# 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
Commented Oct 27, 2016 at 4:22

If the index $$k$$ is meant to be continuous, then you can replace $$\frac{1}{V}\sum_k\to \int dk$$ and your equation follows, up to factors of $$2\pi$$, from the fact that

$$\int dk \ e^{ik(x-x')}=2\pi \delta(x-x').$$

On the other hand, if the index $$k$$ is really discrete, the equation follows from the completeness relation obeyed by the basis eigenstates whose position space wavefunctions are the exponentials $$\frac{1}{\sqrt{V}}e^{ik_nx}$$. This would be the discrete analogue of the above, namely, the completeness relation for the Fourier series instead of Fourier transform.

To establish that, recall that a basis $$|n\rangle$$ must obey completeness:

$$\sum_n |n\rangle \langle n|=\mathbf{1}.$$

This is the equation you are looking for. You have a set of states $$|n\rangle$$ such that

$$\langle x|n\rangle=\dfrac{1}{\sqrt{V}}e^{ik_nx}$$

In that case we have

$$\mathbf{1}=\sum_n |n\rangle \langle n|=\sum_n \int dx dx' |x\rangle \langle x|n\rangle \langle n|x'\rangle \langle x'|= \int dxdx' |x\rangle\langle x'|\frac{1}{V}\sum_n e^{ik_nx}e^{-ik_nx'}$$

But then recall that in the position basis the matrix elements of $$\mathbf{1}$$ are just $$\delta(x-x')$$. Therefore we must have

$$\frac{1}{V}\sum_n e^{ik_nx}e^{-ik_nx'}=\delta(x-x')$$

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.