Following the treatment of Weinberg, chapter 2, we consider $\psi_{p,\sigma}$ as single-particle eigenstates of the 4-momentum. Weinberg says that $\sigma$ labels all other degrees of freedom and we take this label to be discrete for one-particle states. So what exactly is the physical implication of discrete and continuous labeling of other degrees of freedom? And why is discrete labeling physically pertinent to single-particle states?

  • $\begingroup$ Hi 1989189198 - we prefer that each individual question should be asked separately. I've removed the extra questions (beyond the first) from your post, but I would encourage you to make a new post for each of the other two things you wanted to ask. $\endgroup$ – David Z May 25 '13 at 5:51
  • $\begingroup$ That's all right. I actually thought the opposite. But is there any way I can recover the original question? (since then it would be easier to type out the other parts) $\endgroup$ – 1989189198 May 25 '13 at 5:53
  • $\begingroup$ Yep, the revision history of every question and answer is always accessible. Look to the left of your user card (the blue box at the bottom right of the post with your username in it), and you'll see the text "edited # minutes ago" (or in general, "edited <time>"). Click on that to see the post's revision history. $\endgroup$ – David Z May 25 '13 at 6:00

Apart from the spatial translations corresponding to the momentum operator, the other symmetries (that I can think of) that are relevant in particle physics i.e. things like

Spatial Rotations

Phase Transformations

Flavour Transformations

Colour Transformations

are represented by the action of compact Lie groups. The irreducible unitary Hilbert space representations of compact Lie groups are finite dimensional and this is reflected in the discrete labelling.

  • $\begingroup$ 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? $\endgroup$ – 1989189198 May 26 '13 at 11:46
  • $\begingroup$ The idea is that the system we're modelling is invariant under spacetime translations: it doesn't matter if we do our experiment here or three miles north of here (assuming spacetime is flat!). This is the symmetry. To implement this symmetry, we'd perform a translation - a spatial shift. Now the momentum operator $-i\frac{d}{dx}$ represents an infinitesimal version of such a shift, for reasons explained here. $\endgroup$ – twistor59 May 26 '13 at 12:02
  • $\begingroup$ The relationship between these group theory concepts and quantum mechanics should be explained in any good QM book. A good one is Asher Peres "Quantum Theory: Concepts and Methods". You might be able to find a free pdf copy online if you Google for it. $\endgroup$ – twistor59 May 26 '13 at 12:06
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    $\begingroup$ If I understand correctly you're asking why Weinberg says "discrete" as opposed to "discrete and finite"? I think he emphasized "discrete" to distinguish it from "continuous" - there's a continuous spectrum of momentum values, but a discrete spectrum of, for example, spin component measurements. Now all the examples of these "not momentum" symmetries that I know of are represented by compact Lie groups and they would also have the restriction that, in addition to being discrete, the labels would also be finite (like, say, 2 for a spin 1/2 particle $J_z$ component). $\endgroup$ – twistor59 May 26 '13 at 12:21
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    $\begingroup$ I just wanted to note that we have non-compact Lorentz group. This is ok while the small group is compact, but in the massless case it is not ($ISO(2)$). In takes experimental evidence that there is no continuous degrees of freedom for massless particles to say that we pick a very special finite-dimensional rep of the small group. In general, if the translation subgroup of $ISO(2)$ is non-trivially represented ('2d-momentum' non-zero), then the rotation gives you continuous spectrum. $\endgroup$ – Peter Kravchuk May 26 '13 at 12:41

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