The fermion anti-commutation relations are given as $$\{\psi_{\alpha}({\bf x},t),\psi_{\beta}^{\dagger}{(\bf x'},t)\} = \delta_{\alpha,\beta} \, \delta({\bf x} - {\bf x'}).$$ I am interested in determining $\{\psi_{\alpha}({\bf x},t),{\bar \psi}{(\bf x'},t) \psi({\bf x'},t)\}$. Does $\{\psi_{\alpha}({\bf x},t),{\bar \psi}_{\beta} ({\bf x'},t)\}$ simplify to anything? In general you have $\{\psi_{\alpha}({\bf x},t),(\psi^{\dagger}({\bf x'},t) \, \gamma^0)_{\beta}\}$ which is equal to $$\{\psi_{\alpha}({\bf x},t),\psi^{\dagger}_{\rho}({\bf x'},t) \gamma^0_{\rho\beta}\},$$ with the sum over $\rho$ assumed. In the energy representation, for example, it is straightforward to check that the $\gamma^0_{\rho\beta}$ can be taken outside the anti-commutator, but how do you show this in general (if instead of $\gamma^0$ you had, say, $\gamma^1$ then this is not so obvious since the $\gamma_1$ involves the Pauli matrix $\sigma_1$ and the spinor $\psi$ also involves $\sigma$ so it doesn't look easy to see that it would be true in this case)?

  • $\begingroup$ What kind of indices are the $\alpha,\beta$? Are there any of them missing in $\{\psi_{\alpha}({\bf x},t),{\bar \psi}{(\bf x'},t) \psi({\bf x'},t)\}$? $\endgroup$
    – coconut
    Dec 26, 2016 at 10:20
  • $\begingroup$ the $\alpha, \beta$ are components of the spinor $\psi$. $\endgroup$
    – jim
    Dec 26, 2016 at 13:55
  • $\begingroup$ $\gamma^\mu$ is a number. The (anti-)commutator is taken between operators. A $c$-number may always be taken out of any (anti-)commutator. $\endgroup$
    – Prahar
    Dec 26, 2016 at 15:25

1 Answer 1


As you correctly state: \begin{align} \{\psi_{\alpha}({\bf x},t),{\bar \psi}_{\beta} ({\bf x'},t)\} &= \{\psi_{\alpha}({\bf x},t),(\psi^{\dagger}({\bf x'},t) \, \gamma^0)_{\beta}\} = \{\psi_{\alpha}({\bf x},t),\psi^{\dagger}_{\rho}({\bf x'},t) \gamma^0_{\rho\beta}\} \\ &= \psi_{\alpha}({\bf x},t)\psi^{\dagger}_{\rho}({\bf x'},t) \gamma^0_{\rho\beta} + \psi^{\dagger}_{\rho}({\bf x'},t) \gamma^0_{\rho\beta} \psi_{\alpha}({\bf x},t) \\ &= \gamma^0_{\rho\beta}\left(\psi_{\alpha}({\bf x},t) \psi^{\dagger}_{\rho}({\bf x'},t) + \psi^{\dagger}_{\rho}({\bf x'},t) \psi_{\alpha}({\bf x},t)\right) \\ &= \gamma^0_{\rho\beta} \{\psi_{\alpha}({\bf x},t),\psi^{\dagger}_{\rho}({\bf x'},t) \} \\ &= \gamma^0_{\alpha\beta}\delta(\mathbf{x}-\mathbf{x}') \end{align}

After explicitly writing indices on everything, we are just dealing with products of (Grassman) numbers. $\gamma^0_{\alpha\beta}$ commutes with any other element, so it can be taken out.

The commutation relations between $\psi$ and $\bar{\psi}\psi$ should be expressed as commutators, because $\psi$ is a fermion and $\bar{\psi}\psi$ is a boson. Using the equation above and $\{\psi_\alpha(\mathbf{x},t),\psi_\beta(\mathbf{x}',t)\}=0$ we get \begin{align} [\psi_{\alpha}({\bf x},t),{\bar \psi}{(\bf x'},t) \psi({\bf x'},t)] =& [\psi_{\alpha}({\bf x},t),{\bar \psi}_\beta{(\bf x'},t) \psi_\beta({\bf x'},t)] \\ =& \psi_{\alpha}({\bf x},t){\bar \psi}_\beta{(\bf x'},t) \psi_\beta({\bf x'},t) - {\bar \psi}_\beta{(\bf x'},t)\psi_\beta({\bf x'},t) \psi_{\alpha}({\bf x},t) \\ =& -\{\psi_{\alpha}({\bf x},t),{\bar \psi}_\beta{(\bf x'},t)\} \psi_\beta({\bf x'},t) - {\bar \psi}_\beta{(\bf x'},t)\psi_{\alpha}({\bf x},t) \psi_\beta({\bf x'},t) \\ &- {\bar \psi}_\beta{(\bf x'},t)\psi_\beta({\bf x'},t) \psi_{\alpha}({\bf x},t) \\ =& -\delta_{\alpha\beta} \delta(\mathbf{x}-\mathbf{x}')\psi_\beta({\bf x'},t)- {\bar \psi}_\beta{(\bf x'},t) \{\psi_{\alpha}({\bf x},t),\psi_{\beta} ({\bf x'},t)\} \\ =& -\delta(\mathbf{x}-\mathbf{x}')\psi_\alpha({\bf x'},t) \end{align}


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