I am reading "Introductory to Quantum Optics" by Christopher C. Gerry and Peter L. Knight but I don't understand a solution from which you can obtain the matrix elements of an operator in the number basis if you know the diagonal coherent-state matrix elements of that operator.

page 56:

The diagonal elements of an operator $\hat{F}$ in a coherent state basis completely determine the operator. From Eqs. (3.76) and (3.77) we have

$$ \langle\alpha|\hat{F}|\alpha\rangle e^{\alpha^*\alpha} = \sum_n\sum_m \frac{\alpha^{*m}\alpha^n}{\sqrt{m!n!}} \langle m |\hat{F} | n \rangle$$ Treating $\alpha$ and $\alpha^*$ as independent variables it is apparent that

$$ \frac{1}{\sqrt{m!n!}} \left. \left[ \frac{\partial^{n+m} \left( \langle \alpha | \hat{F} | \alpha \rangle e^{\alpha^*\alpha} \right)}{\partial\alpha^{*m}\partial\alpha^n }\right] \right|_{\alpha^*=0 \\ {\alpha=0}} = \langle m | \hat{F} | n \rangle $$

I am a bit confused right now. In the first equation, $n$ and $m$ are just indices but when I take the derivatives in the second equation into account, $n$ and $m$ are both outside of the sum and that doesn't really make sense to me.

$$ \frac{1}{\sqrt{m!n!}} \left. \left[ \frac{\partial^{n+m} \left( \langle \alpha | \hat{F} | \alpha \rangle e^{\alpha^*\alpha} \right)}{\partial\alpha^{*m}\partial\alpha^n }\right] \right|_{\alpha^*=0 \\ {\alpha=0}} = \frac{1}{\sqrt{m!n!}} \left. \left[ \frac{\partial^{n+m}}{\partial\alpha^{*m}\partial\alpha^n} \sum_n\sum_m \frac{\alpha^{*m}\alpha^n}{\sqrt{m!n!}} \langle m |\hat{F} | n \rangle \right] \right|_{\alpha^*=0 \\ {\alpha=0}}$$

As I understood it, $n$ and $m$ only have a real meaning inside of the sums and I don't really know how to apply the derivatives to prove the solution.

  • $\begingroup$ Are $|n\rangle$ and $|m\rangle$ referring to the harmonic-oscillator eigenstates (the "number of photons" basis)? $\endgroup$ Jun 22, 2018 at 20:44
  • $\begingroup$ Yes, they are the harmonic-oscillator eigenstates. $\endgroup$
    – user164828
    Jun 22, 2018 at 20:49
  • $\begingroup$ Then $n$ and $m$ are the number of photons in the states that you're trying to determine the operator's effect on. All this says is that the number of photons in the state is related to the number of derivatives on the left side. $\endgroup$ Jun 22, 2018 at 20:57
  • 2
    $\begingroup$ As a comment as written your reference to "Introductory Quantum Optics" is not useful unless you specify at the least the authors. $\endgroup$ Jun 23, 2018 at 12:47

1 Answer 1


Your confusion is due to a slight abuse of notation, but is fairly common practice. One should not treat $m$ and $n$ to have the same meaning in the first and second equation within your quote. In the first equation, $m,n$ are dummy variables, in the second they are specific choices.

Let us rename the $m,n$ in the second equation to $m',n'$. Now, when performing the derivative, you will pull out precisely the piece of in the sum of the first equation where $m=m',n=n'$, arriving at the desired result. To get this, you also assume $\langle m | \hat{F} |n \rangle$ is independent of $\alpha, \alpha^*$.


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