Normal ordering of passive linear optics

Question

I am trying to normally-order operators in quantum optics. Having normally-ordered expressions is useful when evaluating expectation values in quantum optics, as most of the times we can write the states in terms of coherent states, meaning that the normal ordering allows us to replace operators with c-numbers. For active elements, the paramount example is the squeezing operator, that satisfies the relations

$$\hat{S}(z)=\exp{\{z^*\hat{a}^2-z\hat{a}^{\dagger \; 2}\}} = e^{-\frac{\tanh(|z|)}{2}\hat{a}^{\dagger \; 2}} \left(\dfrac{1}{\cosh{|z|}}\right)^{\hat{n}+1/2}e^{\frac{e^{i\text{arg(z)}}\tanh(|z|)}{2}\hat{a}^{2}},$$

but the same holds for the two mode squeezing operator, that reads:

$$\hat{S}_{2 \; mode}(z)=\exp{\{z^*\hat{a}\hat{b}-z\hat{a}^{\dagger}\hat{b}^{\dagger}\}} = e^{-\frac{\tanh(|z|)}{2}\hat{a}^{\dagger}\hat{b}^{\dagger}}\left(\frac{1}{\cosh{|z|}}\right)^{\hat{n}_a+\hat{n}_b+1}e^{\frac{e^{i\text{arg(z)}}\tanh(|z|)}{2}\hat{a}\hat{b}}.$$

I was wondering if there is a clever way to write a similar expression for passive elements (linear interferometry) such as beam splitter, whose operators usually read $$\exp{\{i\theta(\hat{a}\hat{b}^{\dagger}+\hat{b}\hat{a}^{\dagger})\}}.$$ The point here is that the two ladder operators come always in pairs like $\hat{a}\hat{b}^{\dagger}$ or vice-versa, so while I can normal order an operator with respect to, let's say, $\hat{a}$, the same operator will be anti-normally ordered in the $\hat{b}$. It would be nice to have the same ordering for both modes, for example because, considering the Bloch-Messiah decomposition, we could in principle normally order any second-order nonlinear optical transformation. As a test for this application, we could consider the Bloch-Messiah decomposition of $\hat{S}_{2 \; mode}(z)$: we should be able to reconstruct the expression given above by reshaping the product of three normally-ordered operators. I tried many techniques (e.g. looking at the $\mathfrak{s}\mathfrak{u}(2)$ algebra of the bosonic modes, using coherent states' relations such as $1=\int \frac{d\alpha^2}{\pi} \left|\alpha\right> \left<\alpha\right|$...)

Answer 1

You cannot separate the modes in products like $a^\dagger b$. On the other hand you can normal order in the $\mathfrak{su}(2)$ sense because $$ a^\dagger b\mapsto L_+\, ,\quad a b^\dagger \mapsto L_-\, ,\quad \frac12\left(a^\dagger a - b^\dagger b\right)\mapsto L_z $$ so your exponential $$ e^{i\theta (a b^\dagger + b a^\dagger)}\mapsto e^{i\theta (L_++L_-)} $$ which can be disentangle using the expression \begin{align} e^{i\theta (L_++L_-)}=\left( \begin{array}{cc} \cos (\theta ) & i \sin (\theta ) \\ i \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)&= \left(\begin{array}{cc} 1&\alpha \\ 0 &1 \end{array}\right) \left(\begin{array}{cc}e^{\beta/2}&0 \\ 0 & e^{-\beta/2} \end{array}\right) \left(\begin{array}{cc} 1&0 \\ \gamma &1\end{array}\right)\\ &=e^{\zeta L_+}e^{i\theta L_z} e^{\eta L_-} \end{align}

Answer 2

ZeroTheHero is right when he says that it is probably impossible to separate modes in products like $a^\dagger b$. However, the concept of normal order also applies in the presence of such terms. I am almost certain that there is an equality of the following form $$ \exp\left(i\theta(\hat{a}\hat{b}^\dagger+\hat{b}\hat{a}^\dagger)\right) = :\exp\left(x(\hat{b}^\dagger\hat{a}+\hat{a}^\dagger\hat{b}) + y(\hat{a}^\dagger\hat{a}+\hat{b}^\dagger\hat{b})\right): \tag{eq} $$ where $x$ are $y$ are functions of $\theta$, and $:\ldots:$ denotes normal ordering, as in the following examples $$ :\exp(z\hat{a} + w\hat{b}^\dagger): = \sum_{k,l = 0}^\infty\frac{z^kw^l}{k!l!}(\hat{b}^\dagger)^l(\hat{a})^k,\quad :\exp(z\hat{b}\hat{a}^\dagger): = \sum_{k=0}^\infty\frac{z^k}{k!}(\hat{a}^\dagger)^k(\hat{b})^k. $$ General methods for finding equalities of type (eq) are developed in Berezin's book. Useful information on this subject can also be found in the second chapter of the book by Itzykson and Drouffe.

In this relatively simple example, the parameters $x$ and $y$ can most likely be found by requiring equality (eq) to hold on the vectors $\hat{a}^\dagger|0\rangle$ and $\hat{a}^\dagger\hat{b}^\dagger|0\rangle$, or by solving differential equations obtained by differentiating with respect to $\theta$.

Update. Vectors $(\hat{a}^\dagger\pm\hat{b}^\dagger)|0\rangle$ are eigenvectors of the operator $\hat{a}^\dagger\hat{b}+ \hat{b}^\dagger\hat{a}$, the corresponding eigenvalues ​​are $\pm 1$. Applying equality (eq) to these vectors yields the equations $$ e^{i\theta} = 1 + x + y,\quad e^{-i\theta} = 1 - x + y, $$ from which it follows that $x = i\sin(\theta)$, $y = \cos(\theta)-1$.