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Commutation of Vector-Potential and Field-Operator

  1. Jan 15, 2014 #1

    I am thinking for some hours about the commutation of the field-Operator/(annihilation-Operator): [itex]\Psi[/itex] and the vector-potential: [itex]\vec{A(\vec{r})}[/itex].

    I have noticed in my lecture notes that [itex]\vec{A(\vec{r})}\Psi = \Psi\vec{A(\vec{r})}[/itex].

    But I don't understand why they commute?

    Thank u!
  2. jcsd
  3. Jan 17, 2014 #2


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    What is [itex]\Psi[/itex]? If it is a different field than [itex]\vec A[/itex] (such as a Dirac field or a scalar field), then it commutes with [itex]\vec A[/itex] by the usual quantization rules.
  4. Jan 18, 2014 #3
    So this field-operator [itex]\Psi ^{\dagger}[/itex] creates a particle located at point [itex]\vec{x}[/itex]:

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

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

    $$\vec{A(\vec{x})} :=\sqrt{\frac{2 \pi \hbar c^2}{r}}


    \frac{\hat{e_{\vec{K},\lambda}}}{\sqrt{c |k|} }
    \lbrace e^{i\vec{K} \vec{x}} a^{\dagger} _{\vec{K},\lambda}
    e^{-i\vec{K} \vec{x}} a _{\vec{K},\lambda}

    So why can I say that both operators commute?

    Thank u!!!
  5. Jan 18, 2014 #4


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