1. NPTEL β Physics β Mathematical Physics - 1
Lecture 14
Matrix representing of Operators
While representing operators in terms of matrices, we use the basis kets to compute the
matrix elements of the operator as shown below
< π·1|π₯|π·1 >< π·1|π₯|π·2 > β¦ β¦ β¦ β¦ β¦ β¦ < π·1|π₯|π·π >
π =
(< π·π|π₯|π·1 > β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . < π·π|π₯|π·π >)
Where |π·π >βs are the eigenkets of X
The expectation value of X is
< π > =< π·|π|πΉ >
= βπ,π < πΉ|π·π >< π·π|π₯|π·π >< π·π|πΉ >
If two operators A, B commute then they have same set of eigenkets.
Change of basis
Suppose we have t observables A and B. The ket space is spanned by {|π·
>} and
.
.
.
.
.
.
.
.
{|π₯ >}. e.g. for a spin 1
system, we can have a description in terms of |π Β±> or |π Β±> or
2 π§ π₯
|π π¦Β±> where π π, (π β π₯, π¦, π§) are the components of the spin 2
matrices. Either of
these
different set of base kets span the same space. We are interested in finding out how these
two descriptions are related.
Changing the set of base kets is referred to as a change of basis or a change of
representation. Thus given two sets of basis kets, both satisfying orthonormality
and completeness, there exists an unitary operator U such that
{|π₯ >} = π{|π· >}
Where U is an unitary operator satisfying,
ππβ = 1
and πβ π = 1
Eigenvectors corresponding to different eigenvalues are linearly independent. Let π’, π£ are
eigenvectors of the matrix A corresponding to π1and π2,
then ππ’ + ππ£ = 0 if they are not independent.
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Opearating by A. π, π β 0.
ππ1π’ + ππ2π’ = 0
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(1) and (2) have unique solutions if
Det|
π π
| = 0 β ππ(π β π ).
ππ1 ππ2
Thus π, π = 0
2 1
Hence they are linearly independent.
Summarizing, a change of basis kets is an unitary operation that establishes the
link between the two kets.
Linear Algebra in Quantum Mechanics
We have seen in quantum mechanics that the state of an electron subjected
to some potential is given by a wave function Ξ¨(πβ, π‘) which is obtained from the
solution of a linear second order differential equation, called the Schrodinger
equation. This implies that if Ξ¨1 and Ξ¨2 are the solutions for the Schrodinger
equation, then πΆ1Ξ¨1 + πΆ2Ξ¨2 is also a solution of the equation with πΆ1 and πΆ2 as
arbitrary complex numbers.
It is actually the linearity of Schrodinger equation that allows for the superposed
solution as a valid solution. The commutativity, associativity for Ξ¨ are trivially
satisfied.
Another close link between the content of this chapter to quantum mechanics can
be mentioned as follows. Since all the physical observables correspond to
real values, they must correspond to Hermitian matrices. As an example, the
students can be asked
that operator like the momentum (πΜπ₯ = βπβ ) or the Pauli matrix ππ¦ (
π
ππ₯
contain π(= ββ1),
however are still unitary
and yield real
eigenvalues.
0 β π
π 0
) which
Also for another example, one can show that eigenvectors of ππ¦ corresponding to
the (real) eigenvalues (Β±1) are orthogonal.
Consider a basis set {π’π} for (π = 1, 2 β¦ β¦ . π) in a π π space. The set {π’π} is
orthonormal if < π’π |π’π > = β« πΌ3π π’π
β (πβ)π’π (πβ) = πΏππ
Thus every function Ξ¨(πβ) β π π can be expanded in one (and only one) way in
terms of
π’π(πβ)
Ξ¨(πβ) = βπ πππ’π (πβ) which means that the set ππ is unique and can be found by,
ππ = < π’π |Ξ¨ > = β« π3π π’π
β(πβ)Ξ¨(πβ).
Let π·(πβ) and Ξ¨(πβ) be two functions which are expand using the basis {π’π}
π·(πβ) = βπ ππ π’π(πβ) Ξ¨(πβ) = βπ ππ π’π (πβ)
The inner product of these two functions is expressed as,
< π·|Ξ¨ > = βπ ππ
β
ππ
and < Ξ¨|Ξ¨ > = βπ|ππ|2.
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Change of basis
For a plane wave,
Ξ¨(π₯) =
Ξ¨Μ (p)
=
1
β2πβ ββ
β πππ₯
β« ππ
Ξ¨Μ (π)π
ββ
1
β2Οβ ββ
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β βπππ₯
β« ππ₯
Ξ¨(π₯)π
ββ
< Ξ¨(π₯)|Ξ¨(π₯) > = < Ξ¨Μ (π)|Ξ¨Μ (π) >
This is known as Persevalβs identity.
Projection Operators
A projection operator p is a self adjoint operator satisfying the condition,
π2 = π
Let π1 and π2 be two projection operators, then the question is whether π1 + π2 is also a
projection operator.
(π1 + π2)2 = π1
2 + π2
2 + π1π2 + π2π1
π1 + π2 is also a projection operator if
π1π2 + π2π1 = 0 (1)
That is if they anticommute. The relation between π1 and π2 is more clearly expressed by
left multiplying (1) by π1,
π2 + π1π2π1 = 0
Again right multiplying by π1
π1π2π1 + π2 = 0
Subtracting (2) and (3)
π1π2 β π2π1 = 0
Thus (4) and (1) are simultaneously satisfied if π1π2 = 0
= π2π1.
(2)
(3)
(4)
Thus the necessary and sufficient condition are that the two operators have to be mutually
orthogonal.
Some important properties of Hermitian Matrices
1.Two Hermilian matrices can be simultaneously diagonalised if and only if they
commute. The converse is also true.
2.The angle of rotation π for the rotation described by a proper orthogonal matrix
π π is given by,
2πππ π = π‘π[π π] β 1
3.For a matrix:
a) All eigenvalues π are distinct. There are three mutually orthogonal
eigenvectors.
b) Two of the eigenvalues are equal π1 = π2. The corresponding eigenvector is
unique. But one can always generate an arbitrary orthogonal eigenvector.
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NPTEL β Physics β Mathematical Physics - 1
Thus these two along with the eigenvector corresponding to π3 constitutes the three
eigenvectors.
c) π1 = π2 = π3 any three mutually perpendicular eigenvectors will do.
To summarize this chapter, a little more can be said about the Hermitian matrices and
their relevance to quantum mechanics. To prove that the eigenvalues as real, one may
proceed in the following manner.
π΄|πΌ >= π|πΌ >
Inner product with πΌ > yields
< πΌ|π΄|πΌ >= π < πΌ|πΌ >
The adjoint equation can be denoted as,
< πΌ|π΄ = πβ < πΌ| (remembering A = π΄β )
Again taking an inner product with |πΌ > yields the result π = πβ , or in other words,
the eigenvalues of Hermilian matrices are real.
Likewise, the eigenkets corresponding to different eigenvalues are orthogonal.
π΄|πΌ >= π|πΌ >
π΄|π½ >= π½|π½ >
< πΌ|π΄|π½ >= π½ < πΌ|π½ >
< π½|π΄|πΌ >= πΌ < π½|πΌ >
Because of the Hermilian nature of the eigenvalues,
< πΌ|π΄|π½ >= π½|π΄|πΌ >
Hence (πΌ β π½) < πΌ|π½ > = 0
Thus if πΌ β π½ < πΌ|π½ > = 0
So they are orthogonal. Further eigenvectors of a Hermitian matrix span the space.
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Tutorials
1.Prove each of the following,
a) P is orthogonal if and only if ππ is orthogonal
b) If P is orthogonal, then πβ1is orthogonal.
c) If P and Q are orthogonal, then PQ is orthogonal.
Ans: a) (ππ)π = π.
Thus P is orthogonal if and only if πππ = πΌ if and only if (ππ)πππ = πΌ, that is, if and
only if ππ is orthogonal.
b) We have ππ = πβ1 since P is orthogonal. Thus by part (a) πβ1 is orthogonal. c) We
have ππ = πβ1, ππ = πβ1
2. Prove that momentum operator in quantum mechanics is a Hermitian Operator.
< ππ₯ > = β« βπ₯
Ξ¨β (β ) Ξ¨dx
β πβ β
ββ
β πΞ¨
= - πβ β«
Ξ¨β
ππ₯
ββ ππ₯
= -πβ [Ξ¨β
Ξ¨ β« β β« Ξ¨(π₯) ππ₯]
dΞ¨ β
dx
β β
ββ ββ
0
= iβ β« Ξ¨(π₯) ππ₯
πΞ¨ β
ππ₯
β
ββ
= < ππ₯ >β
3. A particle of mass m is in a one-dimensional box of width a. At
π‘ = 0, the particle is
in the state Ξ¨(π₯, 0) =
3π· +4π·
2 9
β25
, The π· function are orthogonal eigenstates of the
π
Hamitian π»π·π = β2βa π ππ (ππ₯βπ) what will be the measurement of energy, E yield at
π§ = 0. What is the probability of finding this value ?
Ans: First check that πΉ is properly normalized.
< πΉ|πΉ > = 1
Ξ¨ = βπ ππ|π·π >, π2 = , π9 = , ππ = 0
3 4
β25 β25
for π β 2 or 9.
P(πΈ2) = 25
and π(πΈ9) = 25
.
9 16
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In an ensemble of 2500 identical one-dimensional boxes each containing
an identical particle in the same state Ξ¨(π₯, 0). Measurement E at π‘ = 0 , finds
900 particle of each πΈ2 = 4πΈ, and 1600 particles πΈ9 = 81πΈ.
4. It can be seen that πββββ1β, βπβββ2β and πββββ3β are all linearly independent as they do
not lie in a plane. Thus, a unit vector in the direction of βπβββ1β is
π£βββ = = [2, 3, 0] = [0.554, 0.831,
0].
1
πβββββ1β
1
|πββββββ| β3
1
We subtract from πββββ2β, the component of πββββ2β in the
direction of βπ£ββ1β.
π£ββββ2β = βπβββ2ββ< βπ£ββ1β|πββββ2β > βπ’βββ1β
< βπ£β1β|πββββ2β > = 4.16 β< βπ£ββ1β|πββββ2β > βπ£ββ1β = [2.30,
3.45, 0]
Hence, π£ββββ2β = [3.7, β2.45, 0].
Hence calculate the unit vector in the direction of βπ’βββ2β
π’
βββββ
=
2
π£
ββββ
2β
|π£
βββββ
|
2
Finally calculate π£ββββ3β = βπβββ3ββ < βπ£ββ1β|πββββ3β > βπ£
ββ1β
The components of βπβββ3β in the direction of
βπ£β1β and βπ£βββ2β are taken out. That is substract β < βπ£
βββ2β|πββββ3β > βπ£βββ2β
So the corresponding orthonormal set is given by,
πΜ
=
π
π£
β
βπ
|βπ£
β|
π
5. Consider a basis {πβββ1 , βπββ2 } = {( ) , (
)} to represent
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1 β2
2 3
a vector π£ = ( ) in a two
3
5
dimensional space such that,
π£ = π1βπββ1 + π2βπββ2
Find a 2x2 matrix [A], such that
π£ = [π΄].
π’β
where π’β = (π
2
)
π1
Solution: First let us find π’
β
π1 ( ) + π2 ( ) = ( )
1 β2
2 3
3
5