1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Little help with a commutator

  1. Jun 18, 2010 #1
    Hi, could someone give me a hand with the two long commutators on page 25 of Peskin and Schroeder? I'm not sure how to deal with the gradient in the first and the laplacian in the second. Thanx alot
     
  2. jcsd
  3. Jun 18, 2010 #2

    strangerep

    User Avatar
    Science Advisor

    Try integration by parts in the first one, and see how far you can get... :-)
     
    Last edited: Jun 19, 2010
  4. Jun 19, 2010 #3
    sorry, I'm not sure I see what u mean. I understand how to pass from the first to the second expression of the Hamiltonian used by "integrating by parts" the grad term. But my problem is how to handle the commutators. Any more suggestion? I still can't go on, I'm probably missing something easy.
    Thanx
     
  5. Jun 21, 2010 #4

    strangerep

    User Avatar
    Science Advisor

    Hmmm. I'm not sure what you mean by "handle" the commutators. You can
    just move them inside the integral. E.g.,

    [tex]
    \left[A \,,\, \int\!\! B \right] ~\to~ \int [A,B]
    [/tex]

    (If that wasn't your problem, you need to be more explicit...)
     
  6. Jun 21, 2010 #5
    sorry, that was not what I meant. I meant: what is the trick to calculate commutators involving gradient and laplacian operators (for example [pi,del squared phi])? Do I treat them as operators? How do I deal with the fact that they only act on the operator immediately to the right? Is this the right way of doing these things or do I have to integrate something? Thanx
     
  7. Jun 21, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    I think the easiest method may be to use the expressions in terms of ladder operators (2.27) and (2.28) and the commutators in (2.32).

    Alternatively, if you want to do the calculation in the position basis as Peskin & Schroeder do, you'll find the identity [itex][A,BC]=[A,B]C+B[A,C][/itex] very useful. For example,

    [tex] [\phi(\textbf{x},t),(\mathbf{\nabla}\phi(\textbf{x}',t))^2]= [\phi(\textbf{x},t),\mathbf{\nabla}\phi(\textbf{x}',t)]\cdot\mathbf{\nabla}\phi(\textbf{x}',t)+\mathbf{\nabla}\phi(\textbf{x}',t)\cdot[\phi(\textbf{x},t),\mathbf{\nabla}\phi(\textbf{x}',t)] \end{aligned} [/tex]

    Where [itex][\phi(\textbf{x},t),\mathbf{\nabla}\phi(\textbf{x}',t)][/itex] can be easily calculated using (2.27).
     
    Last edited: Jun 21, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Little help with a commutator
  1. Commutator help (Replies: 4)

  2. Help with commutators (Replies: 0)

Loading...