Solve Tricky Commutator: Heisenberg Picture, a_k(t)

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Homework Help Overview

The discussion revolves around evaluating a commutator in the context of the Heisenberg picture, specifically involving the operator \( a_k(t) \) and a sum of products of creation and annihilation operators.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster seeks guidance on how to begin evaluating the commutator and expresses uncertainty about the distribution of terms within the sum. Some participants clarify the distributive property of commutators and suggest pulling scalars outside the commutator.

Discussion Status

Participants are actively engaging with the problem, providing hints and clarifications. There is a productive exchange regarding the manipulation of the commutator and the proper treatment of indices in the summation.

Contextual Notes

There is a mention of potential confusion regarding the use of indices in the summation, indicating a need for careful consideration of dummy indices in the context of the operators involved.

IHateMayonnaise
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Homework Statement



Part of a much larger problem dealing with the Heisenberg picture. I am not remembering how to start evaluating the following commutator:

[tex]\left [ a_k(t),\left(\sum_{k,\ell}a_k^\dagger <k|h|\ell>a_\ell\right)\right][/tex]

Homework Equations



See (a)

The Attempt at a Solution



Just need some help getting started on this one, after that I'm good. What I do know is that when you do the commutator you cannot just lump the first term ([itex]a_k(t)[/itex]) into the sum.. any hints on how to go about breaking this down? Halp!

Thanks

IHateMayonnaise
 
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To start with, the commutator is distributive (i.e. [itex][A,B+C]=[A,B]+[A,C][/itex] ), so you take the sum out front. Also, like any inner product, [itex]\langle k|H|l\rangle[/itex] will be a scalar, and so can be pulled outside the commutator...
 
Thanks for the reply!

So you said that because the commutator is distributive I can take the sum out front. Do you mean I can do this:

[tex] \left [ a_k(t),\left(\sum_{k,\ell}a_k^\dagger a_\ell\right)\right]=\sum_{k,\ell}<k|h|\ell> \left [ a_k(t),a_k^\dagger a_\ell\right][/tex]
 
Yes, exactly...now keep going...simplify the commutator [itex]\left [ a_k(t),a_k^\dagger a_\ell\right][/itex]
 
gabbagabbahey said:
Yes, exactly...now keep going...simplify the commutator [itex]\left [ a_k(t),a_k^\dagger a_\ell\right][/itex]

I think I got it from here, thank you so much for your help!
 
Wait, what does the index [itex]k[/itex] represent in the [itex]a_k(t)[/itex]?...If it is not being summed over you should use a different letter for the dummy index in your sum.

[tex]\left [ a_k(t),\left(\sum_{n,\ell} a_n^\dagger a_\ell\right)\right]=\sum_{n,\ell} \left [ a_k(t),a_n^\dagger a_\ell\right][/tex]
 
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