I am trying to prove the commutation relations between the creation and annihilation operators in field theory. I was already able to show that [itex][a_k, a_q^\dagger]=i\delta(k-q)[/itex]. I want to show that [itex][a_k^\dagger, a_q^\dagger]=0,[/itex] but I am stuck on the last step. I think my confusion has to do with interpreting the delta function and I was hoping someone can help me out. So far I have shown that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]a_k^\dagger = \int \! d^3\!x \: e^{-ikx}\left[ \sqrt{\frac{E_k}{2}} \phi(x) - i \sqrt{\frac{1}{2E_k}} \pi(x) \right],[/tex]

where [itex]\phi(x)[/itex] and [itex]\pi(x)[/itex] are the field and conjugate momentum operators respectively. Based on this I was able to show that

[tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: d^3x \: e^{-i(kx+qy)} \left[ -i \sqrt{\frac{E_k}{E_q}} [\phi(x), \pi(y)] + i \sqrt{\frac{E_q}{E_k}} [\phi(y),\pi(x)] \right].[/tex]

Using the commutation relation [itex][\phi(x), \pi(y)]=i \delta^3(x-y)[/itex] this becomes

[tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: e^{-i (k+q) y} \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

And I believe the integral over y gives us a delta function. If I have done everything correctly then this should be

[tex][a_k^\dagger, a_q^\dagger]=\delta^3(k+q) \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

Since I already know the commutation relations I know that the final expression should equal zero for any arbitrary value of [itex]k[/itex] and [itex]q[/itex]. But my final expression doesn't really say that for [itex]k=-q[/itex]. I don't really understand where this discrepancy is coming from, though. Is there some way to interpret this delta function that I'm not understanding or have I made a computational error somewhere? Thanks

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# Proving [itex][a_k^\dagger, a_q^\dagger]=0[/itex]

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