# Answer to Question #38473 in Quantum Mechanics for sebastian hoffmann

Question #38473

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?

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?

Expert's answer

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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