Perplexing commutator

  • Thread starter waht
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  • #1
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When simplifying this

[tex] \int d^3x' [\pi(x), \frac{1}{2}\pi^2(x') + \frac{1}{2} \phi(x')( -\nabla^2 + m^2)\phi(x')] [/tex]

we know that

[tex] [\pi(x), \pi(x')] = 0 [/tex]

[tex] [\phi(x), \pi(x')] = -i\delta(x-x') [/tex]

how does that simplify to

[tex] \int d^3x' \delta(x-x')( -\nabla^2 + m^2)\phi(x')[/tex]

I know that

[tex] [\pi(x), \pi^2(x')] = 0 [/tex]

but not sure how does the laplacian gets factored out like that and one-half disappears?
 
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Answers and Replies

  • #2
dextercioby
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Have you tried to see what part integration brings you ?
 
  • #3
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By expanding the commutator we get,

[tex]
\frac{1}{2} \int d^3x' \pi(x)\phi(x')( -\nabla^2 + m^2)\phi(x') - \phi(x')( -\nabla^2 + m^2)\phi(x')\pi(x)
[/tex]

the laplacian in the second term is sandwiched between two phis, integration by part doesn't seem to help to factor it out.
 
  • #4
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I think you're better off using [tex][A,BC] = [A,B]C + B[A,C][/tex] and then integration by parts. Also, it's a good time in your life to realize that a commutator behaves very much like a functional derivative (recall QM 101).
 

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