February 18, 2015
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Instantons in Path Integral Formulation of mechanics is based around the propagator: x f e iht / x i In path integral formulation of quantum mechanics we relate the propagator to a sum over all possible paths with a phase: x f e iht/ x i = N Dx[t] e is[x]/
Instantons in Wick Rotation In order to make the integral well defined we perform a Wick Rotation: t T = it e iht/ HT / e S = i dt 1 ( ) dx 2 V (x) 2 dt [ S = dt 1 ( ) dx 2 V (x)] 2 dt [ ( ) 1 dx 2 = dt + V (x)] 2 dt This is equivalent working with an inverted potential.
Instantons in Semi-Classical Solution If we are to solve the Classical equations of motion to an inverted potential we arrive at the usual equation: d 2 x dt 2 V (x) = 0 we also have the conserved energy quantity: E = 1 2 ( ) dx 2 V (x) dt Now assume that x(t) has quantum corrections, we can expand it in. To first order we have the equation: d 2 x (1) dt 2 V (x (0) )x (1) = 0
Instantons in Introducing ness More formally we can perturb around the classical solution x(t): x(t) = x(t) + n c n x n (t) But now let the x n (t) be eigenvalues of the equation: d 2 x n dt 2 V ( x)x n = λ n x n This introduces quantumness to the equations of motion. Integrating over all of these perturbations is equivalent to integrating over all paths. Now we want them to be orthogonal so there is no over counting of perturbations in our integral: T /2 T /2 x n (T )x m (T )dt = δ nm
Instantons in Formal Integration over Modes Now if we expand the potential around the classical solution we get: V (x) = V ( x) (x x)v ( x) + 1 2 V ( x)(x x) 2 +... Now integrating, using the perturbative expansion of x, the first term will give us a factor of e S0/ and the last term will give us: 1 = [ det( t 2 + V ( x)) ] 1/2 λn n This is a problem if there exist any λ n = 0. But since these satisfy the semi-classical equation, these are actually just a deformation of the classical solution so integrating over them corresponds to integrating over the free parameters of the classical solution.
Instantons in Quick Summary Consider the Path Integral formulation of Wick Rotate (make time imaginary t it), causes us to invert the potential Solve the classical equations of motion there, plug into action Integrate over parameter space (zero modes) and multiply by 1 λn (quantum fluctuations).
Instantons in What are Instantons? Instantons are classical solutions to the Wick rotated equations of motion which have non-trivial topology... To illustrate this, we will work out the canonical double well potential.
Instantons in Double Well Double well has potential: V (x) = λ(x 2 a 2 ) 2 Interested in the classical solutions which are topologically nontrivial. In this example that is when a particle will tunnel from one minima to another. Each tunneling event goes as ±a tanh(c(t t )) We can glue solutions together in alternating order to have more general paths. Limit of large separation (quick tunneling): dilute gas approximation.
Instantons in Double Well (cont.) Now when we do the path integral we will have an expression that looks like: a e HT / a = N det ( 2 t + V ) n odd T /2 T /2 tn 1 dt 1... dt n K n e ns 0/ T /2 Since the particle spends most of its time at the bottom of one of the wells which is approximately harmonic, we can approximate the normalization by the harmonic oscillator: N det ( 2 t + V ) ( ω π ) 1/2 e ωt /2
Instantons in Double Well (cont. cont.) The integration time integration gives us a factor of T n n!. Now we get the expression: ( ω 1/2 a e HT / a = e ωt /2 π ) n odd T n n! K n e ns 0/ ( ω ) 1/2 = e ωt /2 1 [e KTe S 0 / e / ] KTe S 0 π 2 And similarly: ( ω 1/2 a e HT / a = e ωt /2 π ) n even T n n! K n e ns 0/ ( ω ) 1/2 = e ωt /2 1 [e KTe S 0 / + e / ] KTe S 0 π 2
Instantons in Double Well (cont.cont.cont.) Examining the exponentials we find: E = ω 2 ± Ke S 0/ It is now important to determine K. As it turns out, these are generally very annoying to calculate so I will just tell you: ( ) 1/2 ( S0 det ( t 2 + ω 2 ) 1/2 ) K = 2π det ( t 2 + V ( x)) This is determined by comparing the exact calculation for the one instanton with the single instanton term in the sum. Furthermore there is a formula for computing the fraction of coefficients: det ( t 2 + ω 2 ) det ( t 2 + V ( x)) = ψ SHO(T /2) (1) ψ 0 (T /2)
Instantons in Uneven Double Well Decay We can also consider the uneven well. These are pictures of the classical solution and zero mode: Since there is a node there will actually be a single mode which has a lower (negative) eigenvalue. This makes K imaginary. We now have that the energy has an imaginary part which is exactly the decay width: Im[E] = S 0 2π K e S 0/ = Γ/2
Instantons in Derrick s Theorem There are no non-trivial topological solutions to the double well s qft equivalent in dimension other than 2. There are no non-trivial matter solutions due to Derrick s Theorem. Take a scalar field theory: L = 1 2 ( µφ) 2 V (φ) [ ] 1 E = d d 1 x 2 ( µφ) 2 + V (φ) = I 1 + I 2 Assume there is time dependent solution of finite energy φ(x). Now define: φ λ (x) = φ(λx). Then the energy changes as: E λ = λ 2 d I 1 + λ d I 2 Want λ = 1 minimize the energy otherwise the solution is unstable
Instantons in Derricks Theorem (cont.) Varying the Energy with respect to λ we find: E λ = (2 d)i 1 d I 2 = 0 λ λ=1 2 E λ λ=1 λ 2 = (d 2)(d 1)I 1 + d(d 1) I 2 Solving these we find: ( 2 d I 2 = d ) I 1 2 E λ λ 2 λ=1 = 2(d 2)I 1 < 0 for d > 2
Instantons in Gauge Instantons We can however have topological solutions to gauge fields. Take SU(N) gauge theory: S = d 4 1 x 2g 2 (F µν) a 2 Require finite energy and finite action. This enforces: ( ) 1 A µ U 1 µ U + O r r 2 These can be classified by their winding number at r = : k = 1 16π 2 d 4 x trf µν F µν = 1 8π 2 d 3 xn µ ɛ µνρσ tr [A ν ρ A σ + 23 ] A νa ρ A σ r= = 1 8π 2 tr [A da + 23 ] A A A r=
Instantons in Winding Number Classification This classification holds because: S = 1 2g 2 d 4 x trf 2 = 1 4g 2 d 4 x tr(f F ) 2 1 2g 2 1 2g 2 d 4 x tr(f F ) = 8π2 g 2 (±k) d 4 trf F This bound is saturated when F is (anti-)self dual. These winding numbers are determined by the types of gauges we can have, that is the class of functions U(x) which maps spatial infinity S 3 for Euclidean space to the gauge group. This tells us that the k are exactly determined by π 3 (G).
Instantons in Pure Gauge Theory For a pure gauge theory, everything else is the same as before Determinant Integrate over Zero modes Classical action Now we can use Index Theorems to determine how many zero modes Can show exact correspondence and orthogonality of zero modes from free parameters Can use Fadeev-Popov method to convert integration over zero modes to integration over parameters
Fermions??? When add fermions to theory, everything is the same except for integration over the fermionic zero modes... There are grassmanian degrees of freedom: N=2 C(R) k Integration over grassmanian variable picks out zero modes from operators in the expectation value of operators ψ(y i ) = ψ i ψ = ψ cl + n ψ n ψ zero f (x)k O(A, ψ, ψ)ψ 1...ψ N ) k N = N DAD ψdψ DK i O(A, ψ, ψ)ψ 1...ψ N e S[A,ψ] i=1 O(A, ψ, ψ 0 e 8π 2 g 2 k
Breaking of U(1) Symmetry Now that we have operators with uneven number of fermions and their conjugates, the symmetry of the Lagrangian: ψ e iϕ ψ is broken. If we have a term in Lagrangian: L = θ 16π 2 trf µν F µν then this anomalous phase can be compensated by: θ θ + 2C(R)kϕ Now we have a discrete symmetry: θ θ + 2π ψ e iϕ ψ ϕ = 2π 2C(R)k Important when up in N=2 SUSY Moduli space
Baryon Decay Consider the Strong Force: SU(3) The k=1 instanton background leads to an integration over 6 zero modes. (it removes 6 quark operators from expectation values) This leads to a 6 (anti-)fermion vertex in our effective action V u 1 Lu 2 Lu 3 Ld 1 Ld 2 Ld 3 L This clearly violates baryon conservation and can lead to proton, neutron decay: p + n e + + ν µ Suppressed by e 8π 2 g 2 k 10 34.
Conclusion In, instantons describe tunneling/decay phenomena by using a semi-classical approximation to equations of motion with imaginary time. In instantons are described by gauge fields with non-trivial winding at infinity. They also lead to a description of tunneling and decay. In QFT when instantons interact with matter they give rise to fermionic zero modes which allow for chiral asymmetric operators which lead to U(1) symmetry breaking and baryon decay.
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