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Im doing this in the context of field operators $$\psi(x)=\sum_k a_k e^{ikx},$$ $$\psi^T(y)=\sum_k a_k^T e^{-iky},$$ and their being defined as the fourier transform of the creation/annihilation operators $a_k,a_k^T$. Specifically, i have to prove: $$[\psi(x),\psi^T(y)]_\zeta=\delta(x-y)$$ With $\zeta=+1$ for bosons and $=-1$ for fermions. We have already proved in class that $$[a_k,a_k'^T]=\delta(k-k').$$ For whatever reason Im really struggling with the first proof. Any tips? And then I was jsut wondering whether it would be simpler to prove the the commutation relations hold for all fourier transforms of operators, not just this case. But im not sure how to do that either, or whether it would actually be simpler, or even if its actually true..

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    $\begingroup$ Just plug in and reduce one integral,... But maybe your momentum sums should be integrals if you are using a $delta$-function for momenta, instead of a Kroncecker delta. $\endgroup$ – Cosmas Zachos Apr 16 at 19:28
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A plug in will work. The operation is very common in quantum field theory. Please refer to any QFT textbook for canonical quantization techniques. $[\psi(x),\psi^{\dagger}(y)]_{\xi} = [\sum_k a_{k}e^{ikx}, \sum_k' a^{\dagger}_{k'}e^{-ik'y}]_{\xi} = \sum_{kk'} e^{ikx-ik'y} [a_k, a^{\dagger}_{k'}]_{\xi} = \sum_{kk'} e^{ikx-ik'y} \delta_{kk'} = \sum_k e^{ik(x-y)} = \delta(x-y)$

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  • $\begingroup$ okay, thanks!! using the linearity of the commutator is obvious now i think about it..but i dont understand the second to last equality? how did you get rid of the delta function? $\endgroup$ – Learn4life Apr 17 at 8:23
  • $\begingroup$ @Learn4life Due to the Kronecker delta, the terms in that summation are all zero except for when $k=k'$. Therefore only terms with $k=k'$ remain and for those terms the Kronecker delta is one and can be omitted. $\endgroup$ – Crimson Apr 26 at 21:17
  • $\begingroup$ Note that the last equation is only valid when the k-values of the summation are infinitessimally close together. $\endgroup$ – Crimson Apr 26 at 21:20

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