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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

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# Commutator of kinetic energy and potential energy

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