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I am working on second quantization of the Dirac field with discrete momentum I was asked to compute the creation/annihilation anticommutator by imposing the anticommutators on $\psi$ i.e.

$$ \{\psi_a(\vec{x}),\psi^{\dagger}_b(\vec{y})\} = \delta^{(3)}(\vec{x}-\vec{y})\delta_{ab}$$

I start with:

$$\psi(\vec{x}) = \sum_{r,\vec{k}} \sqrt{\dfrac{m}{VE_{\vec{k}}}}\bigg[ c_k(\vec{k})\,u_r(\vec{k}) \,e^{-i x \cdot p} + d^{\dagger}_r(\vec{k})\,v_r(\vec{k})\, e^{ix \cdot p} \bigg] $$

Since I am trying to isolate $c$ and $d$ I try to multiply for the expoencial $e^{ix\cdot p'}$ and integrate over the volume, and I will apear an integral like this

$$\iiint_V e^{-i \,x \cdot (p'-p)} \,d^3x$$

Is there any way approximate this? Like doing:

$$\iiint_V e^{-i \,x \cdot (p'-p)} \,d^3x \approx V\delta_{p,\,p'}$$

João Viana
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1 Answers1

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none

Cited from Mahan's book. It may help you.

Yang
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    Hi Fabregas, welcome to Physics SE! As a general rule, both questions and answers should avoid using images of text. Instead, please type whatever answers you'd like to provide, and use MathJax to typeset any mathematical expressions. That way your answers will render properly in all browsers, and will be able to be indexed by search engines. – J. Murray May 03 '20 at 15:53
  • Thank you! It really helped. I noticed that the creation/annihilation operators do not depend on the volume. So I can take V to infinity. Make the delta appear, and convert it back to the discret delta function and make the summation disapear! – João Viana May 03 '20 at 16:08
  • @J.Murray Thank you for your reminder. I won't do that next time :-) – Yang May 03 '20 at 22:49
  • @JoãoViana You're welcome. – Yang May 03 '20 at 22:49