M c W W W ù úû é êë p M c W 2 2 2 p c + + + W = - ( ) 1 tan 2 0 2 3 2 3 3 M c p M c W W g b g and at first we obtain: which we can reduce further with the goal of having only to calculate the ratio p2 /MW c.
    • 2014 MRT Going through patiently with the algebra we obtain: ù úû p c M c M c p c W W b é + ÷ø ö + + - + ÷ø ö êë çè æ çè æ + + ÷ø ö çè æ + + W = - 2 2 2 2 2 2 4 2 2 2 2 3 b 2 2 2 4 2 2 2 2 2 2 2 4 2 3 2 1 tan M c p c M c M c M c p c M c M c p c W W W W W W Reducing ruther and deviding both the numerator and denominator by MW ù ú ú ú û p 1 1 W W é ê ê ê ë ö ÷ ÷ø p æ + + ÷ ÷ø ç çè ö æ ç çè ö æ ö æ b + - + + ÷ ÷ø æ ç çè + ö ÷ ÷ ÷ ø ç ç ç è + ÷ ÷ø ç çè + W = - 2 2 2 p 2 2 3 2 2 2 2 2 3 p b 1 1 1 1 1 tan p M c M c M c M c M c W W W ö ÷ ÷ ÷ ÷ ÷ ÷ ÷ ø æ ç ç ç ç ç ç ç è ü ï ï ï ý ï ï ï þ ì ï ï ï æ - - ÷ ÷ø v v í ï ï ï î + ù ú ú ú ú ú ú û é ê ê ê ê ê ê ë ö ö + ÷ ÷ø æ p M c æ ç çè + ÷ ÷ø ç çè ö + ÷ ÷ø ç çè ö æ ç çè p W W @ - ... p 1 1 arctan 1 1 1 2 2 2 2 2 2 M c M c c c W W 2c4 we get: By dividing by 1+Ö[( p2/ MW c)2 + 1] and using the approximation (1 + x)-1 = 1 – x +… where x =Ö[1 – (| v |/c)2]´[1 + …/(1 + Ö…)] we finally get the Wigner Angle W which is given by: So with only the ratio p2 /MW c and | v | we can calculate this relativistic deflection angle.
    • Finally, by using the following trigonometric identities relating tan W to cos W and/or sin W: = ± sin W - W W = ± - W tan 1 cos W 2 2 1 sin cos = 1 + tan W W W = + W W 1 + W = cos 1 + W W = tan sin 1 tan tan 1 tan 1 tan 2 2 2 2 2 and we find that: such that the boost matrix becomes: ù ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ê ê ê ê ê ê ê ê ë 1 + W 1 1 tan - + W b + 1 tan M c 1 tan p 1 0 1 b p 0 1 0 0 ö ö + W p M c W p + W 1 tan W p M c W p M c b p - - + W + W 1 1 tan W ù - ö ö + é æ æ b - + W ö ö + ÷ ÷ø æ p æ ç çè b - - W ù ú ú ú ú ú ú û é ê ê ê ê ê ê ë + ÷ ÷ø æ æ ç çè + ÷ ÷ø ç çè - + + W ú ú ú ú ú ú û ê ê ê ê ê ê ë + ÷ ÷ø ç çè + ÷ ÷ø ç çè + - + W - W W - + - + W + ÷ ÷ø ç çè - L = 2 2 3 2 2 2 3 3 2 2 3 2 2 2 3 3 2 2 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 2 2 2 2 3 3 2 2 2 3 2 2 2 3 3 2 1 tan 1 1 tan tan 1 1 1 1 1 tan 1 0 1 tan 1 1 1 1 tan 1 1 0 1 1 tan tan 1 tan 1 1 tan 1 1 1 ( ) b b b b b b b b b M c p p M c M c M c M c M c p M c L p W W W W W W W W W ´ Sparing you the beast of a representation this will give… but now you have a few ways of calculating it: either by using the relation W = arctan{– (| v |/c)( p2/ MW c)[1–…]} or the one above by using tan W and its square, with p2 /MW c and b3 = | v |/c given beforehand. 2014 MRT
    • We then obtain for the relativistic one-particle state of spin-1 as seen from observer O: 0 3 3 0 2 2 W p p p p p p L L ¢ = L + L + L + L L L ¢ L L = + = å å D D Y Y Y s s s ( ) ( , 1) [ ( , )] ( , ) [ ( , )] ( , ) + p p = g + + + - b g W L s ¢ ( )( ) (0)(0) (0)( 2 ) ( 3 3 )(0) ( ) ( , ) ¢=- ¢+ p p Y D D D D [ ( ) ( , 1) ( ) ( ,0) ( ) ( , 1) ] as well as the final representation (with cos W and sin W or W = arctan (…) as given above): ù ú ú é ö æ v U Y p Y Y Y 1 [(1 cos ) ( , 1) 2 sin ( ,0) (1 cos ) ( , 1) ] 2 s = +1 s = 0 s = -1 s ¢ = +1 s ¢ = 0 s ¢ = -1 (1) 1 1 (1) 0 1 (1) 3 1 1 1 1 (1) 0 1 0 3 1 1 (1) 0 1 0 1 1 0 0 0 1 1 (1) 0 1 0 2 = W L - + W L + W L + - + + + + + ¢=- ¢+ + ¢=- ¢+ å p p p W p p p p p U p Y Y Y g s s s s s s m m 1 + W (1 cos ) 2 s s ¢ 0 2 0 sinW 0 2 1 - W 0 (1 cos ) 2 0 0 The D( j = 1) s ¢ +1(W) were provided by the following representation: ( ) ( , 1) 1 2 1/ 4 2 - W L - + W L + + W L + û ê ê ë ÷ ÷ ø ç ç è L + = - - p p p c - 2 1 ù é p p j p k p j v v W k & J L = - = - - æ ö çè ˆ ˆ ˆ 1 2 2 2 ˆ ( 1)  2  2014 MRT p M c j j 2 3 2 + = + = y z z ÷ø Jz =  J M v ® W 0 ú úû c c y p D 2 x s [ ] ( ) ( 1) å¢ = j W ¢+ L 0 1 p ê êë ˆ p2 L(v) Lp ˆ | Y( p,s = +1) ñ s | Y(L p,s ¢) ñ O O Suppose that observer O sees a W boson (spin-1 and mass MW ¹ 0) with momentum p2 in the y-direction and spin z-component Jz = s  = +. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the | Y( p,s )ñ state?
    • 2014 MRT Mass Zero Weinberg suggests that we first have to ‘work out’ the structure of the little group and tells us to consider an arbitrary little-group element W m n , with W m n k n = k m, where k m is light-like four-vector, the standard four-momentum for this case, i.e., k m = [0,0,1,1] (c = 1). Acting on a time-like four-vector t m = [0,0,0,1], such that a Lorentz transformation must yeild a four-vector W m n t n whose length and scalar product with W m n k n = k m are the same as those of t m : m W t k t k W t W t t t = = - n r n ( )( ) 1 = = - m ( ) 1 m m m n n m m m r Any four-vector that satisfies the second condition (i.e., (W m n t n )k m = -1) may be written: Wm t = + n [a ,b ,z ,1 z ] n and the first condition (i.e., (W m ) = -1) then yeilds the relation: ( 2 2 ) 2 n t n )(W r m t r z = 1 a + b It follows that the effect of W m n on t n is the same as that of the Lorentz transformation: ù ú ú ú ú û é ê ê ê ê ë + - z a b z a - a 1 0 b - b - = z a b z m n a b 1 0 1 1 S ( , ) This does not mean that W equals S (a ,b ), but it does mean that S -1 (a ,b )W is a Lorentz transformation that leaves t m invariant (i.e., it is therefore a pure rotation!)
    • ù z a j b j a j b j z 1 cos sin sin cos a j j a cos sin b j j b 2014 MRT Also, S m n (a ,b ) like W m n leaves k m invariant, so S -1 (a ,b )W must be a rotation by some angle j around the 3-axis: ù ú ú ú ú û S -1(a ,b )W = R(j ) 1 0 0 0 é ê ê ê ê j j 0 cos sin 0 = 0 0 0 1 ë - 0 sin cos 0 ( ) j j n j Rm where: The most general element of the little group is therefore of the form: S(a ,b )S -1(a ,b )W = S(a ,b )R(j ) hence: When we apply matrix multiplication we get: ú ú ú ú û é ê ê ê ê ë + - + - - - - sin cos - + - = ù ú ú ú ú û 1 0 0 0 é ê ê ê ê ë 0 cos sin 0 - ù ú ú ú ú û é ê ê ê ê ë + - z a b z a - a 1 0 b - b - = z a j b j a j b j z j j j j z a b z j a b cos sin sin cos 1 0 sin cos 0 0 0 0 1 1 0 1 1 W( , , ) W(j ,a ,b ) = S(a ,b )R(j )
    • ù a b a j a b j b 2014 MRT For j , a , and b infinitesimal, the general group element is: ú ú ú ú û é ê ê ê ê ë 0 0 - - - 0 - - m m Wm (j ,a ,b ) =d + ω Þ = m ù ú ú ú ú z a b z a j j a b j j b û é ê ê ê ê ë - - - - cos sin - - - + ù ú ú ú ú û 1 0 0 0 é = ê ê ê ê ë ® j = cos 0 j = sin 1 = 0 0 0 ω sin cos 0 1 0 0 0 0 1 0 0 0 0 1 0 ( 0) a b z a b z n j z m n  W n n n where (note that we are using the result for W(j ,a ,b ) but we use the antisymmetric properties of w m n and the equality –a = a cosj – b sinj and –b = a sinj + b cosj ): From U(1 + w) = 1 + ½(i /) w rs M rs (we consider no translation so that e = 0 ) we see then that the corresponding Hilbert space operator is decoupled to 1 followed by 16 terms: U W i M i M M M M = + åå = + å + + +   s i M M M M = + + + + + [(ω ω ω ω ) 2 M M M M + + + + (ω ω ω ω ) M M M M + + + + (ω ω ω ω ) (ω ω ω ω )] 1 (ω ω ω ω ) 2 ω 1 2 [ ( , , )] 1 33 33 23 23 13 13 03 03 32 32 22 22 12 12 02 02 31 31 21 21 11 11 01 01 30 30 20 20 10 10 00 00 3 3 2 2 1 1 0 0 M M M M + + + +  s s s s s s s s s r rs j a b rs
    • Hence, by inserting the components of the w rs matrix and factoring for a , b and j : U W i M M M M M M M M 00 10 20 30 01 11 21 31 j a b a b a j a = + × + × + × + × × + × - × + × (0 0 0 2 02 12 22 32 03 13 23 33 M M M M M M M M - b × + j × + × + b × + × - a × - b × + ×  i M M M M M M M M M M 10 20 01 21 31 02 12 32 13 23 = + + - - + - + + - - a b a j a b j b a b  i M M M M M M M M M M 10 01 31 13 20 02 32 23 21 12 = + - + - + - + - + - + a a a a b b b b j j  i M M M M M M M M M M [ ( ) ( ) ( )] 2 1 [( ) ( ) ( )] 2 1 ( ) 2 1 0 0 0 ) [ ( , , )] 1 10 01 31 13 20 02 32 23 21 12 = + a - + - + b - + - + j - +  Recalling that M mn = - M nm, we decompose further: U W i M M M M M M M M M M 01 01 13 13 02 02 23 23 12 12 j a b a b j = + - - - - + - - - - + +  i M M M M M 01 13 02 23 12 = + - - + - - + a b j  i M M M M M = + - - + - - + [2 ( ) 2 ( ) 2 ] 2 1 [ ( 2 2 ) ( 2 2 ) (2 )] 2 1 [ ( ) ( ) ( )] 2 [ ( , , )] 1 01 13 02 23 12 a b j  In essence this is an acceptable solution: U[W(j ,a ,b )] = 1+ i [a (-M 01 -M13 ) + b (-M 02 -M 23 ) +j M12 ]  2014 MRT
    • But for the sake of driving things contrary to Weinberg’s notation again, we use temporal-spacial ordering of indices: U W i M M i M M i M 01 13 02 23 12 j a b = + a - - + b - - + j [ ( , , )] 1 ( ) ( )    i K J i K J i J = + - + + - + + a b j 1 ( ) ( ) 1 2 2 1 3    where we used the M mn matrix representation arrived at earlier (especially M12 º J 3 ): ù é K K K 1 2 3 K J J 1 3 2 K - J J ù é 01 02 03 M M M 01 12 13 M M M 02 12 23 M - M M 2 3 1 U[W(j ,a ,b )] = 1+ i Aa + i Bb + i J3j = 1+ i (Aa + Bb + J3j )     So, finally, we have: where: 2014 MRT ú ú ú ú û ê ê ê ê ë - - = ú ú ú ú ú û ê ê ê ê ê ë - - = 0 0 0 0 0 0 0 0 3 2 1 03 13 23 K J J M M M Mmn A = -M 01 -M13 = -K + J and B = -M -M = -K + J 2 1 02 23 1 2 Weinberg defines his A and B as -J2 + K1 and -J1 + K2, respectively.
    • Massless particles are not ‘observed’ (e.g., in any laboratory experiment) to have any continuous degree of freedon like an infinitesimal angle j so this means that states are distinguished by the eigenvalue of the remaining generator: J3 Y(k,e ) =e Y(k,e ) Since the momentum k is in the 3-direction, e gives the component of angular momentum in the direction of motion, or helicity. We are now in a position to calculate the Lorentz transformation properties of general massless particles states. The relation U(W ) = 1 + (i /) A a + (i /) B b + (i /) J 3 j generalises for finite a and b to: a b a b i A + i B U [ S ( , )] = e   j i J j 3 [ ( )] e U R =  and for finite j to: 2014 MRT
    • An arbitrary element W of the little group can be put in the form W(j ,a ,b ) = S(a ,b ) R(j ), so that: 2014 MRT i A i B i J i a + b j ej ( ) ( , ) e e ( , ) e ( , ) 3 e e e U W Y k =    Y k =  Y k and therefore the previously developped equation U(W ) |Y(k,e )ñ = Se ' De' e (W ) |Y(k,e ¢)ñ gives: e e ej i D (W) e e ¢e = d ¢  where j is the angle defined by expressing W as in the equation W(j ,a ,b ) = S(a ,b ) R(j ). The Lorentz transformation rule for a massless particle of arbitrary helicity is now given by the U(L) |Y( p, e )ñ = [N(p)/N(Lp)] Se ' De ' e [W(L, p)] |Y(Lp, e ¢)ñ and N(p) = Ö(k0/p0) equations (adapted from the Mass Positive-Definite case and replacing mj by e ) as: U p p e j ( , ) ( ) ( , ) ( ) e ( , ) 0 0 p e e p i p L L Y = L  Y L with j (L, p) defined by: W(L, p) º L-1(Lp)L L( p) º S[a (L, p),b (L, p)]R[j (L, p)]
    • So, for massless particles, the most general element of the little group is of the form: 1 0 0 0 é ù é g a b g a - a W S R where Sm and R j a b a b j a b n b - b + j j 0 cos sin 0 0 sin cos 0 1 0 Massless particles are not observed to have any continuous degree of freedom like q ; to avoid such a continuum of states, we must require that physical states (called | Y(k, e ) ñ) are eigenvectors of A and B. Since the momentum k is in the 3-direction, e gives the component of angular momentum in the direction in the direction of motion, or helicity. For finite a , b and j : U[S(a ,b )] = ei(a A+ib B)  and U[R(j )] = ei J3j  ù ú ú ú ú û ê ê ê ê ë - = ú ú ú ú û ê ê ê ê ë - - = = 0 0 0 1 ( ) 1 0 1 1 ( , , ) ( , ) ( ) ( , ) j j j g a b g m n | Y» ñ i L 0 ( p) e Suppose that observer O sees a massless particle with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same relativistic particle’s state? An arbitrary element W of the little group can be put in the form W(j,a,b ) = S(a,b ) R(j ), and: D ¢ (W) = ei ( L , p)  where j is the angle defined by expressing W as in W = S Ä R above. The Lorentz transformation rule for a massless particle of arbitrary helicity e is given by (with the basis k = k m = [k, 0,0,k ]): | YO ( p,e ) ñ | YO (L p,e ) ñ 2014 MRT p e » e j  p 0 U(W) Y(k,e ) = eia A +ib B ei J3j  Y(k,e ) = eie j  Y(k,e ) And therefore, with U(W) | Y ñ = Se ' De ' e (W) | Y ñ, we get: e e e j e e d ¢ U p p e j ( , ) ( ) ( , ) ( ) e ( , ) 0 0 p e e p i p L = L L Λ Y  Y since (L p)0 = L0 m p m and can be found with p m = [ p0,0] = [ | p |2, 0 ]. E/c O L(v)
    • for a given L and p we need to fix a convention for the standard Lorentz transformation that take us from k = k m = [k,0,0,k ] to p = p m. This may conveniently be chosen to have the form: 2014 MRT To calculate the little-group element W(L, p) º L-1(L p)L L( p) =W[j (L, p),a (L, p),b (L, p)] = S [a (L, p),b (L, p)] R[j (L, p), p] ù ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ë ö æ p 2 2 + - 0 0 1 0 1 0 0 0 0 1 0 - + ö = ÷ ÷ø æ = ç çè 0 0 1 u u u u u u u u B u 2 2 1 2 2 1 2 2 p k ÷ ÷ø ç çè = k L( p) R(pˆ )B where B ( u = | p | /k ) is a pure boost along the 3-axis (e.g., z-direction): ˆ and R( p) is a pure rotation that carries the 3-axis (e.g., z-direction) into the direction of the unit vector p. ˆ
    • Note that the helicity is Lorentz invariant; a massless particle of a given helicity s looks the same (aside from its momentum) in all inertial frames. Because electromagnetic and gravitational forces obey space inversion symmetry, the massless particles of helicity ±1 associated with electromagnetic phenomena are both called photons. The massless particles of helicity ±2 that are believed to be associated with gravitation are both called gravitons. On the other hand, the supposedly massless particles of helicity ±1/2 that are emitted in nuclear beta decay have no interactions (apart from gravitation) that respect the symmetry of space inversion, so these particles are given different names: neutrino for helicity +1/2 and antineutrino for helicity -1/2. 2014 MRT Even though the helicity of a massless particle is Lorentz invariant, the state itself is not. In particular, because of the helicity-dependent phase factor exp( i e j / ) in: i p 0 ( , ) U p p = L L e j L ( Λ ) Y ( , e ) ( ) e  Y ( p , e ) E c a state formed as a linear superposition of one-particle states with opposite helicities will be changes by Lorentz transformation into a different superposition.
    • For instance, a general one-photon state of four-momenta may be written | Y( p ; e )ñ = e +| Y( p,+1) ñ + e -| Y( p,-1) ñ, where | e + |2 + | e - |2 = 1. The generic case is one of elliptical polarization, with | e ± | both non-sero and unequal. Circular polarization is the limiting case where either e + or e - valishes, and linear polarization is the opposite extreme, with | e + | = | e - |. The overall phase of e + and e - has no physical significance, and for linear polarization may be ajusted so that e - = e + *, but the relative phase is still important. Indeed, for linear Plane polarized gravitons can be defined in a similar way, and here the equation above becomes: 2014 MRT polarizations with e - = e + *, the phase of e + may be identified as the angle between the plane of polarization and some fixed reference direction perpendicular to p. The equation: i p 0 ( , ) U p p = L L e j L ( Λ ) Y ( , e ) ( ) e  Y ( p , e ) E c shows that under a Lorentz transformation Lm n , this angle j rotates by an amount j (L, p). i p 0 2 ( , ) U p p = L L e j L ( Λ ) Y ( , e ) ( ) e  Y ( p , e ) E c and has the consequence that a Lorentz transformation L rotates the plane of polarization by an angle 2j (L, p).
    • Again, now for the calculation based on the data provided: U p p L = L L = L + + L - ( ) ( , ) ( ) e ( , ) e ( , 1) e ( , 1) | Y» ñ col[k ,0, k , 0] = L(v) s i s j  L 0 ( p) e p 0 E/c U p p ( ) ( , ) ( ) e ( , ) e W L W L - W L | YO ( p,s ) ñ | YO (L p,s ) ñ 2014 MRT Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? s j é In this case, the B m p n = 0 n = 1 n = 2 n = 3 m = 0 m = 1 m = 2 m = 3 2 2 1 + k k p p 2 n m ù n are given by (with g = 1/Ö(1 – b 2) and b = v /c ): 2 2 1 - k k p p 2 2 2 1 - 0 0 k k p p 2 1 1 2 2 1 + k k p p 2 0 0 0 0 0 0 0 0 Quantum Fields – Problem 2. O ( , ) 0 0 s s p i B p L Y = L  Y L or, by expansion: where the normalization constant is the same as found in Problem 1, i.e., Ög3. ú ú û ê ê ë ( , ) ( , ) 3 ( , ) 0 0 p p p p i p i p i p Y e  Y e g  Y  Y | k ñ » p B(| p |/k ) R L(ˆp)
    • ö æ 2 i.e., boosting along the photon’s the 2-axis direction and carring it along the direction of p in the 3-axis direction. The result is: 2014 MRT First we calculate the Lorentz transformation along the 2-axis: ù ú ú ú ú ú ú ú ú ú ú ú û 2 2 2 + - 2 2 2 k k 0 1 0 0 é ê ê ê ê ê ê ê ê ê ê ê ë ù 0 1 2 2 2 + - - - k k + = 0 ù ú ú ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ê ê ë 2 2 2 k - é ù é ö 1 0 0 0 j j 0 cos sin 0 0 sin cos 0 2 2 2 2 2 2 + k - +    - - + = ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ë 2 2 2 - + ú ú ú ú û ê ê ê ê ë - = ÷ ÷ ø æ ç ç è = 2 2 2 2 k 1 0 0 k 2 2 2 k 2 2 2 k 0 0 0 1 2 2 1 0 2 1 0 0 0 2 1 0 2 1 2 2 2 k 2 2 2 k k cos90 0 0 1 2 2 2 k 2 2 2 2 1 cos90 0 2 1 sin 90 sin 90 2 1 sin 90 0 2 1 0 cos90 0 2 1 0 2 1 0 0 0 1 2 1 0 2 1 0 2 1 0 0 0 1 ( ) ( ˆ ) 2 2 2 k 2 2 2 2 2 2 2 2 2 2 2 2 3 3 2 k k k k k k k k k k k k k j j k p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p    L p R B ÷ ÷ ø ç ç è = k 2 ( ) 3 ( ˆ 3 ) 2 p L p R p B
    • Then we calculate the Wigner Rotation: W3 L p = L3 - 1 2 1 2 Lp L L p as anticlockwize around the positive 1-axis. We will need this again later. Then we have: é é g b g 0 0 3 3 3 0 1 0 0 0 0 1 0 2 2 2 k 2 2 2 k 2 2 2 k p where the gamma factor is g3 = Ö(1 - b3 2 2 2 + k - k k 1 0 0 ù ù é ù 2 2 2 k 2 2 2 2 2 2 k k 2 2 2 k 2 2 2 2 2 2 2 k 2) (which is a function of the rapidity b 3 = | v3 | /c along the 3-axis). ú ú ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ê ê ë - - + - - - + - + - = ú ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ê ë - - + ú ú ú ú ú û ê ê ê ê ê ë - - L = 3 2 3 3 2 3 3 2 2 3 3 2 3 2 3 2 2 2 3 3 3 3 2 2 1 0 2 1 1 0 0 2 1 0 2 1 0 0 2 1 0 2 1 0 0 0 1 2 1 0 2 1 0 0 0 2 1 0 2 1 0 0 ( ) g k b g k b g k k b g k g k g k k k b g g p p p p p p p p p p p p p p p p p p p L p ( , ) ( ) 3 2 ( ) ´ =- ´ 2014 MRT
    • Then we have with the definition of the basis momentum k k ù ú ú ú ú ú ú ú ú û 2 2 2 + - - é ê ê ê ê ê ê ê ê ë - = 0 ö ö ÷ ÷ ø é ù æ + æ - ç ç è ÷ ÷ ø ç ç è = 0 1 0 0 1 0 1 ù ú ú ú ú ú ú ú ú ú ú ú ú û é é ê ê ê ê ê ê ê ê ê ê ê ê ë 2 2 2 2 2 2 2 ö ÷ ÷ ø 2 2 2 2 2 2 2 2 2 2 2 2 - + ÷ ÷ ø æ ö ö æ æ ö æ 2 2 2 2 2 2 2 2 2 - - ÷ ÷ ø ç ç è ö æ p æ ç ç 2 2 2 2 - + è ÷ ÷ ø ç ç è - ÷ ÷ ø ç ç è + ÷ ÷ ø ç ç è ö ç ç è + = ù ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ë × 2 2 2 + × + - × - + × 2 2 2 2 2 2 2 × - 2 + - × - = ù ú ú ú ú ú û ê ê ê ê ê ë = ú ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ê ë - - + - - - + L = 0 2 k 2 2 2 k p 3 3 2 2 2 2 2 3 2 2 2 2 3 3 2 2 3 3 2 2 2 2 2 2 3 2 2 3 2 2 2 2 3 3 2 2 2 2 2 2 2 3 3 2 2 3 2 3 2 3 2 3 3 2 3 3 2 2 3 3 2 3 2 3 3 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 0 2 2 1 1 0 0 2 1 0 2 2 2 1 ( ) p p p p p p p p p p p p p p p p p p p p p p p p p p p p p k p p p p p p p p p p p L p k b g k g k k k b g k b g k k k k k k k g k g k k k k b g k k b g k k k k k k k k k g k k g k k g k k b g k k b g k k k k b g k g k k g k m m and since | p2 | = p2: ù ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê p p 0 3 2 1 ë - ö ö ÷ ÷ ø æ + æ - ç ç è ÷ ÷ ø ç ç è = ù ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ë L L L L = 2 k 2 2 2 p 3 3 2 2 2 2 2 1 2 3 2 1 2 p p p p p p p b g k g and and thus: 2014 MRT
    • 2 Finally we get the eigenvector for the boosted momentum Lp : 2 2 2 ö æ - + ÷ ÷ ø æ + = - ÷ ÷ ø æ - + ÷ ÷ ø æ L = + p p i j k i j kˆ 1 1 ˆ 2 1 ˆ 2 1 ˆ ˆ 2 1 ˆ 3 2 2 2 3 2 2 2 2 2 2 2 3 3 2 2 2 2 2 2 b k k b k k b g - - ÷ ÷ ø ç ç è ö ç ç è ö ç ç è ö ç ç è p p p p p p p p p k v v ù so that the four-vector (L p )m is given by: é 1 1 ˆ 1 2 ù é p p i j L º L v ˆ 1 1 1 ˆ 2 1 1 2 2 2 2 çè 2 ö 2 2 çè 2 ö 2 2 0 p ö c çè p c p p p ö c çè p p ÷ø - æ - ú ú ú ú ú û ê ê ê ê ê ë ÷ø æ + - ú ú ú ú ú û ê ê ê ê ê ë ÷ø æ º L = - ÷ø - æ k k and x We see that the effect of a Lorentz boost L( p) on the initial unit 2momentum pfollowed by a Lorentz transformation L(v) which 2 3has the effect of creating (as seen from O) two component L(v) vectors pand p, magnitudes being equal to ½[1 – 1/( p/ k)2] p, 31 22 2in the positive 1- and 2-axis directions and a momentum vector p2 pin the negative 3-axis direction: 3 L2( p) W-1(W) y z 3 2 2 1 p ö c çè p c ÷ø - æ = - v v The effect of the Wigner rotation W (L, p) will produce a rotation (negative in this case) around the 1-axis which results in an angle W between the 2-axis but also some dispersion resulting in the overall boost vector Lp º p1 i +p2 j – p3 k. ˆ ˆ ˆ ˆ p3 ˆ ˆ Suppose that observer O sees a photon with momentum p in the y-direction and polarization vector in the z-direction. A second observer O moves relative to the first with velocity v in the z-direction. How does O describe the same photon? ˆ ˆ ˆ 2014 MRT Lp ˆ » p2 ˆ p1 ˆ W
    • We can then go through the matrix multiplication with the order described in Problem 1: ù 1 0 0 0 0 1 0 0 W - W 0 0 cos sin 2014 MRT ù ú ú ú ú ú ú ú ú ú û g b g 0 0 3 3 3 0 1 0 0 0 0 1 0 é ê ê ê ê ê ê ê ê ê ë é ù é + - ù 0 ( ) 1 ( ) 1 0 0 ( ) 1 1 0 0 ( ) 1 + - W - - W cos ( ) 1 cos ( ) 1 0 ( ) 1 ( ) 1 0 0 ( ) 1 1 0 0 W W - - W + W - + ù ú ú ú ú ú û é ê ê ê ê ê ë - g - b g = ú ú ú ú û ê ê ê ê ë W W ú ú ú ú ú ú ú ú ú û ê ê ê ê ê ê ê ê ê ë - - + ú ú ú ú ú û é ê ê ê ê ê ë - - L = L - L = - ( ) 1 0 0 sin cos 2 sin 2 2 sin 2 2 2 0 0 3 3 3 0 1 0 0 0 0 1 0 0 0 0 0 sin cos 0 0 0 1 2 0 2 0 2 2 0 0 ( ) ( ) ( , ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 1 3 2 3 2 1 k k k k k k k k k k k k b g g k k k k k k k k b g g p p p p p p p p p p p p p p p p p p p p L ´ p L p W p
    • 2014 MRT We finally obtain for the boost matrix L3 ´ 2 (Lp) as a function of the deflection angle W: ù ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ë cos sin ( ) 1 W+ W - 0 ( ) 1 ( ) 1 0 0 ( ) 1 cos W+ sin W ( ) 1 - 0 - ( ) 1 + - - - + W - W W - W- W - + - L = sin cos 2 2 ( ) 1 p ( ) 1 2 1 0 0 2 cos cos 2 sin 2 2 2 ( ) 3 2 2 2 3 3 3 2 2 2 3 3 2 2 2 3 3 2 2 2 3 3 2 2 2 2 2 2 3 3 3 2 2 2 3 2 2 2 3 3 2 g k k g b g k k b g k k b g k k b g k k k k b g g k k g k k g p p p p p p p p p p p p p p p L ´ p Since L3 ´ 2 (Lp) is symmetric we can ‘extract’ the [L3 ´ 2 (Lp)]3 2 = [L3 ´ 2 (Lp)]2 3 components: W+ W - 0 = - ( ) 1 - W = - - W cos sin ( ) 1 2 cos sin 2 2 2 2 3 3 3 3 2 2 2 3 3 k g b g k g k k b g p p p p ü ï ï ý ï ï þ ì ï ï 2 1 v v tan 1 arctan 1 í p ï ï î ù ú ú ú ú ú û é ê ê ê ê ê ë ö çè 2 - ÷ø æ Þ W = ù ú ú ú ú ú û é ê ê ê ê ê ë ö çè 2 - ÷ø æ W = k k k k 2 2 2 2 1 2 p p c p c and we obtain: expressed only by calculating the ratio p2 /k .
    • Finally, by using the same trigonometric identities relating tan W to cos W and/or sin W: W = + W sin 1 tan W cos 1 + W W = tan 1 tan 2 2 and such that the boost matrix becomes: ù ú ú ú ú ú ú ú ú ú ú ú ú û é ê ê ê ê ê ê ê ê ê ê ê ê ë 1 + W - - 1 + tan W + 1 ( ) 1 - + W 1 tan W - 1 + tan W + 1 + tan W W 1 1 1 0 ( ) 1 ( ) 1 0 0 ( ) 1 + W - - 2 2 k ( ) 1 k p - + - - + W - + W + W W + W + - L = 3 2 2 ( ) 1 2 2 2 3 3 2 2 0 ( ) 1 2 3 2 2 3 3 ( ) 1 2 2 2 3 3 2 2 2 2 2 3 3 2 2 2 2 2 2 3 2 2 3 3 2 2 3 2 2 2 3 3 2 1 tan tan 2 tan 1 tan 2 2 1 0 0 2 1 tan 1 tan 2 tan 2 tan 1 tan 2 2 ( ) g k g b g k k b g k k b g k b g k k k b g g k k g k k g k p p p p p p p p p p p p p p p L ´ p We can now either use the relation W = arctan{½ (| v |/c)[( p2/k)2 – 1]/(p2/k)} or the one above by using tan W and its square, with p2 /k and b3 = | v |/c given beforehand to calculate things. 2014 MRT
    • The Klein-Gordon Equation When Schrödinger wrote down the non-relativistic equation ( i  ¶t | Y ñ = (  2 /2 m) | Y ñ) now bearing his name, he formulated the corresponding relativistic equation. The equation is derived by inserting the operator substitution E ® i  ¶t and p ® - i  Ñr (with p = [E /c, p] = i  ¶m where ¶m = ¶ /¶ x m = {¶ /¶ c t , Ñ} and ¶ m = ¶ /¶ xm = {¶ /¶ c t , -Ñ}) into the relativistic relation between energy and momentum for a free particle: where mo is the rest mass of the particle. We then obtain the Klein-Gordon equation: o r j r = - Ñ + j ¶ In natural units (i.e. c =  = 1) and the Dirac notation j (x) = á x |j ñ with x = x m = [t , r] , we get: 2014 MRT E2 c2 2 m2c4 = p + o where the D’Alembertian operator is defined as  = (1/c 2) ¶ t - Ñ 2 = hmn ¶ m ¶n = ¶ m ¶m . The amplitude j is a one-component scalar quantity which under an inhomogeneous Lorentz transformation, x m = Lm n x n + a m, transforms as j (x) = j (x) or equivalently as j (x) = j [L-1(x - a)] . 2 ( , ) ( 2c2 2 m2c4 ) ( ,t) t t - ¶   ( 2 ) 0 + mo x j =
    • In order to give a physical interpretation to the Klein-Gordon equation, by analogy with the non-relativistic equation, one might try to define a probability density, r , and a probability current, j, in such a way that a continuity equation holds between them. One is then led to the following expressions for r and j: æ i    and r = j ¶ j - ¶ j j = j × ¶ j - ¶ j × j i = j × ¶ i j - ¶ i j × j ÷ø ÷ ¶ m c which by virtue of the Klein-Gordon equation, satisfy: ( ) = 0 Ñ · j + ¶ r ¶ t The constants appearing in the density and current have been so determined that these expressions reduce to the usual expressions for the Schrödinger theory in the non-relativistic 2014 MRT 2 ( ) 2 2 * * o * 0 0 * * * 2 o ö ç çè ¶ j o m i i m c t t limit. If in the expression for r we substitute for i  ¶t j , Ej we obtain: = E r j *j 2 moc which for E » m oc 2 indeed reduces to the expression for the probability density in non-relativistic quantum mechanics. It is, however, to be noted that in general r may assume negative as well as positive values because the Klein-Gordon equation is of second order in the time variable and therefore j and ¶t j can be prescribed arbitrarily at some time to. In 1934, Pauli and Weisskopf re-interpreted the Klein-Gordon equation as a quan-tum field equation analogous to Maxwell’s equations for the electromagnetic field.
    • The Dirac Equation In 1928, Dirac discovered the relativistic equation which now bears his name while trying to overcome the difficulties of negative probability densities of the Klein-Gordon equation. The Dirac equation has special importance because it describes particles of spin-½, and both electrons and protons have spin-½. Many other elementary particles, including the neutron, the m meson, &c., have spin-½. The reasoning which led Dirac to the Dirac equation was as follows: If we wish to prevent the occurrence of negative probabilities densities, we must then avoid time derivatives in the expression for the probability density r. The wave equation must therefore not contain time derivatives higher than first order. Relativistic covariance, furthermore, requires that there be essentially complete symmetry in the treatment of the spatial and time components. We must therefore also require that only first-order spatial derivatives appear in the wave equation. Thus the Dirac wave function must satisfy a first-order linear differential equation in all four x º x m = [t , r] coordinates. The linearity is required in order that the superposition principle of quantum mechanics hold. Finally, we must also require y (x) that obey the equation: æ + m c ö y x 2 2 o = ÷ ÷ø ( ) 0 2 ç çè  (where = ¶ /¶t + Ñ2) if it is to describe a free particle of rest mass mo, since this equation implies that the energy momentum relation for a free particle p 2 = mo 2014 MRT 2c 2 is satisfied, and that in the correspondence limit, classical relativity still remains a valid concept.
    • Let us therefore assume that y consists of N components y a (a = 1, 2, …, N) where the number N is as yet unspecified; it will turn out to be four. The most general first-order linear equation is then one which expresses the time derivative of one component as a linear combination of all the components as well as their spatial derivatives. Inserting the appropriate dimensional factors (i.e., c and  = h/2p ), the most general equation possible is: ¶ åå = å= 1 0 ( 1,2,..., , 4) 4 ka and b k a are dimensionless constants, independent of the space-time coordinates x 0, x 1, x 2, x 3. A natural way to simplify these equations is to use matrix notation which reduces them to the following equation: m m i im c x p p m y a y by m y 0 p j 2014 MRT y + a ¶ y b y a k i + = = 1 o 3 1 4 1 im c N i.e. c t x N k i N k i ¶ ¶ = = a = a a a a a  Assuming the homogeneity of space-time, the a i 3 ¶ L + = æ = Y L + ¶ o ( )0 In this equation, y is a column matrix of N rows, and a = a i = {a 1, a 2, a 3} and b are both Hermitian matrices of N rows and columns. This is the Dirac equation in its simplest form! The Dirac equation can also be written in Hamiltonian (with H also Hermitian) form: ö ÷ø çè ¶ ¶ = = å 0 ( , ) ( ) 0 1 p i i x ct x  ¶ α Ñ  y y ( b 2 )y H i c moc t i = = - · + ¶ The probability density, r, and a probability current, j, are thus spelled out as: r =y *y and ji = cy *a iy
    • In order to derive the properties of the a and b matrices, we must multiply the equation (1/c)(¶y / ¶ t) + Sa i(¶y / ¶ x i) + (i mc / ) b y = 0 by the operator: i o im c 1 ¶ - i ¶ - o ¶ åå å = = = å= ¶ which has the effect of introducing second derivatives. The terms with ¶ t or mixed derivatives between space and time cancel and we obtain: 2014 MRT ¶ 3 y = a a + a a ¶ y b y a b ba y 2 2 o 1 We have symmetrized the a ia j term, which is permissible since ¶ / ¶ x i and ¶ / ¶ x j commute. To agree with the Klein-Gordon equation, the right-hand side of the equation above must reduce to: Ñ - m c This imposes the following conditions: a b  c t x i i 3 1 ¶ - + + ¶ ¶ ¶ ¶ 1 2 o 2 3 1 3 1 2 2 2 2 2 1 ( ) ( ) i i i i i j i j i j j i x m c im c c t x x   2 2 2 o y y 2  i j j i i j a a + a a = d i i + = i I a b = = a b ba 2 2 1 ( ) 0 ( ) i.e., that the a s as well as any a i and b anticommute, and that the square of all four matrices is unity.
    • Without going into the derivations, if I denotes the unit 2´2 matrix, and s = s i are the Pauli matrices, then the 4´4 matrices are given by: ù é ù é = a s and 0 b s satisfy all our conditions: they are Hermitian and can be seen to anticommute by using the anticommutative properties of the three Pauli matrices s s (i.e., si sj = di j + i eijk sk ). multiplied by a matrix – whereas the time derivatives is not. To eliminate this distinction, let us multiply (1/c)(¶y / ¶ t) + Si a i(¶y / ¶ x 2014 MRT úû êë - = úû êë I I i i i 0 0 0 Let us finally put the Dirac equation in covariant form. When the Dirac equation is written as (1/c)(¶y / ¶ t) + Sa i(¶y / ¶ x i) + (i mc / ) b y = 0, the spatial derivatives are i o i) + (i mo c / ) b y = 0 by b on the left to obtain: - i  b ¶ å= i 0 y - i  ba ¶ y + m cy = o 0 3 1 i i We can make this equation look even more symmetrical by introducing the matrices g m : ïî ïí ì g = b = g i = b a i g m 0 With these definitions, g 0 is Hermitian, with (g 0)2 = +1, and the g s are anti-Hermitian, i.e., (g i)* = -1 with (g i)2 = -1 so that the g matrices satisfy the following commutation rule: {g m ,g n } =g mg n +g ng m = 2hmn (The Clifford Algebra)
    • The Dirac matrices gm are written (N.B., this is one of the many basis available): ù 0 0 1 0 0 0 0 1 1 0 0 0 2014 MRT é = ù é 0 0 1 0 0 0 0 1 g 0 , g 1 ,g 2 andg 3 1 0 0 0 or in a more abridged way: ú ú ú ú û é ê ê ê ê ë - - = ù ú ú ú ú û é ê ê ê ê ë 0 0 0 0 0 0 - - = ù ú ú ú ú û ê ê ê ê ë ú ú ú ú û ê ê ê ê ë - - = 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 1 0 0 i i i i ù úû é - êë ù = = úû é êë - = = 0 0 0 0 0 0 σ σ i i I I g g and g g By extension, a fifth matrix is defined by: ù úû é- = = úû êë 5 0 1 2 3 é 0 ù 5 0 1 2 3 g 5 g g g g g or g g g g g êë - = - = = I I i i I I i 0 0 0 We list a few identities for these gamma matrices: trg mg n = 4hmn and trg mg ng lg s = 4(hmnhls -hmlhns +hmshnl ) where tr is the trace – or the sum of diagonal components – and h00 = -1,h11 = h22 = h33 = +1 and hmn º 0 for m ¹ n . And the following identities follow from the basic Clifford identity: Sum over m : g m p/ g m m m = -2 p/ , g p/ q/ g m = 4 p × q and g p/ q/ r/ g m = -2r/ p/ q/ which are not really used presently but more often in quantum electrodynamics.
    • In terms of the g matrices, the equation -i b ¶0y - i  Si b a i¶iy + mo c y = 0 now reads: where our summation convention has been reintroduced. With this last equation, we have written the Dirac equation in a covariant form where space and time derivatives are treated alike. 2014 MRT i m c i γ ¶ m c ö çè æ- g ¶ + y y m æ + · - = ÷ø ö çè 0 o o = ÷ø m   Feynman has introduced the so-called ‘slash’ notation to simplify the equation even further. For example, he denoted by p the ‘reduced’ quantity: p / = g × p = g m p m = g pm m = g 0 p0 - γ · p where gm is defined by gm = hmn g n . With this notation, natural units (i.e., c =  = 1) and the Dirac notation y (x) = á x |y ñ with x º x m = [t , r] , we get: (-i ¶/ + moc) x y = 0 where ¶ = g m ¶m = g 0 ¶ 0 + g · Ñ.
    • The wave function is now a bi-spinor y that holds 4 components: ù é y æ = y y æ ö 1 2 3 y The Dirac equation describes 4 linear coupled differential equations. The matrices gm sum-up the notion of spin since they correspond to a generalization of the Pauli spin matrices. For this reason, the Dirac equation is convenient to the description of fermions (i.e., half-integer spin particles). In fact, of the 4 degrees of liberty of the bi-spinor, 2 are used to represent the particle in the states of spin ±½ and the 2 other ones, the antiparticle with states of spin ±½. Under a more explicit form where the matrices are developed, i.e., a system of 4 coupled equations, we have: 2014 c p ip cp m c E - y + y + - y = ( ) ( ) 0 x y z c p ip cp m c E + y - y + - y = ( ) ( ) 0 x y z c p ip cp m c E - y + y - + y = ( ) ( ) 0 x y z c p ip cp m c E ( ) ( ) 0 MRT 4 2 1 2 o 3 2 2 1 o 2 2 3 4 o 1 2 4 3 o + y - y - + y = x y z ú ú ú ú ú û ç çè ê ê ê ê ê ë ÷ ÷ø ÷ ÷ø ö ç çè 4 y (Primus, u or spin-±½ ‘Matter’) (Secundus, v or spin-±½ ‘Anti-matter’)
    • The Dirac equation (c a · p + b mo c 2)y = E y may also be written in two-component form by means of the Pauli spin matrices: 2014 MRT é = 0 1 x y i z and the 2´2 identity matrix: With s = [sx ù é - = úû ù é = 0 1 ù é = so that in place of the 4 coupled equations we have: where ,sy ,sz ] it is seen that ù úû é êë 1 0 - = úû êë ù êë 0 1 0 0 1 0 i s , s and s úû êë 1 0 I ù úû é êë - = úû êë I I 0 0 0 0 and b σ σ α σ p c m c E · y + - y = ( ) 0 v u 2 c m c E ( ) 0 o 2 o · - + = u v σ p y y ù úû y é = úû êë ù y é = y u and v êë 3 4 1 2 y y y
    • The equations c s · p yv + ( moc 2 – E )yu = 0 and c s · p yu - ( moc 2 + E )yv = 0 provide a useful starting point for the non-relativistic approximation. From c s · p yv + ( moc 2 – E )yu = 0, 2014 MRT y σ p y 2 c - = · u v E moc which enables us to make an estimate of the relative magnitude of yu and yv. Approx-imating cmv 2 c v y » = 1 mv 2 v 2 u y Consequently, for v /c << 1, the component of interest is y, also known as the “large” ucomponent. Eliminating ybetween the equations c s · p y+ ( mc 2 – E )y= 0 and v v o u c s · p yu - ( moc 2 + E )yv = 0, yu satisfies: u c2 · 2 = E -m c E + m c (σ p) y u ( )( )y u 2 o At non-relativistic energies, we have: 2 o s · p » m v and E – moc 2 » ½m v 2, we get: 2 o 2 E¢ º E -moc << m c or æ ¢ ÷ø ÷ E - m c 2 o E + m c = m c E ¢ + E » 2 m c 2 E ¢ m c ö ç çè 2 o o 2 o 2 o 2 ( )( ) 2 1 y σ p σ p y E¢ = · · ( )( ) u u m 2 o The equation c 2( s · p)2yu = (E – moc 2)(E + moc 2)yu = 0 then becomes:
    • Further yu = [(s · p)(s · p)/2mo]yu reduction of equation is possible with the aid of the identity (s · A)(s · B) = (A · B) + i s · (A ´ B) (which is easily verified from the definitions of the Pauli matrices – See Landau-Lifshitz Quantum Mechanics). Therefore: 2014 MRT (σ · p)(σ · p) = p2 and, substituting in yu = [(s · p)(s · p)/2mo]yu we get: or, with p = -i  Ñ: 2 2m E¢ = p y u y u o E  ¢y = - Ñ y u 2 2 m o u 2 This equation is still in two-component form since yu is a two-component function. But each component of yu satisfies this equation too so that we can drop all subscripts and write: 2 2 y 2y ¢ = - Ñ m o E  which is the time-independent Schrödinger equation for a free particle.
    • The states of positive energy and spin ±½ are represented by the solutions u1 = col[1 0 0 0] and u2 = col[0 1 0 0] whereas the states of negative energy and spin ±½ are represented by the solutions u3 = col[0 0 1 0] and u4 = col[0 0 0 1]. 2014 MRT Using a parity transformation such that g 0y (t ,-r) = y (t ,r), we get: 2 o 0 3 2 o 0 2 0 0 1 0 ù é = g = and g 3 1 1 1 0 0 u 1 0 0 0 u p m c Σ p m c Σ = =- = =+ ú ú ú ú û ê ê ê ê ë ù ú ú ú ú û é ê ê ê ê ë which shows that fermions (i.e., electrons, protons, neutrons) have positive parity whereas: 2 o 0 4 0 0 0 0 ù ú ú ú ú 1 u p m c Σ p m c =- =- =- g = and g =+ û é ê ê ê ê ë - = 3 3 1 2 o 0 1 3 0 0 0 ù ú ú ú ú 1 - 0 u Σ û é ê ê ê ê ë which shows that antifermions (i.e., positrons) have negative parity.
    • We are now in a position to investigate the Lorentz invariance of the Dirac equation and to establish its connection with the representation of the inhomogeneous Lorentz group. Under an inhomogeneous Lorentz transformation x m = Lm n x n + a m with hm n Lm r Ln s = hrs the Dirac equation will be form-invariant if we define: y (x) = S(L)y (x) = S(L)y [L-1(x - a)] and if S(L) satisfies the condition S(L)-1g l m g m ; S(L) is a 4´4 matrix which S(L) = Ll operates on the components of y , i.e., the equation above written explicitly reads: 4 ( ) ( ) [ 1( )] ya x Sab y b x a The form-invariance of the Dirac equation is the statement that in the new frame y obeys the equation: m where ¶m = ¶ /¶ x m . Note that the g matrices remain unaltered under Lorentz transformation. ), we write S(L) 2014 MRT Let us now exhibit the actual form of S for a given Lorentz transformation. For the infinitesimal Lorentz transformation L = 1 + e l , x m = x m + e l m n x n (l mn = -l nm to first order in e as follows: S( 1 + e l) = 1 + e T and [S( 1 + e l)]-1 = S-1( 1 + e l) = 1 - e T . For the infinitesimal situations, the equation S(L)-1g l m g m can be rewritten as: S(L) = Ll å= = L L- - 1 b (-i g ¶m + moc)y (x) = 0 S -1g m S = (1-e T )g m (1+e T ) =g m +e (g mT -Tg m ) = Lm g =g +e l g n n m m n n where T must be such that g m T - Tg m = l m n g n and is given by T = (1/8) l mn (gm gn - gn gm ).
    • As an example of a transformation of single-particle states*, consider a free positive-energy electron (of mass mo) of helicity +1 and momentum p along the positive z-direction. We choose a barred system in such a way that it will coincide with the rest system of the electron (see Figure). In the primed system the electron wave function can be written as: 1 0 ù é - - ( ) ( ) e 1   0 The question is: What is the wave function for the same physical situation in the unbarred system? O v v p3 = O z 2014 MRT The answer Sakurai derives starts with: y x SLorentzy x = - S- i I s 3 E ù 1 0 = 0 1 úû sinh and the Pauli spin matrix: é êë - * J.J. Sakurai, Advanced Quantum Mechanics, 1967 – P. 101. y m p o E c c i m c t i m c t V x 2 o 2 o e 0 ú ú ú ú û ê ê ê ê ë y = Y x = ( ) 1 ( ) where (Majorana representation for gamma matrices and x4 = ict) 2 sinh 0 0 2 cosh 2 sinh 2 cosh 3 3 3 4 4 4 1 Lorentz q s q q s q g g ù úû é êë = - = ´ + and with the angle q given by: 3 m c m c o 2 o 2 2 cosh p q = and q =
    • Then, since q q q q p and 2 cosh = + = + = æ Sakurai obtains: E m c c ö çè 2 3 2 c = - ÷ø s 3 3 sinh 2 o E m c 2 cosh p + mc 1 2 2 2 ( ) cosh 1 2 2 2 o SLorentz s c 3 3 2 o p m c E m c E + m c 2 ( ) 2 é ê ê ê ê ê ê ë = m c E m c E m c = + 2 (1) 2 o 2 o 2 o p + é ê ê ê ê ê 2 3 o m c p u ( ) 1 0 c E m c 0 2 o 2 o m c E + m c 2 ( ) 1 0 0 0 1 0 0 0 2 o + 2 o 2 o 2 o - 1 = ù ú ú ú ú ú û ë + ù ú ú ú ú û é ê ê ê ê ë ù ú ú ú ú ú ú û ù ú ú ú ú û é ê ê ê ê ë m c This result is in complete agreement with u(1)(p), with çp1 ç= çp2 ç= 0 obtained by solving directly the Dirac equation. 2014 MRT
    • As for the spacetime dependence of the wave function, Sakurai points out that: and finds: ù é t t x cosh 3 sinh = q - q ù 1 3 é 1 0 0 E - - - - - - x S x S Lorentz Lorentz i x - i Et   i i Et i m c t i x Et S E m c V    Lorentz i m c t 1 u ( ) e V m c EV c E m c m c V · -   = where he used V = (mc2/E)V that follows from the Lorentz contraction of the normalization volume along the direction of motion. In his own words: Once we know the form of the wave function for a particle at rest, the correct wave function for a moving particle of definite momentum can be cons-tructed just be applying S -1. This operation is sometimes known as the Lorentz boost. = ú ú ú ú ú û ê ê ê ê ê ë + + = ú ú ú ú û ê ê ê ê ë = = = p x p p p p x p u ( ) e e 0 2 e 0 0 ( ) ( ) ( ) e 1 (1) 2 o (1) ( ) 2 o 2 3 o 2 1 1 1 o 3 3 3 3 2 o 2 o y y Y 2 3 o 2 o x m c t m c c p = - 2014 MRT
    • For an infinitesimal rotation through an angle e about the 1-axis, l 23 = -l 32 = +1 with all other l mn = 0. The generator for such a transformation is given by the equation T = (1/8) l mn (gm gn - gn gm ), and is T = ½g2 g3 . With the representation of the a matrix: we get: 2014 MRT úû ù é = ù é T ( R úû ) 1 i = + i 0 1 Σ The S(q ) corresponding to a rotation through the angle q about the 1-axis is therefore: S R T R i q q 2 ( ) Σ = = + ( ) e e Σ sin 2 cos 1 1 1 2 1 q q i = + The mapping |y ñ ® | y ñ for an arbitrary Lorentz transformation is induced by a unitary operator U(L,a) : with U(L,a)y = y x U L a y = x y = U L a y x ( , ) ( , ) ( ) y - S 1 x a = L L - ( ) [ ( )] êë 0 0 i i i s s a 1 1 1 2 2 3 0 2 êë = - = + s s a a 
    • For an infinitesimal quantum transformation, we write: where D, the infinitesimal generator, is determined by using equations x m = x m + e l m n x n and S( 1 + e l) = 1 + e T by equation á x | U(L,a) | y ñ = á x | y ñ = U(L,a)y (x) = S(L)y [L-1(x – a)] : 2014 MRT U a i e D  (L, ) = 1+ so that i e D e T y x e l x 1 (1 ) ( ) é T x x x = + - ¶ ö çè e y e l y (1 ) ( ) ( ) ( ) x m x T x x y e l y D = -i  T -l r x s ¶ ( s r ) ù For an infinitesimal rotation about the 3-axis, T = ½ i Sand only l 21 = -l 12 = +1 are 3 different from zero, so that: D =  3 Σ3 + (r ´ p)2 3 Thus, a Dirac particle has in addition to its orbital momentum, r ´ p, an intrinsic angular momentum S of magnitude /2. It is to be noted that the spin operator ½ S is not a constant of motion, since [H , S] ¹ 0. The same is true for the orbital angular momentum; however, the total angular momentum J = (/2)S + (r ´ p) is a constant of the motion.    = + - ¶ + úû êë + ¶ - + = ÷ø æ + ( ) ( ) ( ) m n n m m
    • We now consider the Dirac equation with electromagnetic coupling. The free-particle Dirac equation (c a · p + b mo c L = 1 m v2 + q v ·A - in which v is the velocity of a particle with positive charge q and rest mass mo, A and j are the vector and scalar potentials, respectively. The fields are then given by: 2014 MRT qj c 2 o B = A and E = - 1 ¶ A Ñ´ -Ñj c ¶ t It will now be assumed that the same modification can be introduced into the free-particle Dirac equation (c a · p + b mo c 2)y = E y so that the proper equation for a particle in a field with vector potential A and scalar potential j is: y [ ( ) b 2 j ]y E = α · cp - qA + moc + q In two-component form, by analogy to the equations c s · p yv + ( moc 2 – E )yu = 0 and c s · p yu - ( moc 2 + E )yv = 0, we set: c q m c q E · - - - = · - + + = ( ) ( ) y j y y ( ) ( ) v u u 2 c q m c q E u v v σ p A σ p A y j y y o 2 o 2)y = E y must now be modified to include effects due to external fields. Classical considerations suggest how this may be accomplished. In the presence of external fields a possible Lagrangian for the system is:
    • As in the free particle case, we shall be interested mainly in the “large” component yu . From the equations s · (c p – qA)yv + ( moc q ( ) 1 ¢ - = · K · j y u (σ π) (σ π)y u 2 E E m c K m c = - π p A 1 2 2 o o j j ¢ = - , = and + ¢- = ¢ + - The identity (s · A)(s · B) = (A · B) + i s · (A ´ B) gives (s · A)2 = p2 + i s · (p ´ p) = [p – (q /c)A]2 – i (q /c) s · (p ´ A + A ´ p). But p ´ A = -i  Ñ ´ A - A ´ p, therefore we get: 2014 MRT 2 + qj)yu = Eyu and s · (c p – qA)yu - ( moc 2 - qj)yv = Eyv, we set: where Our objective now is to obtain an approximation to the equation above to order v 2/c 2. c E m c q E q m c 1 [( ) / 2 ] 2 2 o 2 o 2 o m E q ö · = æ - q q 2 ( )  A σ A p π σ ´ Ñ · - ÷ø çè m c c o and the approximation to (E ¢– qj )yu = (1/2mo)(s · p) K (s · p)yu with K = 1 is: ù é q q ( ) 1  ¢ - = æ - Ñ´ ö çè j y u p A σ A y u ú úû ê êë - · m c ÷ø c m E q o 2 2 o 2 2 2
    • σ π σ π W ¢ - W = ¢- - · ¢ - - ¢ - · ( ) ( ) ( ) ( ) ( ) ( ) 2 2 - - j y j y j y j y 2014 MRT The next higher approximation is obtained by setting: K E¢ - qj 2 m c 2 o = - 1 So that the equation (E ¢– qj )yu = (1/2mo)(s · p) K (s · p)yu now becomes: ù é - · ¢ E q u σ π σ π σ π y u j y j ( ) 1 2 2 ¢ - = · - · ( ) 2 o m c m E q Let y = Wyu where W = 1 + (s · p)2/8mo To order v é ¢- ¢ W W = W σ · π - σ · π E - q σ · π úû - 1 ( ) - 1 1 - ( ) W- 2/c 2, we get: ê êë 2c ( ) ( ) 2 and, to order v ú úû 2 o 2/c 2, W-1 = 1 - (s · p)2/8mo 2c 2. Multiplying the equation above on the left by W-1 and replacing yu by W-1yu one obtains: j y j 1y 2 o 1 2 o 2 ( ) ( ) 2 ù êë m c m E q σ π σ π σ π W · W = · - · ( ) ( ) ( ) 1 4 y y y - · ¢ - - - · W = · ¢ W · E q E q j y j y ( ) 4 - - ( ) ( ) 2 ( ) 2 1 8 2 2 1 8 8 2 2 o 1 2 o 1 o 3 2 o 2 o 1 2 1 o 2 2 o 2 2 o 1 1 σ π σ π σ π σ π m c m c m m m m c m c E q E q m c E q E q
    • Hence, the transformed equation W-1(E ¢ - qj ) W-1yu, to order v é 2 σ π σ π σ π ¢ - = · - · + · ¢- ( ) ( ) ( ) 1 ( ) 1 j y j - · ¢- · + ¢ - · 2 ù ( )( )( ) ( ) ( ) j j y · ¢- - ¢ - · = · - · = · σ π E q E q σ π q σ p q σ p i q σ j j j j j Ñ Ñ Ñ Ñ ´Ñ ( )( ) ( )( ) ( ) i i · ¢ - + ¢- · = · · ¢ - - ¢ - · - σ π E q j E q j σ π σ π σ π E q j E q j σ π ( ) ( ) ( )( ) ( )[( )( ) ( )( )] - · ¢- - ¢ - · · + · ¢ - · σ π E q j E q j σ π σ π σ π E q j σ π [( )( ) ( )( )]( ) 2( )( )( ) = · - · + · ¢ - · q  Ñ Ñ q  ´Ñ E q Substitution of the above simplification into the equation (E ¢ - qj )y = […]y above gives: 2014 MRT Therefore: 2/c 2, becomes: A number of simplifications are possible with the aid of: úû êë 2 2 o 2 2 o 2 2 o 4 3 2 o 2 o 8 4 1 ( ) 8 8 2 m c E q E q m c E q m m c m c E q σ π σ π σ π j j j j Ñ σ · π σ · - σ · σ · π = - · + σ · π ( )( ) ( )( )  2  2 2( )( )( ) 2 2 2 σ π σ π σ π j j j ù é 2 ( ) 1 ( ) 1 q q E q   ¢ - = σ · π - σ · π + Ñ·Ñ - σ · π ´Ñ 2 2 y j j y j úû êë o 2 2 o 4 3 2 o 2 ( ) o 8 8 4 2 m c m c m m c
    • Also, from the relation (s · A)2 = [p – (q /c)A]2 - (q /c) s · Ñ ´ A obtained earlier, we get: and, to order v ( ) 1 · = æ - q q A σ A p π σ ´ Ñ · - ÷ø m m o 2/c 2, we finally have: 2014 MRT 1 ( σ π ) 4 p · = 2 4 1 1 1 ( 1 4 c π p p p ´Ñj ´Ñj Ñ´Ñj Ñj ´ ) Ñj ´ 2 2 2 2 c i c = = - - = - c c c  ö çè m c c 2 o 2 o 2 2 2 1  With q = -e, where e is the electronic charge, we obtain for an electron to order v ( ) 1 where E ¢ = E – moc 2/c ù é e ö çè Ñ´ 2 2 2 4 e e p e  2. This equation, which may be regarded as the Schrödinger equation for an electron interacting with fields describable by the potentials A and j , is the starting point for discussions of atomic and molecular properties. 2 : j j y j y úû - - · - · ê êë - · + ÷ø ¢+ = æ + σ p p A σ A Ñ Ñ Ñ ´ 2 2 o 2 2 o 3 2 o o 2 o 8 8 4 m c m c m c m c c m E e  
    • This significance of the various terms and their energies, indicated to within an order of magnitude, are (the ‘cm-1’ scale is given by the wave number 1/l @ 8000 cm-1): This is the energy associated with the scalar potential energy, j (105 cm-1). This contains the kinetic energy (i.e., p2 /2mo) and interaction term (i.e., (e/2moc)(p · A + A · p) + e2A2/2moc2) with a field represented by a potential vector A (105 cm-1). The interaction terms are responsible or contribute to numerous physical processes among which are absorption, emission and scattering of electromagnetic waves, diamagnetism, and the Zeeman effect. The interaction of the spin magnetic moment (i.e., mS = 2×e/2mo× /2) with a magnetic field B = Ñ ´ A (1 cm-1). Thus, it is the magnetic moment of one Bohr magneton, e /2moc (i.e., mB = 9.2741´10-24 A×m2 or J/T) with the magnetic field. 2014 MRT ej This term appears in the expression of the relativistic energy: m c p c m c p 2 2 2 2 o 2 8 p and is therefore a relativistic correction to the kinetic energy (i.e., p2/2m) (0.1 cm-1). The spin-orbit interaction (10-103 cm-1). More precisely, it is (e /8moc2){s · [p - (e/c)A] ´ E - s · E ´ [p - (e/c)A]} and it arises from the fact that the motion of the magnetic moment gives rise to an electric moment for the particle which then interacts with the electric field. 2 1 2 o ö ÷ø e æp + A çè c m e ö 1 æ p + A σ ·Ñ´A e  2 o m c 4 p 3 2 o 8m c + @ + - + 3 2 o 2 2 o ( ) m c m 4 e - Ñ·Ñj 2 2 o 2 8m c This term produces an energy shift in s-states and is known as the Darwin (1887-1962) term (< 0.1 cm-1). It is thus a correction to the direct point charge interaction due to the fact that in the representation (Foldy-Wouthuysen), the particle is not concentrated at a point but is spread out over a volume with radius whose magnitude is roughly that of a Compton wavelength, /moc. e - σ ·Ñj ´ p 2 2 4moc ej c m - ÷ø çè 2 2 o As a combination, this term represents the interaction of a point charge with the electromagnetic field.
    • The solution in the Coulomb field has important applications, particularly to the energy levels of the hydrogen atom and to the calculation of x-ray spectra due to the K and L electrons of the heavy elements. The atomic number is the number of protons found in the nucleus of an atom and therefore identical to the charge number of the nucleus. It is conventionally represented by the symbol Z. The atomic number uniquely identifies a chemical element. In an atom of neutral charge, the atomic number is also equal to the number of electrons. In the general case of hydrogen-like atoms (e.g., the Hydrogen atom H1 has Z = 1), the energy levels Enj are described by the total angular momentum quantum number j with: 2014 MRT 2 2 E m c Z n j ( ) 2 é ¢+ + - 2 2 2 2 1 2 o 1 ù úû êë = + n j a Z a where n¢ = 0,1,2,….; j = 1/2,3/2,…; a = e 2/4pc @ 1/137 and can be approximated by: ù 2 2 E m c 1 a m c a Z n j 4 » - +  ú úû é ê êë ö + ÷ ÷ ø æ ç ç n j n è - + Z n 1 1 3 2 2 1 2 2 2 o 2 2 o and where the principle quantum number n = n¢ + j + ½ = 1,2,…. The observed fine structure of the levels of hydrogen and hydrogen-like atoms, particularly Helium He+, is in good agreement with the Dirac theory. FIN
    • References / Study Guide C. Harper, Introduction to Mathematical Physics, Prentice Hall, 1976. California State University, Haywood This is my favorite go-to reference for mathematical physics. Most of the complex variable and matrix definitions, &c. are succinctly spelled out and the wave packet discussions served as a primer followed by more in-depth discussion reference of Appendix A of Sakurai’s book below and also the solutions to the wave function (i.e. Radial, Azimutal and Angular) are very well presented and treated in this very readable 300 page volume. J.J. Sakurai, Modern Quantum Mechanics, Revised Edition, Addison-Wesley, 1994. University of California at Los Angeles Only half (Chapters 1-3 or about the first 200 pages) of this book makes for quite a thorough introduction to modern quantum mechanics. Most of the mathematical and angular momentum framework (i.e. orbital, spin and total as well as spherical harmonics discussed earlier) is based on Sakurai’s post-humanous presentation of the subject. S.S. Schweber, An Introduction to Relativistic Quantum Field Theory, Dover, 1989. Brandeis University Quite a voluminous 900 page tome written by a colleague of Hans Bethe. Most of the postulates and rotation invariance discussion, which was really the base or primer for the rest, is based on the first 50 pages of this book. The other 50 pages formed the basis of the relativistic material as well as the Klein-Gordon and Dirac equations. S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press, 2014. Josey Regental Chair in Science at the University of Texas at Austin Soon to become the seminal work on Quantum Mechanics, this ‘course’ from Weinberg fills-in all the gaps in learning and exposition required to fully understand the theory. Including the Historical Introduction, I would definitely suggest Chapter 2-4 and do the problems! Once you do, you will have truly mastered the theory to handle the next set of books from Weinberg – especially QFT I & II. S. Weinberg, The Quantum Theory of Fields, Volume I, Cambridge University Press, 1995. Josey Regental Chair in Science at the University of Texas at Austin Volume 1 (of 3) introduces quantum fields in which Chapter 1 serves as a historical introduction whereas the first 5 sections of Chapter 2 make for quite an elevated introduction to the relativistic quantum mechanics discussed here. Weinberg’s book was crucial in setting up the notation and the Lorentz and Poincaré transformations and bridging the gap for the relativistic one-particle and mass zero states using Wigner’s little group. Very high level reading! M. Weissbluth, Atoms and Molecules, Academic Press, 1978. Professor Emeritus of Applied Physics at Stanford Weissbluth’s book starts by knocking you off your chair with Angular Momentum, Rotations and Elements of Group Theory! The text is mathematical physics throughout but where more than one sentence occurs, the physics (or quantum chemistry) abounds. Chapter 15 starts Part III – One-electron Atoms with Dirac’s equation and in Chapter 16, it offers most of the explanation of the coupling terms generated from the Dirac equation. 2014 MRT