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Commutation of time derrivative

  1. Feb 1, 2014 #1
    Hi I regard,

    $$[\partial_t \Psi, \Psi]=0$$
    but [itex]\Psi[/itex] is a field-operator.

    I don't understand why the commutation of the derrivative of the operator [itex]\Psi[/itex] by itself should be zero?
    THX
     
  2. jcsd
  3. Feb 1, 2014 #2

    WannabeNewton

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    Could you provide a little more context? The commutator shouldn't vanish in general. For example if ##\Psi## is a (free) real Klein Gordon field then from the canonical commutation relations we have ##[\Psi(x,t), \partial_t \Psi (x',t)] = i\delta^{3}(x - x')## so clearly they don't commute in this case.
     
    Last edited: Feb 1, 2014
  4. Feb 1, 2014 #3
    I regard a Klein-Gordon-field. But for the Klein-Gordon-Field, the commutation-relation is $$ [\Psi^\dagger (x,t), \partial_t \Psi (x',t)] = -i\hbar\delta(r-r')$$



    I even know that: $$ [\Psi^\dagger(x,t), \Psi (x',t)] = 0$$
    and $$ [\partial_t\Psi^\dagger(x,t), \partial_t \Psi (x',t)] = 0$$

    and for sure $$ [\partial_t\Psi^\dagger(x,t), \Psi (x',t)] =-i\hbar\delta(r-r') $$



    Maybe with the formula:
    $$[A,f(B)]=i \frac{d}{dB} f(B)$$ ???
     
    Last edited: Feb 1, 2014
  5. Feb 1, 2014 #4

    WannabeNewton

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    I'm not sure what you're asking. If you're talking about complex Klein Gordon fields as opposed to real Klein Gordon fields then ##[\varphi, \pi^{\dagger}] = [\varphi, \partial_t \varphi] = 0## is a trivial statement of second quantization; ##\varphi## and ##\varphi^{\dagger}## are regarded as independent fields and ##\pi^{\dagger} = \partial_t \varphi## is the conjugate momentum of ##\varphi^{\dagger}## not that of ##\varphi##.
     
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