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Particles and Wigner little groups

  1. Mar 27, 2012 #1
    Hello, from Weinberg's Quantum Field Theory book I am confused about the equation (2.5.5). I'll describe the problem briefly here, but in any case, here's that page from Weinberg's book (page 64) http://books.google.am/books?id=h9k...crRhAeM1rWgAQ&redir_esc=y#v=onepage&q&f=false

    Here's the question. First Weinberg shows that [tex] U(\Lambda)\Psi_{p,\sigma} = \Sigma_{\sigma '} C_{\sigma'\sigma}(\Lambda,p) \Psi_{\Lambda p,\sigma} [/tex] (2.5.3)
    ([tex]\Lambda[/tex] is and arbitrary Lorentz transformation, the index p means the state is the eigenvector of momentum with eigenvalue p, sigma represents all the other degrees of freedom)and then says that different particle types correspond to different irreducible representations of Lorentz group, i.e. such matices C, which are not block diagonal (simga, as it turns out later, is the particle spin). And then he goes on to show that such irreducible representations can be obtained from representations of the Wigner little groups, which are groups, containing all the Lorentz transformations that leave some defined "standard" momentum k invariant.
    For that he first writes: We can define the states
    [tex]\Psi_{p,\sigma}[/tex] of momentum p by
    [tex]\Psi_{p,\sigma} \equiv N(p) U(L(p))\Psi_{k,\sigma}[/tex]. (2.5.5)
    and then the logic leading to Wigner little groups is straightforward (k is the standard momentum, belonging to the same type as p, i.e. with the same length as p and k0 has the same sign as p0).
    My question is: Can we really define the states [tex]\Psi_{p,\sigma}[/tex]?? In general on the left hand side of (2.5.5) should be not exactly [tex]\Psi_{p,\sigma}[/tex], but a linear combination of such states with different sigmas, as in (2.5.3). Instead Weinberg somehow defines it with the same sigma on the left and right hand sides in (2.5.5). Isn't it the same thing, as to say that if in one inertial frame some particle has spin projection +1/2, it will have the same spin projection in other inertial frame, related to the previous by Lorentz transformation L(p)? Considering that p varies within a very large range (it is confined only by it's length and the sign of p0), it is very doubtful. Or maybe (2.5.5) is just a trick that makes the rest calculations easier and we'll get the same result in any case (though I couldn't do that)?
  2. jcsd
  3. Mar 27, 2012 #2


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    This is a particularly convenient choice of basis concerning the representation theory of the Poincare group. It's called the Wigner basis. Weinberg shows, how the representation of the little group, acting on the subspace with [itex]p=k[/itex], to the corresponding representation of the full (proper orthochronous) Lorentz group, using this choice of a basis.
  4. Mar 27, 2012 #3
    thanks for the answer. from the pure mathematical point of view I agree, cause then we may consider the index sigma as just a numerical sign to enumerate the basis vectors and we are free to choose the basis vectors as we want. But from the point of view of quantum mechanics, the index sigma has a very definite meaning, it's the value of S_z and for instant takes the values +1/2 or -1/2, and when we write the same index in both sides of (2.5.5) we infer that after making the Lorentz transformation, we will still have an eigenvector of S_z with the same eigenvalue. Right?
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