The canonical commutation relations of the complex scalar field $\phi$ are given by

$$[\phi(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$ $$[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$

Since $[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=[\phi(t,\vec{x}),\pi(t,\vec{y})]^{*}$, the second commutation relation ought to include a minus sign.

Can someone resolve this apparent contradiction?


1 Answer 1


$$[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]$$ $$= \phi^{*}(t,\vec{x}) \pi^{*}(t,\vec{y}) - \pi^{*}(t,\vec{y}) \phi^{*}(t,\vec{x}) $$ $$ =\left( \pi(t,\vec{y}) \phi(t,\vec{x}) \right)^{*}-\left( \phi(t,\vec{x}) \pi(t,\vec{y}) \right)^{*} $$ $$ =[\pi(t,\vec{y}),\phi(t,\vec{x})]^{*}$$ $$ =-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{*}$$

  • $\begingroup$ Can you show that $ =-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{\dagger}=[\phi(t,\vec{x}),\pi(t,\vec{y})]$? $\endgroup$ Sep 18, 2016 at 22:51
  • $\begingroup$ No, because I do not believe that to be true. What I can show is $[\phi^{+}(t,\vec{x}),\pi^{+}(t,\vec{y})]=-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{+}$ - that follows by the same argument as above. Complex conjugation does not make a difference. $\endgroup$
    – Sanya
    Sep 18, 2016 at 23:21
  • $\begingroup$ Well, $[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]=[\phi(\vec{x}),\pi(\vec{y})]=i\delta^{3}(\vec{x}-\vec{y})$ and $[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]=-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{\dagger}$ seem to imply that $-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{\dagger}=[\phi(\vec{x}),\pi(\vec{y})]$!!! Doesn't it? $\endgroup$ Sep 18, 2016 at 23:49
  • $\begingroup$ Why is $[\phi(t,\vec{x}),\pi(t,\vec{y})]=[\phi^{+}(t,\vec{x}),\pi^{+}(t,\vec{y})]$? $\endgroup$
    – Sanya
    Sep 18, 2016 at 23:55
  • 1
    $\begingroup$ Comment to the answer (v2): Consider to mention for clarity what the plus and star notations mean. $\endgroup$
    – Qmechanic
    Sep 19, 2016 at 21:57

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.