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Questions in Weinberg The Quantum Theory of Fields Vol 1

  1. Mar 3, 2012 #1
    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. jcsd
  3. Mar 3, 2012 #2
    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 :smile:

    Ilm
     
  4. Mar 3, 2012 #3
    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?
     
  5. Mar 4, 2012 #4
    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 :smile:
     
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