# Algebra of the complex scalar field - complex conjugate of commutation relations

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?

• The equation after "Since" is not true. Remember that $(AB)^\dagger = B^\dagger A^\dagger$. Commented Sep 18, 2016 at 21:30
• Related post by OP: physics.stackexchange.com/q/280928/2451 Commented Sep 18, 2016 at 22:17

$$[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]$$ $$= \phi^{*}(t,\vec{x}) \pi^{*}(t,\vec{y}) - \pi^{*}(t,\vec{y}) \phi^{*}(t,\vec{x})$$ $$=\left( \pi(t,\vec{y}) \phi(t,\vec{x}) \right)^{*}-\left( \phi(t,\vec{x}) \pi(t,\vec{y}) \right)^{*}$$ $$=[\pi(t,\vec{y}),\phi(t,\vec{x})]^{*}$$ $$=-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{*}$$
• Can you show that $=-[\phi(t,\vec{x}),\pi(t,\vec{y})]^{\dagger}=[\phi(t,\vec{x}),\pi(t,\vec{y})]$? Commented Sep 18, 2016 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. Commented Sep 18, 2016 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? Commented Sep 18, 2016 at 23:49
• Why is $[\phi(t,\vec{x}),\pi(t,\vec{y})]=[\phi^{+}(t,\vec{x}),\pi^{+}(t,\vec{y})]$? Commented Sep 18, 2016 at 23:55