I am thinking for some hours about the commutation of the field-Operator/(annihilation-Operator): $$\Psi$$ and the vector-potential: $$\vec{A(\vec{r})}$$.
I have noticed in my lecture notes that $$\vec{A(\vec{r})}\Psi = \Psi\vec{A(\vec{r})}$$.
But I don't understand why they commute?
1
Expert's answer
2014-01-21T10:49:55-0500
Vector-potential is not a differential operator. Thus, multiplication of vector and scalar values commutes as in linear algebra (commutative property of multiplication of vector and scalar).
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