Magnetic Monopoles, Duality and SUSY.pptx
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A review of Dirac monopoles, T'Hooft Polyakov monopoles, and Olive-Montonen duality in N=4 yang mills Higgs theory.
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Magnetic Monopoles, Duality and SUSY.pptx
- 1. Magnetic monopoles, duality and Supersymmetry MuhammadHassaan Saleem Supervisor: K.S. Narain
- 2. 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. ⟨𝜓|𝑥⟩) and it can also be seen in the momentum space wave function (i.e. ⟨𝜓|𝑝⟩). Another example for the concept of duality would be seen in the harmonic oscillator with the lagrangian. 𝐿 = 𝑝2 2𝑚 − 1 2 𝑚𝜔2 𝑥2 It is easy to see that in the replacement 𝑥 → 𝑝 m𝜔 , 𝑝 → −𝑚𝜔𝑥 the harmonic oscillator is self dual as can be seen below 𝐿 → −𝑚𝜔𝑥 2 2𝑚 − 1 2 𝑚𝜔2 𝑝 𝑚𝜔 2 = − 𝑝2 2𝑚 − 1 2 𝑚𝜔2 𝑥2 = −𝐿 So, the lagrangian changes by an overall factor but that doesn’t change the equations of motion.
- 3. Electromagnetic duality Sourcless Maxwell equations (MEs) are invariant in the transformations 𝐷: 𝐸𝑖 → 𝐵𝑖 , 𝐵𝑖 → −𝐸𝑖 . It can be generalized to 𝐸𝑖 = 𝐸𝑖 cos 𝜃 + 𝐵𝑖 sin 𝜃 , 𝐵𝑖 = −𝐸𝑖 sin 𝜃 + 𝐵𝑖 cos 𝜃 and MEs will still be invariant as can be seen below for the Faraday and Maxwell laws (written in term of indices and 𝜕𝑡 is time derivative); 𝜖𝑖𝑗𝑘𝜕𝑗𝐵𝑘 = 𝜕𝑡𝐸𝑖 𝜖𝑖𝑗𝑘𝜕𝑗𝐸𝑘 = −𝜕𝑡𝐵𝑖 ⇒ 𝜖𝑖𝑗𝑘 𝜕𝑗 cos 𝜃 𝐸𝑘 + sin 𝜃 𝐵𝑘 = −𝜕𝑡 cos 𝜃 𝐵𝑖 − sin 𝜃 𝐸𝑖 ⇒ 𝜖𝑖𝑗𝑘𝜕𝑗𝐸𝑘 ′ = −𝜕𝑡𝐵𝑖 ′ 𝜖𝑖𝑗𝑘 𝜕𝑗 cos 𝜃 𝐵𝑘 − sin 𝜃 𝐸𝑘 = 𝜕𝑡 cos 𝜃 𝐸𝑖 + sin 𝜃 𝐵𝑖 ⇒ 𝜖𝑖𝑗𝑘𝜕𝑗𝐵𝑘 ′ = 𝜕𝑡𝐸𝑖 ′ Introduce 𝐹 𝜇𝜈 with 𝐹0𝑖 = 𝐸𝑖, 𝐵𝑖 = 1 2 𝜖𝑖𝑗𝑘𝐹 𝑗𝑘. So we can visualize 𝐷 transformation as 𝐸𝑖 → 𝐵𝑖 ⇒ 𝐹0𝑖 → 1 2 𝜖𝑖𝑗𝑘𝐹 𝑗𝑘 =∗ 𝐹0𝑖, 𝐵𝑖 → −𝐸𝑖 ⇒ 1 2 𝜖𝑖𝑗𝑘𝐹 𝑗𝑘 → −𝐹0𝑖 = 1 2 𝜖𝑖𝑗𝑘 ∗ 𝐹 𝑗𝑘, ∗ 𝐹 𝜇𝜈 = 1 2 𝜖𝜇𝜈𝛼𝛽𝐹𝛼𝛽 So, we can see that the 𝐷 transformation as 𝐹 𝜇𝜈 =∗ 𝐹 𝜇𝜈. If there are no monopoles, we can take 𝐵𝑖 = 1 2 𝜖𝑖𝑗𝑘 𝐹 𝑗𝑘 = ∇ × 𝐴 𝑖 = 1 2 𝜖𝑖𝑗𝑘 𝜕𝑗𝐴𝑘 − 𝜕𝑘𝐴𝑗 , 𝐸𝑖 = 𝐹0𝑖 = −∇𝜙 − 𝜕𝑨 𝜕𝑡 𝑖 = 𝜕0𝐴𝑖 − 𝜕𝑖𝐴0 ⇒ 𝐹 𝜇𝜈 = 𝜕𝜇𝐴𝜈 − 𝜕𝜈𝐴𝜇
- 4. Electromagnetism in Quantum Mechanics In Quantum Mechanics, the electromagnetic coupling can be realised by minimum coupling scheme 𝑝𝑖 → −𝑖 ∇ − ieA 𝑖 Moreover it is well known that the Schrodinger wave equation (SWE) is invariant in the gauge transformation 𝜓 → 𝑒−𝑖𝑒𝜒 𝜓, 𝐴𝑖 → 𝐴𝑖 − 𝑖 𝑒 𝑒𝑖𝑒𝜒 ∇𝑒−𝑖𝑒𝜒 , 𝑒𝑖𝑒𝜒 ∈ 𝑈(1) In 1931, Dirac [1] tried to add magnetic monopoles without disturbing the coupling [1]. His work gave a remarkable result which gave a reason for the quantization of electrical charge in the presence of magnetic monopoles. We can understand that work by considering a magnetic monopole at 𝑟 = 0 and since we have a magnetic monopole now, it cant the case that 𝐵𝑖 = ∇ × 𝐴 𝑖 everywhere. However, we can still define 𝐴𝑖in patches and in the region where the patches overlap, they should differ by a gauge transformation. So, far away from the monopole (for 𝑟 > 𝑟0 > 0), the magnetic field should look like 𝑩 = 𝑔𝑟 4𝜋𝑟2 Now, we want some 𝑨 to give the above magnetic field by taking it’s curl. We use the polar coordinate system and in this system, 𝑩 (for an arbitrary 𝑨 is) 𝐵𝑟 = 1 𝑟 𝑠𝑖𝑛𝜃 𝜕 sin 𝜃 𝐴𝜙 𝜕𝜃 − 𝜕𝐴𝜃 𝜕𝜙 , 𝐵𝜃 = − 1 𝑟 𝑠𝑖𝑛𝜃 𝜕𝐴𝑟 𝜕𝜙 − sin 𝜃 𝜕(𝑟𝐴𝜙) 𝜕𝑟 , 𝐵𝜙 = 1 𝑟 𝜕(𝑟𝐴𝜃) 𝜕𝑟 − 𝜕𝐴𝑟 𝜕𝜃 Since we want the 𝜙 component of 𝑩 to vanish, we set 𝐴𝜃 = 𝐴𝑟 = 0. Moreover, 𝑩 falls as 1/𝑟2 and thus, 𝑨 falls as 1/𝑟. Therefore, 𝑟𝐴𝜙 should be independent of 𝑟. So, 𝐵𝜃 is zero as well. 𝐵𝑟 = 1 𝑟 𝑠𝑖𝑛𝜃 𝜕 sin 𝜃 𝐴𝜙 𝜕𝜃 = 𝑔 4𝜋𝑟2 ⇒ 𝐴𝜙 = 𝑔 4𝜋𝑟 𝜉 − cos 𝜃 sin 𝜃 Where 𝜉 is arbitrary. [1] P.A.M Dirac (1931)
- 5. Now, we choose 𝑨 in two patches and then them 𝑨𝑁 and 𝑨𝑆 (i.e. for Northern patch and the Southern patch). We choose 𝜉 = 1 for 𝑨𝑁 and 𝜉 = −1 for 𝑨𝑆 and we have; 𝑨𝑁 = 𝑔 4𝜋𝑟 1 − cos 𝜃 sin 𝜃 𝜙, 𝑨𝑆 = − 𝑔 4𝜋𝑟 1 + cos 𝜃 sin 𝜃 𝜙 The overlap region happens to be 2𝜃 = 𝜋 circle and on that circle, the difference of the two potentials is 𝑨𝑁 𝜃 = 𝜋 2 − 𝑨𝑆 𝜃 = 𝜋 2 = 𝑔 2𝜋𝑟 𝜙 = −∇ − 𝑔 2𝜋 𝜙 = −∇𝜒 So, the difference is a gauge transformation. The 𝑈(1) element 𝑒𝑖𝑒𝜒 should be continuous and thus, we have the following condition; 𝑒𝑖𝑒 𝜒 2𝜋 −𝜒 0 = 1 ⇒ 𝑒 𝜒 2𝜋 − 𝜒 0 = 2𝜋𝑛, 𝑛 ∈ ℤ Using the expression for 𝜒, we get 𝑒𝑔 = 2𝜋𝑛, 𝑛 ∈ ℤ Comments A single monopole will ensure the quantization of electrical charge. 𝑒𝜒 = 0 and 𝑒𝜒 = 2𝜋 represent the same 𝑈(1) element and thus, the 𝑈(1) group is compact. So, we have the following two equivalent statements (they are the contrapositive of each other) 𝑀𝑜𝑛𝑜𝑝𝑜𝑙𝑒𝑠 ⇒ 𝐶𝑜𝑚𝑝𝑎𝑐𝑡 𝑈 1 𝑔𝑟𝑜𝑢𝑝 𝑁𝑜 𝐶𝑜𝑚𝑝𝑎𝑐𝑡 𝑈 1 𝑔𝑟𝑜𝑢𝑝 ⇒ 𝑁𝑜 𝑚𝑜𝑛𝑜𝑝𝑜𝑙𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 Compact 𝑈(1) group will be realised when a theory with compact gauge group is spontaneously broken to a group containing 𝑈(1) as a product group.
- 6. The t’Hooft, Polyakov monopole In the Dirac’s 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 t’Hooft, Polyakov monopole. It is nothing but a non abelian Yang Mills Higgs theory in 3 + 1 dimensions. The lagrangian is as follows; ℒ = − 1 4 𝐹 𝜇𝜈 𝑎 𝐹 𝑎 𝜇𝜈 + 1 2 𝐷𝜇𝜙𝑎 𝐷𝜇 𝜙𝑎 − 𝑉 𝜙 , 𝜙 = 𝑣 ≠ 0 The EOM for 𝐹 𝜇𝜈 𝑎 is (also called Gauss constraint) is 𝐷𝜇𝐹 𝑎 𝜇𝜈 = 𝑒𝜖𝑎𝑏𝑐𝜙𝑏𝐷𝜈 𝜙𝑐 We define 𝐹𝜇𝜈 as 𝐹 𝜇𝜈 = 𝜙𝑎 𝐹 𝜇𝜈 𝑎 − 1 𝑒 𝜖𝑎𝑏𝑐 𝜙𝑎 𝐷𝜇𝜙𝑏 𝐷𝜈𝜙𝑐 where 𝜙𝑎 is a unit vector in the direction of 𝜙𝑎 and for 𝜙𝑎 to exist, 𝜙𝑎 ≠ 0 everywhere. Note that 𝐹 𝜇𝜈 is gauge invariant. It can be shown that 𝐹 𝜇𝜈 = 𝜕𝜇(𝜙𝑎 𝐴𝜈 𝑎 ) − 𝜕𝜈 (𝜙𝑎 𝐴𝜇 𝑎 ) − 1 𝑒 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝜇𝜙𝑏 𝜕𝜈 𝜙𝑐 The proof is as follows. The definition of 𝐹 𝜇𝜈 can be expanded and written as follows; 𝐹 𝜇𝜈 = 𝜙𝑎 𝜕𝜇𝐴𝜈 𝑎 − 𝜕𝜈𝐴𝜇 𝑎 + 𝑒𝜖𝑎𝑏𝑐 𝐴𝜇 𝑏 𝐴𝜈 𝑐 − 1 𝑒 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝜇𝜙𝑏 + 𝑒𝜖𝑏𝑑𝑒 𝐴𝜇 𝑑 𝜙𝑒 𝜕𝜈𝜙𝑐 + 𝑒𝜖𝑐𝑓𝑔 𝐴𝜈 𝑓 𝜙𝑔 = 𝜕𝜇 𝜙𝑎 𝐴𝜈 𝑎 − 𝜕𝜈 𝜙𝑎 𝐴𝜇 𝑎 − 1 𝑒 𝜖𝑎𝑏𝑐 𝜙𝑏 𝜕𝜇𝜙𝑏 𝜕𝜈𝜙𝑐 +𝐴𝜇 𝑑 𝜕𝜈𝜙𝑎 𝛿𝑎𝑑 − 𝜕𝜈𝜙𝑐 𝜖𝑎𝑏𝑐 𝜖𝑏𝑑𝑒 𝜙𝑎 𝜙𝑒 − 𝐴𝜈 𝑓 𝛿𝑎𝑓 𝜕𝜇𝜙𝑎 + 𝜕𝜇𝜙𝑏 𝜖𝑎𝑏𝑐 𝜖𝑐𝑓𝑔 𝜙𝑎 𝜙𝑔 + 𝑒𝜖𝑎𝑏𝑐 𝜙𝑎 𝐴𝜇 𝑏 𝐴𝜈 𝑐 − 𝑒𝜖𝑎𝑏𝑐 𝜙𝑎 𝜖𝑏𝑑𝑒 𝜖𝑐𝑓𝑔 𝐴𝜇 𝑑 𝐴𝜈 𝑓 𝜙𝑒 𝜙𝑔
- 7. 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; 𝜕𝜈𝜙𝑎 𝛿𝑎𝑑 − 𝜕𝜈𝜙𝑐 𝜖𝑎𝑏𝑐 𝜖𝑏𝑑𝑒 𝜙𝑎 𝜙𝑒 = 𝜕𝜈𝜙𝑑 + 𝜕𝜈𝜙𝑐 𝛿𝑎𝑑 𝛿𝑐𝑒 − 𝛿𝑎𝑒 𝛿𝑐𝑑 𝜙𝑎 𝜙𝑒 = 𝜕𝜈𝜙𝑑 + 𝜙𝑑 𝜙𝑒 𝜕𝜈𝜙𝑒 − 𝜕𝜈𝜙𝑑 = 0 Where I used the fact that 𝜙𝑎 𝜕𝜈𝜙𝑎 = 1 2 𝜕𝜈 𝜙𝑎 𝜙𝑎 = 1 2 𝜕𝜈 1 = 0 The second bracketed term vanishes by a similar proof. Now we need to prove that 𝜖𝑎𝑏𝑐 𝜙𝑎 𝐴𝜇 𝑏 𝐴𝜈 𝑐 − 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜖𝑏𝑑𝑒 𝜖𝑐𝑓𝑔 𝐴𝜇 𝑑 𝐴𝜈 𝑓 𝜙𝑒 𝜙𝑔 vanishes. For this task, we can manipulate the last term as; 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜖𝑏𝑑𝑒 𝜖𝑐𝑓𝑔 𝐴𝜇 𝑑 𝐴𝜈 𝑓 𝜙𝑒 𝜙𝑔 = 𝛿𝑎𝑓 𝛿𝑏𝑔 − 𝛿𝑎𝑔 𝛿𝑏𝑓 𝜙𝑎 𝜖𝑏𝑑𝑒 𝐴𝜇 𝑑 𝐴𝜈 𝑓 𝜙𝑒 𝜙𝑔 = 𝜖𝑒𝑑𝑏 𝜙𝑒 𝐴𝜇 𝑑 𝐴𝜈 𝑏 where I dropped the vanishing term 𝜖𝑏𝑑𝑒 𝜙𝑎 𝜙𝑏 𝜙𝑒 𝐴𝜇 𝑑 𝐴𝜈 𝑎 and used the fact that 𝜙𝑒 𝜙𝑒 = 1. So, we can now see that 𝜖𝑎𝑏𝑐 𝜙𝑎 𝐴𝜇 𝑏 𝐴𝜈 𝑐 − 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜖𝑏𝑑𝑒 𝜖𝑐𝑓𝑔 𝐴𝜇 𝑑 𝐴𝜈 𝑓 𝜙𝑒 𝜙𝑔 = 𝜖𝑎𝑏𝑐 𝜙𝑎 𝐴𝜇 𝑏 𝐴𝜈 𝑐 − 𝜖𝑎𝑏𝑐 𝜙𝑎 𝐴𝜇 𝑏 𝐴𝜈 𝑐 = 0 So, we have shown that 𝐹 𝜇𝜈 = 𝜕𝜇(𝜙𝑎 𝐴𝜈 𝑎 ) − 𝜕𝜈 (𝜙𝑎 𝐴𝜇 𝑎 ) − 1 𝑒 𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝜇𝜙𝑏 𝜕𝜈 𝜙𝑐
- 8. Now, a magnetic field can be defined as 𝐵𝑖 = 1 2 𝜖𝑖𝑗𝑘𝐹 𝑗𝑘 = 𝜖𝑖𝑗𝑘𝜕𝑗 𝜙𝑎 𝐴𝑘 𝑎 − 1 2𝑒 𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝑗𝜙𝑏 𝜕𝑘𝜙𝑐 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; 𝑔 = ∫ 𝑑𝑆𝑖𝐵𝑖 = − 1 2𝑒 𝑆2 𝑑𝑆𝑖𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝑗𝜙𝑏 𝜕𝑘𝜙𝑐 = − 1 2𝑒𝑣3 𝑆2 𝑑𝑆𝑖𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐 𝜙𝑎 𝜕𝑗𝜙𝑏 𝜕𝑘𝜙𝑐 This integral can be evaluated by choosing the form of 𝜙𝑎 = 𝑣 𝑥𝑎 at infinity (where 𝑥𝑎 = 𝑟 𝑥𝑎 ). We simply get; 1 𝑣3 𝑆2 𝑑𝑆𝑖𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐𝜙𝑎 𝜕𝑗𝜙𝑏 𝜕𝑘𝜙𝑐 = 𝑆2 𝑟2 𝑑Ω 𝑥𝑖𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐𝑥𝑎𝜕𝑗𝑥𝑏𝜕𝑘𝑥𝑐 = 𝑆2 𝑑Ω 𝑥𝑖𝜖𝑖𝑗𝑘𝜖𝑎𝑏𝑐𝑥𝑎 𝛿𝑏𝑗𝛿𝑐𝑘 − 2𝛿𝑏𝑗𝑥𝑘𝑥𝑐 Where I dropped the vanishing terms and used the identity 𝜕𝑖𝑥𝑗 = 1 𝑟 𝛿𝑖𝑗 − 𝑥𝑖𝑥𝑗 Now, using the identities 𝜖𝑖𝑗𝑘𝜖𝑘𝑙𝑚 = 𝛿𝑖𝑙𝛿𝑗𝑚 − 𝛿𝑖𝑚𝛿𝑙𝑗, 𝜖𝑖𝑗𝑘𝜖𝑗𝑘𝑙 = 2𝛿𝑖𝑙 The above integral becomes; 2 𝑆2 𝑑Ω 𝑥𝑖𝑥𝑎 𝛿𝑎𝑖 − 𝛿𝑎𝑖 + 𝑥𝑖𝑥𝑎 = 8𝜋 ⇒ 𝑔 = − 4𝜋 𝑒 Now, I can choose a mapping 𝑓: 𝑆𝜙 2 → 𝑆𝑥 2 such that 𝑆𝜙 2 covers 𝑆𝑥 2 𝑛 times. This will simply change the integral to be multiplied by 𝑛 as we are going around the 𝑆𝜙 2 𝑛 times. So, we get the following condition 𝑒𝑔 = 4𝜋𝑛 𝑛 ∈ ℤ
- 9. Bogomol’nyi bound and BPS solution We calculate the energy of the static solution of YMH theory in the 𝐴0 𝑎 = 0 gauge and we get; 𝐸 = ∫ 𝑑3 𝑥 1 2 𝐵𝑖 𝑎 𝐵𝑖 𝑎 + 1 2 𝐷𝑖𝜙𝑎 2 + 𝑉 𝜙 = ∫ 𝑑3 𝑥 1 2 𝐵𝑖 𝑎 − 𝐷𝑖𝜙𝑎 2 + 𝐵𝑖 𝑎 𝐷𝑖𝜙𝑎 + 𝑉 𝜙 Using zeroth component of Bianchi identity 𝜖𝛼𝛽𝜇𝜈 𝐷𝛽𝐹 𝜇𝜈 𝑎 = 0 and the definition of 𝐵𝑖 𝑎 i.e. 2𝐵𝑖 𝑎 = 𝜖𝑖𝑗𝑘𝐹𝑗𝑘 𝑎 , we conclude 𝜖0𝑖𝑗𝑘 𝐷𝑖𝐹 𝑗𝑘 𝑎 = 𝐷𝑖𝐵𝑖 𝑎 = 0. Moreover, since 𝐵𝑖 𝑎 𝜙𝑎 is gauge singlet, we have 𝐷𝑖 𝐵𝑖 𝑎 𝜙𝑎 = 𝜕𝑖(𝐵𝑖 𝑎 𝜙𝑎 ). So, the second term in the energy integrand is 𝜕𝑖(𝐵𝑖 𝑎 𝜙𝑎 ) and thus, we can use divergence theorem and it becomes; 𝐸 = ∫ 𝑑3 𝑥 1 2 𝐵𝑖 𝑎 − 𝐷𝑖𝜙𝑎 2 + 𝑉 𝜙 + 𝑣 𝑆2 𝑑𝑆𝑖𝐵𝑖 𝑎 𝜙𝑎 ≥ 𝑣 𝑆2 𝑑𝑆𝑖𝐵𝑖 𝑎 𝜙𝑎 = 𝑣𝑔 So, the energy of the static monopole solution (mass of the monopole) has a lower bound known as Bogomol’nyi bound. If we impose the condition 𝐵𝑖 𝑎 = 𝐷𝑖𝜙𝑎 (also known as the Bogomol'nyi equation) and set 𝑉 𝜙 = 0 everywhere, then the energy of the field configuration is just 𝑣𝑔. Since we are setting 𝑉 𝜙 = 0 everywhere, it seems as if we won't be able to get a nonzero ⟨𝜙𝑎 𝜙𝑎 ⟩. However, since we need 𝜙𝑎 𝜙𝑎 = 𝑣2 at spatial infinity, we can impose this as a boundary condition. For BPS solution, if we try to give 𝐴𝑖 𝑎 (𝑥) a non zero vev, it will break Lorentz invariance. Moreover, for the theory to have non zero topological charge, 𝜙𝑎 should vary at infinity and thus, solutions can't be rotationally invariant. However, we can make the solutions invariant under the diagonal subgroup 𝑆𝑂 3 𝑠 × 𝑆𝑂 3 𝑔where 𝑆𝑂 3 𝑔 is the global gauge group. We would not need the detailed form of the BPS solutions though.
- 10. 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 𝜙𝑎 (𝑥) is a BPS solution, so is 𝜙𝑎 (𝑥 + 𝑋) with 𝑋 fixed due to translational invariance. So, ℝ3 is present as a product in the moduli space of BPS monopole. Now, equation of motion 𝐷𝜇𝐹 𝑎 𝜇𝜈 = 𝑒𝜖𝑎𝑏𝑐𝜙𝑏𝐷𝜈 𝜙𝑐 is expanded as; 𝐷𝜇 𝜕𝜇 𝐴𝑎 𝜈 − 𝜕𝜈 𝐴𝑎 𝜇 + 𝑒𝜖𝑎𝑏𝑐𝐴𝑏 𝜇 𝐴𝑐 𝜈 = 𝑒𝜖𝑎𝑏𝑐𝜙𝑏[𝜕𝜈 𝜙𝑐 + 𝑒𝜖𝑐𝑑𝑒𝐴𝑑 𝜈 𝜙𝑒] Set 𝜈 = 0 and working in the 𝐴𝑎 𝜇 = 0 gauge, we get; 𝐷𝑖𝐴𝑎 𝑖 + 𝑒 𝜙, 𝜙 𝑎 = 0 The linearized form of the gauss law for deformations to the BPS solutions (𝛿𝜙𝑎 , 𝛿𝐴𝑖 𝑎 ) is obtained by letting 𝐴𝑖 𝑎 → 𝐴𝑖 𝑎 + 𝛿𝐴𝑖 𝑎 , 𝜙𝑎 → 𝜙𝑎 + 𝛿𝜙𝑎 and realising that only the deformations are allowed to be time dependant. We get 𝐷𝑖𝛿 𝐴𝑎 𝑖 + 𝑒 𝜙, 𝛿𝜙 𝑎 = 0 We can expand the Bogomol’nyi equation as well and then linearize it as follows; 𝜖𝑖𝑗𝑘 𝜕𝑗 𝛿𝐴𝑎 𝑘 + 𝑒𝜖𝑎𝑏𝑐𝐴𝑏 𝑗 𝛿𝐴𝑐 𝑘 = 𝜕𝑖𝛿𝜙𝑎 + 𝑒𝜖𝑎𝑏𝑐 𝐴𝑖 𝑏 𝛿𝜙𝑐 + 𝑒 𝛿𝐴𝑖, 𝜙 𝑎 ⇒ 𝜖𝑖𝑗𝑘𝐷𝑗 𝛿𝐴𝑎 𝑘 = 𝐷𝑖𝛿𝜙𝑎 + 𝑒 𝛿𝐴𝑖, 𝜙 𝑎 The solution to the two red equations above is 𝛿𝐴0 𝑎 = 0, 𝛿𝜙𝑎 = 0, 𝛿𝐴𝑖 𝑎 = 𝐷𝑖 𝜉 𝑡 𝜙𝑎 , 𝜉 𝑡 is arbitrary function of time. If 𝜉 = 0 then the change in 𝐴𝑖 𝑎 is a large gauge transformation (it doesn’t vanish at infinity). The corresponding 𝑈(1) element is 𝑒𝑒𝜙 and since unbroken 𝑈(1) is compact, the coordinate 𝜉 is on a compact 𝑆1 . So, the moduli space of the BPS monopole is; ℝ3 × 𝑆1
- 11. Witten effect We can add a 𝜃 term in the lagrangian without breaking gauge invariance. The term is; ℒ𝜃 = − 𝜃𝑒2 32𝜋2 𝐹 𝜇𝜈 𝑎 ∗ 𝐹 𝑎 𝜇𝜈 The prefactor is for later convenience. First, we can do a 𝑈(1) case (i.e. the QED case). QED case 𝐹𝜇𝜈 ∗ 𝐹 𝜇𝜈 = 1 2 𝜖𝜇𝜈𝛼𝛽 𝐹 𝜇𝜈𝐹𝛼𝛽 = 1 2 𝜖0𝑖𝑗𝑘 𝐹0𝑖𝐹 𝑗𝑘 + 𝜖𝑖0𝑗𝑘 𝐹𝑖0𝐹 𝑗𝑘 + 𝜖𝑖𝑗0𝑘 𝐹𝑖𝑗𝐹0𝑘 + 𝜖𝑖𝑗𝑘0 𝐹𝑖𝑗𝐹𝑘0 = 2𝜖0𝑖𝑗𝑘 𝐹0𝑖𝐹 𝑗𝑘 = 4 𝑬. 𝑩 ⇒ ℒ𝜃 = − 𝜃𝑒2 8𝜋2 𝑬. 𝑩 For a static monopole, we have 𝑬 = ∇𝐴0, 𝑩 = ∇ × 𝑨 + 𝑔 4𝜋 𝒓 𝑟 ⇒ 𝐿𝜃 = ∫ 𝑑3 𝑥 ℒ𝜃 = 𝑒2 𝑔𝜃 8𝜋2 ∫ 𝑑3 𝑥𝐴0𝛿 3 (𝒓) where I used the following fact; 𝑬. 𝑩 = ∇. 𝐴0 ∇ × 𝐴 + 𝑔 4𝜋 𝑟 𝑟 = − 𝑔 4𝜋 𝐴0∇. 𝑟 𝑟 + 𝑆. 𝑇 = −𝑔𝐴0𝛿 3 (𝒓) This lagrangian is nothing but the interaction of a particle at origin with background field. The charge of the particle is; − 𝑒2 𝑔𝜃 8𝜋2 𝑒𝑔=4𝜋 − 𝑒𝜃 2𝜋 So, the 𝜃 term can give electrical charge to the monopoles.
- 12. SU(2) case Consider large gauge transformations that act on 𝐴𝜇 𝑎 . The finite deformation (without infinitesimal parameter) is; 𝛿𝐴𝜇 𝑎 = 1 𝑒𝑣 𝐷𝜇𝜙𝑎 Let 𝒩 be the generator of this gauge transformation. Until now, we don’t have fundamental fields in this theory and the gauge parameter lives on 𝑆1 an thus, 𝑒2𝜋𝑖𝒩 = 1. 𝒩 is the Noetherian conserved charge (as the gauge transformations act on the fields). It can be easily worked out as follows. Expression for 𝒩 𝒩 = ∫ 𝑑3 𝑥 𝜕ℒ 𝜕 𝜕0𝐴𝜇 𝑎 𝛿𝐴𝜇 𝑎 Now, we write the lagrangian as; ℒ = − 1 4 𝐹 𝑎 𝜇𝜈 𝐹 𝜇𝜈 𝑎 − 𝜃𝑒2 32𝜋2 𝐹 𝜇𝜈 𝑎 ∗ 𝐹 𝑎 𝜇𝜈 + 1 2 𝐷𝜇𝜙𝑎 2 So the required derivative is; 𝜕ℒ 𝜕 𝜕0𝐴𝜇 𝑎 = − 1 2 𝐹 𝑏 𝛼𝛽 𝜕𝐹𝛼𝛽 𝑏 𝜕 𝜕0𝐴𝜇 𝑎 − 𝜃𝑒2 32𝜋2 𝜖𝜌𝜎𝛼𝛽 𝐹𝑏 𝛼𝛽 𝜕𝐹 𝜌𝜎 𝑏 𝜕 𝜕0𝐴𝜇 𝑎 From the definition of 𝐹 𝜇𝜈 𝑎 , the derivative appearing above can be calculated as; 𝜕𝐹 𝜌𝜎 𝑏 𝜕 𝜕0𝐴𝜇 𝑎 = 𝛿𝑎 𝑏 [𝛿𝜌 0 𝛿𝜎 𝜇 − 𝛿𝜎 0 𝛿𝜌 𝜇 ] Using this expression and after some elementary simplification, we have; 𝜕ℒ 𝜕 𝜕0𝐴𝑖 𝑎 = −𝐹 𝑎 0𝑖 − 𝜃𝑒2 16𝜋2 𝜖0𝑖𝑗𝑘 𝐹 𝑎 𝑗𝑘 = −𝑔𝑖𝑗 𝐸𝑎 𝑗 − 𝑔𝑖𝑗 𝜃𝑒2 8𝜋2 (𝐵𝑎)𝑗 = 𝐸𝑎 𝑖 + 𝜃𝑒2 8𝜋2 𝐵𝑎 𝑖 Using all the red equations above, we get; 𝒩 = ∫ 𝑑3 𝑥 1 𝑒𝑣 𝐸𝑖 𝑎 + 𝜃𝑒2 8𝜋2 𝐵𝑖 𝑎 𝐷𝑖𝜙𝑎 = 𝑄 𝑒 + 𝜃𝑒 8𝜋2 𝑔 where 𝑄 = 𝑣−1 ∫ 𝑑3 𝑥𝐸𝑖 𝑎 𝐷𝑖𝜙𝑎 𝑔 = 𝑣−1 ∫ 𝑑3 𝑥𝐵𝑖 𝑎 𝐷𝑖𝜙𝑎
- 13. Now, the condition 𝑒2𝜋𝑖𝒩 = 1 implies that 𝒩 = 𝑛𝑒 ∈ ℤ ⇒ 𝑄 𝑒 + 𝜃𝑒𝑔 8𝜋2 = 𝑛𝑒 Using the condition 𝑒𝑔 = 4𝜋𝑛𝑚, we get; 𝑄 𝑒 + 𝜃𝑛𝑚 2𝜋 = 𝑛𝑒 ⇒ 𝑄 = 𝑒𝑛𝑒 − 𝑒𝜃 2𝜋 𝑛𝑚 So, we can see that for a single monopole (i.e. 𝑛𝑚 = 1) there is an extra contribution to the electrical charge of the monopoles due to the non zero 𝜃 term. This contribution is − 𝑒𝜃 2𝜋 This is equal to the electrical charge that we found in the QED case.
- 14. Olive Montonen and 𝑆𝐿(2, ℤ) duality Set 𝜃 = 0 first. Now, there can be electrically charged states (𝑛𝑒, 0) and magnetic monopoles 0, 𝑛𝑚 . The mass of an electrically charged state is the mass of 𝑊 boson i.e. 𝑀𝑊 = 𝑣𝑒. The mass of a magnetic monopole is 𝑀𝑀 = 𝑣𝑔 ≫ 𝑣𝑒 (at weak coupling). In a dual theory where the roles of electrical and magnetic charges are replaced, we should have 𝑒 ↔ 𝑔, 𝑀𝑊 ↔ 𝑀𝑀 Now, the dual theory should be at strong coupling as small 𝑒 means large 𝑔. 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. 𝑉(𝜙) might not remain vanishing when quantum corrections are included. The states don’t match exactly as the monopoles don’t have spin 1 like the 𝑊 bosons. 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 𝑁 = 4 super Yang Mills theory. Now, let 𝜃 ≠ 0. Now, the 𝑌𝑀𝐻 lagrangian is; ℒ = − 1 4 𝐹2 − 𝜃𝑒2 32𝜋2 𝐹 ∗ 𝐹 − 1 2 𝐷𝜇𝜙𝑎 2 Now, we let 𝜙𝑎 → 𝜙𝑎 𝑒 , 𝐴𝜇 𝑎 → 𝐴𝜇 𝑎 𝑒 ⇒ 𝐹 𝜇𝜈 𝑎 → 𝐹 𝜇𝜈 𝑎 𝑒 After this rescaling of field, the lagrangian becomes (with covariant derivatives changed accordingly); ℒ = − 1 4𝑒2 𝐹2 − 𝜃 32𝜋2 𝐹 ∗ 𝐹 − 1 2 𝐷𝜇𝜙𝑎 2
- 15. Now consider this equivalent way of writing the first two terms of the lagrangian = 1 32𝜋2 𝐼𝑚 𝜃 2𝜋 + 4𝑖𝜋 𝑒2 𝐹 + 𝑖 ∗ 𝐹 2 = − 1 8𝑒2 𝐹2 −∗ 𝐹2 − 𝜃 32𝜋2 𝐹 ∗ 𝐹 = 1 4𝑒2 𝐹2 − 𝜃 32𝜋2 𝐹 ∗ 𝐹 where in the last step, I used the fact that 𝐹2 =−∗ 𝐹2 as can be demonstrated ∗ 𝐹2 = 1 4 𝜖𝜇𝜈𝛼𝛽 𝐹𝛼𝛽𝜖𝜇𝜈𝜌𝜎𝐹𝜌𝜎 = − 1 2 𝛿𝜌 𝛼 𝛿𝜎 𝛽 − 𝛿𝜎 𝛼 𝛿𝜌 𝛽 𝐹𝛼𝛽𝐹𝜌𝜎 = −𝐹2 So, the full lagrangian can be written in case of non zero 𝜃 as; ℒ = − 1 32𝜋2 𝐼𝑚 𝜏 𝐹 + 𝑖 ∗ 𝐹 2 − 1 2 𝐷𝜇𝜙𝑎 2 where 𝜏 = 𝜃 2𝜋 + 4𝜋 𝑒2 𝑖 I will use a result from instanton theory. The 𝑛 instantons in this theory are weighed by the factor 𝑒2𝑖𝜋𝑛𝜏 . So, physics is invariant in the transformation 𝜏 → 𝜏 + 1 (corresponding to 𝜃 → 𝜃 + 2𝜋). Moreover, if we set 𝜃 = 0, then the electromagnetic duality (𝑒 → 𝑔) corresponds to 𝑒 → 𝑔 = 4𝜋 𝑒 ⇒ 𝜏 = 4𝜋𝑖 𝑒2 → − 𝑒2 4𝜋𝑖 = − 1 𝜏 We can identify both of these transformations as special cases of a single general transformation 𝜏 → 𝑎𝜏 + 𝑏 𝑐𝜏 + 𝑑 , 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤ, 𝑎𝑑 − 𝑏𝑐 = 1 This transformation is known as the 𝑆𝐿(2, ℤ) transformation and we can identify 𝜏 → 𝜏 + 1 ⇒ 𝑎 = 𝑏 = 𝑑 = 1, 𝑐 = 0 𝜏 → − 1 𝜏 ⇒ 𝑎 = 𝑑 = 0, 𝑐 = −𝑏 = 1 We can see that 𝐼𝑚 𝜏 = 4𝜋 𝑒2 > 0 So, we are concerned with the upper half 𝜏 plane only. We can also define a fundamental region such that any 𝜏 in the upper half plane can be brought into this fundamental region by an appropriate 𝑆𝐿(2, ℤ) transformation. − 1 2 ≤ 𝑅𝑒 𝜏 ≤ 1 2 , 𝜏 ≥ 1
- 16. 𝑺𝑳 𝟐, ℤ action on the states The action of the 𝑆𝐿(2, ℤ) group on the states (𝑛𝑒, 𝑛𝑚) states requires the electrical charge operator 𝑄𝑒 and magnetic charge operator 𝑄𝑒 = 𝑒𝑛𝑒 − 𝑒𝜃 2𝜋 𝑛𝑚, 𝑄𝑚 = 𝑔𝑛𝑚 = 4𝜋 𝑒 𝑛𝑚 Now, 𝜏 → 𝜏 + 1 ⇒ 𝜃 → 𝜃 + 2𝜋 and thus, we have; 𝑄𝑒 → 𝑒 𝑛𝑒 − 𝑛𝑚 − 𝑒𝜃 2𝜋 𝑛𝑚 ⇒ 𝑛𝑒 → 𝑛𝑒 − 𝑛𝑚 = 𝑎𝑛𝑒 − 𝑏𝑛𝑚 𝑛𝑚 → 𝑛𝑚 = 𝑐𝑛𝑒 + 𝑑𝑛𝑚 Similarly for the 𝜏 → −𝜏−1 transformation (for 𝜃 = 0), we have 𝑄𝑒 = 𝑛𝑒𝑒 → 𝑛𝑒𝑔 ⇒ 𝑛𝑒 → 𝑛𝑚 = 𝑎𝑛𝑒 − 𝑏𝑛𝑚 𝑄𝑚 = 𝑛𝑚𝑔 → 𝑛𝑚𝑒 ⇒ 𝑛𝑚 → 𝑛𝑒 = 𝑐𝑛𝑒 + 𝑑𝑛𝑚 So, we have the following 𝑆𝐿(2, ℤ) action on states (𝑛𝑒, 𝑛𝑚) 𝑛𝑒 𝑛𝑚 → 𝑎 −𝑏 𝑐 𝑑 𝑛𝑒 𝑛𝑚 Bogomol’nyi bound revisited The generalised Bogomol’nyi bound (proved at the end) 𝑀2 ≥ 𝑣2 (𝑄𝑒 2 + 𝑄𝑚 2 ) can entirely be written in term of 𝜏 by using the definition of 𝜏 and the expressions for 𝑄𝑒 and 𝑄𝑚.
- 17. Coupling to fermions We can add fermions into this theory with proper interactions with 𝜙𝑎 and 𝐴𝜇 𝑎 fields. Assume that fermions are in the ′𝑟′ representation of the 𝑆𝑈(2) gauge group. The fermion lagrangian is; ℒ𝜓 = 𝑖𝜓𝑛𝛾𝜇 𝐷𝜇𝜓 𝑛 − 𝑖𝜓𝑛𝑇𝑛𝑚 𝑎 𝜙𝑎 𝜓𝑚 Where 𝑇𝑛𝑚 𝑎 are the anti Hermitian generators in the ′𝑟′ representation. The gamma matrices are chosen as; 𝛾0 = −𝑖 0 𝕀2 −𝕀2 0 , 𝛾𝑗 = −𝑖 𝜎𝑗 0 0 −𝜎𝑗 These matrices satisfy the Clifford algebra 𝛾𝜇 , 𝛾𝜈 = 2𝑔𝜇𝜈 𝕀4 as can be verified easily. The Dirac equation can be worked out from the above lagrangian to get (using 𝜙𝑛𝑚 = 𝜙𝑎 𝑇𝑛𝑚 𝑎 ) 𝑖𝛾𝜇 𝐷𝜇𝜓 𝑛 − 𝑖𝜙𝑛𝑚𝜓𝑚 = 0 We seek solutions of the form 𝜓𝑛 𝑡, 𝑥 = 𝑒−𝑖𝐸𝑡 𝜓𝑛(𝑥) and we also set 𝜓𝑛 𝑥 = 𝜒𝑛 + 𝑥 𝜒𝑛 − 𝑥 𝑇 . Using the 𝛾 matrices and the Dirac’s equation, we have; −𝑖𝐸 𝜒𝑛 − −𝜒𝑛 + − 𝜎𝑗 𝐷𝑗𝜒𝑛 + −𝜎𝑗 𝐷𝑗𝜒𝑛 − − 𝑖𝜙𝑛𝑚 𝜒𝑚 + 𝜒𝑚 − = 0 This gives us two equations; 𝑖𝛿𝑛𝑚𝜎𝑗 𝐷𝑗 + 𝜙𝑛𝑚 𝜒𝑚 − = 𝐷𝑛𝑚𝜒𝑚 − = 𝐸𝜒𝑛 + , 𝑖𝛿𝑛𝑚𝜎𝑗 𝐷𝑗 + 𝜙𝑛𝑚 † 𝜒𝑚 + = 𝐷𝑛𝑚 † 𝜒𝑚 + = 𝐸𝜒𝑛 − , If the above two equations have a solution with 𝐸 = 0 then there can be fermionic collective coordinates in the BPS moduli space. In other words, we are looking for the kernels of 𝐷𝑛𝑚 and 𝐷𝑛𝑚 † . It is easy to see that; ker 𝐷𝑛𝑚 † ⊂ ker 𝐷𝑛𝑙𝐷𝑙𝑚 † = {0} In the last equality, I used the fact that 𝐷𝑛𝑙𝐷𝑙𝑚 † is a positive definite operator.
- 18. For the kernel of 𝐷𝑛𝑚, I need a result which is derived from index theorems. The result says that; dim [ker(𝐷𝑛𝑚)] − dim [ker(𝐷𝑛𝑚 † )] = 𝐴(𝑟)𝑛𝑚 Where 𝐴(𝑟) is a number that depends on the representation ‘𝑟′. For our work, the representations we need are 𝟐 and 𝟑. The values of 𝐴(𝑟) for these representations are 𝐴 𝟐 = 1, 𝐴 𝟑 = 2 Fundamental fermions For the fermions in the fundamental representation, we have only one zero mode for 𝑛𝑚 = 1 (since 𝐴 𝑟 = 1 and ker 𝐷𝑛𝑚 † = {0}). We can write the 𝜓 solution as (I will drop the gauge index for some time now) 𝜓(𝑥) = 𝑎0𝜓0(𝑥) + Non zero modes 𝑎0 will be our fermionic collective coordinate. The monopole ground states can be built by considering a ground state |Ω⟩ such that 𝑎0 Ω = 0 and then, another state can be built as 𝑎0 † |Ω⟩. So, the states |Ω⟩ and 𝑎0 † |Ω⟩ are degenerate. Adjoint Fermions For adjoint fermions, 𝐴 = 2 and thus, for 𝑛𝑚 = 1, we have two zero modes. Moreover, since the monopoles with fermions live in the 𝑆𝑈 2 𝑠 × 𝑆𝑈 2 𝑔 group, the fundamental fermions can be singlets because 𝟐 × 𝟐 = 𝟑 + 𝟏 but this is not the case for adjoint fermions because 𝟑 × 𝟐 = 𝟒 + 𝟐. Since there are two fermion zero modes, the fermions can’t have spin 3/2 and thus, the adjoint fermions should live in the 𝟐 representation of 𝑆𝑈 2 𝑠 × 𝑆𝑈 2 𝑔. Therefore, the spins of the zero modes must be ±1/2. So, the 𝜓 solution is expanded with the spins mentioned on the zero modes and the corresponding coefficients as; 𝜓 = 𝑎0,1/2𝜓0 1 2 + 𝑎0,−1/2𝜓0 − 1 2 + Non zero modes So, we have the four monopole states as shown below Ω Spin 0 𝑎0,1/2 † Ω Spin 1/2 𝑎0,−1/2 † Ω Spin − 1/2 𝑎0,1/2 † 𝑎0,−1/2 † Ω Spin 0
- 19. Monopoles in 𝑁 = 4 supersymmetric theories We need monopoles in 𝑁 = 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 𝑁 = 4 supersymmetry lagrangian. I will use only the result that the 𝑁 = 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 𝑎 0, ± 1 2 𝑛† where n = 1,2 Now, the monopole multiplet is as follows (we again start with the |Ω⟩ state and I drop the 0 subscript from the raising operators. Moreover, I replace ±1/2 with ± for brevity). State Spin |Ω⟩ 0 𝑎± 𝑛† |Ω⟩ ±1/2 𝑎− 𝑛†𝑎+ 𝑚† |Ω⟩ 0 𝑎+ 1† 𝑎+ 2† |Ω⟩ 1 𝑎− 1† 𝑎− 2† |Ω⟩ −1 𝑎∓ 1† 𝑎∓ 2† 𝑎± 𝑛† |Ω⟩ ∓1/2 𝑎+ 1† 𝑎+ 2† 𝑎− 1† 𝑎− 2† |Ω⟩ 0
- 20. 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 𝑁 = 4 gauge supermultiplet, there is an exact match. There is another feature of the 𝑁 = 4 theory which is worth mentioning. It is a known fact (I won't go into the details) that the 𝛽 function of 𝑁 = 4 super Yang Mills theory vanishes and thus, the potential of the 𝑁 = 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 𝑁 = 4 supersymmetric gauge theory.
- 21. Proof of generalized Bogomol’nyi bound (𝑄𝑒, 𝑄𝑚) In the presence of electrical fields (keeping 𝐴0 𝑎 = 0 still), we have the following mass for the dyon; 𝑀 = 1 2 ∫ 𝑑3 𝑥 𝐸𝑖 𝑎 𝐸𝑖 𝑎 + 𝐵𝑖 𝑎 𝐵𝑖 𝑎 + 𝐷𝑖𝜙𝑎 2 = 1 2 ∫ 𝑑3 𝑥 𝐸𝑖 𝑎 − cos 𝛼 𝐷𝑖𝜙𝑎 + 𝐵𝑖 𝑎 − sin 𝛼 𝐷𝑖 𝜙𝑎 2 + ∫ 𝑑3 𝑥 cos 𝛼 𝐸𝑖 𝑎 𝐷𝑖𝜙𝑎 + sin 𝛼 𝐵𝑖 𝑎 𝐷𝑖𝜙𝑎 ≥ cos 𝛼 𝑄𝑒 + sin 𝛼 𝑄𝑚 Now, we can make the bound as tight as possible by differentiating the above expression w.r.t 𝛼 and setting it to zero. This gives; tan 𝛼 = 𝑄𝑚 𝑄𝑒 ⇒ sin 𝛼 = 𝑄𝑚 𝑄𝑒 2 + 𝑄𝑚 2 , cos 𝛼 = 𝑄𝑒 𝑄𝑒 2 + 𝑄𝑚 2 This gives us; 𝑀 ≥ 𝑣 𝑄𝑒 2 + 𝑄𝑚 2 1 2 This is the generalised Bogomol’nyi bound in terms of (𝑄𝑒, 𝑄𝑚).
- 22. Acknowledgements • My supervisor, K.S. Narain, for guiding me in my research work and for his comprehensive explanations. • ICTP staff, specially my professors, for teaching and guiding us in these difficult times • My friends at ICTP and in my country (and my family) for keeping in contact with me and encouraging me in hard times.
- 23. Proof of generalized Bogomol’nyi bound (e,g) In the presence of electrical fields (keeping 𝐴𝜇 𝑎 = 0 still), we have the following mass for the dyon; 𝑀 = 1 2 ∫ 𝑑3 𝑥 𝐸𝑖 𝑎 𝐸𝑖 𝑎 + 𝐵𝑖 𝑎 𝐵𝑖 𝑎 + 𝐷𝑖𝜙𝑎 2 = 1 2 ∫ 𝑑3 𝑥 𝐸𝑖 𝑎 − cos 𝛼 𝐷𝑖𝜙𝑎 + 𝐵𝑖 𝑎 − sin 𝛼 𝐷𝑖 𝜙𝑎 2 + ∫ 𝑑3 𝑥 cos 𝛼 𝐸𝑖 𝑎 𝐷𝑖𝜙𝑎 + sin 𝛼 𝐵𝑖 𝑎 𝐷𝑖𝜙𝑎 We proved previously that 𝐵𝑎 𝐷𝑖𝜙𝑎 = 𝜕𝑖(𝐵𝑖 𝑎 𝜙𝑖). We can also prove that 𝐸𝑖 𝑎 𝐷𝑖𝜙𝑎 = 𝜕𝑖 𝐸𝑖 𝑎 𝜙𝑎 as follows. We see that 𝐸𝑖 𝑎 𝐷𝑖𝜙𝑎 = 𝐷𝑖 𝐸𝑖 𝑎 𝜙𝑎 − 𝜙𝑎 𝐷𝑖𝐸𝑖 𝑎 = 𝜕𝑖 𝐸𝑖 𝑎 𝜙𝑎 − 𝜙𝑎 𝐷𝑖𝐸𝑖 𝑎 as 𝐸𝑖 𝑎 𝜙𝑎 is gauge singlet. Now, we prove that 𝐷𝑖𝐸𝑖 𝑎 = 0. Use the zeroth component of Gauss’ constraint to get; 𝐷𝑖𝐹𝑖0 𝑎 = 𝑒𝜖𝑎𝑏𝑐 𝜙𝑏 𝐷0𝜙𝑐 = 0 The last equality follows because 𝜙𝑎 is time independent and 𝐴0 𝑎 = 0. So, 𝐷𝑖𝐹𝑖0 𝑎 = −𝐷𝑖𝐹0𝑖 𝑎 = −𝐷𝑖𝐸𝑖 𝑎 = 0 So, we get; 𝑀 ≥ ∫ 𝑑3 𝑥 cos 𝛼 𝜕𝑖 𝐸𝑖 𝑎 𝜙𝑎 + sin 𝛼 𝜕𝑖 𝐵𝑖 𝑎 𝜙𝑎 = 𝑣[cos 𝛼 𝑒 + sin 𝛼 𝑔] Now, we can make the bound as tight as possible by differentiating the above expression w.r.t 𝛼 and setting it to zero. This gives; tan 𝛼 = 𝑔 𝑒 ⇒ sin 𝛼 = 𝑔 𝑒2 + 𝑔2 , cos 𝛼 = 𝑒 𝑒2 + 𝑔2 This gives us; 𝑀 ≥ 𝑣 𝑒2 + 𝑔2 1 2