Perturbation Theory: Deciphering Missing Lines of Explanation

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SUMMARY

This discussion focuses on the application of perturbation theory in quantum mechanics, specifically regarding the Heisenberg representation of interacting fields. The equation presented, \phi(\vec{x}, t) = U^{-1}(t) \phi_{a}(\vec{x}, t) U(t), illustrates the transformation of the free field \phi_a under the unitary operator U(t). The participants clarify the derivation of the time derivative expression \frac{\partial}{\partial t} \phi_{a} = \frac{\partial U}{\partial t} \phi U^{-1} + U\phi\frac{\partial U^{-1}}{\partial t} + U\frac{\partial\phi}{\partial t}U^{-1} and emphasize the importance of commutators in this context.

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  • Understanding of quantum mechanics and perturbation theory
  • Familiarity with the Heisenberg representation and unitary operators
  • Knowledge of commutators and their significance in quantum field theory
  • Basic proficiency in mathematical derivations involving operators and fields
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Students and researchers in quantum mechanics, particularly those focusing on quantum field theory and perturbation methods. This discussion is beneficial for anyone seeking to deepen their understanding of the Heisenberg representation and its mathematical foundations.

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:

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

(where \phi_a is the free field before the interaction.

Why is it that we can write:

\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}

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:

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

(where \phi_a is the free field before the interaction.

Why is it that we can write:

\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}

where the square brackets in the third equality are commutators?

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

Thanks.

<br /> \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 /> ...(*)<br />
where the last term of RHS involves the derivative of time with respect to the field \phi whose equation of motion is well known, the Heisenberg's EoM.

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

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

1 = UU^{-1}

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

1 = UU^{-1}

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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