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Let $a_1, \ldots, a_N$ be boson operators, $[a_i, a_j^\dagger ] = \delta_{ij}$. One often considers Bogoliubov transformation $a_i \to \sum_j (A_{ij} a_j + B_{ij} a_j^\dagger)$, where the matrices $A=(A_{ij}), B=(B_{ij})$ satisfy $$AA^\dagger - BB^\dagger = I, \quad AB^T - BA^T=0,$$ to preserve the commutation relation.

Question: Can we find a unitary operator $U$ (acting on the usual bosonic Hilbert space) that satisfies $U a_i U^\dagger = \sum_j(A_{ij} a_j + B_{ij} a_j^\dagger)$?

I think finding $U$ is an entirely different issue than checking the invariance of the commutation relation.

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There is a general argument for the existence of the unitary operator. From the Stone-von Neumann theorem, if you know where $a$ is mapped, then there is only one unitary operator that can represent the transformation (up to a global phase).

You can construct this operator more explicitly. One way is to realize that your transformations form a path connected Lie group. Formally, you can view it as the subgroup of matrices of size $2N\times2N$ of the form: $$ \begin{pmatrix} A & B\\ B^* & A^* \end{pmatrix} $$ with your additional constraints: $$ A A^\dagger -BB^\dagger =1\\ AB^T-BA^T =0 $$ You can consider a path from the identity to your transformation, namely $A_t,B_t$ with $t\in[0,1]$ and: $$ \begin{align} A_0&=1 & B_0&=0 \\ A_1&=A& B_1&=B \end{align} $$ This gives a varying $a_t$ satisfying the ODE: $$ \dot a_t= \alpha_ta_t+ \beta_ta_t^\dagger $$ with the matrices $\alpha_t,\beta_t$ defined by: $$ \frac{d}{dt} \begin{pmatrix} A_t & B _t\\ B _t ^* & A _t ^* \end{pmatrix} = \begin{pmatrix} \alpha_t& \beta_t\\ \beta_t^* & \alpha_t^* \end{pmatrix} \begin{pmatrix} A_t & B _t\\ B _t ^* & A _t ^* \end{pmatrix} $$ You want to recognize the ODE as a Heisenberg equation from a suitably chosen quadratic Hamiltonian: $$ \dot a_t =-i[a,H] $$ Using the CCR’s you can check that a suitable candidate (unique up to additional multiple of identity) is: $$ H_t= i(\alpha_t)_{ij}a_i^\dagger a_j+\frac{i}{2}(\beta_t)_{ij}a_i^\dagger a_j^\dagger-\frac{i}{2}(\beta_t^*)_{ij}a_i a_j $$ $H_t$ is hermitian and gives the correct equations of motion using the relations (taking the derivatives of the constraints): $$ \alpha_t+\alpha_t^\dagger =0 \\ \beta_t-\beta_t^T=0 $$ Your unitary operator is given by the time ordered exponential: $$ U=\mathcal T \exp\left(\int_0^1dt H_t\right) $$ Actually, it turns out that the exponential is surjective on the group, so you can choose Choose a path with constant $\alpha_t,\beta_t$. This gives a constant $H_t$ and the time ordered exponential is a simple exponential.

Physically, you are constructing the unitary operator from harmonic oscillators and squeeze operators. The process is explicit in the case $N=1$.

Hope this helps.

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  • $\begingroup$ Thanks for your answer! I found the abstract argument based on Stone-von Neumann theorem interesting. Could you elaborate more why knowing where $a_i$'s are mapped implies that there exists a unique unitary operator that realizes that transformation? Maybe one transforms $x_i \sim a_i + a_i^\dagger$ and $p_i \sim a_i - a_i^\dagger$ to obtain the "canonical commutation relation"? $\endgroup$
    – Laplacian
    Mar 11, 2023 at 11:18
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    $\begingroup$ Yes exactly, and since your Hilbert space is constructed as an irreducible representation, this guarantees existence and uniqueness. Note that the exact mathematical statement is not about the CCR $[x,p] = i$, but rather its exponentiated form (see Weyl form of CCR for more). $\endgroup$
    – LPZ
    Mar 11, 2023 at 13:01
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The general case (more than one mode) is complicated. A formula is given in eqs 2.17, 2.18 in https://arxiv.org/pdf/1608.03289.pdf, unfortunately the author's notation is rather obscure and ius defined only in their cited book.

There is a discussion in my online notes http://people.physics.illinois.edu/stone/bose_bogoliubov.pdf. It is too long to copy here.

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  • $\begingroup$ Thanks for your answer! I will take time to read your notes. $\endgroup$
    – Laplacian
    Mar 11, 2023 at 11:14

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