Questions in Weinberg The Quantum Theory of Fields Vol 1

[wphysics]

I have a few questions about the contents in Weinberg, the Quantum Theory of fields Vol1.

First one : At the end of page 77(Sec 2.6), we got -ζ_σ=ζ_σ±1. From this, we could refer that ζ_σ should be proportional to (-1)^σ. But, in the next page, Weinberg concluded that ζ_σ =ζ(-1)^(j-σ). As I said before, -σ in the exponential of (-1) can be easily understood. But, I don't know why ζ_σ should be proportional to (-1)^j.

Second one : It is about the equation below the equation (3.1.21) (page 112, sec 3.2)
The integral variable of this equation is dα, which is Ʃ_(n1,σ1,n2,σ2,...)∫d^3p1 d^3p2 ...
The paragraph below this equation explains how we get '0' in some cases by using the contour integration about E_α. But, in my opinion, dα does not include dE_α. How can we consider the contour integration about E_α ?

Third one : In the last paragraph of page 156(Sec 3.7), Weinberg said we note that k^l Y_lm(k) is a simple polynomial function of the three ..., so in order for M_... the coefficients M^j_l's'n',lsn must go as k^(l+1/2)k'^(l+1/2) when k and k' go to zero.
I think k^2l Y_lm(k) is also a simple polynomial function, so why did he choose 'l' specially? In addition, I don't understand how we get k^(1/2)k'^(1/2) in k^(l+1/2)k'^(l+1/2) either.

 

[Ilmrak]

I have a few questions about the contents in Weinberg, the Quantum Theory of fields Vol1.

First one : At the end of page 77(Sec 2.6), we got -ζ_σ=ζ_σ±1. From this, we could refer that ζ_σ should be proportional to (-1)^σ. But, in the next page, Weinberg concluded that ζ_σ =ζ(-1)^(j-σ). As I said before, -σ in the exponential of (-1) can be easily understood. But, I don't know why ζ_σ should be proportional to (-1)^j.

It's an arbitrary chose, since [itex]j[/itex] is constant (it labels the irreducible representation your particle transforms with) [itex](-1)^j[/itex] can be reabsorbed in [itex]\xi[/itex] (which can still depends on [itex]j[/itex]).

To answer the other points I'd need to read those contents again, I'll do this as soon as I'll have some more free time

Ilm

 

[wphysics]

First of all, thank you for your reply.

My further question about your answer.

Because we choose phase is proportional to (-1)^j, we can get Eq(2.6.26) in Page 80.
As you said, if this choice is arbitrary, we might choose (-1)^(2j) or something else (of course I am not sure...)
Then, Eq(2.6.26) is not true, but has a different proportional constant.
But, as far as I know, Eq(2.6.26) can be verified in experiments..

Why do we choose (-1)^j specifically?

 

[Ilmrak]

We choose [itex](-1)^j[/itex] so that [itex](-1)^{j-\sigma}[/itex] is real.

This chose is the most convenient, but is not necessary. Suppose we chose instead [itex]\xi(-1)^\sigma = \xi e^{i \pi \sigma}[/itex]. Then

[itex]
T^2 \Psi_{p, \sigma} =
T \xi e^{i \pi \sigma} \Psi_{\Pi p, -\sigma} =
\xi^* e^{-i \pi \sigma}T \Psi_{\Pi p, -\sigma} =
\xi^* e^{-i \pi \sigma} \xi e^{-i \pi \sigma} \Psi_{p, \sigma} =
e^{-2i \pi \sigma} \Psi_{p, \sigma} =
(-1)^ {2 \sigma} \Psi_{p, \sigma} = (-1)^ {2 j} \Psi_{p, \sigma}
[/itex]

where I used the fact that if [itex]2j[/itex] is odd (even) [itex]2\sigma=2(j - m), m \in \{0, ... , 2j\}[/itex] is odd (even) too.

Ilm

P.s. sorry I still didn't look for an answer to the other two questions