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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?

nightmarish
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1 Answers1

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

Sanya
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  • Can you show that $ =-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{\dagger}=[\phi(t,\vec{x}),\pi(t,\vec{y})]$? – nightmarish Sep 18 '16 at 22:51
  • 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. – Sanya Sep 18 '16 at 23:21
  • 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? – nightmarish Sep 18 '16 at 23:49
  • Why is $[\phi(t,\vec{x}),\pi(t,\vec{y})]=[\phi^{+}(t,\vec{x}),\pi^{+}(t,\vec{y})]$? – Sanya Sep 18 '16 at 23:55
  • Because both of them are equal to $i\delta^{3}(\vec{x}-\vec{y})$. – nightmarish Sep 19 '16 at 00:01
  • As $[\phi^{}(t,\vec{x}),\pi^{}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$, complex conjugation yields $[\phi^{+}(t,\vec{x}),\pi^{+}(t,\vec{y})]=-i\delta^{(3)}(\vec{x}-\vec{y})$ – Sanya Sep 19 '16 at 00:04
  • I don't think so! – nightmarish Sep 19 '16 at 00:51
  • @failexam I updated my answer. – Sanya Sep 19 '16 at 21:06
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    Comment to the answer (v2): Consider to mention for clarity what the plus and star notations mean. – Qmechanic Sep 19 '16 at 21:57
  • @Qmechanic sometimes, I get confused myself ... – Sanya Sep 20 '16 at 14:12
  • @failexam I am sorry about the confusion. Please tell me what the difference between your stars and daggers/pluses is... – Sanya Sep 20 '16 at 14:13
  • @failexam in what I have written so far, * was conjugate transpose and + simple transposition (restricted to the cases where that actually makes sense) – Sanya Sep 20 '16 at 14:17