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May
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comment Physical Interpretation of Lorentz-transformed Single Particle states being linear
Is it that we simply take the inner product of eq. 2.5.8 with itself (taking the most general indices, of course)? Many thanks for the help so far...
May
26
comment Physical Interpretation of Lorentz-transformed Single Particle states being linear
May I ask one more thing if you don't mind? Referring to a question of yours : physics.stackexchange.com/questions/24766/… . Do you know how we can explicitly get 2.5.13 from 2.5.12? I've read the answers and they only mention that the choice of basis in 2.5.12 enables us to get 2.5.13. But how exactly do we "choose the basis" by assuming 2.5.12 and how do we go from there to arrive at 2.5.13?
May
26
accepted Why are non-momentum DoFs of single-particle states discretely labeled?
May
26
comment Why are non-momentum DoFs of single-particle states discretely labeled?
@PeterKravchuk: Thanks for the comment. Unfortunately I'm not qualified enough yet to understand your answer. If at all possible, could you express your answer in simpler terms? (I've recently read up about symmetries and their representations, and I'm somewhat new to this topic)
May
26
comment Why are non-momentum DoFs of single-particle states discretely labeled?
let us continue this discussion in chat
May
26
comment Why are non-momentum DoFs of single-particle states discretely labeled?
I just read your newer response; will I find the answer to the question in my comment above in Peres, then?
May
26
comment Why are non-momentum DoFs of single-particle states discretely labeled?
Ah... my bad. I read your answer as "corresponding to p, the other symmetries are...", whereas you meant "spacial translations, corresponding to p..."; sorry. But the problem is I can't find the link between what you said and what I read, maybe due to insufficient mathematical background. I still don't understand how the idea of irreducible representations of these symmetry group being finite dimensional, leads to the idea of there being finite degrees of freedom. (More like -countable-, since the labeling is only discrete)
May
26
comment Why are non-momentum DoFs of single-particle states discretely labeled?
Thanks! But what exactly do you mean by "symmetries corresponding to momentum operator"? I'm not very clear as to what is the correspondence between degrees of freedom and symmetries of momentum. Also, where can I get a sufficient background on the underlying mathematics (as in any source), provided I'm only interested in the results that would matter for their applications in QFT, not their proofs or any other mathematical asides?
May
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comment Physical Interpretation of Lorentz-transformed Single Particle states being linear
All right thanks, I've accepted your answer. As an aside, how much of Weinberg have you completed and how long did it take?
May
26
accepted Physical Interpretation of Lorentz-transformed Single Particle states being linear
May
26
comment Physical Interpretation of Lorentz-transformed Single Particle states being linear
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?
May
26
comment Irreducible Representations Of Lorentz Group
+1: I had parallel doubts and this answer helped me a lot. (also thanks to the OP for coincidentally coming up with similar doubts!)
May
26
revised Why do single particle states furnish a rep. of the inhomogeneous Lorentz group?
added 59 characters in body
May
26
revised Why do single particle states furnish a rep. of the inhomogeneous Lorentz group?
added 808 characters in body
May
26
comment Why do single particle states furnish a rep. of the inhomogeneous Lorentz group?
@Vibert: That's what I thought. But $U$ matrices map the states $\psi$ to Lorentz-transformed states, so then it should be the $C_{\sigma \sigma'}$ matrices that furnish a representation of the Lorentz group. I'm confused why Weinberg says that the states $\psi$ are the ones furnishing such a representation. (See the edited question)
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25
revised Physical Interpretation of Lorentz-transformed Single Particle states being linear
edited title
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25
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