Commutation of Vector-Potential and Field-Operator

[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

 

[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.

 

[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!

 

[Avodyne]

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