# Commutation of Vector-Potential and Field-Operator

1. Jan 15, 2014

### Abigale

Hello,

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?

Thank u!
Abby

2. Jan 17, 2014

### Avodyne

What is $\Psi$? If it is a different field than $\vec A$ (such as a Dirac field or a scalar field), then it commutes with $\vec A$ by the usual quantization rules.

3. Jan 18, 2014

### Abigale

So this field-operator $\Psi ^{\dagger}$ creates a particle located at point $\vec{x}$:

$$\Psi ^{\dagger} (\vec {x}) :=\sum_{\lambda} <\lambda|\vec{x}> a^{\dagger} _{\lambda}$$

And this Operator $\vec{A(\vec{x})}$ (Vector-Potential) creates a photon with momentum $\hbar \vec K$ an polarisation $\hat{e_{\vec{K},\lambda}}$:

$$\vec{A(\vec{x})} :=\sqrt{\frac{2 \pi \hbar c^2}{r}} \sum_{\vec{K},\lambda} \frac{\hat{e_{\vec{K},\lambda}}}{\sqrt{c |k|} } \cdot \lbrace e^{i\vec{K} \vec{x}} a^{\dagger} _{\vec{K},\lambda} + e^{-i\vec{K} \vec{x}} a _{\vec{K},\lambda} \rbrace$$

So why can I say that both operators commute?

Thank u!!!

4. Jan 18, 2014

### Avodyne

Is the particle created by $\Psi^\dagger$ a photon, or is it some other kind of particle?