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As in this question, let $\psi_{p,\sigma}$ be a single-particle 4-momentum eigenstate, with $\sigma$ being a discrete label of other degrees of freedom.

Weinberg discusses the effect of a homogenous Lorentz transformation $U(\Lambda, 0)$ or $U(\Lambda)$ on these states, and concludes that $U(\Lambda)\psi_{p,\sigma}$ is a linear combination of $\psi_{\Lambda p,\sigma'}$.

$$U(\Lambda)\psi_{p,\sigma} = \sum_{\sigma'} C_{\sigma'\sigma}(\Lambda, p)\psi_{\Lambda p,\sigma'}$$

Again, is there any physical information that we can extract from this? (I realize that $\psi_{\Lambda p,\sigma'}$ represent physical states after Lorentz transformation).

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Yes, indeed, first of all it tells you that a particle of momentum $p$ in one frame looks like a particle of momentum $\Lambda p$ in another frame which is related to the first by Lorentz transformation $\Lambda$. That means $p$ is a Lorentz vector and hence gives you more certainty that it indeed captures momentum of the particle.

If you read a few more pages in the book, you will see that $\sigma$ would correspond to spin for a massive particle and helicity for a massless particle. So, the above equation is telling us that not only do you find a different momentum for the particle but also that you might find the particle carrying a different spin (or more concretely the probabilities to find the particle in different spin states would be changed) when you do a Lorentz transformation.

Compare this with a usual case one encounters in QM. We have a massive motionless particle as our physical system. And we now want to study how the physical state transforms under a rotation $$U(R)\psi_{p=0,\sigma} = C_{\sigma',\sigma}(R,0)\psi_{p=0,\sigma'}$$ Here $\sigma$ labels spin of the particle and $U(R)$ forms a representation of rotation group which is generated by the angular momenta $J_1, J_2, J_3$.

So, this equation is telling you that transformation of single particle states in relativistic quantum mechanics is a straightforward union of the transformations that we expect from our earlier study of relativity and quantum mechanics.

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  • $\begingroup$ Thanks! After reading your answer I'm getting the rough picture. If possible, could you somewhat more explicitly demonstrate the point you made in the last paragraph? Also, one more point, and this may seem daft on my part, but if $U$ is a representation already, shouldn't it simply act on $\psi_p$ to give only $\psi_{\Lambda p}$, because the latter state is Lorentz transformed? Why does the presence of other degrees of freedom induce this weird linear combination as the image of $U$, instead of merely just one state? $\endgroup$
    – 1989189198
    Commented May 26, 2013 at 7:57
  • $\begingroup$ First paragraph of my answer captures the transformation that we expect from relativity and third paragraph captures what we expect from QM. Now it is easy to see that the equation in your question captures the union of both which is what we expect from a relativistic quantum mechanics. $\endgroup$
    – user4235
    Commented May 26, 2013 at 9:28
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    $\begingroup$ Regarding your other question, in general there can be many quantum states which correspond to a single particle having a fixed momentum $p$. So, Weinberg distinguishes all these states which have the same momentum by the label $\sigma$ which he expects to be discrete from the experience gained from experiments. Now, he goes on to show that a homogeneous Lorentz transformation takes a particular state $\psi_{p,\sigma}$ to a state having momentum $\Lambda p$. But we still don't know what is the label $\sigma$ of this new state. So we have to allow for a generic possibility like RHS of above eqn $\endgroup$
    – user4235
    Commented May 26, 2013 at 9:33
  • $\begingroup$ All right thanks, I've accepted your answer. As an aside, how much of Weinberg have you completed and how long did it take? $\endgroup$
    – 1989189198
    Commented May 26, 2013 at 10:53
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    $\begingroup$ The operator $U(W)$ for any particular $W$ is unitary, but the selected answer's point number 2 shows that the matrix corresponding to $U(W)$ in a particular basis may not be unitary. So, Weinberg chooses the basis as $\Psi_{k,\sigma}$ where $k$ is a fixed standard momentum and you vary $\sigma$ to get all basis vectors. Now, in this basis, you can check from (2.5.8) using (2.5.12) that the matrices $D(W)$ corresponding to operators $U(W)$ are all unitary matrices. $\endgroup$
    – user4235
    Commented May 27, 2013 at 3:20

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