Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In quantum field theory, one defines a particle as a unitary irreducible representations of the Poincaré group. The study of these representations allows to define the mass and the spin of the particle. However, the spin is not defined the same way for massive particles (where the eigenvalue of the Pauli-Lubanski vector squared are $-m^2 s(s+1)$ where $$s = -S, -S +1, \cdots, S,$$ and $S$ is the spin of the particle (and $m$ the mass)) and massless particles (where the helicity has eigenvalues $\pm \lambda$ and $S=\left|\lambda\right|$ - not to mention continuous spin representations).

With this definition, the "spin" $S$ that appears in both cases doesn't seem to be exactly the same thing. The eigenvalue that labels the irreducible representations are not from the same operator in the massive and the massless case ... however, it's tempting (for me) to see the maximum value of these eigenvalues as the same physical quantity if both cases.

So I wonder if there is a more general definition that would embrace both massive and massless particles (even less practicable)?

share|improve this question
1  
You can think of spin as residual angular momentum in the rest frame. –  Antillar Maximus Mar 16 '12 at 16:01
3  
@AntillarMaximus Well it does not work for a photon (or any massless particle), as it has no rest frame. –  Georg Sievelson Mar 17 '12 at 0:11
    
Related: physics.stackexchange.com/q/1/2451 –  Qmechanic Jul 25 '13 at 13:09
1  
Ultimately the only reason we think of the angular momentum of a gyroscope, the spin of an electron, and the spin of a photon as related to one another is that they couple to one another. The fact that they couple means that only their sum is conserved; they're not conserved individually. –  Ben Crowell Sep 1 '13 at 1:19

1 Answer 1

up vote 9 down vote accepted

In general, quantum numbers are labels of irreducible representations of the relevant symmetry group, not primarily eigenvalues of an otherwise simply defined operator.

But for every label that has a meaningful numerical value in every irreducible representation, one can define a Hermitian operator having it as an eigenvalue, simply by defining it as the sum of the projections to the irreducible subspaces multiplied by the label of this representation. It is not clear whether such an operator has any practical use.

This also holds for the spin. However, one can define the spin in a representation independent way, though not via eigenvalues.

The spin of an irreducible positive energy representation of the Poincare group is $s=(n-1)/2$, where $s$ is the smallest integer such that the representation occurs as part of the Foldy representation in $L^2(R^3,C^n)$ with inner product defined by
$~~~\langle \phi|\psi \rangle:= \displaystyle \int \frac{dp}{\sqrt{p^2+m^2}} \phi(p)^*\psi(p)$.
The Poincare algebra is generated by $p_0,p,J,K$ and acts on this space as follows. $p$ is multiplication by $p$,
$~~~p_0 := \sqrt{m^2+p^2}$,
$~~~J := q \times p + S$,
$~~~K := \frac{1}{2}(p_0 q + q p_0) + \displaystyle\frac{p \times S}{m+p_0}$,
with the position operator $q := i \hbar \partial_p$ and the spin vector $S$ in a unitary irreducible representation of $so(3)$ on the vector space $C^n$ of complex vectors of length $n$, with the same commutation relations as the angular momentum vector.

The Poincare algebra is generated by $p_0,p,J,K$ and acts on this space irreducibly if $m>0$ (thus givning the spin $s$ representation), while it is reducible for $m=0$. Indeed, in the massless case, the helicity
$~~~\lambda := \displaystyle\frac{p\cdot S}{p_0}$,
is central in the universal envelope of the Lie algebra, and the possible eigenvalues of the helicity are $s,s-1,...,-s$, where $s=(n-1)/2$. Therefore, the eigenspaces of the helicity operator carry by restriction unitary representations of the Poincare algebra (of spin $s,s-1,...,0$), which are easily seen to be irreducible.

The Foldy representation also exhibits the massless limit of the massive representations.

Edit: In the massless limit, the formerly irreducible representation becomes reducible. In a gauge theory, the form of the interaction (multiplication by a conserved current) ensures that only the irreducible representation with the highest helicity couples to the other degrees of freedom, so that the lower helicity parts have no influence on the dynamics, are therefore unobservable, and are therefore ignored.

share|improve this answer
    
Well, thanks. Do you have any reference about that ? I found a 1956 article from L. Foldy, but it only deals with massive particles. Aside from that, I understand it is probably a too much blurry question, but do you have some physical explanation of why the quantity that happens to be the spin of the particle in all cases is the same ? –  Georg Sievelson Mar 16 '12 at 22:49
    
I don't think he ever considered the massless case, though it is clear that he gets a representation if m=0. I figured out the details for myself. So if you need a reference, cite the section ''Particle positions and the position operator'' of Chapter B1: The Poincare group of my <A HREF="mat.univie.ac.at/~neum/physfaq/physics-faq.html">; theoretical physics FAQ</A> –  Arnold Neumaier Mar 16 '12 at 22:54
    
''the same'' makes no sense. It is usually called by the same name, as it contributes to the total angular momentum. But sometimes it is said that a photon has no spin byt only helicity.... –  Arnold Neumaier Mar 16 '12 at 22:56
    
Well, I try to understand to what extent they are similar. But the suppression of the 0-helicity mode in the massless limit somehow fulfils my need of a cloudy physical explanation. Thanks. –  Georg Sievelson Mar 17 '12 at 0:10
    
I added to my answer another comment on the massless limit, to clarify why the suppression occurs. –  Arnold Neumaier Mar 18 '12 at 9:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.