2
$\begingroup$

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:

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

$\endgroup$
0

1 Answer 1

1
$\begingroup$

none

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

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$
    – J. Murray
    May 3, 2020 at 15:53
  • $\begingroup$ 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! $\endgroup$ May 3, 2020 at 16:08
  • $\begingroup$ @J.Murray Thank you for your reminder. I won't do that next time :-) $\endgroup$
    – Fabregas
    May 3, 2020 at 22:49
  • $\begingroup$ @JoãoViana You're welcome. $\endgroup$
    – Fabregas
    May 3, 2020 at 22:49

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.