Questions in Weinberg The Quantum Theory of Fields Vol 1

1. Mar 3, 2012

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.

2. Mar 3, 2012

Ilmrak

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

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

3. Mar 3, 2012

wphysics

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?

4. Mar 4, 2012

Ilmrak

We choose $(-1)^j$ so that $(-1)^{j-\sigma}$ is real.

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

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

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

Ilm

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