# 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