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Bose–Einstein Condensation and Superfluidity

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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 cre- ation 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 Hˆint 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.
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