Quantum Field Theory III

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1 Quantum Field Theory III Prof. Erick Weinberg March 9, 0 Lecture 5 Let s say something about SO(0. We know that in SU(5 the standard model fits into 5 0(. In SO(0 we know that it contains SU(5, in two ways. If we look at the Dynkin diagram and remove a node, we can get SU(5. Another way to see it is that SU(0 is the transformation that fixes the sum x i, but for SU(5 it conserves z i which also have 0 components but it is more restrictive. Let s look at the correspondence. The 0 representation of SO(0 corresponds to 5 5, but it is real and we don t want it. So we need to look at the spinor representation. There are 5 commuting generators so the spinor representation should be written as (±/, ±/, ±/, ±/, ±/. But these are reducible into representations of odd number of + and even number of +. These are 6 and 6 respectively. Let s look at 6. The representation with one + has 5 permutations, the one with three + has 0 permutations, and the one with 5 + has only one case. Under the generators of SU(5 these do not mix with each other, so it reduces into 5 0. The bar or no bar on 5 doesn t matter, just a matter of choice. We can see that 0 is the 0 we want from anomaly considerations. So all the standard model particle fits into 6 exactly, which is economical and attractive. Now we need to break this symmetry. We can imagine But we can also imagine SO(0 SU(5 U( SU(5... ( SO(0 SO(6 SO(4 = SU(4 SU( L SU( R ( where SU( L is the weak coupling that breaks chirality. This looks like we have deeper chiral symmetry which is broken in standard model. Among the infinite amount of papers that are generated, there are also more ways to break the group down.. Solitons Now we change subject completely and talk about solitons. Solitons are localized solutions of classical field equations. By localized we mean that if it starts at some finite region in space it stays there. Originally they were required to retain their form even after scattering. This is the soliton in mathematical sense occurring in the solution in nonlinear differential equations. Let s start with some classical field theory in + dimensions. The Lagrangian is L = ( µϕ V (ϕ, V (ϕ = m ϕ + 4 ϕ4 + 4 v4 = 4 (ϕ v (3

2 where we defined v = m /. The minima of the potential are at v and v. We know from symmetry breaking that there are two vacuum states and elementary excitation has mass m. If you are in a vacuum state and are asked which vacuum you are in, there is no way to tell, because physics looks the same. The classical field equation is ϕ t ϕ x + V ϕ = 0 (4 The static solution is just when the time derivative is zero, and it is equivalent to minimizing the potential energy [ ] U[ϕ(x] = dx ϕ + V (ϕ (5 Now we need to look for a solution. It had better have finite energy and go to one vacuum or another at infinity. Let s consider a configuration which connects one vacuum to the other at different infinities. Now if we start from this configuration and do a variation continuously to lower the energy, we will hit a minimum, which must be different from either vacuum solution. So we should have a solution which connects two vacuua and we call this a kink. Now mathematicians will say the space of functions is not compact and we can t use this argument, but it turns out that here it is fine. Let s multiply the equation by ϕ 0 = ϕ ϕ + ϕ V ϕ = d [ ] dx ϕ + V So the term in brackets is independent of x, and we can shift it to be zero, which we already did when introducing v. So we have dϕ dx = ± dϕ V, = dx (7 V and we can integrate the equation. Let s call the point when ϕ = 0 by x 0, and integrate from there ϕ dϕ x 0 v ϕ = dx = x x 0 (8 x 0 (6 If we do the integral on the left then we will find that [ ] m ϕ = v tanh (x x 0 (9 The size of the kink is on the order of /m which can be seen from the equation directly. So ϕ is on the order of m v m 4 /. The classical energy of this solution is E cl = M cl ( m 4 m3 m = ( m 3 3 = m m (0 3 The term in the final bracket is dimensionless in + dimensions. In the limit of weak coupling we have /m, so this mass, or energy, is much much larger than the elementary excitations m. Now we can consider solutions of many kinks. But we need to have the above integral relation which tells us that we can only go to vacuum at x. But we can have time-dependent solution, which are

3 kinks moving around. When they collide a kink will dissipate with an antikink. The number of kinks minus the number of antikinks will be conserved and we can define the topological current J µ = v ɛµν ν ϕ ( This is obviously conserved because of the ɛ factor. We can similarly define the topological charge Q top = dx J 0 = dx ɛ 0 ϕ = [ϕ( ϕ( ] = ± ( v v But this is not due to a symmetry! Another example which will be important is The classical field equation is V = m4 [ cos ( ] m ϕ ( 0 = ϕ + m3 sin m ϕ Some people call this Sine-Gordon equation. This is a classical theory where there is solition solution. Let s look at the vacuum solutions. It occurs at minimum of the potential, which is at (3 (4 ϕ = πm N = Nv (5 where N is an integer. So we have many vacuum solutions. If we consider ϕ 0 then we have [ V = m4 ] m ϕ + 4!m 4 ϕ = m ϕ 4! ϕ4 + O(ϕ 6 (6 so the elementary excitation around ϕ = 0 has mass m. Now if we proceed as before, we can find the kink solution which connects the vacuum N to N + as ϕ kink = Nv + v π tan ( e m(x x 0 (7 with classical mass M cl = 8m 3 /. All these discussions are classical, but we know that there is nothing as classical field theory in nature. So question is when is the classical theory like above meaningful? Suppose we want to consider the kink as a particle and think of its position, then we need its Compton wavelength to be much less than the size of the particle c M cl m 3 (8 m so this is true in weak coupling. What about the value of the field? How do we measure it? We can t measure the value of a field at some point, because it will oscillate wildly. Let s define a smeared ϕ L (x over a smeared length L. We will expect that ( ϕ ( (d / (9 L 3

4 where d is the space-time dimension and when d = it does not depend on L, up to logarithm terms. So at very small distances we expect large fluctuations, but it should be smooth when L is comparable to the characteristic size of the classical solution ( ϕ L quantum (ϕ L v = m (0 So now let s do quantum field theory. We do QFT about a vacuum using ϕ(x, t = ϕ vac + η(x, t ( and we write the Lagrangian as a classical piece plus a quadratic piece and an interaction piece L = L class + L quad + L int ( The first is just the vacuum energy. The second is [ L = dx ( µη ] V (ϕ vac η (3 We find the normal modes of η using the classical equation of motion and write η(x, t in a sum of modes η(x, t = j [c j (tf j (x + h.c.] (4 where f j are classical functions and c j are quantum mechanical operators. Then the Hamiltonian will look like that of harmonic oscillators and the energy is E = E cl + ( + n j ω j +... (5 j Even if n j = 0 for all j we have divergence when we add / for sufficiently many times. So we add a term δe = j ω j (6 to the Lagrangian to cancel this zero point energy. There is also divergence due to loop corrections so we add a counterterm L ct = (δm ϕ (7 Now we write ϕ(x, t = ϕ kink + η(x, t. This is equivalent to taking the classical kink as a background. Now the normal modes are solutions to the equation ] [ d dx + V (ϕ kink f j (x = ωj kink f j (x (8 and again we have divergences in the energy. But we have already added the vacuum energy and the loop counterterms and we don t have any freedom left in our theory to fix these divergences. These divergences had better cancel. 4

5 Let s try the kink solution [ ] m ϕ kink = v tanh (x x 0 (9 so the potential derivative is V (ϕ kink = m ( 3 cosh (m(x x 0 / (30 So the classical normal mode equation looks like a Schrödinger equation with a potential like the above. Note that the ϕ kink satisfies ϕ kink V ϕ = 0 = ϕ kink V ϕ kink ϕ = 0 (3 So this tells us that ϕ kink is a zero-mode solution with ω = 0. This is the lowest energy solution to the Schrödinger equation and it has no nodes. All the other solutions should have higher energies and more nodes. There is another node which has ω = 3m / and the function is f = sinh(m(x x 0/ cosh (m(x x 0 / After that there is continuum with ω = k + m and { [ ] m(x f k = e ikx 3m tanh x0 m k 3 [ ]} m(x x0 imk tanh (3 (33 We are really interested in the behavior at x ± which is f k 4(m k + 8m k e i[kx±δ(k/] (34 where δ(k = tan ( 3 mk k m (35 We need to find the branch of δ so we define δ(0 + = π and δ(+ = 0. Note we have δ( k = δ(k and there is a discontinuity at 0. Now the energy of the kink is E kink = M cl + ( ω kink j ω vac j + δm dx ( ϕ kink ϕ vac The sums are badly divergent. Let s define the world to have length a and use periodic boundary conditions, so we have a cut off on momentum. We also assume that space has only N points, which means there are only N modes. Now in vacuum instead of continuum of k we should have k n a = πn where n = ±, ±,... for the vacuum solutions. But for the kinks we have k n a + δ(k n = πn where n = ±, ±3,.... This is because for the kink we have two discrete states below the continuum. Now we should have k kink n k vac n = a δ(kvac n + O 5 ( a (36 (37

6 Now the vacuum zero point energies become (0 ( m + 3 m N [ m + (kn kink + m ] (kn vac + m n= = (0 ( m + 3 m m + [ ] (38 k vac (kn vac + m n δ(k n [(k n + m ]a + O(/a Now if we take the continuum limit we have instead of the sum kδ(k dk π k + m = [ d dk π dk [ k + m δ(k] k + m dδ ] dk 0 (39 And the vacuum energy is 3 ω π m 6 dk k + m 6 m 3 π m + m (40 Now we have got rid of most of the divergences but we still have a logarithmic divergence from the above integral. It is cancelled by the counterterm. So in the end we have ( M kink = m ( m + O π m m (4 Now we are just quantizing things at the kink background, instead of the kink itself. But let s look at the spectrum. The first excited state is like a one particle state, and the continuum is like particles scattering off the kink. However the ω = 0 mode doesn t correspond to any excitations. It turns out that this is the mode correspond to the position of the kink. 6

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