In my lecture notes on second quantization it is written that the creation field operator is given by

$|\Psi\rangle^{\dagger}_s (r) = \frac{1}{\sqrt{V}} \sum_{k} e^{-i k r} \hat{a}^{\dagger}_{ks}$

They then go on to write that this expression can be inverted, to give $\hat{a}^{\dagger}_{k} = \frac{1}{\sqrt{V}}\int{d^3r e^{i k r}}|\Psi\rangle^{\dagger} (r)$

However, I honestly don't see how this is done. I'm afraid this is a bit of a trivial question probably, but to me the inversion is not obvious at all. Could someone assist me in seeing that this is true?

  • 1
    $\begingroup$ You have to use $\int d^3 r e^{i k r } e^{ i k' r} = V \delta_{k+k',0}$. Also, I think you have the wrong sign in one of the two exponents in your equations. $\endgroup$ – Prahar Jan 15 '15 at 18:11
  • $\begingroup$ You are correct, that was indeed wrong. $\endgroup$ – user129412 Jan 15 '15 at 19:30

As Prahar has already mentioned, the sign in the exponential of your second expression is wrong.

Mulplying by $e^{-iqr}$ and integrating over space your first expression you have $$ \int d^3 r e^{-iqr} | \Psi \rangle_s^\dagger(r) = \int d^3 r e^{-iqr} \frac{1}{\sqrt{V}} \sum_k e^{ikr} a_{ks}^\dagger $$ now exchanging integral and sum you obtain $$ \int d^3 r e^{-iqr} | \Psi \rangle_s^\dagger(r) = \frac{1}{\sqrt{V}} \sum_k a_{ks}^\dagger \int d^3r e^{i(k-q)r}. $$ Are you able to conclude from here?

  • $\begingroup$ Yes, I see now. You get a delta function from the integral, picking out the q term in the sum. That does indeed give the desired result. Thank you! $\endgroup$ – user129412 Jan 15 '15 at 19:33

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