Commutator of kinetic energy and potential energy

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SUMMARY

The discussion focuses on the transformation of the Hamiltonian \( H_0 = T + V \) in the Dirac picture of quantum mechanics, specifically within the context of second quantization. The user seeks to understand how to manipulate operators in the Dirac picture, particularly when dealing with the potential energy term \( V \) and its relation to the kinetic energy operator \( T \). The conversation highlights the importance of equal-time operators and the implications of non-relativistic Hamiltonians on the treatment of these operators.

PREREQUISITES
  • Understanding of Dirac picture in quantum mechanics
  • Familiarity with second quantization techniques
  • Knowledge of Hamiltonian mechanics, specifically kinetic and potential energy operators
  • Experience with operator algebra in quantum field theory
NEXT STEPS
  • Study the implications of equal-time operator commutation relations in quantum mechanics
  • Explore Dyson's equations and their application in non-relativistic quantum field theory
  • Research the role of time variables in the formulation of quantum mechanical equations
  • Examine the conditions under which potential energy operators commute with kinetic energy operators
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This discussion is beneficial for theoretical physicists, quantum mechanics students, and researchers working on quantum field theory, particularly those interested in the Dirac picture and operator transformations.

Tipi
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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)

O_I=e^{iH_0t}Oe^{-iH_0t}.

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

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)

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

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}

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 e^{iH_0t}e^{-iH_0t}, but I will impose equal-time \psi 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 referring to the title of this post!).

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

Thanks,

TP
 
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Tipi said:
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 e^{iH_0t}e^{-iH_0t}, but I will impose equal-time \psi operators that have different position arguments. Is it the right thing to do ?
Yes. I don't understand why you think this is a problem.
 
Avodyne said:
Yes. I don't understand why you think this is a problem.
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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