Calculating Field moments of two mode squeezed state

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?

• Could you post the reference ? – Gold Feb 13 '19 at 16:21
• Sorry about that, I forgot! Paper link is now added. – user41178 Feb 13 '19 at 17:53

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.

• So this is very confusing. What looks like an annihilation operator is in actual fact a quadrature operator. – flippiefanus Feb 14 '19 at 4:18
• I agree that the notation employed can be confusing. But at least the authors properly defined it in the paper, so that in the end it is possible to understand what they want to say. – Gold Feb 14 '19 at 15:11
• Hello, I waited until I went through the whole thing to reply. Thank you so much, this is exactly what I needed! – user41178 Feb 14 '19 at 21:56