My problem

Find $O_\phi|\psi\rangle$, where the state $|\psi\rangle$ is defined on a composite space $\mathcal H_A\otimes \mathcal H_B$ as

$$|\psi\rangle = \left(\bigotimes_{k=1}^N|\alpha_k'\rangle\right)_A \otimes |0\rangle_B $$

where $|\alpha_k'\rangle$ is a coherent state with $\alpha_k'=\sum_j t_{kj}\alpha_j$ and $|0\rangle_B=\otimes_{j=1}^N |0\rangle_j$ the empty ket on $\mathcal H_B$, while $O_\phi$ for $\phi\in \mathbb R^N$ is the operator $$O_\phi=\exp\left[i\sum_{j,\ k=1}^N \phi_j\left(t_{jk}a_k^\dagger b_j + t_{jk}^* b_j^\dagger a_k\right)\right]$$ where $a_k$ and $b_j$ are annihilation operators for $A$ and $B$ respectively and $t_{jk}$ a complex phase.


$O_\phi$ looks a lot like a product of beam splitters acting locally with parameters $\phi_j$ and coupling the modes of $A$ and $B$, but after the $a_j$'s have been updated via the same rule that defines the $\alpha_k'$'s. Indeed, the state on $A$ is defined by the displacement operator $$D_{\alpha'}=\exp\left(\sum_{j, \ k=1}^N \left(t_{jk}\alpha_j a_k^\dagger - t_{jk}^* \alpha_j^* a_k\right) \right). $$

I know how to compute the action of a simple beam splitter on a product of coherent state (see this answer for an explicit calculation), but I'm a little stuck with this more involved operator.

  • $\begingroup$ On phase space, the coherent state becomes a Gaussian and the beamsplitter retains its form. The operation of the beamsplitter on the coherent state can then be evaluated as an integration over a Gaussian function. Perhaps it would help to do it that way. $\endgroup$ Commented Dec 25, 2023 at 3:36

1 Answer 1


Similarly to the answer you linked we can rewrite the state as $\vert \psi\rangle=\exp\left(-\frac{1}{2}\sum_k \vert\alpha'_k\vert^2\right)\exp\left(\sum_k \alpha'_k a^\dagger_k\right)\vert 0\rangle$. We can also see that $O_\phi^\dagger\vert 0\rangle=\vert 0\rangle$ (if you don't immediately see this, try to explicitly write the power series of the exponential that defines $O_\phi$ and every after the 1 should vanish because an annihilator acts on the vacuum). This means we have:

$O_\phi\vert \psi\rangle=O_\phi\exp\left(-\frac{1}{2}\sum_k \vert\alpha'_k\vert^2\right)\exp\left(\sum_k \alpha'_k a^\dagger_k\right)O_\phi^\dagger\vert 0\rangle=\exp\left(-\frac{1}{2}\sum_k \vert\alpha'_k\vert^2\right)\exp\left(\sum_k \alpha'_k O_\phi a^\dagger_k O_\phi^\dagger \right)\vert 0\rangle$.

So let's try to compute $O_\phi a^\dagger_k O_\phi^\dagger$. $ O_\phi$ is an exponential, so we basically have the form $e^X Y e^{-X}$. To simplify this, we do the following: First define "superoperators", i.e. operators acting on operators as: $L_X Y:=XY$, $R_X Y:=YX$ and $\mathrm{ad}_X Y:=[X,Y]$. In fact, $\mathrm{ad}_X=L_X-R_X$ (just apply it to $Y$ and insert definitions!).

Now we have the following: $e^X Y e^{-X}=L_{e^X} R_{e^{-X}} Y=e^{L_X} e^{-R_X}Y$. Now $[L_X,R_X]=0$ (to see that, just apply that to an arbitrary $Y$), so we can pull both exponentials together without problems (see Baker–Campbell–Hausdorff formula). This gives $e^{L_X} e^{-R_X}Y=e^{L_X-R_X}Y=e^{\mathrm{ad}_X}Y$. Okay, in our case $X=i\sum_{j,l}\phi_j(t_{jl}a_l^\dagger b_j+ t^*_{jl}b_j^\dagger a_l)$ and $Y=a_k^\dagger$ (I had to change the summation index so I don't have $k$ twice). So what is $\mathrm{ad}_X Y$ for example?

Well, by definition $\mathrm{ad}_X Y=[X,Y]=i\sum_{j,l}\phi_j(t_{jl}[a_l^\dagger b_j,a_k^\dagger]+ t^*_{jl}[b_j^\dagger a_l,a_k^\dagger])=i\sum_{j,l}\phi_jt^*_{jl}[b_j^\dagger a_l,a_k^\dagger]$ where the first term in the sum vanished because the commutator is just zero. Now $[b_j^\dagger a_l,a_k^\dagger]=b_j^\dagger[ a_l,a_k^\dagger]=b_j^\dagger\delta_{lk}$, so $\mathrm{ad}_X Y=i\sum_{j}\phi_jt^*_{jk}b^\dagger_j$.

To compute $e^{\mathrm{ad}_X}Y$ we need more than just the first power. We need all powers. But let's continue with the second and maybe we see a pattern. $(\mathrm{ad}_X)^2 Y=i\sum_{j}\phi_jt^*_{jk}\mathrm{ad}_Xb^\dagger_j$. A similar calculation to before should lead to something like $i^2\sum_{j,j'}\phi_{j'}t_{j'j}\phi_jt^*_{jk}a^\dagger_{j'}$. I hope I didn't mess up anything here. When you do the computation yourself you can check if I did a mistake. The thing is: Now our result is again something proportional to an $a_n^\dagger$ (in a sum to be fair). Like our $Y$! So what will happen is that all the odd powers of $\mathrm{ad}_X$ will lead to some sum of $b_m^\dagger$ and the even powers will lead to some sum of $a_n^\dagger$. This will just alternate. You should be able to split the power series $e^{\mathrm{ad}_X}Y$ into two terms (the odd and the even). You should already know computations like this, for example when exponentiating Pauli matrices. Give it a try!

Edit: Here is the rest of the schematic calculation: Call $\mathbf{a}$ the vector $(a_1,a_2,...)$. Similarly $\mathbf{b}=(b_1,b_2,...)$, $\mathbf{\alpha}'=(\alpha'_1,\alpha'_2,...)$. Further let $A$ be the matrix with entries $A_{kj}=\phi_j t^*_{jk}$. With this the calculation above amounts to $\mathrm{ad}_X \mathbf{a}^\dagger=iA\mathbf{b}^\dagger$ (where on the left hand side the operator $\mathrm{ad}_X$ gets applied componentwise while on the right hand side we have matrix-vector multiplication). Analogously you should find $\mathrm{ad}_X \mathbf{b}^\dagger=iA^*\mathbf{b}^\dagger$. This means that (here $\alpha'\cdot ...$ will just be the regular dot product):

$\exp\left(\sum_k \alpha'_k O_\phi a^\dagger_k O_\phi^\dagger \right)=\exp\left(\alpha'\cdot (O_\phi \mathbf{a}^\dagger O_\phi^\dagger) \right)=\exp\left(\alpha'\cdot e^{\mathrm{ad}_X} \mathbf{a}^\dagger \right)\\=\exp\left(\sum_n \frac{1}{(2n)!} \alpha'\cdot(\mathrm{ad}_X)^{2n}\mathbf{a}^\dagger+ \sum_n \frac{1}{(2n+1)!} \alpha'\cdot(\mathrm{ad}_X)^{2n+1}\mathbf{a}^\dagger \right)\\=\exp\left(\sum_n \frac{1}{(2n)!} \alpha'\cdot(-AA^*)^{n}\mathbf{a}^\dagger+ \sum_n \frac{1}{(2n+1)!} \alpha'\cdot(-AA^*)^{n} iA\mathbf{b}^\dagger \right)$

I basically just inserted the power series of the exponential and then split it in even and odd terms. The even and odd powers of $\mathrm{ad}_X$ could be evaluated with $\mathrm{ad}_X \mathbf{a}^\dagger=iA\mathbf{b}^\dagger$ and $\mathrm{ad}_X \mathbf{b}^\dagger=iA^*\mathbf{b}^\dagger$. The resulting operator looks ugly but under all the clutter it is just an exponential of a linear combination of creation operators. When this acts on the vacuum it will just give a tensor product of coherent states depending on the corresponding coefficients in the linear combination (and some factor in front). So the result will not be completely crazy. I'm afraid that it won't get much simpler, at least not without any additional information on the $\phi_j$ and the $t_{jk}$. Maybe you think about what properties these should fulfill and further simplify the result? Especially the $AA^*$ seems like it shouldn't be the end of the story so that we get some $\sin$ and $\cos$ out of the odd and even power series.

Nonetheless, I hope that helped you a bit! Please go through the calculation yourself to verify that I didn't do any mistakes.


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