Little help with a commutator

  • Thread starter Rick89
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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
 

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  • #2
strangerep
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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.
Try integration by parts in the first one, and see how far you can get... :-)
 
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  • #3
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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
 
  • #4
strangerep
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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.
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...)
 
  • #5
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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
 
  • #6
gabbagabbahey
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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).
 
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