0
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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 '16 at 10:20
  • $\begingroup$ the $\alpha, \beta$ are components of the spinor $\psi$. $\endgroup$ – jim Dec 26 '16 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 '16 at 15:25
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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