How many polarisation states does a spin $s$ particle have in $d$ dimensions?

Consider Wigner's classification of massive particles in $d$ dimensions. One is interested in the irreducible representations of the (double cover of the) little group of the reference momentum $p=(m,\boldsymbol 0)$, to wit, of $\mathrm{Spin}(d-1)$. Let $s$ be the spin (weight) of one such representations $R_s$.

What is $n_s(d)\equiv\mathrm{dim}\ R_s$?

In other words, how many polarisation states does a spin $s$ particle have? In $d=4$ we know that $n_s(4)=2s+1$. I would like to know how this is generalised to higher $d$.

We have for example the trivial representation, which has $n_0(d)=1$, and the fundamental representation, which has $n_1(d)=d-1$. I would expect that the first non-trivial representation should be somewhere in between; i.e., it should have $1\le n_{1/2}(d)\le d-1$. Is this correct? If the answer to the question above is unknown, I would settle for the simpler question

What is $n_{1/2}(d)$? (i.e., what is the dimension of the first non-trivial irreducible representation of $\mathrm{Spin}(d-1)$?)

Important note: here I am asking about particles, not fields. Therefore, I am not interested in representations of the Lorentz Group, only on those of the Orthogonal Group. I know that the spinor representation of $\mathrm{Spin}(1,d-1)$ has $2^{\lfloor d/2\rfloor}$ components, but that is not (in principle) relevant to my question. In a sense, I am thinking of QFT à la Weinberg: particles come first, and fields later. At this point, I only care about the former; once I understand them, I will care about the latter.

• I believe that if $s$ is even, we have $n_s(d)=\frac{(d+s-4)!}{(d-3)! s!}(d+2 s-3)$ (and I think I could prove this, but I'm not sure). I also *believe* that if $s$ is odd, we have $n_s(d)=\frac{2 (d+s-7/2)!}{(d-3)! (s-1/2)!}$ (but I have no good proof of that, so it is quite possible this is just wrong). If this is correct, we have $n_{1/2}(d)=2$ for all $d$, but at this point I have no reason to believe this is correct at all. – AccidentalFourierTransform Jan 23 '18 at 15:32
• On- & off-shell DOF of a massless & massive field with spin $s$ in $D$ spacetime dimensions are listed in my Phys.SE answer here. – Qmechanic Jan 23 '18 at 15:46
• Yes, your integral spin expression comports with the standard textbook case that @Qmechanic displays in his answer. – Cosmas Zachos Jan 23 '18 at 17:12
• @CosmasZachos Unless I misunderstood Qmechanic's answer, it is about the components of fields. Here I am asking about particles. They are different objects, and the number of components of the former need not (and usually do not) agree with the number of spin states of the latter. – AccidentalFourierTransform Jan 23 '18 at 17:29
• His answer is about physical d.o.f. of fields, normally associated to particle states. That's what the logic of his formula follows: the respective little groups SO(d-2) and SO(d-1), respectively. My sense is any decent appendix of a supergravity book should settle the issues you might have. Essentially, supergravitists are the only ones to apply all these formulas freely, and the stringency of the matchings involved leaves little room for ambiguity. Trust me, the coincidence of multiplicities noted above is not a random coincidence. – Cosmas Zachos Jan 23 '18 at 17:35

Let me give you a few basic facts about of representation theory of $SO(N)$ groups (set $N=d-1$ for the situation at hand; also, we should be talking about the universal cover $Spin(N)$, but to be more general let's pretend that we are working with complex Lie algebra $so(N,\mathbb{C})$ -- the representation theory over $\mathbb{C}$, in which one is interested in this case, is independent of signature, and is equivalent to that of $so(N,\mathbb{C})$). There are two different cases, $N=2n$ and $N=2n+1$. In the former case the Lie algebra of $SO(N)$ is of $D_n$ type and in the latter of $B_n$ type; the representation theory is slightly different in these two cases.

The first thing to know is that $SO(N)$ is reductive for any $N$ -- that is, all its finite dimensional representations decompose as a direct sum of irreducible representations, so you only need to know what are the irreducible representations. The second thing to know is that for $N>3$ ($d>4$) there is no single "spin" which labels the representations of $SO(N)$, but instead you need to specify the weight, which is a vector of $n$ numbers (recall $N=2n$ or $N=2n+1$). That is just how the life works, and there is no way around this. In particular, for $N>3$ spinor and vector representations are not naturally members of the same family labeled by a single spin, so there is no reason to expect that dimension of spinor representation will be smaller than the dimension of vector representation (and this is in fact false).

These $n$ numbers can be taken to be the so-called Dynkin labels, and we can write them as a list $[\lambda_1,\ldots,\lambda_n]$. The $\lambda_i$ are all independent non-negative integers. To show you the basic idea, let us consider $SO(5)$ and $SO(6)$ with $n=2$ and $n=3$ respectively. For $SO(5)$ the simplest representations, which one usually calls spin-$j$, are $[j,0]$. These are rank-$j$ traceless totally symmetric tensors. The spinor representation is, however, $[0,1]$, so it is not a part of this family. For $SO(6)$ we also have the family $[j,0,0]$ and two chiral spinor representations $[0,1,0]$ (left-handed) and $[0,0,1]$ (right-handed). As you can see, spinors are again not a part of the family. More generally, for $N=2n+1$ the last label is the "spinor label", and for $N=2n$ the last two labels are the "spinor labels".

Now, let me stress that you cannot set $j=\frac{1}{2}$ above, because the labels must be integers. You can in $SO(3)$ because there we have the vector $[\lambda_1]$ and $\lambda_1$ is the spinor label and the traceless-symmetric "spin-$j$" family is $[2j]$, a low-dimensional peculiarity. (There is also a similar peculiarity in $SO(4)$ where the traceless-symmetric irreps are $[j,j]$.) Furthermore, there are other representations besides those mentioned above (you can set all $\lambda_i$ to whatever you want), but they have more complicated interpretations ($p$-forms, mixed-symmetry tensors, spin-tensors, etc).

Important thing to know is, however, that the Fermi-Bose statistics is determined by the parity of $\lambda_n$ ($N=2n+1$) or $\lambda_n+\lambda_{n-1}$ ($N=2n$). Thus to get a fermionic representation it is crucial to set at least one of spinor labels to a non-zero value.

Finally, let me answer directly your question about the dimensions of representations. Hopefully, it is clear by now that you cannot get away with just one number $s$, and instead you need all the $n$ labels $\lambda_i$. Correspondingly, the answer is quite complicated. (Consult any representation theory text for the formula.) The simple special cases are:

• The irreducible spinor representations are $2^n$ for $N=2n+1$ and $2^{n-1}$ for $N=2n$. In the latter case I am talking about Weyl spinors and a Dirac spinor is a sum of left and right Weyl spinors, hence not irreducible.
• Traceless-symmetric spin-$j$ representation has the dimension $$\dim [j,0,\ldots,0] = \frac{\Gamma(j+N-2)(2j+N-2)}{\Gamma(j+1)\Gamma(N-1)} = \frac{(j+N-3)!(2j+N-2)}{j!(N-2)!}.$$
• Let me stress that physical massive one-particle states are finite-dimensional representations of $SO(d-1)$ (cf. Weinberg), and hence all of the above applies. There is no special representation theory for "particles". It is the same for particles and fields, its just that fields are representations of $SO(d)$ and particles are representations of $SO(d-1)$ or $SO(d-2)$, depending on the mass. – Peter Kravchuk Jan 24 '18 at 12:44
• Another thing is that the spinors I'm talking about are the only representations which can be reasonably called spin-$\frac{1}{2}$. Namely, these are the only representations which can only have angular momentum $\hbar/2$ in any projection. Any other representation is either scalar, or will allow for larger angular momentum projections. – Peter Kravchuk Jan 24 '18 at 12:48
• I think your first comment should be a bullet in an edit of your fine answer, as it seems to be the OP's sticking point. Also, perhaps, a link to @Qmechanic 's answer, since the mere dimension formulas cannot lie. Throop rules! – Cosmas Zachos Jan 24 '18 at 15:35
• @CosmasZachos, thanks for pointing out the mistake in the formula -- I accidentally put an extra gamma function in the first expression.. – Peter Kravchuk Jan 24 '18 at 18:23
• Hi @PeterKravchuk, thank you for your answer! It is taking me some time to fully digest it. If you don't mind, I'd like to ask you some questions: 1) To me, the "spin" $j$ (or $s$) is the eigenvalue of the quadratic Casimir $J_{ij}J_{ij}$, where $J_{ij}$ are the generators of the corresponding representation. In particular, $J^2=j(j+N-2)$. How does this fit into your answer? e.g., why can't it be a half-integer? 2) In terms of Dynkin labels, what is the representation that describes the electron for higher $d$? 3) What do you exactly mean by the "spinor" representation? Is it the Clifford one? – AccidentalFourierTransform Jan 27 '18 at 2:36