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strangerep

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Try integration by parts in the first one, and see how far you can get... :-)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.

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Thanx

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strangerep

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Hmmm. I'm not sure what you mean by "handle" the commutators. You cansorry, 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.

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...)

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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).

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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