Calculating Field moments of two mode squeezed state

Question

I am reading through a paper (EDIT: Paper is here) and I actually want to rigorously go through their calculations. I am having some issues, For a two mode squeezed state given by:

$$|\psi\rangle = \sum_{n=0}^{\infty}\sqrt{\frac{N_S^n}{(N_S+1)^{n+1}}}|n\rangle_S |n\rangle_I$$

I am trying to calculate $\langle a_{S_m}^2 \rangle$. Which according to the paper is:

$$ \frac{2 N_S + 1}{4} $$

The state given above is the result of an SPDC output, creating pairs of signal (denoted by S) and idler (denoted by I) photons. $N_S$ is the mean signal photon number.

It has been awhile since I have done these calculations and I just cannot seem to get the above answer. I don't know if $a_{S_m}^2=a_{S_m}a_{S_m}$ or if $a_{S_m}^2=a_{S_m}^{\dagger}a_{S_m}$. If its the former, then the calculation would yield zero due to Fock basis orthogonality.

$$\langle \psi |a_{S_m}a_{S_m}| \psi \rangle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sqrt{\frac{N_S^{m+n}}{(N_S+1)^{n+m+2}}}\langle m |_S \langle m |_I a_{S_m}a_{S_m} | n \rangle_S | n \rangle_I$$

$$\langle \psi |a_{S_m}a_{S_m}| \psi \rangle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sqrt{\frac{N_S^{m+n}}{(N_S+1)^{n+m+2}}}\langle m+2 |_S \langle m |_I| n \rangle_S | n \rangle_I = 0$$

If it's the latter, then we obtain:

$$\langle \psi |a^{\dagger}_{S_m}a_{S_m}| \psi \rangle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sqrt{\frac{N_S^{m+n}}{(N_S+1)^{n+m+2}}}\langle m |_S \langle m |_I a^{\dagger}_{S_m}a_{S_m} | n \rangle_S | n \rangle_I$$

$$\langle \psi |a^{\dagger}_{S_m}a_{S_m}| \psi \rangle = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sqrt{\frac{N_S^{m+n}}{(N_S+1)^{n+m+2}}}(n)\langle m |_S \langle m |_I | n \rangle_S | n \rangle_I$$

$$\langle \psi |a^{\dagger}_{S_m}a_{S_m}| \psi \rangle = \sum_{n=0}^{\infty}\sqrt{\frac{N_S^{2n}}{(N_S+1)^{2n+2}}}(n) = \frac{1}{N_S+1}\sum_{n=0}^{\infty}n\left( \frac{N_S}{N_S+1} \right)^n$$

Which I do not know how to evaluate. Am I on the right track here?

Answer 1

Let $a_S$ and $a_I$ denote the annihilation operators on the $S$ and $I$ parts of the state respectively. We define hermitian operators $a_{S_m}$ and $a_{I_m}$ for $m=1,2$ by means of the equations

$$a_S = a_{S_1}+ia_{S_2},\quad a_{I}=a_{I_1}+ia_{I_2}\tag{1}$$

Here we focus on the $S$ part. Taking the adjoint of the first equation in (1) we get

$$a_S^\dagger=a_{S_1}-i a_{S_2}\tag{2}$$

By summing and subtracting (1) and (2) we may invert the relation to get

$$a_{S_1}=\frac{a_S+a_S^\dagger}{2},\quad a_{S_2}=\frac{a_S-a_S^\dagger}{2i}\tag{3}$$

Hence employing commutation relations $[a_S,a_S^\dagger]=1$ we get the squares

$$a_{S_1}^2=\frac{1}{4}(1+a_S^2+(a_S^\dagger)^2+2a_S^\dagger a_S),\quad a_{S_2}^2=-\frac{1}{4}(-1+a_S^2+(a_S^\dagger)^2-2a_S^\dagger a_S)\tag{4}$$

So these are the operators we wish to find the mean value. For that we rewrite the state in a better form

$$|\psi\rangle=\beta\sum_n \alpha^n |n\rangle_S|n\rangle_I,\quad \beta = \frac{1}{\sqrt{N_S+1}},\quad \alpha=\sqrt{\frac{N_S}{N_S+1}}\tag{5}$$

We turn to the means. Trace over the $I$ part. This amounts to forming $|\psi\rangle\langle \psi|$ and summing over the $I$ index:

$$\rho_S=\operatorname{Tr}_{I}|\beta|^2 \sum_{nm}(\alpha^\ast)^m \alpha^n |n\rangle_S|n\rangle_I \langle m|\langle m|=|\beta|^2\sum_n |\alpha|^{2n} |n\rangle\langle n|.\tag{6}$$

The mean of any operator $O$ acting just on $S$ is thus $$\langle O\rangle=\operatorname{Tr} (\rho_S O)$$

Notice that if $\rho_S O$ has only offdiagonal elements the mean vanishes. Indeed that's obviously the case for $a_S^2$ and $(a_S^\dagger)^2$ as seen by (6). Just the term with $1$ and the term $a_S^\dagger a_S$ remains. The term with $1$ gives $1$ by normalization of the state. But when $a_S^\dagger a_S$ acts on $|n\rangle$ it gives $n|n\rangle$ since its the number operator. Hence

$$\langle a_S^\dagger a_S\rangle=|\beta|^2\sum_n n |\alpha|^{2n}\tag{7}$$

This sum can be found with

$$\sum n x^n = x\frac{d}{dx}\sum x^n=x\frac{d}{dx}\frac{1}{1-x}=\frac{x}{(1-x)^2}\tag{8}$$

In that case combining (8) and (7) and nserting $\alpha,\beta$ as defined in (5)

$$\langle a_S^\dagger a_S\rangle=|\beta|^2\frac{|\alpha|^2}{(1-|\alpha|^2)^2}=N_S\tag{9}$$

Combining it all we get

$$\langle a_{S_1}^2\rangle=\frac{1}{4}(1+2N_S)\tag{10}$$

The case $m=2$ can be dealt with in similar manner.