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Weinberg QFT - Inner product relations, Standard momentum, Invariant integrals

by Lakshya
Tags: quantum field theory
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Lakshya
#1
May2-12, 09:31 AM
P: 72
Weinberg in his 1st book on QFT writes in the paragraph containing 2.5.12 that we may choose the states with standard momentum to be orthonormal. Isn't that just true because the states with any momentum are chosen to be orthonormal by the usual orthonormalization process of quantum mechanics?

Similarly, doesn't 2.5.13 come directly from the unitarity of the operators U? What is Weinberg trying to say?

Then later when he considers inner product relations between arbitrary momenta, then if p and p' are in the same class, then they would have same k. So, if one defines a k', that might not be a standard momentum. So, is Weinberg also allowing for non-standard momenta? But then he considers N(L^-1(p)p')=1 while writing the second equation between 2.5.13 and 2.5.14. This strongly suggests he is only taking standard momenta. Moreover, if I vary p' even in the different class, according to his definition k' is varying. But k' is a fixed standard momentum in a class. I again don't get what is going on.

Plus, I don't get anything he says between 2.5.14 and 2.5.17 (excluding 2.5.14 and 2.5.17). Can somebody explain what he is trying to say? That would be great.
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Bill_K
#2
May3-12, 07:05 AM
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P: 4,160
Yikes! Most writers get paid by the word. I think Weinberg must get paid by the subscript.

What he's trying to say is actually quite simple and is usually dismissed in other QFT books in a line or two. Given a field φ(x) you want to Fourier transform it:

φ(x) = (2π)-2 ∫ eipx c(p) d4p

Taking into account the mass shell condition p2 + m2 = 0 we know that c(p) = δ(p2 + m2) b(p). This can be used to integrate out one of the four p's and thereby reduce the four components of c(p) to three. Then everything will be expressed in terms of a 3-momentum p instead of the 4-momentum p.

Naturally this can't be done in a Lorentz covariant fashion. In particular there will be one 4-momentum that corresponds to the 3-momentum p = 0. This is Weinberg's "standard" 4-momentum, p = (0, 0, 0, m). [Table 2.1].

Well, δ(p2 + m2) = (2E)-1 [δ(p0 - E) + δ(p0 + E)], so integrating out p0:

φ(x) = (2π)-2 ∫ (2E)-1 [eipx b(p) + eipx b(-p)] d3p

That's half of it. The other half is to replace b(p), b(-p) with rescaled operators a, a* that obey nicer commutation relations:

a(p) = (4πE)-1/2 b(p), a*(p) = (4πE)-1/2 b(-p)

for which

[a(p), a*(p')] = δ(p-p')

The bottom line is

φ(x) = (2π)-3/2 ∫ (2E)-1/2 [eipx a(p) + e-ipx a*(p)] d3p

The sole purpose of all of this is to get the non-Lorentz-covariant (2E)-1/2 factor!
Lakshya
#3
May3-12, 08:21 AM
P: 72
Thanks for your reply. BTW, what is Weinberg's k' for different cases? It creates many different problems as I pointed out. Please clarify it also. Does his k' has anything necessarily to do with his previously defined standard momentum? If yes, then what if p and p' belong to the same class? How can there be two standard momentum for the same class?


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