A beam splitter transforms incoming mode operators $\hat{a}_{i}, \hat{b}_{i}$ to outgoing operators $\hat{a}_{o}, \hat{b}_{o}$ as $$ \begin{aligned} &\hat{a}_{o}=\sqrt{\eta} \hat{a}_{i}+i \sqrt{1-\eta} \hat{b}_{i} \\ &\hat{b}_{o}=i \sqrt{1-\eta} \hat{a}_{i}+\sqrt{\eta} \hat{b}_{i} \end{aligned} $$

$\text { where } \eta=\cos ^{2} \theta$ I want to find for the incoming state $\left|\alpha_{i}\right\rangle \otimes\left|\beta_{i}\right\rangle$, which is a product of two coherent states what will be the outgoing states?

First I have showed that: $ \hat{T}=\exp \left[-i \theta\left(\hat{a}^{\dagger} \hat{b}+\hat{a} \hat{b}^{\dagger}\right)\right]$

is equivalent to the transformation matrix of the beam splitter using the below expansion:

$$ \exp \left(\mathrm{i} \theta \sigma_{1}\right)=\cos \theta+\mathrm{i} \sigma_{1} \sin \theta $$

Then expanded the coherent states as follows:

$$ |\alpha\rangle=e^{-\frac{|\alpha|^{2}}{2}} \sum_{n=0}^{\infty} \frac{|\alpha|^{n}}{\sqrt{n !}}|n\rangle $$

and obtained: $\left(\hat{I} \cos \theta+i\left(\hat{a} \hat{b}^{+}+\hat{b} \hat{a}^{+}\right) \sin \theta\right)e^{-|\alpha|^{2} / 2} \cdot e^{-|\beta|^{2} / 2} \sum_{m,n} \frac{|\beta|^{m}|\alpha|^{n}}{\sqrt{ nm !}}(|m\rangle \otimes|n\rangle)$

and further obtained:

$e^{-|\alpha|^{2} / 2} \cdot e^{-|\beta|^{2} / 2} \sum_{m,n} \frac{|\beta|^{m}|\alpha|^{n}}{\sqrt{ nm !}}(|m, n\rangle \cos \theta+i(|m-1, n+1\rangle+|m+1, n-1\rangle) \sin \theta)$

But couldn't continue further at this point. Thanks in advance.


1 Answer 1


The input coherent state is : $$|\alpha\rangle\otimes|\beta\rangle = \exp\left(-\frac{|\alpha|^2+|\beta|}{2}\right)\exp\left(\alpha \hat a_i^\dagger + \beta \hat b_i^\dagger\right)|0\rangle$$

Then, since $\hat T$ is unitary and $\hat T^\dagger|0\rangle=|0 \rangle $ we have :

\begin{align} \hat T|\alpha\rangle\otimes|\beta\rangle &= \exp\left(-\frac{|\alpha|^2+|\beta|}{2}\right) \exp\left(\hat T\left(\alpha \hat a_i^\dagger + \beta \hat b_i^\dagger\right)\hat T^\dagger\right)|0\rangle \\ &=\exp\left(-\frac{|\alpha|^2+|\beta|}{2}\right) \exp\left(\alpha \hat a_o^\dagger + \beta \hat b_o^\dagger\right)|0\rangle \\ &=\exp\left(-\frac{|\alpha|^2+|\beta|}{2}\right) \exp\left( (\alpha\cos(\theta)-i\beta\sin(\theta))\hat a_i^\dagger + (\beta\cos(\theta) - i\alpha\sin(\theta)) \hat b_i^\dagger\right)|0\rangle \\ &=|\alpha\cos(\theta)-i\beta\sin(\theta)\rangle\otimes|\beta\cos(\theta) - i\alpha\sin(\theta) \rangle \end{align}

  • 1
    $\begingroup$ Thanks a lot, but I have tried to replicate your work and realized that it should be $|\alpha\rangle \otimes|\beta\rangle=\exp \left(-\frac{|\alpha|^{2}+|\beta|^2}{2}\right) \exp \left(\alpha \hat{a}_{i}^{\dagger}+\beta \hat{b}_{i}^{\dagger}\right)|0\rangle\otimes|0\rangle$ for second line, can you check it? $\endgroup$
    – asd.123
    May 10, 2022 at 16:48
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    $\begingroup$ Yes, I took the liberty to abuse notation and write $|0\rangle$ in place of $|0\rangle \otimes |0\rangle$ for the ground state. $\endgroup$ May 10, 2022 at 17:19
  • $\begingroup$ @SolubleFish, Thanks for this brilliant answer, but can you describe how you put T and T^\dagger inside exponential? $\endgroup$ Jul 26, 2023 at 15:36
  • 1
    $\begingroup$ If $A$ is a matrix or an operator, its exponential is defined by the Taylor series : $\exp(\hat A) = \sum_{n=0}^{+\infty} \frac{1}{n!} \hat A^n$. Since we have $\left(\hat A^n\right)^\dagger =\left(\hat A^\dagger\right)^n$, we get $\exp\left(\hat A^\dagger\right) = \exp\left(\hat A^\dagger\right)$. (Note that this work for any analytic function whose Taylor coefficients are real.) $\endgroup$ Aug 5, 2023 at 15:27
  • $\begingroup$ You used Displacement operator $D(\alpha) = exp(\alpha a^\dagger)$ to convert $|\alpha>$ into $|0>$, but $D(|\alpha>) = exp(\alpha a^\dagger - \alpha^* a)$. what happened to annihilation operator? $\endgroup$ Nov 10, 2023 at 9:24

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