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 $|\psi\rangle$?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.