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?

  • 1
    $\begingroup$ 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 $\endgroup$
    – mike stone
    Commented Nov 10, 2022 at 13:39
  • $\begingroup$ I found this paper useful at one point. Not sure that it really gets at what you're looking for, but it might help. $\endgroup$
    – march
    Commented Nov 10, 2022 at 17:40
  • $\begingroup$ 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. $\endgroup$
    – skdys
    Commented Jan 6, 2023 at 21:19
  • $\begingroup$ @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. $\endgroup$ Commented Jan 7, 2023 at 0:42
  • $\begingroup$ 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. $\endgroup$
    – skdys
    Commented Jan 8, 2023 at 19:03

1 Answer 1


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 $\mathcal{N}$.)


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