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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?

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1 Answer 1

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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