Derive explicit expression of squeezed vacuum state in the Fock basis

Question

I'm learning quantum optics, and I'm starting to manage boson algebra.

In particular, as a pure exercise, I would like to express a squeezed vacuum state in the Fock basis, which, according to Weedbrook et al. is given by: $$|0,r\rangle = \frac{1}{\sqrt{\cosh(r)}}\sum_n\frac{\sqrt{(2n)!}}{2^nn!}\tanh r^n|2n\rangle$$ Where $r$ is the parameter associated with the squeezing operator $$S(\xi)=\exp\left\lbrace\frac{1}{2}\left(\xi {a^\dagger}^2-\xi^*a^2\right)\right\rbrace ; \xi=re^{i\theta}$$ This expression can easily be obtained using disentangling equations and assuming $\xi$ to be real. However, I asked myself if the same expression could be obtained by using simple algebra (like BCH formula), indeed: $$|0,\xi\rangle=S(\xi)|0\rangle$$ Using BCH formula $e^{A+B} = e^Ae^Be^{-\frac{1}{2}[A,B]}$, I found that: $$S(\xi)=\exp\left(\frac{1}{2}\xi {a^{\dagger}}^2\right)\exp\left(-\frac{1}{2}\xi^* {a}^2\right)\exp\left(-\frac{1}{4}|\xi|^2(\hat{n}+1)\right)$$ Where the last term has been obtained from: $$-\frac{1}{2}\bigg[\frac{1}{2}\xi {a^{\dagger}}^2,-\frac{1}{2}\xi^* {a}^2\bigg] = -\frac{1}{4}|\xi|^2[{a^{\dagger}}^2,a^2]$$ by applying different times the identity $\hat{n}=a^{\dagger}a=aa^{\dagger}-1$. Indeed: $$[{a^{\dagger}}^2,a^2]=a^\dagger a^\dagger aa -aa a^\dagger a^\dagger=a^\dagger aa^\dagger a-a^\dagger a-aaa^\dagger a^\dagger =a^\dagger aaa^\dagger -2a^\dagger a-aaa^\dagger a^\dagger \\=aa^\dagger aa^\dagger -aa^\dagger -2a^\dagger a-aaa^\dagger a^\dagger =-2(aa^\dagger +a^\dagger a)=-2(\hat{n} + 1)$$

If I now apply this decomposition of $S(\xi)$ to the ground state, thanks to the properties of the number and annihilation operator, I have that: $$S(\xi)|0\rangle=\exp\left({-\frac{1}{4}|\xi|^2}\right)\exp\left(\frac{1}{2}\xi {a^{\dagger}}^2\right)|0\rangle=\exp\left({-\frac{1}{4}|\xi|^2}\right) \sum_n\frac{\xi^n}{n!2^n}{a^{\dagger}}^{2n}|0\rangle\\=\exp\left({-\frac{1}{4}|\xi|^2}\right) \sum_n\frac{\xi^n}{n!2^n}\sqrt{(2n)!}|2n\rangle$$ The first equality follows from the expansion of the two exponentials in Taylor series, and noticing that $a^n$ produces the 0 vector when acting on $|0\rangle$.

This expression is different to the one given in the literature, but I can't figure out where I'm going wrong with this derivation. Can you help me?

Answer 1

I’m willing to bet your commutator $[\hat a^\dagger \hat a^\dagger,\hat a\hat a]$ is wrong.

This is because \begin{align} \hat K_+=\frac{1}{2} \hat a^\dagger \hat a^\dagger \, ,\qquad \hat K_-=\frac{1}{2}\hat a\hat a\, ,\qquad \hat K_0=\frac{1}{4}(\hat a^\dagger \hat a +\hat a\hat a^\dagger) \tag{1} \end{align} close under commutation on the algebra $\mathfrak{su}(1,1)$ so that $[\hat K_+,\hat K_-]$ is a multiple of $\hat K_0$, which is not proportional to $\hat n+1$ but rather to $2\hat n+1$.

A useful reference for the realization (1) and for this type of stuff is the book by Perelomov:

Perelomov A. Generalized Coherent States and Their Applications, Springer.

As a bonus the squeezing transformation is basically an $SU(1,1)$ transformation, with matrix elements given in explicit form in

Ui H. Clebsch-Gordan formulas of the SU (1, 1) group. Progress of Theoretical Physics. 1970 Sep 1;44(3):689-702.