Answer to Question #38473 in Quantum Mechanics for sebastian hoffmann

Question

Hi,

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?

Answer 1

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).