Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A QFT, what are the transformation rules for

  1. Oct 29, 2016 #1
    I'm taking an introductory course in QFT. During quantization of the Dirac field, my textbook gives a lot of information on how annihilation and creation operators act on vacuum, but nothing about how they act on non-vacuum states. I need these to compute
    $$
    \int \frac{d^3 p}{(2\pi)^3} \sum_s ( {a^s_ {{\vec{p}}}}^\dagger a^s_ {{\vec{p}}} - {b^s_ {{\vec{p}}}}^\dagger b^s_ {{\vec{p}}} ) |\vec{k},s \rangle,
    $$
    I have searched google, but I couldn't find anything after about 1 hour of searching.

    Are you able to tell me how the annihilation and creation operators from Dirac theory act on non-vacuum?
     
  2. jcsd
  3. Oct 29, 2016 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Take the state you get when you apply a creation operator to the vacuum, and then apply another creation operator to it. What do you get? Notice that to answer this question, you need to figure out what operator corresponds to applying two creation operators in succession. Does your textbook talk about that?
     
  4. Oct 29, 2016 #3
    I am following Peskin. As far as I can tell by reading though Dirac quantization, it doesn't.
     
  5. Oct 29, 2016 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    I assume it talks about what state you get when you apply a creation operator to the vacuum. What state is that, and have you tried applying a second creation operator to that state?
     
  6. Oct 29, 2016 #5
    Yes
     
  7. Oct 29, 2016 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    So what did you get?
     
  8. Oct 29, 2016 #7
    I didnt recognize the operator you're advertising :/
     
  9. Oct 29, 2016 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Um, what? You said "yes" when I asked if you had tried applying a creation operator to the vacuum state, and then taking the state you obtained from that and applying a creation operator to it a second time. I am asking you to give the details of what happened when you did that. I can't help you unless I can see the actual mathematical steps that you're doing.
     
  10. Oct 29, 2016 #9
    Nothing happens magically. I didn't know how to go on from there.
     
  11. Oct 29, 2016 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    From where?

    You haven't posted any math that you personally have done. I can't help you if I can't see what you've already tried. I don't care if you don't think it got you anywhere; I need to see what you've tried.
     
  12. Oct 30, 2016 #11
    I wrote up
    $$
    b^\dagger_s(\vec{p}) | \vec{k}, r \rangle = (2 E(\vec{k}))^{1/2}b^\dagger_s(\vec{p})b^\dagger_r(\vec{k}) | 0 \rangle
    $$
    I thought that perhaps i could get an equation by considering the commutation relation, but the operators commute, so that doesn't help. I didn't know how to go on from here.
     
  13. Oct 30, 2016 #12

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    What if you take the state ##| \vec{k}, r \rangle## and apply an annihilation operator to it? The sum you said you were trying to compute in the OP applies annihilation operators to a non-vacuum state, then creation operators to the result.

    Also, you are applying two different creation operators. What about if you apply the same creation operator twice? Or, even better, apply the corresponding annihilation operator, i.e., take ##b^\dagger_r(\vec{k}) | 0 \rangle## and apply ##b_r(\vec{k})## to it. That's what the sum in your OP does. (And similarly for ##a## and ##a^\dagger##.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: QFT, what are the transformation rules for
  1. What is QFT (Replies: 60)

Loading...