Perturbation Theory: Deciphering Missing Lines of Explanation

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Discussion Overview

The discussion revolves around the application of perturbation theory in quantum field theory, specifically focusing on the Heisenberg representation of interacting fields. Participants are trying to understand the derivation of certain expressions related to time derivatives of fields and the use of commutators in this context.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the derivation of a specific expression involving the time derivative of the interacting field and its relation to commutators.
  • Another participant provides an equation that breaks down the time derivative of the free field and attempts to clarify the grouping of terms into a commutator.
  • A third participant requests a derivation of an identity related to the time derivative of the inverse of the unitary operator, indicating its usefulness.
  • One participant suggests taking the time derivative of the identity \(1 = UU^{-1}\) as a method to derive the needed expression.
  • Another participant acknowledges the brevity of the suggested method and expresses gratitude for the clarification.

Areas of Agreement / Disagreement

The discussion contains multiple viewpoints and attempts to clarify expressions, but no consensus is reached on the overall understanding of the derivation or the implications of the discussed identities.

Contextual Notes

Participants reference specific mathematical identities and expressions without fully resolving the underlying assumptions or providing complete derivations, leaving some steps and implications unclear.

vertices
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Again, I am having difficulty deciphering my class notes - in this case there are missing lines of explanation. If we consider a system of particles that approach and interact, the Heisenberg representation of the interacting field is:

[tex]\phi(\vec{x} , t) = U^{-1} (t) \phi_{a} (\vec{x} , t) U(t)[/tex]

(where [tex]\phi_a[/tex] is the free field before the interaction.

Why is it that we can write:

[tex]\frac{\partial}{\partial t} \phi_{a}= \frac{\partial}{\partial t} U \phi U^{-1}=[\frac{\partial}{\partial t} UU^{-1},\phi_{a}]+iU[H,\phi]U^{-1}[/tex]

where the square brackets in the third equality are commutators?

I don't understand where the third expression comes from?

Thanks.
 
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vertices said:
Again, I am having difficulty deciphering my class notes - in this case there are missing lines of explanation. If we consider a system of particles that approach and interact, the Heisenberg representation of the interacting field is:

[tex]\phi(\vec{x} , t) = U^{-1} (t) \phi_{a} (\vec{x} , t) U(t)[/tex]

(where [tex]\phi_a[/tex] is the free field before the interaction.

Why is it that we can write:

[tex]\frac{\partial}{\partial t} \phi_{a}= \frac{\partial}{\partial t} U \phi U^{-1}=[\frac{\partial}{\partial t} UU^{-1},\phi_{a}]+iU[H,\phi]U^{-1}[/tex]

where the square brackets in the third equality are commutators?

I don't understand where the third expression comes from?

Thanks.

[tex] \frac{\partial \phi_a}{\partial t} = <br /> \frac{\partial U}{\partial t} \phi U^{-1} <br /> + U\phi\frac{\partial U^{-1}}{\partial t} <br /> + U\frac{\partial\phi}{\partial t}U^{-1} <br /> ...(*)[/tex]
where the last term of RHS involves the derivative of time with respect to the field [tex]\phi[/tex] whose equation of motion is well known, the Heisenberg's EoM.

For the first two terms of eq(*), note that,
[tex]\frac{\partial U^{-1}}{\partial t} = -U^{-1}\frac{\partial U}{\partial t} U^{-1}[/tex]
then you will see why they can be grouped into
[tex] \left[ <br /> \frac{\partial U}{\partial t}U^{-1} , \phi_a<br /> \right][/tex]
 
ismaili said:
[tex]\frac{\partial U^{-1}}{\partial t} = -U^{-1}\frac{\partial U}{\partial t} U^{-1}[/tex]

I do not recall this identity .. can you provide a brief derivation/proof/justification? It seems quite useful ...
 
Take time derivative of both sides of the equality

[tex]1 = UU^{-1}[/tex]

Eugene.
 
meopemuk said:
Take time derivative of both sides of the equality

[tex]1 = UU^{-1}[/tex]

Eugene.

That'll do it ... and it certainly was brief. :redface: Thanks!
 
Thank you ever so much ismaili - spent ages trying to see this!
 

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