Solve Tricky Commutator: Heisenberg Picture, a_k(t)

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In summary, the conversation is about evaluating a commutator in the context of the Heisenberg picture. It is mentioned that the commutator is distributive and that the inner product can be pulled outside. The conversation also includes a reminder to use a different letter for the dummy index in the sum.
  • #1
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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  • #2
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...
 
  • #3
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]
 
  • #4
Yes, exactly...now keep going...simplify the commutator [itex]\left [ a_k(t),a_k^\dagger a_\ell\right][/itex]
 
  • #5
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!
 
  • #6
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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FAQ: Solve Tricky Commutator: Heisenberg Picture, a_k(t)

What is a commutator in the Heisenberg picture?

A commutator in the Heisenberg picture is a mathematical operation that describes how two operators, such as position and momentum, interact with each other over time. It is defined as the difference between the product of the two operators and the product of the operators in reverse order.

How do you solve a tricky commutator in the Heisenberg picture?

To solve a tricky commutator in the Heisenberg picture, you need to use the Heisenberg equation of motion, which describes how operators change over time. By applying this equation and using the properties of commutators, you can simplify the expression and solve for the desired operator.

What is the significance of the Heisenberg picture in quantum mechanics?

The Heisenberg picture is one of the two main representations in quantum mechanics, the other being the Schrödinger picture. In the Heisenberg picture, operators are time-dependent, while in the Schrödinger picture, operators are time-independent. The Heisenberg picture is particularly useful for studying the time evolution of operators and observables in quantum systems.

Can the Heisenberg picture be used for all quantum systems?

Yes, the Heisenberg picture can be used for all quantum systems, as long as they are described by operators. This includes systems with a finite or infinite number of particles, as well as systems with discrete or continuous spectra.

Are there any limitations to using the Heisenberg picture?

The Heisenberg picture has some limitations, such as the fact that it cannot be used for systems with time-dependent Hamiltonians. Additionally, it may be more difficult to visualize or interpret physical quantities in the Heisenberg picture compared to the Schrödinger picture. However, it is still an important and widely used tool in quantum mechanics.

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