# Can spin be infinite?

Can spin of a particle or a group of particles become infinity? Explain plz. Is there any representation for spins like dot(for $S=0$) and arrow (for $S=1$)? If so what for $S= \infty$?

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please don't use capslock (I changed the question title) –  Mark Eichenlaub Apr 28 '11 at 4:52
ok.thanx for changing the name.sorry 4 my mistake.!! –  chaos Apr 28 '11 at 14:15

Compact group $SU(2)$ really has only finite-dimensional unitary irreducible representations, but formally it is not enough to close the question, because there are unitary irreducible infinite-dimensional representations of spin group $SL(2,C)$ of four-dimensional relativistic Lorentz group and they were used in some models, below is a cite from: N.N. Bogolubov, A. A. Logunov, A.I. Oksak, I. Todorov, General principles of quantum field theory, Springer, 1989. Appendix I for chapter 9

The concept of an infinite-component field (ICF for short) is the result of abandoning the "technical" requirement that the representations of the Lorentz group according to which the fields transform (say, in the Wightman formalism) be finite-dimensional. This idea turned up at the earliest stages of quantum field theory: in 1932, Majorana gave an example of an infinite-dimensional wave equation $(i \Gamma^\mu \partial_\mu – M) \psi(x) = 0$ without negative-energy solutions of non-negative square mass, that is, without "antiparticles".

Running ahead (see §1.3), it should be noted, however, that the description of composite systems by means of ICF's has met with difficulties which, it would seem, require a weakening of the postulate of (strict) locality.

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Dear @Alex 'qubeat'. Sorry for this nitpicking comment, but the statement that the compact group $SU(2)$ has only finite-dimensional irreducible representations is not correct. There are non-unitary infinite-dimensional irreducible representations of the Lie group $SU(2)$. Of course, in physics one usually requires unitarity. –  Qmechanic May 7 '11 at 21:13
@Qmechanic: why does the unitarian trick fail on these irreducible representations? –  wnoise May 7 '11 at 21:41
@Qmechanic: Anyway, I would like to see an example of representation you are talking about. –  Alex 'qubeat' May 8 '11 at 10:38
@Qmechanic: I believe your statement is not correct. By usual Weylish arguments one can always unitarize a representation of the compact group (just impose arbitrary Hilbert structure and then define new scalar product by averaging over the group action). By Peter-Weyl theorem then every such representation decomposes into sum of unitary irreducibles. What you had in mind was probably the rep. theory of $SL(2, \mathbb{R})$ which indeed admits non-unitary reps. But this is precisely because the group is not compact. –  Marek May 8 '11 at 11:03
@wnoise and @Marek. The standard Weyl unitarian trick for compact groups $G$ and representations $\rho:G\to B(H)$ into bounded operators on a Hilbert space $H$, only preserves the strong operator topology. Thus if the representation is continuous in, e.g., the weak operator topology, the averaging procedure may destroy this notion of continuity. –  Qmechanic May 8 '11 at 20:55

I am not aware of any theory involving a notion of infinite spin. Generally, spin is a quantum number that takes (typically) small integer or half integer values. In principle, you can have a system with as high a spin as you would like, but that's not infinite. So I'd say the answer is no.

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There is an exotic phenomenon in in general relativity regarding large "spin"/angular momentum. It does not have much to do with your question, so I'll add it as a comment: A massive rotating body will drag space along around itself. This "frame-dragging" has been measured for the earth, but it is a very small effect. However, around a quickly rotating black hole it can get so strong that something in orbit would have to move with light speed against it, just to stay where it is (relative to the surrounding ~flat space)! With a lot of handwaving, one could speak of an infinity here :-) –  jdm May 5 '11 at 6:47
the angular momentum of such a rotating black hole is perfectly finite, though--with a maxiumum magnitude proportional to $M^{2}$ –  Jerry Schirmer May 7 '11 at 23:09
@Jerry: Thanks for adding the clarification. I was not claiming that the angular momentum was infinite :-), it was more of an interesting anecdote (hence a comment and not an answer). –  jdm May 9 '11 at 15:37
But intrinsic angular momentum,L=nh/(2pi),which is equal to spin.So when n becomes infinity,L is also infinity.So my question is ,is there any physical significance for this statement rather than a theoretical statement? –  chaos Jun 2 '11 at 2:34

The three-dimensional spin group is the compact group $SU(2)$, which has only finite-dimensional representations. Hence, the spin is always a non-negative integer or half-integer.

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Dear @Oluf. The answer (v1) contains a wrong statement. There exist infinite-dimensional unitary representations of the compact group $SU(2)$. Take, e.g., infinitely many copies of the trivial or adjoint representation. –  Qmechanic May 7 '11 at 11:16
@Qmechanic: Sorry, I meant to say irreducible representation. And of course it is not obvious that a particle has to transform in a irreducible representation, so I guess my answer is a bit too naive. By the way, doesn't infinitely many copies of the trivial representation still just contain a single state? –  Olof May 7 '11 at 11:40
@Oluf. In usually terminology, the trivial (=singlet) group representation is $1$-dim, and the adjoint (=triplet) group representation is $3$-dim. Hope this answers your comment. –  Qmechanic May 7 '11 at 12:09
@Qmechanic: Sure. I'm just saying that a tensor product of even an infinite number of singlets still is a singlet. Or am I missing something? –  Olof May 7 '11 at 12:20
@Oluf. No, you're right, the tensor product of two singlet representations is again a singlet representation. In my comment above, I was alluding to direct sums (as opposed to tensor products) of singlet representations. –  Qmechanic May 7 '11 at 12:34