The Klein-Gordon field as harmonic operators

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I am reading through 'An Introduction to QFT' by Peskin & Schroeder and I am struggling to follow one of the computations.
I follow writing the field [tex]\phi[/tex] in Fourier space
ϕ(x,t)=∫(d^3 p)/(2π)^3 e^(ip∙x)ϕ(p,t)

And the writing the operators [tex]\phi[/tex](x) and pi(x) as
ϕ(x)=∫(d^3 p)/(2π)^3 1/√(2ϖ_p )(a(p)+a(-p)†)e^(ip∙x)
π(x)=∫(d^3 p)/(2π)^3 (-i)√(ϖ_p/2) (a(p)-a(-p)† )e^(ip∙x)

But I struggle to follow the commutator of the two its given as
[ϕ(x),π(x) ]
=∫(d^3 p)(d^3 p')/(2π)^6 (-i)/2 √(ϖ(p')/ϖ(p)) ([a(-p)†,a(p')]-[a(p), a(-p')†])e^(ip∙x)

which I understand how to get to, I know the commutation relation
[a(p),a(p')†]=(2π)^3 δ^((3))(p-p')

Is subbed into the previous commutator and that the final result is
[ϕ(x),π(x) ]= iδ^((3))(x-x')

I just cant work through the steps I someone could go through it and really dumb it down it would be mega appreciated as Peskin & Schroder stress that working through such computations is essential for future formalisms. p.s sorry if you can't read the equations I couldn't get the code to work today, just let me know and I can give it another go.
 

Answers and Replies

  • #2
naima
Gold Member
938
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You just have to use, at the end, the fact that
\int exp ip(x-x') dp gives a dirac of (x-x')
 

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