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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 U(\Lambda)\Psi_{p,\sigma} = \Sigma_{\sigma '} C_{\sigma'\sigma}(\Lambda,p) \Psi_{\Lambda p,\sigma} (2.5.3)
(\Lambda 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
\Psi_{p,\sigma} of momentum p by
\Psi_{p,\sigma} \equiv N(p) U(L(p))\Psi_{k,\sigma}. (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 \Psi_{p,\sigma}?? In general on the left hand side of (2.5.5) should be not exactly \Psi_{p,\sigma}, 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)?
Here's the question. First Weinberg shows that U(\Lambda)\Psi_{p,\sigma} = \Sigma_{\sigma '} C_{\sigma'\sigma}(\Lambda,p) \Psi_{\Lambda p,\sigma} (2.5.3)
(\Lambda 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
\Psi_{p,\sigma} of momentum p by
\Psi_{p,\sigma} \equiv N(p) U(L(p))\Psi_{k,\sigma}. (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 \Psi_{p,\sigma}?? In general on the left hand side of (2.5.5) should be not exactly \Psi_{p,\sigma}, 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)?