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 $\pi$).


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

  • $\begingroup$ right but I thought the OP asked for the other direction,i.e. how to construct the resolution of identity given an overcomplete set of states. $\endgroup$ Jul 15, 2020 at 22:50
  • 1
    $\begingroup$ Given the overcomplete basis of coherent states, the above answer shows exactly what the resolution of the identity is. Both the directions are proved by this calculation. However, if the question is about a general overcomplete basis, and a general prescription for the resolution of identity is wanted, I'm afraid I don't know the answer. $\endgroup$
    – Arkya
    Jul 16, 2020 at 0:03
  • 1
    $\begingroup$ I just didn’t understand the question as equivalent to showing the example of the coherent did resolve the identity, more like given an overcomplete set how to construct the identity. Your answer is correct of course, just not how I understood the question. $\endgroup$ Jul 16, 2020 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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