# Ground state of Bogoliubov quasi-particles (simpler version)

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 $$|\psi\rangle$$?

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.
• Can one derive the squeezing operator $S$ from first principles? May 15 at 13:11
• @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 May 15 at 13:17
• @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. May 15 at 14:00