# Field operator anti-commutator relation

For the field operators (fermions)

$$\hat{\Psi}^\dagger_\sigma(x) = \dfrac{1}{\sqrt{V}}\sum_k e^{-ikx}~\hat{a}^\dagger_{k,\sigma}$$

$$\hat{\Psi}_\sigma(x) = \dfrac{1}{\sqrt{V}}\sum_k e^{ikx}~\hat{a}_{k,\sigma}$$

I want to prove the following anti-commutator relationship: $$\left\{\hat{\Psi}_{\sigma}(x), \hat{\Psi}^\dagger_{\sigma^\prime}(x^\prime)\right\} = \delta(x-x^\prime)\delta_{\sigma,\sigma^\prime}$$

I have \begin{align}\left\{\hat{\Psi}_{\sigma}(x), \hat{\Psi}^\dagger_{\sigma^\prime}(x^\prime)\right\} &= \dfrac{1}{V}\sum_{k,k^\prime} e^{-ikx}e^{-ik^\prime x^\prime} \{\hat{a}_{k,\sigma}, \hat{a}^\dagger_{k,\sigma}\}\\ &=\dfrac{1}{V}\sum_{k,k^\prime} e^{-ikx}e^{-ik^\prime x^\prime} \delta(k-k^\prime)\delta_{\sigma,\sigma^\prime} \\ &= \dfrac{1}{V}\sum_{k} e^{-ik(x-x^\prime)}\delta_{\sigma,\sigma^\prime}\end{align}

But I don't know how to show $$\dfrac{1}{V}\sum_{k} e^{-ik(x-x^\prime)}=\delta(x-x^\prime)\,.$$ I would be thankful for your help!

• – udrv Oct 27 '16 at 4:22

Part of the problem, may lie in the fact that this is expressed in term of summations instead of integrals over $k$. If you had integrals, you'd end up with $$\int \exp[-ik(x-x')] dk = 2\pi\delta(x-x') ,$$ which we know from the Fourier properties of a Dirac delta function.