# Ground state of Bogoliubov quasi-particles

Consider a set of boson/fermion creation and annihilation operators satisfying the canonical (anti-)commutation rules (CCR/CAR):

$$[a_i, a_j]_\eta = [a^\dagger_i, a^\dagger_j]_\eta = 0, \quad [a_i, a^\dagger_j]_\eta = \delta_{ij} \quad (i,j = 1,...,N)$$

where $$\eta$$ is the statistical sign ($$+1$$ for boson, and $$-1$$ for fermion), $$[A,B]_\eta = AB - \eta BA$$, and $$N$$ is the dimension of one-particle Hilbert space. A general canonical transformation mixes the creation and annihilation operators to produce a new set of bosons/fermions $$\{b_j\}_{j=1}^N$$, which I call the Bogoliubov quasi-particles:

$$b_i = \sum_{j=1}^N (u_{ij} a_j + v_{ij} a^\dagger_j) , \quad b^\dagger_i = \sum_{j=1}^N (v^*_{ij} a_j + u^*_{ij} a^\dagger_j)$$

and the $$\{b_i\}$$ operators should also satisfy the CCR/CAR:

$$[b_i, b_j]_\eta = [b^\dagger_i, b^\dagger_j]_\eta = 0, \quad [b_i, b^\dagger_j]_\eta = \delta_{ij} \quad (i,j = 1,...,N)$$

One can show that to preserve the CCR/CAR, the matrices $$U = \{u_{ij}\}$$, $$V = \{v_{ij}\}$$ should satisfy the requirement

$$U V^{\mathsf{T}} = \eta V U^{\mathsf{T}} \quad \text{and} \quad U U^\dagger - \eta V V^\dagger = 1$$

Question: Let $$|0_a\rangle, |0_b\rangle$$ be the vacuum states of the $$a, b$$ particles respectively, i.e.

$$\forall i = 1,...,N: \quad a_i |0_a\rangle = 0, \quad b_i |0_b\rangle = 0$$

How to express $$|0_b\rangle$$ in terms of states and operators of the $$a$$ particles?

• It's a bit more than an "educated guess," but it it does require that the $U$ matrix be invertible, which is not globally true. For the Bosonic case look at arxiv.org/abs/1608.03289 eq 2.17 2.18 Commented Nov 10, 2022 at 13:39
• I found this paper useful at one point. Not sure that it really gets at what you're looking for, but it might help. Commented Nov 10, 2022 at 17:40
• You can check this paper by Kita journals.jps.jp/doi/abs/10.1143/JPSJ.65.908. You can find the same calculation also in his book: link.springer.com/10.1007/978-4-431-55405-9 The result is similar to the one in Mike Stone's note. Commented Jan 6, 2023 at 21:19
• @skdys Could you please specify at which chapter the derivation of the ground state of quasi-particles is given? I had a brief look but did not find it. Commented Jan 7, 2023 at 0:42
• In the book, Eq. 8.4 is the definition of a many-body wavefunction $\vert {\Phi} \rangle$ as a function of a matrix $\phi$. In Sec. 8.3, the standard BdG equations are derived, and finally, after Eq. 8.46, the connection between this matrix $\phi$ and the usual coherence factor u and v is shown. I believe that if you reorder the steps (maybe with the help of the paper), you can get to a formal mathematical derivation if that is your objective. To me, this looks pretty formal already from a physicist's point of view. Commented Jan 8, 2023 at 19:03

The following is summarized from two notes (both can be found here) by Prof. Michael Stone (@mikestone). Confusing typos in these notes are corrected.

From a general perspective, the two set of operators $$\{a\}, \{b\}$$ are two realizations of the CCR/CAR. By the Stone-von Neumann theorem, there is a unitary operator $$\mathcal{U}$$ that acts on the many-body Hilbert space such that

$$b_i = \mathcal{U} a_i \mathcal{U}^\dagger \ \Rightarrow \ b^\dagger_i = \mathcal{U} a^\dagger_i \mathcal{U}^\dagger$$

Then we find $$|0_b\rangle \propto \mathcal{U}|0_a\rangle$$, since

$$a_i |0_a\rangle = \mathcal{U}^\dagger b_j \mathcal{U} |0_a\rangle = 0$$

However, as noted by Prof. Stone, the general form of $$\mathcal{U}$$ is very complicated. Thus we take another approach.

Assume that $$U$$ is invertible (this is not an "innocent assumption"). Then $$b_i |0_b \rangle = 0$$ is equivalent to

$$0 = U^{-1}_{ij} b_j |0_b\rangle = (a_i - S_{ij} a^\dagger_j) |0_b \rangle, \quad S \equiv - U^{-1} V$$

Using the constraint $$U V^\mathsf{T} = \eta V U^\mathsf{T}$$, one can show that

$$S^\mathsf{T} = \eta S$$

i.e. the matrix $$S$$ is symmetric for bosons, and anti-symmetric for fermions. Then we construct the bilinear

$$Q \equiv \frac{1}{2} \sum_{i,j} S_{ij} a^\dagger_i a^\dagger_j$$

and use the BCH formula to calculate

$$\begin{equation*} e^Q a_i e^{-Q} = a_i + [Q,a_i] + \frac{1}{2}[Q,[Q,a_i]] + \cdots \end{equation*}$$

Fortunately, the commutator

$$[Q, a_i] = - S_{ij} a^\dagger_j$$

commutes with $$Q$$ (for both bosons and fermions). Therefore we simply get

$$\begin{equation*} e^Q a_i e^{-Q} = a_i + [Q,a_i] = a_i - S_{ij} a^\dagger_j \end{equation*}$$

and $$b_i |0_b \rangle = 0$$ is equivalent to

$$e^Q a_i e^{-Q} |0_b\rangle = 0$$

This can be satisfied by all $$a_i$$ if and only if

$$e^{-Q} |0_b \rangle \propto |0_a \rangle$$

The "if" is obvious. To show "only if", suppose that $$|\psi\rangle \equiv e^{-Q} |0_b \rangle \ne |0_a \rangle$$; then $$e^Q$$, which consists of only $$\{a^\dagger_i\}$$, cannot annihilate $$|\psi\rangle$$. Finally, up to a normalization constant $$\mathcal{N}$$, the vacuum $$|0_b\rangle$$ is

$$|0_b\rangle = \mathcal{N} e^Q |0_a \rangle$$

(When I have time I will update on finding $$\mathcal{N}$$.)