Commutator of field operator with arbitrary functions

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SUMMARY

The discussion centers on the commutation relation in Quantum Field Theory (QFT), specifically the field operator \(\hat{\phi}\) and its conjugate momentum \(\pi\). The established commutation relation is given by \([\phi(x,t),\pi(y,t)] = i\delta(x-y)\). A participant confirms that the commutator of the field operator \(\phi\) or momentum operator \(\pi\) with the exponential function \(e^{i k \cdot x}\) is zero, as this function acts as a scalar and commutes with all operators.

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eudo
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In QFT, the commutation relation for the field operator \hat{\phi} and conjugate momentum is
<br /> [\phi(x,t),\pi(y,t)] = i\delta(x-y)<br />
Maybe this is obvious, but what would the commutator of \phi or \pi and, say, e^{i k\cdot x} be?
 
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It's obviously 0, because ##\exp(\mathrm{i} k \cdot x)## is just a number, which commutes with all operators.
 
Of course... Thanks
 

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