I'm trying to prove that displacement operators are orthonormal in quantum mechanics, e.g.: $$\text{Tr}\{D^{\dagger}(\alpha)D(\beta)\} =\pi \delta^2(\alpha - \beta)$$ I used the completeness property of coherent states to write: $$\text{Tr}\{D^{\dagger}(\alpha)D(\beta)\} = \frac{1}{\pi}\int \langle\xi|D^{\dagger}(\alpha)D(\beta)|\xi\rangle d^2\xi $$ Using the fact that $D^{\dagger}(\alpha)=D(-\alpha)$ and the group property of displacement operators i get: $$\int \langle\xi|D^{\dagger}(\alpha)D(\beta)|\xi\rangle d^2\xi = e^{i\Im\{\alpha\beta^*\}}\int\langle\xi|\xi+\beta-\alpha\rangle d^2\xi$$ Now using the scalar product formula for coherent states I have that: $$\langle\xi|\xi+\beta-\alpha\rangle = e^{-|\xi|^2/2} e^{-|\xi|^2/2 -|\beta-\alpha|^2/2 -\Re\{\xi^*(\beta-\alpha)\}} e^{(\xi+\beta-\alpha)\xi^*}$$ where I expanded the second exponential whose argument was $|\xi+\beta-\alpha|^2$. Using some algebra I get the following expression: $$\langle\xi|\xi+\beta-\alpha\rangle = e^{ -|\beta-\alpha|^2/2 -\xi(\beta-\alpha)^*} $$ If I put this result into the integral I get the following: $$\text{Tr}\{D^{\dagger}(\alpha)D(\beta)\} = \frac{1}{\pi}e^{i\Im\{\alpha\beta^*\}}e^{ -|\beta-\alpha|^2/2}\int e^{ -\xi(\beta-\alpha)^*} d^2\xi $$

Now, I'm stuck here and I cannot really see how to take the delta function out from this integral. Can you give me an hint?

  • $\begingroup$ It seems to me that you should have $\frac12(\xi^*\gamma-\xi\gamma^*)$ in the exponent in the integral (with $\gamma=\beta-\alpha$). $\endgroup$
    – Adam
    Commented May 24, 2018 at 9:10

2 Answers 2


Let us first correct a little mistake I spotted and then finish the proof. Recheck the following part: $$\langle \xi | \xi \beta - \alpha \rangle = e^{-|\beta-\alpha|^2/2 - i\Im(\xi(\beta-\alpha)^*)}$$ After that everything goes through. You can split $\xi = x + iy$ with $x,y\in \mathbb{R}$ and $(\beta-\alpha)^* = a - ib$ with $a=\Re( \beta-\alpha)$ and $b=\Im( \beta-\alpha)$. Then the integral becomes $$\sim \int e^{-i\Im((x+iy)(a-ib))}dxdy=\int e^{-iya+ixb}=4\pi^2\delta(a)\delta(b)=4\pi^2\delta^2(\beta-\alpha)$$ (I believe the factors of $2\pi$ are correct but check them too.) That should solve your issues.

  • $\begingroup$ Thank you so much for finding my mistake! I have checked my computations different times without noticing that I forgot to put a 1/2 factor when expanding $\Re\{\xi^*(\beta-\alpha)\}$ $\endgroup$
    – steg
    Commented May 24, 2018 at 10:07

You can prove it by putting together a few different statements:

  1. $\boxed{D(\alpha) = \exp(\alpha a^\dagger-\bar\alpha a) = \exp(\alpha a^\dagger)\exp(-\bar\alpha a) e^{-\frac12|\alpha|^2} = \exp(-\bar\alpha a)\exp(\alpha a^\dagger) e^{\frac12|\alpha|^2}}$

  2. $\boxed{\operatorname{tr}(D(\alpha))=\pi \delta^2(\alpha)}$

    One way to show this is expanding the trace with coherent states and exploiting the normally ordered expression for $D(\alpha)$, that is: $$\operatorname{tr}(D(\alpha)) = \int\frac{d^2\beta}{\pi} \langle\beta|D(\alpha)|\beta\rangle = \int\frac{d^2\beta}{\pi} e^{-\frac12|\alpha|^2} e^{\alpha\bar\beta-\bar\alpha \beta} = \pi \delta^2(\beta),$$

  3. $\boxed{D(\alpha)D(\beta) = D(\alpha+\beta) e^{\frac12(\alpha\bar\beta - \bar\alpha\beta)}}$

    This is a simple matter of merging together a product of exponentials via $e^A e^B = e^{A+B +\frac12 [A,B]}$ (which holds whenever $[A,B]$ commutes with $A$ and $B$, as is the case here).

It follows that $$\operatorname{tr}(D^\dagger(\alpha)D(\beta)) = \operatorname{tr}(D(-\alpha)D(\beta)) \\= e^{-\frac12(\alpha\bar\beta-\bar\alpha\beta)} \operatorname{tr}(D(\beta-\alpha)) = \pi \delta^2(\beta-\alpha).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.