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.


1 Answer 1


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.

  • $\begingroup$ Great, thank you very much. I was unaware that all expectation values needed to be normal ordered. $\endgroup$
    – nuLab
    Jun 4, 2020 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.