# Prove orthonormality of displacement operators: ${\rm Tr}[D^\dagger(\alpha)D(\beta)]=\pi\delta^2(\alpha-\beta)$

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?

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

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.

• 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 $\Re\{\xi^*(\beta-\alpha)\}$
– 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 $$\operatorname{tr}(D^\dagger(\alpha)D(\beta)) = \operatorname{tr}(D(-\alpha)D(\beta)) \\= e^{-\frac12(\alpha\bar\beta-\bar\alpha\beta)} \operatorname{tr}(D(\beta-\alpha)) = \pi \delta^2(\beta-\alpha).$$