# Derive explicit expression of squeezed vacuum state in the Fock basis

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?

• presumably you mean $(2n)!$. You essentially have a different parametrization of your squeezing transformation. The first step in reconciling the expressions is to convert $\xi$ to $r$ or vice-versa. Aug 28 '18 at 17:03
• Seem to me that you've used the disentangling incorrectly. Perhaps you can have a look at the answer in physics.stackexchange.com/questions/416105/… Aug 29 '18 at 6:06
• You say "using the BCH formula" but I don't see a simple way to use BCH to get what you have. Can you spell out the details of how you get the RHS? Aug 29 '18 at 12:38
• Check the Wiki page for the BCH formula, you are not using it correctly. Note that $[A,B]$ doesn't commute with $A$ or $B$, unlike the application of BCH to e.g. unitary displacement operators which you might have seen before. Aug 30 '18 at 8:02
• @MarkMitchison You are right. I was applying the BCH formula without veryfing if $[A,B]$ commutes with $A$ and $B$ (lot of simple Boson algebra in the last few days). That was my fault. Thank you so much for the spot!
– steg
Aug 30 '18 at 8:30

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 $$2\hat n+1$$.
As a bonus the squeezing transformation is basically an $$SU(1,1)$$ transformation, with matrix elements given in explicit form in