Commutator of kinetic energy and potential energy

In summary, the Dirac operator in the second quantification is the same as the Heisenberg operator. However, the free hamiltonian will be different depending on whether you write it in terms of the operators in the interaction picture or in terms of the same operators in the Heisenberg picture (neglecting spins).
  • #1
Tipi
48
0
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 referring to the title of this post!).

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

Thanks,

TP
 
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  • #2
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 [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 ?
Yes. I don't understand why you think this is a problem.
 
  • #3
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
 

What is a commutator?

A commutator is a mathematical operation that determines how two quantities, such as kinetic energy and potential energy, interact with each other. It is denoted by [A, B] and is defined as the difference between the products AB and BA.

Why is the commutator of kinetic energy and potential energy important?

The commutator of kinetic energy and potential energy is important because it allows us to understand the relationship between these two types of energy in a physical system. It helps us to analyze the dynamics of the system and calculate important quantities, such as the total energy and the rate of change of energy.

What does a non-zero commutator between kinetic energy and potential energy indicate?

A non-zero commutator between kinetic energy and potential energy indicates that these two forms of energy do not commute, meaning that they do not have a well-defined simultaneous measurement. This can have significant implications in quantum mechanics, where the commutator plays a crucial role in determining the uncertainty in measurements.

How does the commutator of kinetic energy and potential energy relate to Heisenberg's uncertainty principle?

Heisenberg's uncertainty principle states that it is impossible to know the exact values of certain pairs of physical quantities, such as position and momentum, simultaneously. The commutator of kinetic energy and potential energy is related to this principle as it represents the uncertainty in the simultaneous measurement of these two quantities in a quantum mechanical system.

Can the commutator of kinetic energy and potential energy be used to find the total energy of a system?

Yes, the commutator of kinetic energy and potential energy can be used to find the total energy of a system. This is because the total energy of a system is equal to the sum of the kinetic energy and potential energy, and the commutator allows us to calculate the rate of change of energy, which is related to the total energy of the system.

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