Bose–Einstein Condensation and Superfluidity

Bose–Einstein Condensation and Superﬂuidity

In these notes I explain the approximate ground state of a Bose–Einstein condensate, the spectrum of quasi-particle excitations of that ground state, and then explain why the BEC can ﬂow without resistance. For simplicity, I assume zero temperature these notes, but if I have time in class I shall explain what happens at ﬁnite temperatures.

Let me start with the Landau–Ginzburg theory with Hamiltonian

1 λ Hˆ µNˆ = d3x ψˆ† ψˆ µψˆ†ψˆ + ψˆ†ψˆ†ψˆψˆ (1) − 2m ∇ ·∇ − 2 Z where ψˆ(x) is the atom-annihilation quantum field while ψˆ†(x) is the atom-creation field. The Landau–Ginzburg theory with local ψˆ†ψˆ†ψˆψˆ interactions — which correspond to zero- distance two-body repulsion V (x y) = λδ(3)(x y) between the atoms — is a good 2 − − approximation to reality for a low-density ultra-cold BEC of heavy atoms, and it’s a good starting point for qualitative understanding of the superfluid liquid helium. Later in these notes we shall get to a better description of superfluid helium, but for the moment let’s stick with the Landau–Ginzburg theory.

In the undergraduate StatMech textbooks, Bose–Einstein condensation is often described

as all the atoms having exactly zero momenta, or in terms of occupation numbers nk for

momentum modes k, the BEC has n0 = N while all the other nk = 0,

N aˆ† naive BEC = 0 vac . (2) | i √N ! | i

However, this naive ground state has unphysical long-distance correlations between the creation and annihilation ﬁelds at distant points in space,

G(x y) = ψˆ†(x)ψˆ(y) ψˆ†(x) ψˆ(y) = n 0 0 = n for any x y. (3) − − − × − D E D ED E In real-life such correlations can never happen for any macroscopic number of atoms, so let’s make a better starting approximation for the ground state, namely the coherent state of the

1 k = 0 mode,

coherent = e−N/2 exp(√Naˆ†) vac . (4) | i 0 | i In QFT terms, this coherent state obeys

ψˆ(x) coherent = φ(x) coherent for φ(x) = const, (5) | i | i where the speciﬁc value of the φ(x)= ψˆ(x) obtains from the classical ﬁeld theory of the Landau–Ginzburg theory: D E

1 λ H[φ,φ∗] = d3x φ 2 + φ 4 µ φ 2 . (6) 2m |∇ | 2 | | − | | Z For a negative chemical potential µ the lowest-energy state of this classical Hamiltonian is φ(x) 0 — which corresponds to the vacuum state of the quantum ﬁeld theory, — while for ≡ positive chemical potential µ> 0 the lowest-energy state has a non-zero value of the scalar ﬁeld φ, namely µ φ 2 =n ¯ = . (7) | | s λ The phase of φ is arbitrary, as long as it is the same at all x (in the non-moving BEC), so

without loss of generality we assume φground = √n¯s. In the quantum theory, this corresponds to a constant non-zero ground-state expectation value

ψˆ(x) = √n¯s = const. (8) D E The simplest quantum state with this expectation value ψˆ is the coherent state (5). However, interactions of this expectation value with the fluctuations of the quantum fields around this coherent state changes the ground states of the k = 0 modes and they no longer 6 correspond to nk = 0. To see how this works, consider the shifted quantum fields

δψˆ(x) = ψˆ(x) ψˆ = L−3/2 eik·xaˆ − k D E k=06 ∗ X (9) and δψˆ†(x) = ψˆ†(x) ψˆ = L−3/2 eik·xaˆ† , − k D E kX=06 and let’s rewrite the Hamiltonian (1) in terms of these shifted ﬁelds. Term-by-term in the

2 Hamiltonian density, we have

ψˆ† ψˆ = δψˆ † δψ,ˆ (10) ∇ ·∇ ∇ ·∇ † † † † ψˆ ψˆ = (√n¯s + δψˆ )(√n¯s + δψˆ)=n ¯s + √n¯s δψˆ + δψˆ + (δψˆ )(δψˆ), (11) † † † 2 2 ψˆ ψˆ ψˆψˆ = (√n¯s + δψˆ ) (√n¯s + δψˆ) 2 3/2 ˆ† ˆ ˆ† 2 ˆ† ˆ ˆ 2 =n ¯s + 2¯ns δψ + δψ +n ¯s (δψ ) + 4(δψ )(δψ)+(δψ) † † † 2 2 + 2√n¯s(δψˆ )(δψˆ + δψˆ)(δψˆ) +(δψˆ ) (δψˆ) , (12)

hence for λn¯s = µ,

λ λn2 (ψˆ†)2(ψˆ)2 µψˆ†ψˆ = s + 0 (δψˆ† + δψˆ) 2 − − 2 × 1 ˆ† 2 ˆ† ˆ 1 ˆ 2 (13) + λn¯s 2 (δψ ) + (δψ )(δψ) + 2 (δψ) ˆ† ˆ† ˆ ˆ 1 ˆ† 2 ˆ 2 + λ√n¯s(δψ )(δψ + δψ)(δψ) + 2 λ(δψ ) (δψ) .

Note the organization of the RHS here according to net powers of the shifted ﬁelds δψˆ† and δψˆ. Reorganizing the whole Landau–Ginzburg Hamiltonian along the similar lines, we get

Hˆ µNˆ = const + Hˆ + Hˆ (14) − free int where

1 Hˆ = d3x δψˆ† δψˆ + λn¯ (δψˆ†)(δψˆ) + 1 (δψˆ)2 + 1 (δψˆ†)2 free 2m ∇ ·∇ s 2 2 Z k2 (15) = + λn¯ aˆ† aˆ + 1 λn¯ aˆ aˆ +a ˆ† aˆ† 2m s k k 2 s k −k k −k k=06 X comprises the quadratic (and bilinear) terms in the shifted ﬁelds, while

ˆ 3 ˆ† ˆ† ˆ ˆ 1 ˆ† 2 ˆ 2 Hint = d x λ√n¯s(δψ )(δψ + δψ)(δψ) + 2 λ(δψ ) (δψ) (16) Z comprises the cubic and the quartic terms. Physically, the Hˆfree describes the free quanta of the shifted fields — i.e., of the quantum fields’ fluctuations around their ground-state expectation values, — while the Hînt describes the interactions between such quanta.

3 Our next task is to diagonalize the Hˆfree; this should give us the leading approximation to the excitation spectrum as well as the next approximation to the ground state (next after the coherent state). The better approximations after that would obtain by perturbation theory in Hˆint, but I won’t do it in these notes. Instead, diagonalizing just the free Hamiltonian for the ﬂuctuations would be interesting enough.

In terms of atomic creation and annihilation operators, the Hamiltonian (15) has form

ˆ † 1 † † H = Akaˆkaˆk + 2 Bk aˆkaˆ−k +a ˆkaˆ−k , (17) k X

for some real Ak = A−k and Bk = B−k, so it may be diagonalized via a Bogolyubov transform of the creation and annihilation operators:

ˆ † bk = cosh(tk) aˆk + sinh(tk) aˆ−k , × × (18) † † ˆb = cosh(tk) aˆ + sinh(tk) aˆ . k × k × −k for appropriate real parameters tk = t−k. ˆ ˆ† Lemma 1: For any real tk = t−k, the bk and bk operators obey the same bosonic † commutation relations as thea ˆk anda ˆk operators,

† † † ˆ ˆ ˆ ˆ ˆ ˆ ′ bk, bk′ = 0, bk, bk′ = 0, bk, bk′ = δk,k . (19)

Lemma 2: For any Hamiltonian of the form (17) with real Ak = A−k, real Bk = B−k and Bk < Ak, there is a Bogolyubov transform (18) with | |

1 Bk tk = artanh , (20) 2 Ak

which leads to

ˆ ˆ† ˆ H = ω(k) bkbk +constant (21) Xk for ω(k) = A2 B2 . (22) k − k q 4 ˆ Clearly, the ground state of the Hamiltonian (21) is the state annihilated by all the bk operators,

k, ˆb ground = 0, (23) ∀ k | i

ˆ† while the excited states obtain by acting with the bk operators on the ground state,

excited = ˆb† ˆb† ground , E E = ω(k ) + + ω(k ). (24) | i k1 ··· kn | i excited − ground 1 ··· n

Physically, we may interpret such excitations as containing n quasiparticles of respective ˆ† ˆ energies ω(k1),...ω(kn). Thus, the operators bk create quasiparticles, the operators bk annihilate those quasiparticles, and the ground state deﬁned by eq. (23) is the quasiparticle vacuum.

ˆ† ˆ Lemma 3: The quasiparticle creation and annihilation operators bk and bk are related † to the atomic creation and annihilation operatorsa ˆk anda ˆk by a unitary operator transform,

ˆ ˆ ˆ ˆ ˆb† = e+F aˆ† e−F , ˆb = e+F aˆ e−F , (25) k × k × k × k × for the antihermitian operator

1 † † Fˆ = tk aˆ aˆ aˆ aˆ . (26) 2 k −k − k −k k X

Consequently, the state

ˆ ground = e+F coherent (27) | i | i

ˆ is annihilated by all the quasiparticle annihilation operators bk, so it’s the ground state of the Hamiltonian (21).

Unlike the coherent state of the BEC which has all the atoms in the k = 0 mode, the | i ground state (27) also has a lot of atoms paired in (+k, k) modes. In fact, experiments − with the Bose–Einstein condensates of ultra-cold atoms show more atoms in such k pairs ±

5 than the atoms in the k = 0 mode itself. Lemma 4: For the state (27), the net number of atoms in k = 0 modes is 6

2 Nk = ground Nˆk ground = sinh (tk). (28) =06 h | =06 | i Xk=06

Lemma 5: The quasiparticle vacuum state (23) has zero net mechanical momentum, while the quasiparticles have deﬁnite momenta k, thus

ˆ † ˆ† ˆ Pnet = kaˆkaˆk = kbkbk . (29) k k X X

I shall prove the Lemmas 1–5 later in these notes, but first let me put them in the specific context of the Bose–Einstein condensate. Let’s go back to the Landau–Ginzburg theory and the Hamiltonian (14) for the fluctuation fields. — or rather the free part of that Hamiltonian. In terms of the atomic creation and annihilation operators, the free part (15) of that Hamiltonian is

k2 Hˆ = + λn¯ aˆ† aˆ + 1 λn¯ aˆ aˆ +a ˆ† aˆ† , (30) free 2m s k k 2 s k −k k −k k=06 X

which is a special case of (17) with

k2 A = + λn¯ , B = λn¯ , (31) k 2m s k s hence

1 2λn¯sm for small k, tk = artanh ∞ (32) 2 2λn¯ m + k2 −→ 2 0 for large k. while

k2 2 2 2 k λn¯s ω(k) = + λn¯s (λn¯s) = k + . (33) s 2m − × 4m2 m r

6 Graphically,

ω(k) (34)

2 atoms knocked out of BEC, ω k ≈ 2m

phonons, ω c k ≈ s × k

Note that at high quasiparticle momenta k we have ω(k) k2/2m while t(k) 1 and • ≈ ≪ hence ˆb aˆ and ˆb† aˆ† . In other words, the quasiparticle created by the ˆb† and k ≈ k k ≈ k k ˆ annihilated by the bk is approximately a free atom, or rather an atom kicked out of the BEC condensate and given a high momentum k.

On the other hand, for low (but non-zero) quasiparticles momenta •

ω(k) k λn¯ /m k c (35) ≈ × s ≡ × s p while t(k) is large, hence ˆb† (ˆa† +ˆa ) which means that it creates a quantum of k ∝ k −k δψ∗(x)+ δφ(x) δn (x), a quantum of the condensate density wave. In other words, ∝ s ˆ† ˆ the quasiparticle created by the bk (and annihilated by the bk) is a phonon; indeed,

the quasiparticle’s velocity cs = λn¯s/m is the speed of sound in the BEC.

Finally, at the intermediate momentap k, the quasiparticles interpolate between the • phonons and the atoms kicked out of the BEC.

7 Thus far, we have ignored the interactions between the ﬂuctuation ﬁelds δψˆ(x) and δψˆ†(x) and hence between the quasiparticles. In quasiparticle terms, the interaction Hamil- tonian

ˆ 3 ˆ† ˆ† ˆ ˆ 1 ˆ† 2 ˆ 2 Hint = d x λ√n¯s(δψ )(δψ + δψ)(δψ) + 2 λ(δψ ) (δψ) (16) Z comprises terms of the form

ˆbˆbˆb, ˆb†ˆbˆb, ˆb†ˆb†ˆb, ˆb†ˆb†ˆb†, and ˆbˆbˆbˆb, ˆb†ˆbˆbˆb, ˆb†ˆb†ˆbˆb, ˆb†ˆb†ˆb†ˆb, ˆb†ˆb†ˆb†ˆb†. (36)

In particular, the ˆb†ˆb†ˆb† and the ˆb†ˆb†ˆb†ˆb† terms do not annihilate the ground state of the | i free Hamiltonian, so it suﬀers perturbative corrections. Fortunately, there is a way to recast all such corrections in terms of the unitary operator transform,

ˆ true ground state = eF coherent (37) | i | i

for 1 † † Fˆ = tk(â aˆ aˆ aˆ ) 2 k −k − k −k Xk=06 + perturbative corrections involving terms of the form (38) (âaâˆ aˆ†aˆ†aˆ†), (âaââˆ aˆ†aˆ†aˆ†aˆ†),... − − Consequently, we may redefine the quasiparticle creation and annihilation operators as

ˆ ˆ ˆ ˆ ˆb† = e+F aˆ† e−F , ˆb = e+F aˆ e−F , (25) k × k × k × k ×

Fˆ ˆ ˆ† in terms of the perturbatively corrected unitary transform e , so that the bk and the bk obey ˆ the bosonic commutation relations and all the bk annihilate the true ground state. Thus, we still have the quasiparticle picture of the excited states of the complete Hamiltonian, although the relation ω(k) between quasiparticles energy and momentum suﬀers perturbative corrections. Nevertheless, qualitatively the ω(k) function remains as on the diagram (34), so the low-momentum quasiparticles are phonons, the high-momentum quasiparticles are atoms kicked out of the BEC, and the intermediate-momentum quasiparticles interpolate between the two.

8 Superfluid Helium

Thus far we have used the Landau–Ginzburg theory which is good for the BEC of ultra- cold atoms but becomes inaccurate for the superﬂuid liquid helium. The problem stems from higher average momenta of the helium atoms and hence shorter De Broglie wavelength which becomes comparable to the range of the interatomic forces. Consequently, we may longer approximate the two-body potential as V (x y) = λδ(3)(x y), so instead of the 2 − − Landau–Ginzburg Hamiltonian we should use

1 1 Hˆ µNˆ = d3x ˆ ψ† ψˆ µψˆ†ψˆ + d3x d4y V (x y) ψˆ†(x)ψˆ†(y)ψˆ(y)ψˆ(x). − 2m ∇ ·∇ − 2 2 − × Z Z Z (39) To find the ground state of this Hamiltonian, we start with the classical field limit and minimize the classical Hamiltonian 1 1 H[φ,φ∗] = d3x φ 2 µ φ 2 + d3x d4y V (x y) φ(x) 2 φ(y) 2. (40) 2m |∇ | − | | 2 2 − ×| | ×| | Z Z Z Again, the minimum obtains for φ(x) = const, specifically µ φ 2 =n ¯ = , any constant phase of φ, (41) | | s λ where

def 3 λ = d x V2(x) > 0, (42) Z or in terms of the Fourier transform of the two-atom potential

3 −ik·x W (k) = d x V2(x)e , (43) Z λ = W (0). (44)

Given the classical ground-state expectation value of the condensate field φ, we go back to the quantum field theory and shift the quantum field just as we did before,

δψˆ(x) = ψˆ(x) ψˆ = L−3/2 eik·xaˆ − k D E k=06 ∗ X (9) and δψˆ†(x) = ψˆ†(x) ψˆ = L−3/2 eik·xaˆ† , − k D E kX=06 Next, we re-express the Hamiltonian (39) in terms of the shifted quantum ﬁelds and re-

9 arrange the terms according to the net power of the shifted ﬁelds. Just as we had earlier for the Landau–Ginzburg theory, we end up with

Hˆ µNˆ = const + Hˆ + Hˆ (14) − free int

where Hˆfree comprises the quadratic (and bilinear) terms while Hˆint comprises the cubic and the quartic terms which we treat as perturbations.

Lemma 6:

1 Hˆ = d3x δφˆ† δψˆ free 2m ∇ ·∇ Z 3 3 † 1 1 † † + d x d y V2(x y) ψˆ (x)ψˆ(y) + ψˆ(x)ψˆ(y) + ψˆ (x)ψˆ (y) − × 2 2 (45) Z Z k2 = +n ¯ W (k) aˆ† aˆ + 1 n¯ W (k) aˆ aˆ +ˆa† aˆ† . 2m s k k 2 s k −k k −k k=06 X

Again, this Hamiltonian can be diagonalized by a Bogolyubov transform, exactly as we did it for the LG theory in lemmas 1–5, except for the new values of

k2 A = +n ¯ W (k) and B =n ¯ W (k). (46) k 2m s k s

Consequently, we end up with

ˆ ˆ† ˆ Hfree = ω(k)bkbk +const (47) Xk=06

so the ground state is the quasiparticle vacuum, while the quasiparticles have deﬁnite momenta k and energies

k2 2 2 2 k n¯s ω(k) = +¯nsW (k) (λn¯s) = k + W (k) . (48) s 2m − × 4m2 m r

10 For the helium atoms, the W (k) drops oﬀ at large momenta,

W (k)

hence the energy-momentum relation ω(k) for the quasiparticles — or equivalently, the wavenumber-frequency dispersion relation for the waves of small ﬂuctuations — has a dip:

ω(k) (49)

ω = csk

ω k = vc

2 k ω = 2m k Again, this curve shows that the low-momenta quasiparticles are phonons while the high momenta quasiparticles are helium atoms knocked out from the BEC. But now we also have unexpectedly-low energy quasiparticles at intermediate momenta; they are called the rotons

11 for historical reason. The rotons have much larger phase space than the phonons, so at temperatures T 1 K the rotons dominate the quasiparticle gas, which is the normal-ﬂuid ∼ component of the ﬁnite-temperature liquid Helium II.

The most important feature of the ω(k) curve (49) is the positive lower bound on the energy-to-momentum ratio,

k : ω(k) > v k for a positive v . (50) ∀ c ×| | c

We shall see momentarily that it is this lower bound which gives rise to the superﬂuidity.

Superfluidity

Consider a flowing superfluid; for simplicity, let it flow with a uniform velocity v. Clas- sically, this flow is described by

φ(x) = √n¯ exp(imv x) (51) s × · while quantum mechanically, we have a coherent pile up of atoms into the k = mv mode, while other atoms form pairs with momenta k = mv k . Altogether, we have a 1,2 ± red quantum state very much like the state of the superfluid at rest, except all atom’s momenta are shifted by mv. In other words, the state of the flowing superfluid obtains from the ground state of the superfluid at rest via Galilean boost of velocity v.

The excitation spectrum of the moving superﬂuid also obtains via Galilean boost of the Hamiltonian,

Hˆ ′ = Hˆ + v Pˆ + 1 v2M . (52) · net 2 net

For simplicity, let’s ignore the interactions between the quasiparticles and focus on their free Hamiltonian. In quasiparticle terms,

ˆ ˆ† ˆ Hfree = const + ω(k) bkbk , (53) Xk ˆ ˆ† ˆ Pnet = k bkbk , (54) Xk

12 hence for a moving superﬂuid

Hˆ ′(v) = const′ + ω(k)+ v k ˆb† ˆb . (55) · k k k X

(In this formula, the quasiparticle momenta k are in the rest frame of the superﬂuid rather than the lab frame. In the lab frame, they have p = k + mv.)

Before we apply this formula to the liquid Helium II, consider the ideal gas. For the ideal gas at rest, the Hamiltonian has form (53) where the ‘quasiparticles’ are simple the atoms and ω(k)= k2/2m. Consequently, the uniformly ﬂowing gas has

k2 Hˆ ′(v) = const + + v k ˆb† ˆb (56) 2m · k k Xk

where the frequencies k2 ω′(k) = + v k (57) 2m ·

are positive for some modes k and negative for other modes. In particular, for k in opposite direction from the gas ﬂow v and of magnitude k < 2mv, the frequency ω′(k) is negative.

Now, while a harmonic oscillator with a positive frequency ω has a unique ground state, the oscillator with a negative frequency has energy spectrum unlimited from below. Which means that any perturbation — however small it might be — would cause transitions building up the number of quanta while lowering the energy. For the ideal gas, this means spontaneous build up of atoms with ω′(k) < 0 — i.e., with lab-frame velocities

k vk = v + < v , (58) m | |

by taking them out of the coherent motion with the gas. In other words, any interactions with the outside world (for example, the walls of the pipe the gas flows through) would spontaneously knock the atoms out of the coherent flow of the gas and slow them down. Such slowed-down atoms would dissipate the net energy and the net momentum of the flowing gas; it is this dissipation that we experimentally observe as resistance to the flow.

13 Now consider the superfluid Helium with ω(k) as on the diagram (49). Unlike the ideal gas with ω k2, the superfluid has ω k at low momenta, and for any quasiparticle ∝ ∝ | | momenta ω(k) > v k for some positive v . Consequently, as long as the Helium flows at c ×| | c speed v less than the critical speed vc, we have

ω′(k) = ω(k) + v k > 0 for all k. (59) ·

Indeed,

ω(k) + v k > ω(k) vk > ω(k) v k 0. (60) · − − c ≥

Consequently, there are no negative-energy quasiparticles — like the slowed-down atoms — so there are no microscopic transitions lowering the superfluid’s net energy. This means no energy dissipation, and that’s why the superfluid flows without resistance, hence the name superfluidity.

Proofs of the Lemmas

Lemma 1: the bosonic commutation relations (19) for the quasiparticle creation and annihilation operators. Starting from the bosonic commutation relations

† † † ′ [âk, aˆk′ ] = 0, [âk, aˆk′ ] = 0, [âk, aˆk′ ] = δk,k (61) for the operators creating and annihilating the atoms, and treating eqs. (18) as the definitions ˆ ˆ† of the bk and bk operators, we immediately calculate

† [ˆb , ˆb ′ ] = cosh(tk) sinh(tk′ ) [ˆa , aˆ ′ ]= δ ′ k k × k −k k,−k † ′ ′ + sinh(tk) cosh(tk) [ˆa−k, aˆk′ ]= δ−k,k × − (62) = δk′ k cosh(tk) sinh(tk′ ) sinh(tk) cosh(tk′ ) = sinh(tk′ tk) ,− × − − ′ ′ = 0 because t = tk when k = k. k −

ˆ† ˆ† In the same way, [bk, bk′ ] = 0.

14 Finally,

† † [ˆb , ˆb ′ ] = cosh(tk) cosh(tk′ ) [ˆa , aˆ ′ ]= δk k′ k k × k k , † ′ ′ + sinh(t−k) sinh(t−k ) [ˆa−k, aˆ ′ ]= δ−k,−k × −k (63) 2 2 2 2 = δk k′ cosh (tk) sinh (t k) = cosh (tk) sinh (tk)=1 , × − − − = δk,k′ .

Quod erat demonstrandum.

Lemma 2: bringing the Hamiltonian (17) to the form (21). Let’s start by expressing ˆ† ˆ † the product bkbk in terms of thea ˆ anda ˆ operators. Applying both deﬁnitions (18), we immediately obtain

ˆ† ˆ 2 † † † bkbk = cosh (tk)âkaˆk + cosh(tk) sinh(tk)(âkaˆ−k +a ˆ−kaˆk) (64) 2 † † + sinh (tk)(â−kaˆ−k =a ˆ−kaˆ−k + 1).

Likewise,

ˆ† ˆ 2 † † † b−kb−k = cosh (t−k)â−kaˆ−k + cosh(t−k) sinh(tk)(â−kaˆk +a ˆkaˆ−k) (65) 2 † † + sinh (t−k)(âkaˆk =a ˆkaˆk + 1).

Assuming t−k = tk, we may combine

ˆ† ˆ ˆ† ˆ 2 2 † † bkbk + b−kb−k = cosh (tk) + sinh (tk) = cosh(2tk) (âkaˆk +â−kaˆ−k) × (66) † † + 2 cosh(tk) sinh(tk) = sinh(2tk) (â aˆ +a ˆ aˆ ) + const. × k −k −k k Now let’s plug this formula into a Hamiltonian of the form (21) for some ωk and require that the result matches the original Hamiltonian (17). Assuming ω−k ωk, we obtain ≡ 1 Hˆ = ω ˆb† ˆb = ω (ˆb† ˆb + ˆb† ˆb ) k k k 2 k k k −k −k Xk Xk (67) † 1 † † = ωk cosh(2tk)âkaˆk + 2 ωk sinh(2tk) aˆkaˆ−k +a ˆ−kaˆk + const. Xk Xk This formula must match (up to a constant) the original Hamiltonian (17), so we need to

15 choose the parameters ωk = ω−k and tk = t−k such that

ωk cosh(2tk) = Ak and ωk sinh(2tk) = Bk . (68)

These equations are easy to solve, and the solution exists as long as Ak = A−k, Bk = B−k, and Ak > Bk , namely | |

1 Bk 2 2 tk = artanh and ωk = Ak Bk . (69) 2 Ak − q Quod erat demonstrandum.

Lemma 3: By the Hadamard Lemma,

ˆ ˆ e+F aˆ e−F =a ˆ + [F,ˆ aˆ ] + 1 [F,ˆ [F,ˆ aˆ ]] + 1 [F,ˆ [F,ˆ [F,ˆ aˆ ]]] + (70) × k × k k 2 k 6 k ···

† and likewise for thea ˆk. Speciﬁcally, for

ˆ 1 † † F = tp aˆpaˆ−p aˆpaˆ−p , (26) 2 p − X we have

† † † [F,ˆ aˆ ] = tk[ˆa aˆ , aˆ ] = +tkaˆ (71) k − k −k k −k and ˆ † † [F, aˆ−k] = +tk[ˆakaˆ−k, aˆ−k] = +tkaˆk . (72)

Consequently, multiple commutators yield

n aˆk for even n, [F,ˆ [F,...ˆ [F,ˆ aˆ ] ]] = (tk) , (73) k ··· n times × † ( aˆ−k for odd n.

and therefore

n n ˆ ˆ (t ) (t ) e+F aˆ e−F = k aˆ + k aˆ† k n! k n! −k × × even n × × X oddXn † (74) = cosh(tk) aˆ + sinh(tk) aˆ × k × −k ˆ = bk .

16 In exactly the same way

† † n aˆk for even n, [F,ˆ [F,...ˆ [F,ˆ aˆ ] ]] = (tk) , (75) k ··· n times × ( aˆ−k for odd n.

and therefore

n n ˆ ˆ (t ) (t ) e+F aˆ† e−F = k aˆ† + k aˆ k n! k n! −k × × even n × × X oddXn † (76) = cosh(tk) aˆ + sinh(tk) aˆ × k × −k ˆ† = bk .

This establish the unitary transform (25) of the creation and annihilation operators.

The unitary transforms (25) automatically preserve the commutation relations between the operators, so instead of going through the algebra of proving the Lemma 1, we may simply use

† ˆ † ˆ ˆ ˆ ˆ ˆ F −F F ′ −F ′ [bk, bk′ ] = e [ˆak, aˆk′ ]e = e δk,k e = δk,k (77)

ˆ ˆ ˆ† ˆ† and likewise for the [bk, bk′ ] and [bk, bk′ ].

In the same matter, ground = eFˆ coherent being the quasiparticle vacuum automati- | i | i cally follows from the unitary operator transform (25) and the fact that the coherent state | i is annihilated by all thea ˆ operators with k = 0. Indeed, k 6

ˆ ˆ ˆ ˆ ˆ ˆb ground = eF aˆ e−F eF coherent = eF aˆ coherent = eF 0 = 0. (78) k | i k × | i k | i ×

. . . Q E D

Lemma 4: The operator Fˆ — and hence its exponential eFˆ — creates and annihilates the atoms in k pairs independently from all the other pairs. Consequently, the BEC ground ± state ground = eFˆ coherent can be written as a direct product of independent states of | i | i

d C (t) = (n + 1)C (t) nC − (t) (83) dt n n+1 − n 1

subject to initial conditions Cn(0) = δn,0. Instead of going through the long song and dance of solving the equations (83), let me simply give you the solutions

(tanh(t))n C (t) = (84) n cosh(t)

and verify that they indeed solve the equations (83):

d (tanh(t))n (tanh t)n−1 1 sinh t = n (tanh t)n dt cosh(t) cosh t × cosh2 t − × cosh2 t 1 = n(tanh t)n−1 (1 tanh2 t) (tanh t)n+1 (85) cosh t × − − (tanh t)n−1 (tanh t)n+1 = n (n + 1) , cosh t − cosh t quod erat demonstrandum.

18 Now, given the quantum state (80) of the k modes of the atoms, we may calculate the ± expectation value of the atom number in these two modes as

∞ 2 N k = (2n) C (tk). (86) ± × n nX=0

Speciﬁcally, for the Cn coeﬃcients as in eq. (84),

2 2 2n 2 tanh (tk) 2 N k = n(tanh tk) = = 2 sinh (tk). (87) ± 2 2 2 2 cosh tk × n cosh tk × (1 tanh (tbk)) X − Finally, combining all such k pairs of modes, we get the net (average) number of atoms in ± all the k = 0 modes as 6 1 N = N = sinh2(t ). (28) k=06 2 ±k k k k X X . . . Q E D

Lemma 5: the net momentum operator is

ˆ † Pnet = kaˆkaˆk . (88) Xk

Using eqs. (64) and (65) from the proof of Lemma 2 and t−k = tk, we immediately see that

† † 2 2 † † ˆb ˆb ˆb ˆb = cosh (tk) sinh (tk)=1 (ˆa aˆ aˆ aˆ ). (89) k k − −k −k − × k k − −k −k Consequently, for the momentum operator (88) we have

Pˆ = k aˆ† aˆ = ( k) aˆ† aˆ × k k − × −k −k Xk Xk 1 = k (ˆa† aˆ aˆ† aˆ ) 2 × k k − −k −k k X (90) 1 = k (ˆb† ˆb ˆb† ˆb ) 2 × k k − −k −k Xk = k ˆb† ˆb . × k k Xk Quod erat demonstrandum.

19 Alternatively, we may use the unitary operator transform of Lemma 3 and the fact that the Fˆ operator commutes with the net momentum Pˆ . Indeed, for every k mode

Pˆ aˆ aˆ =a ˆ (Pˆ k)â =a ˆ aˆ Pˆ (91) k −k k − −k k −k and likewise ˆ † † † ˆ † † † ˆ Paˆkaˆ−k =a ˆk(P + k)â−k =a ˆkaˆ−kP, (92) hence Pˆ Fˆ = FˆPˆ . Consequently, the BEC ground state ground = eFˆ coherent has the | i | i same momentum as the coherent state, namely zero, and the quasiparticles created by the | i ˆ† ˆ bk and annihilated by the bk carry the same definite momenta k as the atoms created by the † aˆk and annihilated by thea ˆk.

Lemma 6: the ﬁnite-range potential V (x y) for the helium atoms. Consider the net 2 − potential operator

1 Vˆ = d3x d3y V (x y) ψˆ†(x)ψˆ†(y)ψˆ(y)ψˆ(x). (93) 2 2 − × Z Z In terms of the shifted ﬁelds δψˆ(x)= ψˆ(x) √n¯ and δψˆ†(x)= ψˆ†(x) √n¯ , we have − s − s ˆ† ˆ† ˆ ˆ 2 3/2 ˆ† ˆ† ˆ ˆ ψ (x)ψ (y)ψ(y)ψ(x)=n ¯s +n ¯s δψ (x) + δψ (y) + δψ(x) + δψ(y) † † +n ¯s δψˆ (x)δψˆ(x) + δψˆ (y)δψˆ(y)

† † +n ¯s δψˆ (x)δψˆ(y) + δψˆ (y)δψˆ(x) (94)

† † +n ¯s δψˆ (x)δψˆ (y) + δψˆ(y)δψˆ(x) + cubic + quartic.

The terms on the ﬁrst two lines here depend only on the x or only on the y, so when we plug them into the potential operator (93), we may immediately integrate over the other space position to obtain

[@any ﬁxed y] d3x V (x y) = [@any ﬁxed x] d3y V (x y) = W (0). (95) 2 − 2 − Z Z Consequently, integrating over the expansion (94) in the context of the potential (93) and

20 making use of the x y symmetry, we obtain ↔ Vˆ =n ¯ W (0) d3x 1 n¯ + √n¯ δψˆ†(x)+ δψˆ(x) + δψˆ†(x)δψˆ(x) s × × 2 s s Z n¯ + 2 d3x d3y V (x y) 2δψˆ†(x)δψˆ(y) + δψˆ†(x)δψˆ†(y) + δψˆ(y)δψˆ(x) 2 × 2 − × Z Z + cubic + quartic. (96)

Now consider the other non-derivative term in the Helium’s Hamiltonian

Hˆ = Kˆ + Vˆ µN,ˆ (97) net − namely the chemical potential term,

µNˆ = µ d3x ψˆ†(x)ψˆ(x) − − Z (98) = µ d3x n¯ + √n¯ δψˆ(x)+ δψˆ†(x) + δψˆ†(x)δψˆ(x) − s s Z If we generalize the µ = λn¯s formula of the Landau–Ginzburg theory to the

µ = W (0) n¯ , (99) × s then the chemical potential term (98) cancels the top line of the two-body potential (96) (except for the constant part), hence

n¯ Vˆ µNˆ = 2 d3x d3y V (x y) 2δψˆ†(x)δψˆ(y) + δψˆ†(x)δψˆ†(y) + δψˆ(y)δψˆ(x) − 2 2 − × Z Z + constant + cubic + quartic . (100) Thus altogether,

Hˆ = constant + Hˆfree + Hˆinteractions (101)

where 1 Hˆ = d3x δψˆ†(x) δψˆ(x) free 2m ∇ ·∇ Z n¯ + 2 d3x d3y V (x y) 2δψˆ†(x)δψˆ(y) + δψˆ†(x)δψˆ†(y) + δψˆ(y)δψˆ(x) . 2 2 − × Z Z (102) This completes the proof of the ﬁrst part of the Lemma 6 — the top two lines of the eq. (45).

21 To prove the second part of the Lemma (the bottom line of eq. (45)) we simply Fourier transform from the shifted creation and annihilation ﬁelds to the creation and annihilation operators for modes k = 0, 6

ˆ† −3/2 +ikx † ˆ −3/2 −ikx δψ (x) = L e aˆk , δψ(x) = L e aˆk . (103) k k X=06 X=06 Consequently,

d3x d3y V (x y) δψˆ†(x)δψˆ(y) = 2 − × Z Z 3 3 −3 ikx−ik′y † = d x d y V2(x y) L e aˆkaˆk′ − × ′ × (104) Z Z Xk,k ′ † −3 3 3 ikx−ik y = aˆkaˆk′ L d x d y V2(x y) e ′ × − × Xk,k Z Z where

′ L−3 d3x d3y V (x y) eikx−ik y = 2 − × Z Z ′ = L−3 d3y d3(z = x y) V (z) eik(y+z)−ik y − 2 × Z Z (105) ′ = d3z V (z)eikz L−3 d3y eiky−ik y 2 × Z boxZ = W (k) δk k′ , × , hence

d3x d3y V (x y) δψˆ†(x)δψˆ(y) = W (k) aˆ† aˆ . (106) 2 − × × k k Z Z Xk In the same way we obtain

d3x d3y V (x y) δψˆ†(x)δψˆ†(y) = 2 − × Z Z ′ 3 3 −3 ikx−ik y † † = d x d y V2(x y) L e aˆkaˆ−k (107) − × ′ × Z Z Xk,k = W (k) aˆ† aˆ† × k −k Xk

22 and likewise

d3x d3y V (x y) δψˆ(x)δψˆ(y) = W (k) aˆ aˆ . (108) 2 − × × −k k Z Z Xk

Combining all these formulae with the gradient term in the Hamiltonian (102),

1 1 k2 Kˆ = d3x ψ† ψ = d3x δψ† δψ = aˆ† aˆ , (109) 2m ∇ ·∇ 2m ∇ ·∇ 2m k k Z Z Xk

we ﬁnally assemble all quadratic terms to

k2 Hˆ = + W (k)¯n aˆ† aˆ + 1 W (k)¯n aˆ† aˆ† +a ˆ aˆ . (110) free 2m s k k 2 s k −k −k k k X Quod erat demonstrandum.