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Commutator of kinetic energy and potential energy

  1. Sep 3, 2009 #1
    Hi,

    I am working with the Dirac picture in the second quantification. An operator in this picture is defined as (where some constants are 1)

    [tex]O_I=e^{iH_0t}Oe^{-iH_0t}[/tex].

    Now, it is evident that the hamiltonian [tex]H_0 = T + V[/tex] is the same in Heisenberg or Dirac picture since the exponential [tex]e^{iH_0t}[/tex] commute with the free hamiltonian. This mean the free hamiltonian will be the same, no matter you write it in terms of the operators [tex]\psi_I(x,t)[/tex] in the interaction picture or in terms of the same operators [tex]\psi(x)[/tex] in the Heisenberg picture (neglecting spins):

    [tex]H_0 = \int dx_1\psi^\dag(x_1)T(x_1)\psi(x_1)+\frac{1}{2}\int dx_1dx_2\psi^\dag(x_1)\psi^\dag(x_2)V(x_1,x_2)\psi(x_2)\psi(x_1)[/tex]

    Now I am trying to retreive the same expression with operators in Dirac picture. If I transform this hamiltonian, I have still [tex]H_0[/tex] on the left, and some new exponentials on the right :

    [tex]H_0 = \int dx_1e^{iH_0t}\psi^\dag(x_1)T(x_1)\psi(x_1)e^{-iH_0t}+\frac{1}{2}\int dx_1dx_2e^{iH_0t}\psi^\dag(x_1)\psi^\dag(x_2)V(x_1,x_2)\psi(x_2)\psi(x_1)e^{-iH_0t}[/tex]

    In the first term, we see the operators in Dirac picture. It's in the second term that I'm not sur what to do, precisely with the operators with x labeled by 2. I can insert identities [tex]e^{iH_0t}e^{-iH_0t}[/tex], but I will impose equal-time [tex]\psi[/tex] operators that have different position arguments. Is it the right thing to do? If yes, I will have V in the interaction picture and I think this is not normal (it could be normal if V commute with H_0, i.e. T commute with V : the question refering to the title of this post!).

    Any reference or help on these precise questions will be appreciated.

    Thanks,

    TP
     
  2. jcsd
  3. Sep 3, 2009 #2

    Avodyne

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    Science Advisor

    Yes. I don't understand why you think this is a problem.
     
  4. Sep 4, 2009 #3
    I think I was wondering about relativity. I should normally introduce as many time variable as space coordinate variables. But maybe your right, since I work with an explicitly non-relativistic hamiltonian.

    Maybe you can see more clearly my concerns knowing I'm trying to obtain the movement equation for the green function (Dyson's eqs in differential form) which, in my references, include as many time variable as coordinate variable.

    Cheers,

    TP
     
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