Review of Seiberg Witten duality.pptx
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A review of the work of Seiberg and Witten in 1994.
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Review of Seiberg Witten duality.pptx
- 1. Review of Seiberg Witten duality MuhammadHassaan Saleem Supervisor: Dr. Bushra haider
- 2. Objective and structure We want to work out the form of low energy π = 2 SYM βπ=2 = 1 16π πΌπ β« π2 π β±β² β² π ππΌ π πΌ + β« π2 π π2 π π β±β² π where β± π is known as the prepotential. We essentially need to find the prepotential. The work is divided as follows: 1) Review of π© = 2 SYM theory and the topics needed to understand it. 2) Review of Olive Montonen duality (1977) 3) Review of Seiberg Witten duality (1994)
- 3. π© = 2 SYM
- 4. SUSY algebra The SUSY algebra is Where ππΌπ½ are called central charges and they are antisymmetric in their indices. Moreover, πππ πΌ π½ = ππΌπΎ π ππ πΎπ½ β (π β π) where ππ = (1, π) and ππ = (1, βπ).
- 5. Massless multiplets Take π π = (πΈ, 0,0, πΈ) and we get π1 πΌ , π1 π½ = 0 , π2 πΌ , π2 π½ = 4πΈ. So π1 πΌ = π1 πΌ = 0 (Positivity of Hilbert space). Define ππΌ = 1 4πΈ π2 πΌ , (πβ )πΌ = 1 4πΈ π2 πΌ β ππΌ , ππ½ = 1 The states πβ 1 πβ 2 β¦ πβ π |πβ© are built and they are 2π in number if πΌ = 1,2, β¦ π. N=1 multiplets π = 0 β 0, 1 2 β β 1 2 , 0 (chiral multiplet) π = 1 2 β 1 2 , 1 β β1, β 1 2 gauge multiplet N=2 multiplets π = 0 β 0, 1 2 , 1 2 , 1 β β1, β 1 2 , β 1 2 , 0 (gauge multiplet) π = β 1 2 β β 1 2 , 0,0, 1 2 β β 1 2 , 0,0, 1 2 (hypermultiplet)
- 6. Massive multiplets (π© = 2) Let π π = (π, π) and then , we have ππΌ πΌ , ππ½ π½ β = 2ππΏπΌπ½πΏπΌπ½ , ππΌ πΌ , ππ½ π½ = ππΌπ½ππΌπ½ , ππΌ πΌ β , ππ½ π½ β = ππΌπ½ πβ πΌπ½ Since ππΌπ½ is anti symmetric, we can write it as ππΌπ½ = 0 π βπ 0 , π > 0 Define π΄πΌ = 1 2 ππΌ 1 + ππΌπ½ ππ½ 2 β , π΅πΌ = 1 2 ππΌ 1 β ππΌπ½ ππ½ 2 β β π΄πΌ, π΄π½ β = 2π β π πΏπΌπ½, π΅πΌ, π΅π½ β = 2π + π πΏπΌπ½ For positivity of Hilbert space, 2π β π β₯ 0. If 2π = π then it is called a BPS multiplet (or short multiplet).
- 7. Chiral and vector superfield In superspace (π₯π , π, π), a general field is πΉ π₯π , π, π = π π₯ + ππ π₯ + ππ π₯ + πππ π₯ + πππ π₯ + πππ ππ£π π₯ + ππππ π₯ + ππππ π₯ + πππππ(π₯) While the SUSY transformation is; πΉ β πβπ ππ+ππ πΉππ ππ+ππ , πΏπΉ = π ππ + ππ πΉ, ππΌ = βπ π πππΌ β ππΌπ½ π ππ½ ππ, ππΌ = π π πππΌ + ππ½ ππ½πΌ π ππ Define π·πΌ = π πππΌ + πππΌπ½ π ππ½ ππ, π·πΌ = π πππΌ + πππ½ ππ½πΌ π ππ, π·πΌ, ππ½ = 0 Chiral superfield π·πΌπ = 0 chiral , π·πΌπ = 0 (anti β chiral) π·πΌ π = 0, π·πΌπ¦π = π·πΌ π₯ π + πππ½ ππ½πΌ π ππΌ = 0 β π π, π¦π = π§ π¦ + 2ππ π¦ β πππ(π¦) Vector superfield Letβs impose πΉ = πΉ and to accommodate π£π β π£π + πππΌ gauge, letβs have πΉ β πΉ + π + π gauge (Wess Zumino gauge). Then we have π = πππ ππ£π + πππππ β πππππ + 1 2 πππππ Non abelian gauge is ππ β ππΞ©β ππ πβπΞ©β but the form of vector superfield is the same.
- 8. π© = 1 SYM The following actions are supersymmetric β« π4 π₯β« π2 ππ2 π πΉ, β« π4 π₯β« π2 π π , π·πΌπ = 0, π2 π = 1 2 ππππ Now, π πΌ = β 1 4 π·2 (πβπ π·πΌππ ) Is chiral as π·3 = 0. Moreover, π πΌ β ππΞ π πΌπβπΞ as usual for non abelian theories. The gauge invariant term is π‘π(ππΌ π πΌ). βπ=1 = πΌπ π 32π β« π2 π π‘π ππΌ π πΌ = π‘π β 1 4 πΉππ πΉ ππ β ππππ π·ππ + 1 2 π2 + Ξπ2 32π2 πΉππ β πΉ ππ Where π = Ξ 2π + π 4π π2 , π·πππ½ = ππππ½ β π 2 π£π, ππ½ , πΉ ππ = πππ£π β ππ π£π β π 2 π£π, π£π , β πΉ ππ = 1 2 ππππΌπ½πΉπΌπ½
- 9. π© = 1 SYM with matter The chiral field is ππ invariant under the gauge transformation ππ β ππΞ π π ππ , Ξ = Ξπ ππΊ π , DπΌΞ = 0 which implies that πβ π2ππ π is gauge invariant and thus, the matter lagrangian is βπππ‘π‘ππ = β« π2 π π2 π ππ2ππ π = π·ππ§ β π·π π§ β ππππ π·ππ + ππ + ππ 2 π§ππ β ππ 2 πππ§ + ππ§ππ§ where π·ππ§ = πππ§ β π 2 ππ£π π ππΊ π π§ So, the total π© = 1 lagrangian is βπ©=1 = βπππ’ππ + βπππ‘π‘ππ = πΌπ π 32π β« π2 π π‘π ππΌ π πΌ + β« π2 π π2 π ππ2ππ π = π‘π β 1 4 πΉππ πΉ ππ β ππππ π·ππ + 1 2 π2 + Ξπ2 32π2 πΉππ β πΉ ππ + π·ππ§ β π·π π§ β ππππ π·ππ + ππ + ππ 2 π§ππ β ππ 2 πππ§ + ππ§ππ§
- 10. π© = 2 SYM In π© = 2 theory, the gauge representation is adjoint with the representation matrices ππ . Using this, we can write the terms in π© = 1 lagrangian as; π·ππ§ π β π·π π§ π = π‘π π·ππ§ β π·π π§ , πππ πππ·π π π = π‘π ππππ·π π , ππππ = π‘π [ππ] π§πππ ππ π π ππ = π‘π π§ π, π , πππ§ = π‘π π, π π§ , π§πππ ππ π π π§π = π‘π π π§, π§ Making these arrangements, the π© = 1 lagrangian becomes; β = π‘π β 1 4 πΉππ πΉ ππ β ππππ π·ππ + 1 2 π2 + Ξπ2 32π2 πΉππ β πΉ ππ + π·ππ§ β π·π π§ β ππππ π·ππ + ππ + ππ 2π§ π, π β ππ 2 π, π π§ + ππ π§, π§ It is symmetric under the exchange π β π (and also under the π symmetry) so it is π© = 2 supersymmetric. So, the π© = 2 lagrangian is; βπ©=2 = πΌπ π 32π β« π2 π π‘π ππΌ π πΌ + β« π2 ππ2 π π‘π πβ π2ππ π
- 11. Low energy π© = 2 SYM SUSY breaking The field equations for ππ and ππ give; ππ = 0, ππ = βπ π§, π§ π This gives the scalar potential π = ππ + 1 2 ππ2 = 1 2 π π§, π§ 2 For SUSY to break, the potential should vanish as ππΌ πΌ Ξ© 2 + ππΌ πΌ Ξ© 2 = 4 Ξ© π0 Ξ© β₯ 0 β ππΌ πΌ Ξ© = 0 β πΌ if Ξ© π0 Ξ© = 0 β π = 0 Low energy effective lagrangian In π© = 2 lagrangian, ππΏππ can be replaced by an arbitrary function πππ(π§) and ππ can be replaced by an arbitrary function πΎπ(π§) (details in the sigma model). πππ π§ and πΎπ(π§) can be expressed in terms of prepotential as πππ π§ = π2 4ππ β±ππ π§ , πΎ π§, π§ = π2 8ππ π§π πβ± π§ ππ§π β π§π πβ± π§ ππ§π This gives the following effective lagrangian βπ©=2 eff = 1 16π Im β« π2 π β±ππ π ππΌπ π πΌ π + β« π2 π π2 π ππ2ππ π β±π(π)
- 13. What is duality? Duality is the presence of two different perspectives of a single problem or of a single physical system. The main examples of the presence of duality would include the wave particle duality. A physical system can be viewed in the position space wave function (i.e. β¨π|π₯β©) and it can also be seen in the momentum space wave function (i.e. β¨π|πβ©). Another example for the concept of duality would be seen in the harmonic oscillator with the lagrangian. πΏ = π2 2π β 1 2 ππ2 π₯2 It is easy to see that in the replacement π₯ β π mπ , π β βπππ₯ the harmonic oscillator is self dual as can be seen below πΏ β βπππ₯ 2 2π β 1 2 ππ2 π ππ 2 = β π2 2π β 1 2 ππ2 π₯2 = βπΏ So, the lagrangian changes by an overall factor but that doesnβt change the equations of motion.
- 14. Electromagnetic duality Sourcless Maxwell equations (MEs) are invariant in the transformations π·: πΈπ β π΅π , π΅π β βπΈπ . It can be generalized to πΈπ = πΈπ cos π + π΅π sin π , π΅π = βπΈπ sin π + π΅π cos π and MEs will still be invariant as can be seen below for the Faraday and Maxwell laws (written in term of indices and ππ‘ is time derivative); πππππππ΅π = ππ‘πΈπ πππππππΈπ = βππ‘π΅π β ππππ ππ cos π πΈπ + sin π π΅π = βππ‘ cos π π΅π β sin π πΈπ β πππππππΈπ β² = βππ‘π΅π β² ππππ ππ cos π π΅π β sin π πΈπ = ππ‘ cos π πΈπ + sin π π΅π β πππππππ΅π β² = ππ‘πΈπ β² Introduce πΉ ππ with πΉ0π = πΈπ, π΅π = 1 2 πππππΉ ππ. So we can visualize π· transformation as πΈπ β π΅π β πΉ0π β 1 2 πππππΉ ππ =β πΉ0π, π΅π β βπΈπ β 1 2 πππππΉ ππ β βπΉ0π = 1 2 ππππ β πΉ ππ, β πΉ ππ = 1 2 ππππΌπ½πΉπΌπ½ So, we can see that the π· transformation as πΉ ππ =β πΉ ππ. If there are no monopoles, we can take π΅π = 1 2 ππππ πΉ ππ = β Γ π΄ π = 1 2 ππππ πππ΄π β πππ΄π , πΈπ = πΉ0π = ββπ β ππ¨ ππ‘ π = π0π΄π β πππ΄0 β πΉ ππ = πππ΄π β πππ΄π
- 15. Electromagnetism in Quantum Mechanics In Quantum Mechanics, the electromagnetic coupling can be realised by minimum coupling scheme ππ β βπ β β ieA π Moreover it is well known that the Schrodinger wave equation (SWE) is invariant in the gauge transformation π β πβπππ π, π΄π β π΄π β π π ππππ βπβπππ , ππππ β π(1) Consider a magnetic monopole at π = 0 and since we have a magnetic monopole now, it cant the case that π΅π = β Γ π΄ π everywhere. We can still define π΄π in patches and in the region where the patches overlap, they should differ by a gauge transformation. For π > π0 > 0, the magnetic field should look like; π© = ππ 4ππ2 Now, we want some π¨ to give the above magnetic field by taking itβs curl. We use the polar coordinate system and in this system, π© (for an arbitrary π¨ is) π΅π = 1 π π πππ π sin π π΄π ππ β ππ΄π ππ , π΅π = β 1 π π πππ ππ΄π ππ β sin π π(ππ΄π) ππ , π΅π = 1 π π(ππ΄π) ππ β ππ΄π ππ Since we want the π component of π© to vanish, we set π΄π = π΄π = 0. Moreover, π© falls as 1/π2 and thus, π¨ falls as 1/π. Therefore, ππ΄π should be independent of π. So, π΅π is zero as well. π΅π = 1 π π πππ π sin π π΄π ππ = π 4ππ2 β π΄π = π 4ππ π β cos π sin π Where π is arbitrary. [1] P.A.M Dirac (1931)
- 16. Now, we choose π¨ in two patches and then them π¨π and π¨π (i.e. for Northern patch and the Southern patch). We choose π = 1 for π¨π and π = β1 for π¨π and we have; π¨π = π 4ππ 1 β cos π sin π π, π¨π = β π 4ππ 1 + cos π sin π π The overlap region happens to be 2π = π circle and on that circle, the difference of the two potentials is π¨π π = π 2 β π¨π π = π 2 = π 2ππ π = ββ β π 2π π = ββπ So, the difference is a gauge transformation. The π(1) element ππππ should be continuous and thus, we have the following condition; πππ π 2π βπ 0 = 1 β π π 2π β π 0 = 2ππ, π β β€ Using the expression for π, we get ππ = 2ππ, π β β€ Comments ο A single monopole will ensure the quantization of electrical charge. ο ππ = 0 and ππ = 2π represent the same π(1) element and thus, the π(1) group is compact. So, we have the following two equivalent statements (they are the contrapositive of each other) πππππππππ β πΆππππππ‘ π 1 ππππ’π ππ πΆππππππ‘ π 1 ππππ’π β ππ ππππππππ π πππ’π‘ππππ Compact π(1) group will be realised when a theory with compact gauge group is spontaneously broken to a group containing π(1) as a product group.
- 17. The tβHooft, Polyakov monopole In the Diracβs treatment of the monopole, the interior of the monopole is not studied. In order to study the interior of a monopole, we study a monopole configuration called tβHooft, Polyakov monopole. It is nothing but a non abelian Yang Mills Higgs theory in 3 + 1 dimensions. The lagrangian is as follows; β = β 1 4 πΉ ππ π πΉ π ππ + 1 2 π·πππ π·π ππ β π π , π = π£ β 0 The EOM for πΉ ππ π is (also called Gauss constraint) is π·ππΉ π ππ = ππππππππ·π ππ We define πΉππ as πΉ ππ = ππ πΉ ππ π β 1 π ππππ ππ π·πππ π·πππ where ππ is a unit vector in the direction of ππ and for ππ to exist, ππ β 0 everywhere. Note that πΉ ππ is gauge invariant. It can be shown that πΉ ππ = ππ(ππ π΄π π ) β ππ (ππ π΄π π ) β 1 π ππππ ππ ππππ ππ ππ The proof is as follows. The definition of πΉ ππ can be expanded and written as follows; πΉ ππ = ππ πππ΄π π β πππ΄π π + πππππ π΄π π π΄π π β 1 π ππππ ππ ππππ + πππππ π΄π π ππ ππππ + πππππ π΄π π ππ = ππ ππ π΄π π β ππ ππ π΄π π β 1 π ππππ ππ ππππ ππππ +π΄π π ππππ πΏππ β ππππ ππππ ππππ ππ ππ β π΄π π πΏππ ππππ + ππππ ππππ ππππ ππ ππ + πππππ ππ π΄π π π΄π π β πππππ ππ ππππ ππππ π΄π π π΄π π ππ ππ
- 18. The terms that we actually want are the red terms. The extra terms can be shown to be zero. For example, the first bracketed expression is; ππππ πΏππ β ππππ ππππ ππππ ππ ππ = ππππ + ππππ πΏππ πΏππ β πΏππ πΏππ ππ ππ = ππππ + ππ ππ ππππ β ππππ = 0 Where I used the fact that ππ ππππ = 1 2 ππ ππ ππ = 1 2 ππ 1 = 0 The second bracketed term vanishes by a similar proof. Now we need to prove that ππππ ππ π΄π π π΄π π β ππππ ππ ππππ ππππ π΄π π π΄π π ππ ππ vanishes. For this task, we can manipulate the last term as; ππππ ππ ππππ ππππ π΄π π π΄π π ππ ππ = πΏππ πΏππ β πΏππ πΏππ ππ ππππ π΄π π π΄π π ππ ππ = ππππ ππ π΄π π π΄π π where I dropped the vanishing term ππππ ππ ππ ππ π΄π π π΄π π and used the fact that ππ ππ = 1. So, we can now see that ππππ ππ π΄π π π΄π π β ππππ ππ ππππ ππππ π΄π π π΄π π ππ ππ = ππππ ππ π΄π π π΄π π β ππππ ππ π΄π π π΄π π = 0 So, we have shown that πΉ ππ = ππ(ππ π΄π π ) β ππ (ππ π΄π π ) β 1 π ππππ ππ ππππ ππ ππ
- 19. Now, a magnetic field can be defined as π΅π = 1 2 πππππΉ ππ = ππππππ ππ π΄π π β 1 2π ππππππππ ππ ππππ ππππ If we try to calculate the surface integral of the magnetic field over a sphere at large distance, we will not get contribution from the first term in the expression for magnetic field as it is a curl. So, we get; π = β« ππππ΅π = β 1 2π π2 πππππππππππ ππ ππππ ππππ = β 1 2ππ£3 π2 πππππππππππ ππ ππππ ππππ This integral can be evaluated by choosing the form of ππ = π£ π₯π at infinity (where π₯π = π π₯π ). We simply get; 1 π£3 π2 πππππππππππππ ππππ ππππ = π2 π2 πΞ© π₯ππππππππππ₯ππππ₯ππππ₯π = π2 πΞ© π₯ππππππππππ₯π πΏπππΏππ β 2πΏπππ₯ππ₯π Where I dropped the vanishing terms and used the identity πππ₯π = 1 π πΏππ β π₯ππ₯π Now, using the identities ππππππππ = πΏπππΏππ β πΏπππΏππ, ππππππππ = 2πΏππ The above integral becomes; 2 π2 πΞ© π₯ππ₯π πΏππ β πΏππ + π₯ππ₯π = 8π β π = β 4π π Now, I can choose a mapping π: ππ 2 β ππ₯ 2 such that ππ 2 covers ππ₯ 2 π times. This will simply change the integral to be multiplied by π as we are going around the ππ 2 π times. So, we get the following condition ππ = 4ππ π β β€
- 20. Bogomolβnyi bound and BPS solution We calculate the energy of the static solution of YMH theory in the π΄0 π = 0 gauge and we get; πΈ = β« π3 π₯ 1 2 π΅π π π΅π π + 1 2 π·πππ 2 + π π = β« π3 π₯ 1 2 π΅π π β π·πππ 2 + π΅π π π·πππ + π π Using zeroth component of Bianchi identity ππΌπ½ππ π·π½πΉ ππ π = 0 and the definition of π΅π π i.e. 2π΅π π = πππππΉππ π , we conclude π0πππ π·ππΉ ππ π = π·ππ΅π π = 0. Moreover, since π΅π π ππ is gauge singlet, we have π·π π΅π π ππ = ππ(π΅π π ππ ). So, the second term in the energy integrand is ππ(π΅π π ππ ) and thus, we can use divergence theorem and it becomes; πΈ = β« π3 π₯ 1 2 π΅π π β π·πππ 2 + π π + π£ π2 ππππ΅π π ππ β₯ π£ π2 ππππ΅π π ππ = π£π So, the energy of the static monopole solution (mass of the monopole) has a lower bound known as Bogomolβnyi bound. If we impose the condition π΅π π = π·πππ (also known as the Bogomol'nyi equation) and set π π = 0 everywhere, then the energy of the field configuration is just π£π. Since we are setting π π = 0 everywhere, it seems as if we won't be able to get a nonzero β¨ππ ππ β©. However, since we need ππ ππ = π£2 at spatial infinity, we can impose this as a boundary condition. For BPS solution, if we try to give π΄π π (π₯) a non zero vev, it will break Lorentz invariance. Moreover, for the theory to have non zero topological charge, ππ should vary at infinity and thus, solutions can't be rotationally invariant. However, we can make the solutions invariant under the diagonal subgroup ππ 3 π Γ ππ 3 πwhere ππ 3 π is the global gauge group. We would not need the detailed form of the BPS solutions though.
- 21. Moduli space of BPS solution Moduli space is the space of fixed energy and fixed topological charge solutions. The coordinates on the moduli space are called moduli or collective coordinates. If ππ (π₯) is a BPS solution, so is ππ (π₯ + π) with π fixed due to translational invariance. So, β3 is present as a product in the moduli space of BPS monopole. Now, equation of motion π·ππΉ π ππ = ππππππππ·π ππ is expanded as; π·π ππ π΄π π β ππ π΄π π + ππππππ΄π π π΄π π = πππππππ[ππ ππ + ππππππ΄π π ππ] Set π = 0 and working in the π΄π 0 = 0 gauge, we get; π·ππ΄π π + π π, π π = 0 The linearized form of the gauss law for deformations to the BPS solutions (πΏππ , πΏπ΄π π ) is obtained by letting π΄π π β π΄π π + πΏπ΄π π , ππ β ππ + πΏππ and realising that only the deformations are allowed to be time dependant. We get π·ππΏ π΄π π + π π, πΏπ π = 0 We can expand the Bogomolβnyi equation as well and then linearize it as follows; ππππ ππ πΏπ΄π π + ππππππ΄π π πΏπ΄π π = πππΏππ + πππππ π΄π π πΏππ + π πΏπ΄π, π π β πππππ·π πΏπ΄π π = π·ππΏππ + π πΏπ΄π, π π The solution to the two red equations above is πΏπ΄0 π = 0, πΏππ = 0, πΏπ΄π π = π·π π π‘ ππ , π π‘ is arbitrary function of time. If π = 0 then the change in π΄π π is a large gauge transformation (it doesnβt vanish at infinity). The corresponding π(1) element is πππ and since unbroken π(1) is compact, the coordinate π is on a compact π1 . So, the moduli space of the BPS monopole is; β3 Γ π1
- 22. Witten effect We can add a π term in the lagrangian without breaking gauge invariance. The term is; βπ = β ππ2 32π2 πΉ ππ π β πΉ π ππ The prefactor is for later convenience. First, we can do a π(1) case (i.e. the QED case). QED case πΉππ β πΉ ππ = 1 2 ππππΌπ½ πΉ πππΉπΌπ½ = 1 2 π0πππ πΉ0ππΉ ππ + ππ0ππ πΉπ0πΉ ππ + πππ0π πΉπππΉ0π + ππππ0 πΉπππΉπ0 = 2π0πππ πΉ0ππΉ ππ = 4 π¬. π© β βπ = β ππ2 8π2 π¬. π© For a static monopole, we have π¬ = βπ΄0, π© = β Γ π¨ + π 4π π π β πΏπ = β« π3 π₯ βπ = π2 ππ 8π2 β« π3 π₯π΄0πΏ 3 (π) where I used the following fact; π¬. π© = β. π΄0 β Γ π΄ + π 4π π π = β π 4π π΄0β. π π + π. π = βππ΄0πΏ 3 (π) This lagrangian is nothing but the interaction of a particle at origin with background field. The charge of the particle is; β π2 ππ 8π2 ππ=4π β ππ 2π So, the π term can give electrical charge to the monopoles.
- 23. SU(2) case Consider large gauge transformations that act on π΄π π . The finite deformation (without infinitesimal parameter) is; πΏπ΄π π = 1 ππ£ π·πππ Let π© be the generator of this gauge transformation. Until now, we donβt have fundamental fields in this theory and the gauge parameter lives on π1 an thus, π2πππ© = 1. π© is the Noetherian conserved charge (as the gauge transformations act on the fields). It can be easily worked out as follows. Expression for π© π© = β« π3 π₯ πβ π π0π΄π π πΏπ΄π π Now, we write the lagrangian as; β = β 1 4 πΉ π ππ πΉ ππ π β ππ2 32π2 πΉ ππ π β πΉ π ππ + 1 2 π·πππ 2 So the required derivative is; πβ π π0π΄π π = β 1 2 πΉ π πΌπ½ ππΉπΌπ½ π π π0π΄π π β ππ2 32π2 ππππΌπ½ πΉπ πΌπ½ ππΉ ππ π π π0π΄π π From the definition of πΉ ππ π , the derivative appearing above can be calculated as; ππΉ ππ π π π0π΄π π = πΏπ π [πΏπ 0 πΏπ π β πΏπ 0 πΏπ π ] Using this expression and after some elementary simplification, we have; πβ π π0π΄π π = βπΉ π 0π β ππ2 16π2 π0πππ πΉ π ππ = βπππ πΈπ π β πππ ππ2 8π2 (π΅π)π = πΈπ π + ππ2 8π2 π΅π π Using all the red equations above, we get; π© = β« π3 π₯ 1 ππ£ πΈπ π + ππ2 8π2 π΅π π π·πππ = π π + ππ 8π2 π where π = π£β1 β« π3 π₯πΈπ π π·πππ π = π£β1 β« π3 π₯π΅π π π·πππ
- 24. Now, the condition π2πππ© = 1 implies that π© = ππ β β€ β π π + πππ 8π2 = ππ Using the condition ππ = 4πππ, we get; π π + πππ 2π = ππ β π = πππ β ππ 2π ππ So, we can see that for a single monopole (i.e. ππ = 1) there is an extra contribution to the electrical charge of the monopoles due to the non zero π term. This contribution is β ππ 2π This is equal to the electrical charge that we found in the QED case.
- 25. Olive Montonen and ππΏ(2, β€) duality Set π = 0 first. Now, there can be electrically charged states (ππ, 0) and magnetic monopoles 0, ππ . The mass of an electrically charged state is the mass of π boson i.e. ππ = π£π. The mass of a magnetic monopole is ππ = π£π β« π£π (at weak coupling). In a dual theory where the roles of electrical and magnetic charges are replaced, we should have π β π, ππ β ππ Now, the dual theory should be at strong coupling as small π means large π. Moreover, this proposal of a dual theory is based on the classical spectrum as was first given by Olive and Montonen in 1977. However, there are some problems pointed by them. ο Potential π(π) might not remain vanishing when quantum corrections are included. ο The states donβt match exactly as the monopoles donβt have spin 1 like the π bosons. ο The duality is hard to test because the dual theory is at strong coupling. We will see that the first two problems are solved by incorporating the theory in π = 4 super Yang Mills theory. Now, let π β 0. Now, the πππ» lagrangian is; β = β 1 4 πΉ2 β ππ2 32π2 πΉ β πΉ β 1 2 π·πππ 2 Now, we let ππ β ππ π , π΄π π β π΄π π π β πΉ ππ π β πΉ ππ π π After this rescaling of field, the lagrangian becomes (with covariant derivatives changed accordingly); β = β 1 4π2 πΉ2 β π 32π2 πΉ β πΉ β 1 2 π·πππ 2
- 26. Now consider this equivalent way of writing the first two terms of the lagrangian = 1 32π2 πΌπ π 2π + 4ππ π2 πΉ + π β πΉ 2 = β 1 8π2 πΉ2 ββ πΉ2 β π 32π2 πΉ β πΉ = 1 4π2 πΉ2 β π 32π2 πΉ β πΉ where in the last step, I used the fact that πΉ2 =ββ πΉ2 as can be demonstrated β πΉ2 = 1 4 ππππΌπ½ πΉπΌπ½ππππππΉππ = β 1 2 πΏπ πΌ πΏπ π½ β πΏπ πΌ πΏπ π½ πΉπΌπ½πΉππ = βπΉ2 So, the full lagrangian can be written in case of non zero π as; β = β 1 32π2 πΌπ π πΉ + π β πΉ 2 β 1 2 π·πππ 2 where π = π 2π + 4π π2 π I will use a result from instanton theory. The π instantons in this theory are weighed by the factor π2ππππ . So, physics is invariant in the transformation π β π + 1 (corresponding to π β π + 2π). Moreover, if we set π = 0, then the electromagnetic duality (π β π) corresponds to π β π = 4π π β π = 4ππ π2 β β π2 4ππ = β 1 π We can identify both of these transformations as special cases of a single general transformation π β ππ + π ππ + π , π, π, π, π β β€, ππ β ππ = 1 This transformation is known as the ππΏ(2, β€) transformation and we can identify π β π + 1 β π = π = π = 1, π = 0 π β β 1 π β π = π = 0, π = βπ = 1 We can see that πΌπ π = 4π π2 > 0 So, we are concerned with the upper half π plane only. We can also define a fundamental region such that any π in the upper half plane can be brought into this fundamental region by an appropriate ππΏ(2, β€) transformation. β 1 2 β€ π π π β€ 1 2 , π β₯ 1
- 27. πΊπ³ π, β€ action on the states The action of the ππΏ(2, β€) group on the states (ππ, ππ) states requires the electrical charge operator ππ and magnetic charge operator ππ = πππ β ππ 2π ππ, ππ = πππ = 4π π ππ Now, π β π + 1 β π β π + 2π and thus, we have; ππ β π ππ β ππ β ππ 2π ππ β ππ β ππ β ππ = πππ β πππ ππ β ππ = πππ + πππ Similarly for the π β βπβ1 transformation (for π = 0), we have ππ = πππ β πππ β ππ β ππ = πππ β πππ ππ = πππ β πππ β ππ β ππ = πππ + πππ So, we have the following ππΏ(2, β€) action on states (ππ, ππ) ππ ππ β π βπ π π ππ ππ Bogomolβnyi bound revisited The generalised Bogomolβnyi bound (proved at the end) π2 β₯ π£2 (ππ 2 + ππ 2 ) can entirely be written in term of π by using the definition of π and the expressions for ππ and ππ.
- 28. Coupling to fermions We can add fermions into this theory with proper interactions with ππ and π΄π π fields. Assume that fermions are in the β²πβ² representation of the ππ(2) gauge group. The fermion lagrangian is; βπ = ππππΎπ π·ππ π β ππππππ π ππ ππ Where πππ π are the anti Hermitian generators in the β²πβ² representation. The gamma matrices are chosen as; πΎ0 = βπ 0 π2 βπ2 0 , πΎπ = βπ ππ 0 0 βππ These matrices satisfy the Clifford algebra πΎπ , πΎπ = 2πππ π4 as can be verified easily. The Dirac equation can be worked out from the above lagrangian to get (using πππ = ππ πππ π ) ππΎπ π·ππ π β ππππππ = 0 We seek solutions of the form ππ π‘, π₯ = πβππΈπ‘ ππ(π₯) and we also set ππ π₯ = ππ + π₯ ππ β π₯ π . Using the πΎ matrices and the Diracβs equation, we have; βππΈ ππ β βππ + β ππ π·πππ + βππ π·πππ β β ππππ ππ + ππ β = 0 This gives us two equations; ππΏππππ π·π + πππ ππ β = π·ππππ β = πΈππ + , ππΏππππ π·π + πππ β ππ + = π·ππ β ππ + = πΈππ β , If the above two equations have a solution with πΈ = 0 then there can be fermionic collective coordinates in the BPS moduli space. In other words, we are looking for the kernels of π·ππ and π·ππ β . It is easy to see that; ker π·ππ β β ker π·πππ·ππ β = {0} In the last equality, I used the fact that π·πππ·ππ β is a positive definite operator.
- 29. For the kernel of π·ππ, I need a result which is derived from index theorems. The result says that; dim [ker(π·ππ)] β dim [ker(π·ππ β )] = π΄(π)ππ Where π΄(π) is a number that depends on the representation βπβ². For our work, the representations we need are π and π. The values of π΄(π) for these representations are π΄ π = 1, π΄ π = 2 Fundamental fermions For the fermions in the fundamental representation, we have only one zero mode for ππ = 1 (since π΄ π = 1 and ker π·ππ β = {0}). We can write the π solution as (I will drop the gauge index for some time now) π(π₯) = π0π0(π₯) + Non zero modes π0 will be our fermionic collective coordinate. The monopole ground states can be built by considering a ground state |Ξ©β© such that π0 Ξ© = 0 and then, another state can be built as π0 β |Ξ©β©. So, the states |Ξ©β© and π0 β |Ξ©β© are degenerate. Adjoint Fermions For adjoint fermions, π΄ = 2 and thus, for ππ = 1, we have two zero modes. Moreover, since the monopoles with fermions live in the ππ 2 π Γ ππ 2 π group, the fundamental fermions can be singlets because π Γ π = π + π but this is not the case for adjoint fermions because π Γ π = π + π. Since there are two fermion zero modes, the fermions canβt have spin 3/2 and thus, the adjoint fermions should live in the π representation of ππ 2 π Γ ππ 2 π. Therefore, the spins of the zero modes must be Β±1/2. So, the π solution is expanded with the spins mentioned on the zero modes and the corresponding coefficients as; π = π0,1/2π0 1 2 + π0,β1/2π0 β 1 2 + Non zero modes So, we have the four monopole states as shown below Ξ© Spin 0 π0,1/2 β Ξ© Spin 1/2 π0,β1/2 β Ξ© Spin β 1/2 π0,1/2 β π0,β1/2 β Ξ© Spin 0
- 30. Monopoles in π = 4 supersymmetric theories We need monopoles in π = 4 supersymmetry in order to solve two major problems in the Olive Montonen proposal. We don't need to go into the details of the derivation of the π = 4 supersymmetry lagrangian. I will use only the result that the π = 4 supersymmetric yang mills Higgs theory contains only one multiplet (without going into spins higher than 1) with contains one gauge field, six scalars and two Weyl fermions (two Dirac fermions) and all of them are in the adjoint representation since they are in the same multiplet as the gauge field. Now, since there are two Dirac fermions in this theory, the number of fermion zero modes double and thus, we have the following creation operators π 0, Β± 1 2 πβ where n = 1,2 Now, the monopole multiplet is as follows (we again start with the |Ξ©β© state and I drop the 0 subscript from the raising operators. Moreover, I replace Β±1/2 with Β± for brevity). State Spin |Ξ©β© 0 πΒ± πβ |Ξ©β© Β±1/2 πβ πβ π+ πβ |Ξ©β© 0 π+ 1β π+ 2β |Ξ©β© 1 πβ 1β πβ 2β |Ξ©β© β1 πβ 1β πβ 2β πΒ± πβ |Ξ©β© β1/2 π+ 1β π+ 2β πβ 1β πβ 2β |Ξ©β© 0
- 31. We can see that in this multiplet, we have states which have spin 1. Moreover, if we compare the spin content of this multiplet with the π = 4 gauge supermultiplet, there is an exact match. There is another feature of the π = 4 theory which is worth mentioning. It is a known fact (I won't go into the details) that the π½ function of π = 4 super Yang Mills theory vanishes and thus, the potential of the π = 4 theory won't receive quantum corrections. Therefore, the first two problems in the Olive Montonen proposal that I mentioned are solved by incorporating 4 dimensional YMH theory in the π = 4 supersymmetric gauge theory.
- 33. Low energy π© = 2 SYM The scalar potential for π© = 2 theory is; π π§, π§ β π§, π§ 2 Now, π = 0 for SUSY to remain intact β π§, π§ = 0 For ππ(2) gauge, π§ = 1 2 ππ + πππ ππ Gauge invariance renders π1 = π2 = 0 and π§, π§ = 0 renders π1 = π2 = 0. Let π = π3 + ππ3 and ππ(2) Weyl transformation renders π βΌ βπ. So, π2 is gauge invariant (i.e. π‘π π§2 is gauge invariant). Define π’ = π‘π π§2 , π§ = 1 2 ππ3 Moduli space is the complex π’ plane. π£π 3 , π3 and π3 remain massless due to β¨π§β© and thus, low energy theory concerns only one gauge boson. The lagrangian is π(1) lagrangian now and is given as; βπ©=2 πππ = 1 16π β« π2 π β±β²β² π ππΌ π πΌ + β« π2 ππ2 π πβ±β² π
- 34. β±β²β²(π§) as metric The kinetic terms from the low energy lagrangian are βΌ πΌπ β±β²β² π§ πππ§ 2 , βΌ βπΌπ β±β²β² π§ πΉππ πΉ ππ which suggests that β±β²β²(π§) is a metric. The metric on the moduli space is ππ 2 = πΌπ β±β²β² π ππππ = πΌπ π π ππππ Let Ξ¨ be π© = 2 chiral field. Then the one loop prepotential is; β± Ξ¨ = π 2π Ξ¨2 πΌπ Ξ¨2 Ξ©2 For large π, Ξ¨ βΌ π and β± π βΌ π 2π π2 πΌπ π2 Ξ©2 π β β β π π = π π πΌπ π2 Ξ©2 + 3 π β β Since π(π) doesnβt depend on π, it is holomorphic β πΌπ π is harmonic function β it doesnβt have local minima β πΌπ π isnβt positive definite. So, π(π) and thus, π should be defined on certain patches of π’ plane.
- 35. Seiberg Witten duality Define ππ· = β±β²(π) and thus, ππ 2 = πΌπ β±β²β² π ππππ = πΌπ (πππππ·) The metric is symmetric is π and ππ·. We can now define ππ· = β±β² π and the dual prepotential is defined by Legendre transformation β±π· ππ· = β± π β πππ· β π = ββ±π· β² (ππ·) The duality transformation (called π) is π: π β ππ· = β±β² π , β± π β β±π· ππ· β π: ππ· π β 0 β1 1 0 ππ· π This implies the following term invariant πΌπ β« π2 ππ2 ππβ±β² π β πΌπ β« π2 ππ2 πππ·β±π· β² ππ· = πΌπ β« π2 ππ2 ππβ±β² π The Bianchi identity ππΌπ½ππ β πΉ ππ = 0 identifies as Im π·πΌππΌ = 0 in superspace. Imposing this constraint using a lagrange multiplier ππ· in calculating partition function, we get; π = β« π·π exp iβ = β« π·π π·ππ· exp i 16π Im β« π4 π₯ β« π2 π β±β²β² π ππΌ π πΌ + 1 2 β« π2 π π2 π ππ·π·πΌππΌ = β« π·ππ· exp i 16π Im β« π4 π₯ π2 π β 1 β±β²β² π ππ· πΌ ππ· πΌ This gives us a dual coupling ππ· ππ· = β 1 π π = β 1 β±β²β² π = β ππ πππ· = β±π· β²β² (ππ·)
- 36. This gives us the following lagrangian under duality transformation βπ=2 eff β 1 16π β« π2 π β±π· β²β² ππ· (ππ·)πΌ (ππ·)πΌ+β« π2 ππ2 π ππ·β±π· β² ππ· So the lagrangian is invariant under duality transformation. πΊπ³(π, β€) duality The forms of ππΏ(2, β€) matrices are π = 0 1 β1 0 , π π = 1 π 0 1 , π β β€ π is already realised and π(π) gives ππ· β ππ· + ππ, π β π β πππ· β ππ·π β πππ· β ππ·π, β±β²β² π = πππ· ππ = β±β²β² π + π β βπ=2 β βπ=2 + π 16π Im β« π2 π ππ· πΌ ππ· πΌ β β« π4 π₯ βπ=2 β β« π4 π₯ βπ=2 + π 16π Im β« π4 π₯ π2 πππ· πΌ ππ· πΌ = β« π4 π₯ βπ=2 β b 16π β« π4 π₯πΉππ β πΉ ππ = β« π4 π₯ βπ=2 β 2πππ where π = 1 32π2 β« π4 π₯πΉππ β πΉ ππ is Yang Mills instanton number. Since πππ is used in calculation of functional integrals, π and π β 2πππ are equivalent and thus, π(π) is a symmetry transformation. So, Seiberg Witten duality also extends to ππΏ(2, β€) group.
- 37. Moduli space structure Singularity at infinity As π β β β π’ β β, we have; ππ· = πβ±(π) ππ = π ππ π 2π π2 πΌπ π2 Ξ©2 = π π π πΌπ π2 Ξ©2 + 1 Now, π’ β π2ππ π’ β π β πππ π = βπ and ππ· β βππ· + 2π So, we have; ππ· π β β1 2 0 β1 ππ· π = πβ ππ· π Other singularities The R charges are as follows π§ β π2ππ π§, π πΌ β πππ π πΌ, π β πππ π β π2 π = πβ2ππ π2 π β Ξ¨ β e2ππ Ξ¨ β β± Ξ¨ β Ξ¨2 β π4ππ β±(Ξ¨) The instanton corrections for β±(Ξ¨) are given as; β± Ξ¨ πΌππ π‘πππ‘πππ = Ξ¨2 π ππ Ξ©2 Ξ¨2 2π The β± Ξ¨ β π4ππ β±(Ξ¨) in realised now if π = 2ππ/8 π β β€+ (i.e. β€8 symmetry) i.e. π§2 β βπ§2 and hence π’ β βπ’ is a symmetry. So, at least three non zero points should be singularities or π’ = 0, β should be singularities. Later case implies πβ = π0 (contour deformation) and thus, π2 is a good coordinate on whole of complex plane (which is not true). So, we should have three singularities π’ β β, π’, βπ’.
- 38. Other singularities At π’ = 0, the gauge group restores and everything becomes massless. We can try to interpret other singularities the same way. Singularities canβt be due to gauge boson becoming massless (it requires tπ π2 to become dimensionless but itβs dimension is 2). The trick is to consider massless monopoles. Massive states have to be in short representations and thus, mass is proportional to π. For π© = 2, π = πππ + ππ·ππ Thus, ππ· = 0 at singularities. Coupling π© = 2 SYM to hypermultiplets, we have SQED with beta function; Ξ πππ· πΞ = ππ· 3 8π2 Which gives ππ· πππ· πππ· = ππ· π πππ· 4ππ ππ· 3 = β π π β ππ· = β π π πΌπ ππ· = β ππ πππ· β π βΌ π + π π ππ· πΌπ ππ· Now, ππ· = 0 at π’ = π’ , we have ππ· = π(π’ β π’) near π’ = π’ where π is a constant and thus, near π’ = π’, we have; π π’ = π + π π π π’ β π’ πΌπ (π’ β π’) For π’ β π’ β π2ππ (π’ β π’), we have; ππ· β ππ·, π β π β 2ππ· β ππ· π β 1 0 β2 1 ππ· π = ππ’ ππ· π Using πβ = ππ’πβπ’, we have; πβπ’ = β1 2 β2 3
- 39. Dyon interpretation For dyon ππ, ππ becoming massless, π = πππ + ππ·ππ should be invariant under monodromy action. In other words π = ππ ππ ππ· π β ππ ππ π ππ· π = π β ππ ππ π = ππ ππ For ππ’ Μ, (1 0) is a left eigenvector and for πβπ’ , (1 β1) is left eigenvector (actually (πππ πππ) with integer π is a left eigenvector but they arenβt stable as ππ and ππ have to be relatively prime). For ππ ππ to be a left eigenvector, the monodromy matrix is 1 + 2ππππ 2ππ 2 β2ππ 2 1 β 2ππππ This shows that (for π β β€) πβ βπ ππ’πβ π = 1 β 4π 8π2 β2 1 + 4π corresponds to ππ, ππ = (β2π, 1) Moreover, πβ βπ πβπ’πβ π = β1 β 4π 8π2 + 2π + 8 β2 3 + 4π corresponds to ππ, ππ = (β2π β 1,1).
- 40. Solution for π π’ , ππ·(π’) Consider π2 π ππ§2 β π π§ π π§ = 0 (1) Regular singular point π§ = π§0 is such that π§ β π§0 2 π(π§) is analytic at π§ = π§0. Theorem: Non trivial constant monodromies correspond to singular points of π(π§) which are regular singular points. Following Seiberg and Witten, π’ = 1 and thus, the singular points are π’ = Β±1, π’ β β. The terms 1 π§ β 1 , 1 π§ + 1 in π(π§) correspond to third order poles at infinity. So, the general π(π§) is (following the convention) π π§ = β 1 4 1 β π1 2 π§ + 1 2 + 1 β π2 2 π§ β 1 2 β 1 β π1 2 β π2 2 + π3 2 π§ + 1 π§ β 1 , ππ > 0 The transformation π π§ = π§ + 1 1βπ1 2 π§ β 1 1βπ2 2 πΊ π§ + 1 2 equation (1) becomes; π 1 β π π2 πΊ π ππ2 + π β π + π + 1 π ππΊ π ππ β πππΊ π = 0 where π is an independent variable and 2π = 1 β π1 β π2 + π3, 2π = 1 β π1 β π2 β π3, π = 1 β π1.
- 41. The solutions to this equation are hypergeometric functions πΉ π, π, π; π = π=0 β π π π π π π ππ π! Where π π are Pochhammerβs symbol π 0 = 1, π π+1 = π π + 1 β¦ (π + π) The two solutions that we pick are πΊ1 π = βπ βπ πΉ π, π + 1 β π, π + 1 β π; 1 π πΊ2 π = 1 β π πβπβπ πΉ π β π, π β π, π + 1 β π β π; 1 β π Now, πΊ1 has trivial monodromy at infinity and πΊ2 has trivial monodromy at π = 1. Finding π, π, π For large π§ π π§ = β 1 4 1 β π3 2 π§2 β 4π§2 π2 π π§ ππ§2 + 1 β π3 2 π π§ = 0 β π π§ = π§1Β±π3 2 π3 β 0 , π π§ = π§ πΌπ π§ (π3 = 0) The large π’ expression for ππ· resembles the π3 = 0 case. So, set π3 = 0. Around π§ = 1 π π§ = β 1 4 1 β π2 2 π§ β 1 2 β π2 π π¦ ππ¦2 + 1 β π2 2 4π¦2 π π¦ = 0, π¦ = π§ β 1 ππ· is linear in π’ β 1 around π’ = 1 and thus, the second derivative should vanish and this, π2 = 1. β€2 symmetry implies π1 = 1. β π = β 1 2 , π = β 1 2 , π = 0
- 42. β π π§ = πΊ π§ + 1 2 β πΊ1 π’ = π π’ + 1 2 1 2 πΉ 1 2 , β 1 2 , 1; 2 π’ + 1 , πΊ2 π’ = β π’ β 1 2 πΉ 1 2 , 1 2 , 2; 1 β π’ 2 The solutions for π and ππ· which give correct asymptotics are ππ· π’ = βππΊ2 π’ = π π’ β 1 2 πΉ 1 2 , 1 2 , 2; 1 β π’ 2 π π’ = β2ππΊ1 π’ = 2 π’ + 1 1 2πΉ 1 2 , β 1 2 , 1; 2 π’ + 1 We can find ππ·(π) now and then, β±(π) is determined by integrating ππ· π = β±β²(π) with respect to π which gives you low energy lagrangian.
- 43. Further developements Extensions ο N. Seiberg and E. Witten added hypermultiplets in 1994. ο A first step was taken to generalize to ππ(π) group in 1995 by A. Klemm et al. ο Low energy action was calculated by Klemm et al. for ππ(3) in 1995. ο A. Brandhuber et al. calculated the low energy action for ππ(2π) in 1995. ο U. H. Danielsson et al. wrote the weak coupling limit for arbitrary gauge groups and wrote the solution for ππ(2π + 1) groups in 1995. ο P.C. Argyres et al. worked out the description of metric on moduli space for ππ(π) theory in 1995. ο U.H. Danielsson and B. Sundborg address the same problem with some simple groups with matter multiplets in 1996. Tests ο E. Witten and C. Vafa proposed a test at strong coupling in 1994. ο A. Lossev, N. Nekrasov and S.L. Shatashvilli proposed a test of Seiberg Witten solution by instanton calculations in 1999. ο The test proposed in 1999 was performed by N. Nekrasov (and the low instanton results matched the literature values) in 2003.
- 44. Proof of generalized Bogomolβnyi bound (ππ, ππ) In the presence of electrical fields (keeping π΄0 π = 0 still), we have the following mass for the dyon; π = 1 2 β« π3 π₯ πΈπ π πΈπ π + π΅π π π΅π π + π·πππ 2 = 1 2 β« π3 π₯ πΈπ π β cos πΌ π·πππ + π΅π π β sin πΌ π·π ππ 2 + β« π3 π₯ cos πΌ πΈπ π π·πππ + sin πΌ π΅π π π·πππ β₯ cos πΌ ππ + sin πΌ ππ Now, we can make the bound as tight as possible by differentiating the above expression w.r.t πΌ and setting it to zero. This gives; tan πΌ = ππ ππ β sin πΌ = ππ ππ 2 + ππ 2 , cos πΌ = ππ ππ 2 + ππ 2 This gives us; π β₯ π£ ππ 2 + ππ 2 1 2 This is the generalised Bogomolβnyi bound in terms of (ππ, ππ).
- 45. Acknowledgements I would like to thank ο My family for their support in these difficult times. ο My supervisor and other teachers for guiding me about various things regarding the thesis. ο My friends for their support. ο People with whom I had useful discussions, specially Dr. K.S. Narain.
- 46. Proof of generalized Bogomolβnyi bound (e,g) In the presence of electrical fields (keeping π΄π π = 0 still), we have the following mass for the dyon; π = 1 2 β« π3 π₯ πΈπ π πΈπ π + π΅π π π΅π π + π·πππ 2 = 1 2 β« π3 π₯ πΈπ π β cos πΌ π·πππ + π΅π π β sin πΌ π·π ππ 2 + β« π3 π₯ cos πΌ πΈπ π π·πππ + sin πΌ π΅π π π·πππ We proved previously that π΅π π·πππ = ππ(π΅π π ππ). We can also prove that πΈπ π π·πππ = ππ πΈπ π ππ as follows. We see that πΈπ π π·πππ = π·π πΈπ π ππ β ππ π·ππΈπ π = ππ πΈπ π ππ β ππ π·ππΈπ π as πΈπ π ππ is gauge singlet. Now, we prove that π·ππΈπ π = 0. Use the zeroth component of Gaussβ constraint to get; π·ππΉπ0 π = πππππ ππ π·0ππ = 0 The last equality follows because ππ is time independent and π΄0 π = 0. So, π·ππΉπ0 π = βπ·ππΉ0π π = βπ·ππΈπ π = 0 So, we get; π β₯ β« π3 π₯ cos πΌ ππ πΈπ π ππ + sin πΌ ππ π΅π π ππ = π£[cos πΌ π + sin πΌ π] Now, we can make the bound as tight as possible by differentiating the above expression w.r.t πΌ and setting it to zero. This gives; tan πΌ = π π β sin πΌ = π π2 + π2 , cos πΌ = π π2 + π2 This gives us; π β₯ π£ π2 + π2 1 2
- 47. Questions?