Physics 127c: Statistical Mechanics

Weakly Interacting Bose Gas

Bogoliubov Theory

The Hamiltonian is X X + 1 + + H = εkb bE + u(˜ q)bE b bE0 bE. (1) kE k 2 kE+Eq kE0−Eq k k kE kEkE0qE We will look at the weakly interacting system at low temperatures. Already, without interactions, we have Bose condensation b+b = N = N. 0 0 0 (2) Interactions will reduce this, but if they are weak the depletion of the condensate will be a small fraction (N − N0)/N  1, or X0 D E b+b  N, kE kE (3) kE P0 with denoting the sum over all kE 6= 0. With the macroscopic occupation of the zero mode we can neglect the quantum fluctuations relative to the mean (cf. the classical treatment of electromagnetism)

b → N 1/2,b+ → N 1/2 0 0 0 0 (4) where N0 is just a number (called a c-number to distinguish it from an operator). Equation (4) is now substituted into the Hamiltonian, followed by expansion in (N −N0)/N. The kinetic energy term is unchanged, since there is no contribution from the kE = 0 state. The potential energy can be N 1/2 V = V + V +··· written as the sum of terms at successive order (decreasing powers of 0 ) 0 1 with 1 V = u(˜ 0)N 2 (5) 0 2 0 N X0 V = 0 u(˜ q)(bE +b+ + b b + b+ b + b+b ) +˜u( )(b+b + b+ b ). 2  qE −Eq qE −Eq −Eq −Eq qE qE 0 qE qE −Eq −Eq (6) 2 qE

(The first order term V1 is zero by momentum conservation.) It’s convenient to combine all the u(˜ 0) terms

N X0 1 2 0 + + 1 2 u(˜ 0)N + u(˜ 0)(b bqE + b b−Eq ) ' u(˜ 0)N (7)  0  qE −Eq  2 2 qE 2 leaving X0 0 N0 + + + + V = u(˜ q)(bE b + bqEb−Eq + b b−Eq + b bqE) (8) 2  qE −Eq −Eq qE 2 qE so that we can write the Hamiltonian as H = const + H1 with H1 the part to be solved

X0 1 + + + + H1 = [εq + n0u(˜ q)E ](bqE bqE + b−Eq b−Eq ) + n0u(q)(b˜ qE b−Eq + bqEb−Eq ) (9) 2 qE

n = N / V 0 qE −Eq where 0 0 is the condensate density and since the first term in 1 mixes and we have explicitly written the kinetic energy terms from qE and −Eq.

1 + The momentum state creation and annihilation operators bqE ,bqE are no longer ladder operators of the + Boson Hamiltonian. We look for new creation and annihilation operators αqE ,αqE that do have this property, i.e. operators satisfying

+ [αqE,αqE0 ] = δqEqE0 (10a) + + [αqE ,αqE0 ] = 0 = [αqE,αqE0 ] (10b) and in terms of which the Hamiltonian has the form X + H1 = Eq αqE αqE + const. (11) qE

+ We look for αqE ,αqE in the form

+ αqE = uq bqE + vq b−Eq , (12) + + αqE = uq bqE + vq b−Eq , (13) with uq ,bq depending only on |qE| and real. This type of transformation, preserving the commutation rules, is + called a canonical transformation. This from of the ansatz is motivated by the way bqE and b−Eq appear in H1. + We can also argue that αqE should create momentum h¯qE—which a linear combination of creating at qE and + destroying at −Eq accomplishes. Note that αqE no longer adds a particle to the system! This is OK because the condensate soaks up any deficit. Sorry about the notation: uq is not related to u(˜ q)E ! We first check the commutation rules

+ 2 + 2 + [αqE,αqE ] = uq [bqE,bqE ] + vq [b−Eq ,b−Eq ] (14a) 2 2 = uq − vq . (14b)

So we set 2 2 uq − vq = 1 (15) and can then invert Eq. (12)

+ bqE = uq αqE − vq α−Eq , (16a) + + bqE = uq αqE − vq α−Eq. (16b)

These expressions can then be used to evaluate H1 in terms of the α

X0 2 H1 = [(εq + n0u(˜ q))vE q − n0u(˜ q)uE q vq ] qE X0 1 2 2 + + + [(εq + n0u(˜ q))(uE q + vq ) − 2n0u(˜ q)uE q vq ](αqE αqE + α−Eq α−Eq ) 2 qE X0 1 2 2 + + + [−2(εq + n0u(˜ q))uE q vq + n0u(˜ q)(uE q + vq )](αqE α−Eq + αqEα−Eq ). (17) 2 qE

We now choose uq ,vq to eliminate the last term

2uq vq n u(˜ q)E = 0 . (18) 2 2 uq + vq εq + n0u(˜ q)E

2 2 2 Since uq − vq = 1 this is most easily solved by introducing

uq = cosh φq ,vq = sinh φq (19) and then Eq. (18)is n0u(˜ q)E tanh 2φq = (20) εq + n0u(˜ q)E and ε + n u(˜ q)E 2 2 q 0 uq + vq = cosh 2φq = (21) Eq

n0u(˜ q)E 2uq vq = sinh 2φq = (22) Eq with q 2 2 Eq = (εq + n0u(˜ q))E − (n0u(˜ q))E (23) s   h¯ 2q2 h¯ 2q2 = 2n u(q)˜ + . (24) 2m 0 2m

With these results H1 can be simplified

X0 X0 1 + H1 =− (εq + n0u(˜ q)E − Eq ) + Eq αq αq (25) 2 qE qE giving the desired form. We recognize that Eq is the excitation energy spectrum of the new Bosons. Notice that Eq is linear at small q p Eq ' n0u(˜ 0)/mhq¯ (26) p but crosses over to the free particle quadratic form for q & mn0u(˜ 0)/h¯. The linear spectrum at small q is crucial to the phenomenon of superfluidity. + Since αqE ,αqE are creation and annihilation operators and ladder operators of the Hamiltonian, we can calculate physical quantities as we did for the harmonic oscillator. For example the ground state of the interacting system |ψ0i satisfies

+ αqE |ψ0i = hψ0| αqE = 0 for all qE. (27)

Condensate Depletion The condensate depletion, or the number particles excited out of the zero momentum state by the interactions in the ground state, is

X0 D E 0 + N = N − N0 = ψ0 bqE bqE ψ0 (28) qE X0 + + = ψ0 (uq αq − vq α−q )(uq αq − vq α−q ) ψ0 (29) qE   X0 X0 ε + n u(˜ q)E 2 1 q 0 = vq = − 1 . (30) Eq qE qE 2

3 Since the term in () becomes unity when Eq ' εq , i.e. except for the small range of q (for weak interactions) where the spectrum is linear, we can replace u(˜ q)E by u(˜ 0) = g, say. Then converting the qE sum to an integral (noting that the qE = 0 term gives negligible contribution) Z !  ∞ 1 h¯ 2q2/2m + n g N 0 = 4π dq q2 p 0 − 1 . (31) ( π)3 2 2 2 2 2 0 2 (h¯ q /2m)(2n0g + h¯ q /2m) p 2 2 Substituting y = h¯ q /2mn0g gives

  / Z   N 0 1 1 2mn g 3 2 ∞ y2 + 1 = 0 y2 − 1 dy. (32a) 2 2 4 2 1/2 N0 4π n0 h¯ 0 (y + 2y ) √ The integral is 2/3, so that   / N 0 8 n a3 1 2 = 0 (33) N0 3 π where a = mg/4πh¯ 2 is known as the scattering length. Note that the depletion of the condensate is small for weak interaction, g small, but the expansion in g is nonanalytic. This shows us that we could not get the results by simple perturbation theory: the canonical transformation goes beyond this.

Other Thermodynamic Properties Since we have the excitation energy spectrum, and can write any physical operator in terms of the corre- + + sponding creation and annihilation operators αqE ,αqE (via their expressions in terms of bqE ,bqE) we can now calculate the thermodynamic properties at nonzero temperature. In this, we neglect the interaction between the new Bosons, which would arise from the terms V3 and V4 containing 3 and 4 α operators. This is OK for weak interactions, except very near the superfluid transition temperature, where we expect the universal fluctuation behavior characteristic of superfluids. This universal behavior derives from the interactions. The linear spectrum at small q corresponds to phonon excitations (you can check that the slope is just the speed of sound in the interacting gas.) The low temperature thermodynamics will have the familiar power law behavior for these excitations, namely a T 3 specific heat. The normal fluid density will grow as T 4.

Further Reading Now would be a good time to review Lecture 15 and problem 3 of Homework 7 for Ph127a where the physics of superfluidity was discussed. In particular, the importance of the linear spectrum εk ∝ k (rather 2 than εk ∝ k ) and the notion of the normal fluid density were discussed. Pathria §10.2-10.7 discusses the weakly interacting Bose gas.

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