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This is a simplified version of one of my previous questions. Let $b_1, b_2$ be two boson operators; their vacuum is denoted as $|0\rangle$, i.e. $b_i |0\rangle = 0$. We can make a canonical Bogoliubov transformation:

$$ \beta_1 = u b_1 + v b^\dagger_2, \quad \beta_2 = u b_2 + v b^\dagger_1 $$

where $u,v$ are two real positive numbers. $u^2 - v^2 = 1$ to make sure that $\beta_i$ are still bosons, i.e. satisfy the canonical commutation relations. Now I want to find the vacuum $|0_\beta\rangle$ of $\beta_1, \beta_2$, i.e. $\beta_i |0\rangle = 0$. I know the answer is (up to a normalization constant)

$$ |0_\beta\rangle = e^Q |0\rangle, \quad Q = -\frac{v}{u} b^\dagger_1 b^\dagger_2 \tag{1} $$

which I found by an educated guess (see an alternative way to get it in this answer).

To get $|0_\beta\rangle$ more naturally, I attempt to use projections: we should be able to get $|0_\beta\rangle$ starting from any state $|\psi\rangle$ that is not orthogonal to $|0_\beta\rangle$ by projecting out all states with number of $\beta_i$ greater than 0. In this spirit, we should have

$$ |0_\beta\rangle \propto \prod_{n=1}^\infty (\beta^\dagger_1 \beta_1 - n) (\beta^\dagger_2 \beta_2 - n) |\psi\rangle \tag{2} $$

This method works for fermions (see Eqs. (1.33), (1,34) in D-wave Superconductivity by Xiang et al), for which one can simply choose $|\psi\rangle$ as the vacuum of the original fermions. But for bosons things are much more complicated due to the infinite product; and one cannot choose $|\psi\rangle = |0\rangle$, since $\langle 0_\beta | 0 \rangle = \langle 0 | \exp(Q^\dagger) | 0 \rangle = 0$ (as noted by @Quantum Mechanic; since $Q^\dagger$ only contains annihilation operators).

Question: How to derive Eq. (1) from Eq. (2), and how may we choose a convenient $|\psi\rangle$?

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The Bogoliubov transformation is achieved by a two-mode squeezing operator that achieves $$S b_i S^\dagger=\beta_i.$$ Using this and the definition of the vacuum $b_i|0\rangle=0$, one finds $$0=S b_i|0\rangle=S b_i S^\dagger S|0\rangle=\beta_i S|0\rangle.$$ We thus learn that the null eigenstate of the operator $\beta_i$ is simply the squeezed vacuum state $S|0\rangle$. Up to proportionality constants and braiding relations, that is what's given by $e^Q|0\rangle$.

That's the easiest method for me. In terms of your projections method, one might have to worry about the state where the projections start: you need some proof that $|0\rangle$ has nonzero overlap with $|0_\beta\rangle$ because otherwise the projection will vanish.

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  • $\begingroup$ Can one derive the squeezing operator $S$ from first principles? $\endgroup$ May 15, 2023 at 13:11
  • $\begingroup$ @ZhengyuanYue Yes! It must be an element of SU(1,1) because it performs an SU(1,1) transformation on the operators $b_i$. The Lie algebra of su(1,1) and its Lie group SU(1,1) are essential here $\endgroup$ May 15, 2023 at 13:17
  • $\begingroup$ Can this be generalized to more than 2 bosons? You may refer to my "complete" version of the question and see if you have additional comments to the existing ones there. $\endgroup$ May 15, 2023 at 13:26
  • $\begingroup$ @ZhengyuanYue I'm not sure if there's a direct way, but my easy way to generalize is to do a ("linear" or "beam splitter") transformation $Ua_iU^\dagger=\sum_j u_{ij}a_j$ either before or after the $S$ (Bogoliubov) transformation and then see how that changes the $\beta_i$, the $S$ operator, etc. $\endgroup$ May 15, 2023 at 14:00

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