Can anyone help me to find the mean value of the displacement operator $$D(\alpha) = \exp( \alpha a^\dagger -\alpha^* a)$$ for a Coherent State $\left|\beta\right> = D\left(\beta\right)\left|0\right>$?

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    $\begingroup$ Hi! Welcome to QCSE! This question seems unmotivated right now. Please provide some details of what you are asking and what you know. $\endgroup$ – Mark S Oct 1 '19 at 0:30

So you want to calculate $\left<\beta |D\left(\alpha\right)|\beta\right>$ where $\left|\beta\right>$ is a coherent state and $D\left(\alpha\right)$ is the displacement operator.

The easiest way of doing this is to take $\left<\beta |D\left(\alpha\right)|\beta\right> = \left<0|D^\dagger\left(\beta\right) D\left(\alpha\right)D\left(\beta\right)|0\right>$ and write this in terms of the exponential of creation and annihilation operators $D(\alpha) = \exp( \alpha a^\dagger -\alpha^* a)$, which I'll denote as $D(\alpha) = \exp(Y)$.

At this point, note that $D^\dagger(\beta) = \exp( \beta^* a -\beta a^\dagger) = D\left(-\beta\right) = \exp(-X)$ and from here, we can use the Baker–Campbell–Hausdorff formula. In this case, $$\left[X,Y\right] = \left[\beta a^\dagger -\beta^* a,\alpha a^\dagger -\alpha^* a\right] = \beta\alpha^*\left[a, a^\dagger\right] - \beta^*\alpha\left[a, a^\dagger\right] = \beta\alpha^* - \beta^*\alpha$$ as $\left[a, a^\dagger\right] = 1$.

As $\beta\alpha^* - \beta^*\alpha$ is just a constant, this commutes with both $\beta a^\dagger -\beta^* a$ and $\alpha a^\dagger -\alpha^* a$ and so, all the higher order terms in BCH are $0$.

This gives that $e^{-X}e^Y = e^{-X+Y-\frac{1}{2}\left[X, Y\right]}$ and further that $$e^{-X}e^Ye^X = e^{-X+Y-\frac{1}{2}\left[X, Y\right] + X + \frac{1}{2}\left[-X+Y-\frac{1}{2}\left[X, Y\right],X\right]} = e^{Y - \left[X, Y\right]}.$$

While this still looks complicated, $\left[X, Y\right] = \beta\alpha^* - \beta^*\alpha$ is still a constant, so we can rewrite this as $$\left<0|e^{-X}e^Ye^X|0\right> = \left<0|e^{-\left[X, Y\right]}e^Y|0\right> = e^{-\left[X, Y\right]}\left<0|e^Y|0\right>.$$

At this point, we can write $e^Y\left|0\right> = D\left(\alpha\right)\left|0\right> = e^{-\frac{1}{2}\left|\alpha\right|^2}e^{\alpha a^\dagger}\left|0\right>$ (which can also be shown using BCH) to get $$\left<\beta |D\left(\alpha\right)|\beta\right> = e^{-\beta\alpha^* + \beta^*\alpha}\left<0|e^{-\frac{1}{2}\left|\alpha\right|^2}e^{\alpha a^\dagger}|0\right> = e^{-\beta\alpha^* + \beta^*\alpha - \frac{1}{2}\left|\alpha\right|^2}$$

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