Problem with Solutions for Quantum Field Theory I - Assignment 4 | PHY 396K, Assignments of Physics

Download Problem with Solutions for Quantum Field Theory I - Assignment 4 | PHY 396K and more Assignments Physics in PDF only on Docsity! PHY–396 K. Problem set #4. Due October 2, 2007. 1. When an exact symmetry of a quantum field theory is spontaneously broken down, it gives rise to exactly massless Goldstone bosons. But when the spontaneously broken symmetry was only approximate to begin with, the would-be Goldstone bosons are no longer exactly massless but only relatively light. The best-known examples of such pseudo-Goldstone bosons are the pi-mesons π± and π0, which are indeed much lighter then other hadrons. The Quantum ChromoDynamics theory (QCD) of strong interactions has an approximate chiral isospin symmetry SU(2)L×SU(2) ∼= Spin(4) which would be exact if the two lightest quark flavors u and d were exactly massless; in reality, the current quark masses mu and md do not exactly vanish but are small enough to be treated as a perturbation. Exact or approximate, the chiral isospin symmetry is spontaneously broken down to the ordinary isospin symmetry SU(2) ∼= Spin(3), and the 3 generators of the broken Spin(4)/Spin(3) give rise to 3 (pseudo) Goldstone bosons π± and π0. QCD is a rather complicated theory, so it is often convenient to describe the physics of the spontaneously broken chiral symmetry in terms of a simpler effective theory such as the linear sigma model. This model has 4 real scalar fields; in terms of the unbroken isospin symmetry, we have an isosinglet σ(x) and an isotriplet π˜(x) comprising π1(x), π2(x) and π3(x) (or equivalently, π0(x) ≡ π3(x) and π±(x) ≡ ( π1(x)± iπ2(x) ) / √ 2). The Lagrangian L = 12(∂µσ) 2 + 12(∂µπ˜)2 − λ8 (σ2 + π˜2 − f2)2 + βσ (1) is invariant under the SO(4) rotations of the four fields, except for the last term which we take to be very small. (In QCD β ∼ mu+md2f 〈 ΨΨ 〉 which is indeed very small because the u and d quarks are very light.) In class, we discussed this theory for β = 0 and showed that it has SO(4) spontaneously broken to SO(3) and hence 3 massless Goldstone bosons. In this exercise, we let β > 0 but β  λf3 to show how this leads to massive but light pions. 1 (a) Show that the scalar potential of the linear sigma model with β > 0 has a unique minimum at 〈π˜〉 = 0 and 〈σ〉 = f + βλf2 + O(β2). (2) (b) Expand the fields around this minimum and show that the pions are light while the σ particle is much heavier. Specifically, M2π ≈ (β/f) while M2σ ≈ λf2. 2. The rest of this homework is about the Bogolyubov transform and the superfluid helium. Let us start with some kind of annihilation and creation operators âk and â † k which satisfy the bosonic commutation relations [âk, âk′ ] = [â † k, â † k′ ] = 0, [âk, â † k′ ] = δk,k′ . (3) Let us define new operators b̂k and b̂ † k according to b̂k = cosh(tk)âk + sinh(tk)â † −k , b̂ † k = cosh(tk)â † k + sinh(tk)â−k (4) for some arbitrary real parameters tk = t−k. (a) Show that the b̂k and the b̂ † k satisfy the same bosonic commutation relations as the âk and the â†k. The Bogolyubov transform — replacing the ‘original’ creation and annihilation operators â†k and âk with the ‘transformed’ operators b̂ † k and b̂k — is useful for diagonalizing quadratic Hamiltonians of the form Ĥ = ∑ k Akâ † kâk + 1 2 ∑ k Bk ( âkâ−k + â † kâ † −k ) (5) where for all momenta k, Ak = A−k, Bk = B−k, and Ak > |Bk|. 2