σ 2 + π = 0 while σ satisfies a cubic equation λf 2, σ 3 +f + β = 0 the second derivatives of the potential are = λ(σ 2 f 2 )δ ij, π i π j

Transcription

1 PHY 396 K. Solutions for problem set #4. Problem 1a: The linear sigma model has scalar potential V σ, π = λ 8 σ + π f βσ. S.1 Any local minimum of this potential satisfies and V = λ π V σ = λ σ + π f = 0 π σ + π f σ β = 0, S. which together imply π = 0 while σ satisfies a cubic equation σσ f β λ = 0. S.3 For small β > 0 this equation has three real solutions, namely σ 1 β λf, σ f + β λf, σ 3 +f + β λf. S.4 Furthermore, at π = 0 the second derivatives of the potential are V σ = λ3σ f, V π i π j = λσ f δ ij, v σ π i = 0, S.5 which means that the stationary point σ 1 is the local maximum of V σ, π, the stationary point σ is a saddle point, and only the σ 3 is a minimum. In other words, V σ, π has a unique minimum at = 0, σ = σ 3, exactly as in eq.. Q.E.D. π 1

2 Problem 1b: Let us shift the σ field by its vacuum expectation value σx = σ + σx. Expanding the scalar potential into powers of the π and σ fields, we have V π, σ = λ π + σ + σ f β σ + σ 8 S.6 = const + V + V 3 + V 4 where V 4 = λ 8 π + σ, V 3 V = λ σ σ π + σ = λ 4 σ f π + λ 4 3 σ f σ = β σ π + β σ + λ σ σ, S.7 and the last equality follows from eq. S. for the σ. Lagrangian into powers of π x and σx and write Consequently, we expand the while L = L free + L int + const S.8 where L free = L kin V π, σ [ ] 1 = µ σ β σ + λ σ σ + [ 1 µπ β ] σ π S.9 describes the free π and σ fields, and L int = V 3 π, σ V 4 π, σ S.10 describes their interactions. The particles masses follow from the quadratic Lagrangian S.9: M π = β σ β f, and M σ = M π + λ σ λf M π. Q.E.D.

3 Problem a: Combining definitions 4 with commutation relations 3, we immediately calculate [ˆb k, ˆb k ] = cosht k sinht k δ k, k sinht k cosht k δ k,k = 0 S.11 where the second equality follows from t k = t k for k = k. Likewise, [ˆb k, ˆb k ] = 0. Finally, [ˆb k, ˆb k ] = cosht k cosht k δ k,k sinht k sinht k δ k, k = δ k,k cosh t k sinh t k = 1. S.1 In other words, the ˆb k and ˆb k operators satisfy the same bosonic commutations relations [ˆb k, ˆb k ] = 0, [ˆb k, ˆb k ] = 0, [ˆb k, ˆb k ] = δ k,k. S.13 as the original â k and â k operators. Q.E.D. Problem b: A straightforward calculation shows that k ω kˆb kˆb k ω k cosht k â kâk + 1 k ω k sinht k â kâ k + â kâk + const. S.14 Therefore, the Hamiltonian 5 can be diagonalized in terms of the transformed creation / annihilation operators 4 if and only if we can solve for ω k and t k such that ω k cosht k = A k and ω k sinht k = B k. S.15 The latter equations are solvable whenever A k > B k and the solution is t k = 1 artanh B k A k and ω k = A k B k. S.16 Q.E.D. 3

4 Problem c: In terms of the momentum modes ã k and ã k of the shifted fields ˆϕx and ˆϕ x, the free Hamiltonian 8 becomes Ĥ free [ k M + λn ã kãk + λn ã k ã k + ã kã k ]. S.17 This Hamiltonian has form 5 in term of the shifted annihilation and creation operators ã k = â k + Nδ bp,0 and ã k = â k + Nδ bp,0, which obviously satisfy the same bosonic commutation relations as the un-shifted operators â k and ã k. Therefore, we may diagonalize this Hamiltonian by means of the Bogolyubov transform ˆbk = cosht k ã k + sinht k ã k, ˆb k = cosht k ã k + sinht k ã k, 4 for suitable parameters t k = t k. Specifically, according to eqs. S.16, we let t k = 1 artanh λn k M + λn = 1 4 log and then the free Hamiltonian S.17 takes form 9 for 1 + 4λnM k, S.18 ω k = k λn M + λn λn = k M + k 4M. 10 Q.E.D. Problem : The key to this exercise is the multiple commutator formula: For any two operators ˆF and Ĝ, e + ˆF Ĝe ˆF = 1 n! [ ˆF, [ ˆF, Ĝ] ] n times n=0 S.19 = Ĝ + [ ˆF, Ĝ] + 1 [ ˆF, [ ˆF, Ĝ]] [ ˆF, [ ˆF, [ ˆF, Ĝ]]] +. Now, let ˆF be as in eq. 11 and let Ĝ be one of the ã k or ã b p. The simple commutators [ ˆF, Ĝ] 4

5 follow directly from the bosonic commutation relations: [ ˆF, ã k ] = t k ã k and [ ˆF, ã k ] = t kã k. S.0 Consequently, [ ˆF, [ ˆF, ã k ] ] n times = t k n { ãk for even n, ã k for odd n, S.1 and therefore eq. S.19 yields e + ˆF ã k e ˆF = t k n ã n! k + even n odd n t k n ã n! k = cosht k ã k + sinht k ã k S. = ˆb k. And since ˆF is anti-hermitian, we also have e + ˆF ã k e ˆF = e + ˆF ã k e ˆF = ˆb k, S.3 and hence the bosonic commutation relations S.13 follow from the relations 3 via unitary equivalence S. and S.3. Finally, for the quantum state Ω = e + ˆF coh we have ˆbk Ω = e + ˆF ã k e ˆF e + ˆF coh = e + ˆF ã k coh = 0 because ã k coh = 0 for all momenta k. Therefore, from the point of view of quasi-particles created by the ˆb k operators and annihilated by the ˆb k, the Ω is the vacuum state. 5

6 Problem d: In terms of atomic creation and annihilation operators ˆP tot k â kâk = k ã kãk, S.4 k 0 hence [ ˆP tot, ã k ] = +k ã k and [ ˆP tot, ã k ] = k ã k. S.5 Consequently, for any t k coefficients, the Bogolyubov-transformed operators 4 operators satisfy [ ˆP tot, ˆb k ] = +k ˆb k and [ ˆP tot, ˆb k ] = k ˆb k, S.6 which means that the quasiparticle created by the ˆb k operator and annihilated by the ˆb k has definite momentum k. Also, straightforward algebra shows that for any k 0 and t k = t k ˆb kˆb k ˆb kˆb k = ã kãk ã kã k. S.7 Consequently, ˆP tot 0 k ã kãk ã kãk ã kã k k 0 ˆb kˆb k ˆb kˆb k k 0 S.8 0 k ˆb kˆb k. In other words, the total mechanical momentum of the superfluid helium is simply the net momentum of all the quasi-particles. 6

7 Problem e: Combining eqs. S.18 and 4 we immediately see that for large momenta k, and therefore t k λnm k 1 = cosht k 1, sinht k t k 1, S.9 ˆb k ã k + t kã k ã k = â k. S.30 Thus, quasi-particles with large momenta are approximately atoms. On the other hand, for small momenta k but k 0, t k 1 4 4λnM log k 1 = cosht k, sinht k 1et k 1 4 4λnM k 1, S.31 and therefore ˆb k 1 4 4λnM k ã k + ã k. S.3 At the same time, the density fluctuations in the superfluid are measured by the operators δnx ˆΨ x ˆΨx n = n ˆϕ x + ˆϕx + ˆϕ x ˆϕx. S.33 Approximating this formula by its leading first-order terms and then taking a Fourier transform, we find δn k Hence, the physical meaning of eq. S.3 is n ã k + ã k. S.34 for small k, ˆb k coeff δn k, S.35 or in other words, for small momenta, the quasiparticle created by the ˆb k wave in the superfluid, i.e. a phonon. operator is the density 7

8 Problem 3f: Non-relativistic mechanics classical or quantum is invariant under Galilean boosts, which act on coordinates, momenta and energy according to t = t, x i = x i + vt, p i = p i + m i v, H = H + v P total + const. S.36 A superfluid flowing at a uniform velocity v is related to the same superfluid at rest via a Galilean boost. Consequently, the lab-frame Hamiltonian of the flowing superfluid is Ĥ µ ˆN lab frame = Ĥ µ ˆN fluid frame + v ˆP fluid frame + const = Ĥ free + v ˆP fluid frame + Ĥint + const S.37 where lab frame Ĥfree = Ĥ free + v ˆP fluid frame ω k + v k ˆb kˆb k. S.38 The quasiparticle momenta k in this formula are in the fluid s rest frame. Eq. S.38 applies to a flow of an ideal gas just as well as a flow of a superfluid, the only difference being in the dispersion relations ωk. For the ideal gas, ω = k /M, and therefore for any non-zero velocity v, there are modes k for which ω k + v k = k Mv M 1 Mv < 0. S.39 For such modes, excitations created by the ˆb k have negative energy, which means that any perturbation of the flowing gas can create such negative-energy excitations and slow down the flow. For the superfluid however, ω k v c k for all modes, hence ω k + v k v c v k 0 S.40 for all the excitation modes, provided v v 0. Consequently, all excitations in the flowing superfluid have positive lab-frame energies, so they don t get spontaneously created and the flow persists forever without dissipation, thus superfluidity. 8