Weinberg QFT  Inner product relations, Standard momentum, Invariant integralsby Lakshya Tags: quantum field theory 

#1
May212, 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 nonstandard 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. 



#2
May312, 07:05 AM

Sci Advisor
Thanks
P: 3,864

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} ∫ e^{ip·x} c(p) d^{4}p Taking into account the mass shell condition p^{2} + m^{2} = 0 we know that c(p) = δ(p^{2} + m^{2}) 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 3momentum p instead of the 4momentum p. Naturally this can't be done in a Lorentz covariant fashion. In particular there will be one 4momentum that corresponds to the 3momentum p = 0. This is Weinberg's "standard" 4momentum, p = (0, 0, 0, m). [Table 2.1]. Well, δ(p^{2} + m^{2}) = (2E)^{1} [δ(p^{0}  E) + δ(p^{0} + E)], so integrating out p^{0}: φ(x) = (2π)^{2} ∫ (2E)^{1} [e^{ip·x} b(p) + e^{ip·x} b(p)] d^{3}p 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(pp') The bottom line is φ(x) = (2π)^{3/2} ∫ (2E)^{1/2} [e^{ip·x} a(p) + e^{ip·x} a*(p)] d^{3}p The sole purpose of all of this is to get the nonLorentzcovariant (2E)^{1/2} factor! 



#3
May312, 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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