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I'm currently trying to calculate the expectation value

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

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

\begin{equation} \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\}. \end{equation}

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

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

whilst the creation and annihilation operators satisfy the following anti-commutation relation \begin{equation} \{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}, \end{equation}

and all other anti-commutators are zero.

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

\begin{equation} \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}). \end{equation}

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.

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Expectation values in QFT need to be normal ordered in general. So the divergencefree expression you should compute is \begin{equation} \langle \psi(p,s)| :\bar{\psi}(p,s)\Gamma_{\rho} \psi(x) : | \psi(p,s)\rangle. \end{equation}

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.

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  • $\begingroup$ Great, thank you very much. I was unaware that all expectation values needed to be normal ordered. $\endgroup$ – nuLab Jun 4 at 16:00

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