Confusion (2) from Weinberg's QFT.(Little group))

[kof9595995]

There're two questions:
(1)In page 64, he says
[itex]L^{-1}({\Lambda}p){\Lambda}L(p)[/itex]...(2.5.10) ((L(p) is defined in (2.5.4))
belongs to the subgroup(of lorentz group) that leaves the standard momentum k invariant, i.e. [itex]W^{\mu}_{\;\nu}k^{\nu}=k^{\mu}[/itex]...(2.57), and this subgroup is called the little group.
My question is about the converse of the statement: can all little group elements be written as (2.5.10)?In other words, can we use (2.5.10) as an alternative definition of little group?If no, what are the exceptions?If yes, how do I prove it?

(2)Suppose little group can be defined by (2.5.10), does the matrix of [itex]W(\Lambda,p)[/itex]depend on the particular choice of the "standard boost“ of L(p)?
Take an example from page 68, the standard boost given in (2.5.24) (which is a pure boost without rotation), can be written as [itex]L(p)=R(\hat {\mathbf{p}})B(|\mathbf{p}|)R^{-1}(\hat {\mathbf{p}})[/itex], but I find we may just as well define L(p) to be [itex]L(p)=R(\hat {\mathbf{p}})B(|\mathbf{p}|)[/itex] and still we'll get the same p from the standard momentum k, but would W still be the same if I choose L(p) to be the latter way(and this L(p) is later used for mass 0 particles, see (2.5.44))? I checked for pure rotation these two L(p) indeed give the same W, but I don't know how to prove it for a general [itex]\Lambda[/itex]

 

[Petr Mugver]

(1) I think the answer is yes: given a W belonging to the little group associated to a certain k, then choose

[itex]p = k[/itex]
[itex]\Lambda = L(k)WL^{-1}(k)[/itex]

Note that L(k) belongs to the little group, and so does \Lambda, so

[itex]L^{-1}(\Lambda p)\Lambda L(p)=
L^{-1}(k)L(k)WL^{-1}(k)L(k)=W[/itex]

(2) I have to think it a bit better, but I think that W depends on the particular L(p): note that L(p) might be an "ugly" correspondence between R^4 and SO(1,3) (for example, discontinuous).

 

[kof9595995]

(1) I think the answer is yes: given a W belonging to the little group associated to a certain k, then choose

[itex]p = k[/itex]
[itex]\Lambda = L(k)WL^{-1}(k)[/itex]

Note that L(k) belongs to the little group, and so does \Lambda, so

[itex]L^{-1}(\Lambda p)\Lambda L(p)=
L^{-1}(k)L(k)WL^{-1}(k)L(k)=W[/itex]

(2) I have to think it a bit better, but I think that W depends on the particular L(p): note that L(p) might be an "ugly" correspondence between R^4 and SO(1,3) (for example, discontinuous).

(1)Thanks, that's a neat construction
(2)I think the other way:if W depends on L(p), then its representation D probably also depends on L(p), then according to (2.5.11） [itex]U(\Lambda)\Psi_{p,\sigma}=(\frac{N(p)}{N({\Lambda}p)})\sum\limits_{\sigma'}D_{{\sigma'}\sigma}(W({\Lambda},p))\Psi_{{\Lambda}p,\sigma'}[/itex]
the transformation will depend in L(p), but it's physically obvious it should only depend on [itex]\Lambda[/itex]

 

[kof9595995]

Nah, now I think you're right about (2), I found a counter example already. But then I'm even more confused, considering what I argued in post3, we'll conclude while W depends on
L(p), it's representation doesn't. I can hardly imagine this.

 

[Petr Mugver]

Ok I think I've got point (2). Call [itex]\Omega\in\mathbb{R}^4[/itex] one of the disjoint subsets of space-time, such that every two points [itex]p,q\in\Omega[/itex] can be connected by a proper ortochronous homogeneous Lorentz transformation (POHLT). Every such [itex]\Omega[/itex] is good for what follows, except the trivial one [itex]\Omega=\{0\}[/itex]. (So no matter if we consider massive or non-massive particles, or even non-physical states with negative mass and/or energy). Choose once and for all a [itex]k\in\Omega[/itex] and for every [itex]p\in\Omega[/itex] define two POHLTs L(p) and M(p) such
that

[itex](A)\qquad\qquad L(p)k = M(p)k = p\qquad\qquad\forall p\in\Omega[/itex]

I want to show that if

[itex](B)\qquad\qquad L^{-1}(\Lambda p)\Lambda L(p)=M^{-1}(\Lambda p)\Lambda M(p)[/itex]

for every [itex]p\in\Omega[/itex] and for every [itex]\Lambda\in POHLT[/itex] then

[itex]L(p) = M(p)\qquad\qquad\forall p\in\Omega[/itex]

Rewrite (B) as

[itex]\Lambda^{-1}M(\Lambda p)L^{-1}(\Lambda p)\Lambda=M(p)L^{-1}(p)[/itex]

for every bla bla, and choose [itex]\Lambda = M^{-1}(p)[/itex], obtaining

[itex]L^{-1}(p)M(p)=M(k)L^{-1}(k)\equiv C[/itex]

where C is a POHLT independent of p. So

[itex]M(p)=CL(p)[/itex]

Applying this last equation to k we have

[itex]L(p)k=CM(p)k[/itex]

and using (A):

[itex]Cp=p\qquad\qquad\forall p\in\Omega[/itex]

So every p is eigenvector of C with eigenvalue 1, and since, as it is easy to see, the omega's we consider all contain a basis of R^4, we must have

[itex]C=id[/itex]

and hence the thesis.

 

[kof9595995]

Sorry for my late reply, been occupied for a while. There're some points I want to address:
(1)I don't see why my question is equivalent to the proposition you attempted to prove, I wanted to show if L(p) is multiplied by any little group element M(so L(p)M will also transform k to p), the representation D will be the same.
(2) You can' just let [itex]\Lambda[/itex] be [itex]M^{-1}(p)[/itex],we want [itex]\Lambda[/itex] to be any possible transformation when L and M are chosen
(3)I actually think I've figured out the crux of the problem, the point is [itex]\sigma[/itex] is defined by L(p) via (2.5.5), for example, if we choose L(p) to be (2.5.24), then [itex]\sigma[/itex] would be the angular momentum along 3-axis, while if we choose L(p) to be (2.5.44), it'd be helicity, and you see these two definitely transform differently under a general Lorentz transformation

 

[Petr Mugver]

Sorry for my late reply, been occupied for a while. There're some points I want to address:
(1)I don't see why my question is equivalent to the proposition you attempted to prove, I wanted to show if L(p) is multiplied by any little group element M(so L(p)M will also transform k to p), the representation D will be the same.
(2) You can' just let [itex]\Lambda[/itex] be [itex]M^{-1}(p)[/itex],we want [itex]\Lambda[/itex] to be any possible transformation when L and M are chosen
(3)I actually think I've figured out the crux of the problem, the point is [itex]\sigma[/itex] is defined by L(p) via (2.5.5), for example, if we choose L(p) to be (2.5.24), then [itex]\sigma[/itex] would be the angular momentum along 3-axis, while if we choose L(p) to be (2.5.44), it'd be helicity, and you see these two definitely transform differently under a general Lorentz transformation

(1) + (3): It is equivalent because I showed that if instead of L(p) you use any other M(p) (for example the one you said, M(p) = L(p)M, with M in the little group) then when you calculate [itex]D_{\sigma\sigma'}(M^{-1}(\Lambda p)\Lambda M(p))[/itex] you get a different matrix (because D is a faithful representation of the little group), and hence a different representation. Now, if the group under consideration has only one (up to equivalences) representation of that dimensionality, then what you get is an equivalent representation, or, in other words, a simple renaming of the complete orthonormal basis of the Hilbert space. But the construction of Weinberg has many of theese arbitrary choices: this one, but also the one you pointed out about the definition of the sigma index, and also the choice of the standard ket k, the particular choice of D, etc.
All theese possible different choices lead to equivalent representations of the kind
U'(\Lambda) = U U(\Lambda)U^-1, and this is not "unphysical", since it's just a renaming of the kets.

(2): I can, because to proove that a function of \Lambda is different from another function of \Lambda, I just need to find one of such \Lambda's for which the two functions are different.

 

[kof9595995]

ah..I see, I thought you wanted to prove W does not depend on the particular L(p), but actually you were proving the contrary. Thanks a lot.

 

[Petr Mugver]

ah..I see, I thought you wanted to prove W does not depend on the particular L(p), but actually you were proving the contrary. Thanks a lot.

Thanks to you, I also learned a lot thinking about theese things.