160 The one SU 2 instanton solution centered at z \u03bc can be written explicitly

160 the one su 2 instanton solution centered at z μ

This preview shows page 160 - 163 out of 174 pages.

160
Image of page 160
The one SU (2) instanton solution centered at z μ can be written explicitly in com- ponents as A cl = 2 g η aμν ( x - z ) ν ( x - z ) 2 + R 2 , (10.55) where the symbol η aμν is anti-symmetric in μ, ν , defined as η aij = aij , i, j = 1 , 2 , 3 , η ai 4 = - η a 4 i = δ ai . (10.56) Note that the one anti-instanton solution would be obtained by replacing η aμν with ¯ η aμν , defined by ¯ η aij = η aij , ¯ η ai 4 = - η ai 4 . The Euclidean action S E is related to the Lorentzian one by S = iS E . The one instanton solution has Euclidean action S cl E = 8 π 2 /g 2 . In background field gauge, the action for the quantum fluctuations is S E = S cl E + Z d 4 x 1 2 ( D μ A 0 ) 2 + g abc F μν a A 0 A 0 + ( D μ Φ) * ( D μ Φ) + ¯ ψγ μ D μ ψ + ( D μ ¯ η a )( D μ η a ) + · · · (10.57) where we have exhibited only quadratic terms in the quantum fluctuations, which is all we need to the one-loop computation. Note that the term proportional to ( D μ A 0 μ ) 2 from expanding the Yang-Mills Lagrangian is canceled by the gauge fixing term. As before, we will write M A , M Φ , M ψ and M η for the kinetic operators on A 0 , Φ, ψ , and ( η, ¯ η ) respectively. The result of the one-loop Gaussian integral is then e - 8 π 2 /g 2 Z d (zero modes) det 0 M ψ det 0 M η p det 0 M A det 0 M Φ , (10.58) where we have separated out the integration over zero modes of A 0 , Φ , ψ, η in the instanton background: these modes are annihilated by the respective kinetic operator M ; det 0 is the determinant taken over modes of nonzero eigenvalues only, defined using a suitable (gauge invariant) regulator. Explicitly, the kinetic operators are M A A 0 = - D 2 A 0 - 2 g abc F bμν A 0 , M ψ = γ μ D μ , M 2 ψ = D 2 - i 2 γ μν F aμν T a , M Φ = M η = - D 2 . (10.59) Here T a are the generators of the SU (2) gauge group. Let us first calculate D 2 in the 161
Image of page 161
background of an instanton with z = 0 and R = 1, D 2 = ( μ - igT a A cl )( μ - igT b A cl,μ b ) = - 2 iT a η aμν { x ν r 2 + 1 , ∂ μ } - 4 T a T b η aμν η b μ ρ x ν x ρ ( r 2 + 1) 2 = + 4 r 2 + 1 T a η aμν ( - ix ν μ ) - 4 r 2 ( r 2 + 1) 2 T 2 , (10.60) where r 2 = x μ x μ , T 2 = T a T a . Note that the SO (4) rotation generator - ix [ μ ν ] can be split into the SU (2) × SU (2) generators L a 1 = - i 2 η aμν x μ ν , L a 2 = - i 2 η aμν x μ ν . (10.61) They obey L a 1 L a 1 = L a 2 L a 2 L 2 = - 1 8 ( x μ ν - x ν μ ) 2 . (10.62) Writing the Laplacian in terms of L 2 also, we have D 2 = 1 r 3 r r 3 ∂r - 4 L 2 r 2 - 8 r 2 + 1 T a L a 1 - 4 r 2 ( r 2 + 1) 2 T 2 . (10.63) This expression can be generalized to the kinetic operator acting on fields of nonzero spin, M = - 1 r 3 r r 3 ∂r + 4 L 2 r 2 + 8 r 2 + 1 T a L a 1 + 4 r 2 ( r 2 + 1) 2 T 2 + 16 ( r 2 + 1) 2 T a S a 1 , (10.64) where S a 1 together with S a 2 are the SU (2) × SU (2) generators associated with the intrinsic spin of the field. The fermion ψ transforms in the representation ( 1 2 , 0) (0 , 1 2 ) of the SU (2) × SU (2) SO (4) rotation (Euclidean version of Lorentz) group, whereas A 0 μ transforms in the representation ( 1 2 , 1 2 ). Explicitly, acting on the spinor, S 1 and S 2 are S a 1 = - i 8 η aμν γ μν , S a 2 = i 8 η aμν γ μν , S 2 1 = 3 4 1 - γ 5 2 , S 2 2 = 3 4 1 + γ 5 2 .
Image of page 162
Image of page 163

You've reached the end of your free preview.

Want to read all 174 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes