# Unexpected divergence in expectation value

I'm currently trying to calculate the expectation value

$$$$\langle\psi(p,s)|\bar{\psi}(x)\Gamma_\rho \psi(x)|\psi(p,s)\rangle,$$$$

where $$\Gamma_\rho$$ is understood to be some unspecified string of gamma matrices and $$\psi(x)$$ denotes the fermionic field operator

$$$$\psi(x)=\int\frac{d^3\vec{p}}{(2\pi)^3} \frac{1}{\sqrt{2 E_{\vec{p}}}} \sum_{s=\pm \frac{1}{2}}\bigg\{a_{\vec{p},s}u(p,s) e^{ip\cdot x} + b^{\dagger}_{\vec{p},s}v(p,s) e^{-ip\cdot x}\bigg\}.$$$$

Here $$a_{\vec{p},s}$$ and $$b^{\dagger}_{\vec{p},s}$$ are the annihilation and creation operators for particle $$\psi$$ and antiparticle $$\bar{\psi}$$ respectively, whilst $$u(p,s)$$ and $$v(p,s)$$ are the particle and antiparticle Dirac spinors. The single particle state with momentum $$p$$ and helicity $$s$$ is defined as

$$$$|\psi(p,s)\rangle =\sqrt{2E_{\vec{p}}}a^{\dagger}_{\vec{p},s}|0\rangle,$$$$

whilst the creation and annihilation operators satisfy the following anti-commutation relation $$$$\{a_{\vec{p},r},a^{\dagger}_{\vec{q},s}\}= \{b_{\vec{p},r},b^{\dagger}_{\vec{q},s}\}=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q})\delta_{rs},$$$$

and all other anti-commutators are zero.

After working through and simplifying as much as possible, I arrive at the following answer

$$$$\langle\psi(p,s)|\bar{\psi}(x)\Gamma_\rho \psi(x)|\psi(p,s)\rangle = \bar{u}(p,s)\Gamma_\rho u(p,s) + \int d^3 \vec{q} \bigg(\frac{E_{\vec{p}}}{E_{\vec{q}}}\bigg) \sum_{r=\pm\frac{1}{2}} \bigg\{\bar{v}(q,r)\Gamma_\rho v(q,r)\bigg\} \delta^{(3)}(\vec{0}).$$$$

Now, naively I would expect that since the external states are all particles (rather than antiparticle) that I would have no $$v(p,s)$$ spinors in my final answer and something that looks like the first term only. Additionally, there is a divergence due to the presence of the $$\delta^{(3)}(\vec{0})$$ term.

How do I treat this extra term? Does it vanish or have I made a mistake and it shouldn't be there at all? I know that a similarly divergent term arises in the free Dirac Hamiltonian, and it is dealt with using normal ordering. I suspect that there is something akin to that going on here.

Expectation values in QFT need to be normal ordered in general. So the divergencefree expression you should compute is $$$$\langle \psi(p,s)| :\bar{\psi}(p,s)\Gamma_{\rho} \psi(x) : | \psi(p,s)\rangle.$$$$
The $$a$$'s and $$a^{\dagger}$$'s in your expression are already in order, but the $$b$$'s and $$b^{\dagger}$$'s are not, which is what leads to the divergence.