Normalization of overcomplete coherent states as basis

Question

In a complete orthonormal basis $|x\rangle$, we often use the completeness relation:

$$\sum_{n=0}^\infty | x \rangle \langle x | = \mathbb{I}$$

if the basis is continuous we use the natural extension

$$\int | x \rangle \langle x | dx = \mathbb{I}.$$

This makes sense only if the choice of basis is complete. What if it is overcomplete? As an example, consider the overcomplete basis of coherent states. How does one construct the identity from these? I have seen the normalization

$$\frac{1}{\pi}\int | \alpha \rangle \langle \alpha | d^2\alpha = \mathbb{I}$$

(the $d^2$ implies integration over the real and imaginary parts of $\alpha$ separately.)

How does one derive the $\frac{1}{\pi}$ factor?

I thought you could get it by computing $$Tr(\int | \alpha \rangle \langle \alpha | d^2\alpha) = \int Tr(| \alpha \rangle \langle \alpha |) d^2\alpha = \int \langle \alpha | \alpha \rangle d^2\alpha,$$

but this of course fails since it becomes infinite (I was hoping to get $\pi$).

Answer 1

You can use the expansion of a coherent state in terms of the SHO states:

$$|\alpha\rangle=e^{-|\alpha|^2/2}\sum_l {\alpha^l\over \sqrt{l!}}|l\rangle$$

Now, we can evaluate the left hand side of the identity in question in between two SHO states, say $\langle m|$ and $|n\rangle$, as follows

\begin{align} \int \langle m|\alpha\rangle\langle \alpha|n\rangle d^2 \alpha &= {1\over \sqrt{m!n!}} \int d^2\alpha \quad \alpha^m \bar{\alpha}^n e^{-|\alpha|^2}\\ &= {1\over \sqrt{m!n!}} {\partial\over\partial J^m} {\partial\over\partial \bar{J}^n} \left(\int d^2\alpha \quad e^{-|\alpha|^2+J\alpha+\bar{J}\bar{\alpha}}\right)\bigg|_{J=\bar{J}=0} \\ &= {1\over \sqrt{m!n!}} {\partial\over\partial J^m} {\partial\over\partial \bar{J}^n} \left( \int d^2\alpha \quad e^{-|\alpha-J|^2}e^{J\bar{J}}\right) \bigg|_{J=\bar{J}=0}\\ &= {1\over \sqrt{m!n!}} (\sqrt{\pi})^2 {\partial\over\partial J^m} {\partial\over\partial \bar{J}^n}e^{J\bar{J}} \bigg|_{J=\bar{J}=0}\\ &= {\pi\over \sqrt{m!n!}} {\partial\over\partial J^m} {\partial\over\partial \bar{J}^n} \sum_{r} {(J \bar{J})^r \over r!}\\ &= {\pi\over \sqrt{m!n!}} \delta_{mn} n!\\ &= \pi \end{align} where I have made use of the usual gaussian integral $\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$. Thus, I have shown above that $$\langle m|\left({1\over\pi} \int |\alpha\rangle\langle \alpha| d^2 \alpha \right)|n\rangle =\delta_{mn}$$ or, equivalently, $${1\over\pi} \int |\alpha\rangle\langle \alpha| d^2 \alpha = \mathbb{I} $$