# The Pauli-Lubanski vector for massless particles

Homework Helper
Gold Member
Does anyone know how to prove that $W^2 = 0$ for massless particles, where W is the Pauli-Lubanski vector

$$W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\rho \sigma} P_\nu$$
??

The books I have either give no proof, or they rely on the expression obtained in the massive case,

$$W^2 = - m^2 \vec{J}^2$$

However, the derivation of this formula relies entirely on working in the rest frame of the particle, so it makes no sense to infer anything about the massless limit from this result.

I have found that, in general,

$$W^2 = - \frac{1}{2} P^2 M_{\mu \nu} M^{\mu \nu} + M_{\mu \nu} M^{\sigma \nu} P^\mu P_{\sigma}$$

In the massless limit, the first term is zero but I can't see why the second term is also zero. I have considered working in the frame

$$p_\mu = (E,0,0,E)$$

but I still don't get zero. The antisymmetry of the Lorentz generators M does not help.

Hans de Vries
Gold Member
There is only a trivial reason: (edit: the reason you gave I see)
The definition of $P_\mu P^\mu=m^2$ as the first Casimir invariant of the
Poincaré group in the massive case.

The Pauli Lubanski vector transforms as a spin vector and the square
of a spin vector is in principle never zero. In the rest frame we have
per definition $W_0=0$ and $W^2=W_x^2+W_y^2+W_z^2$.

Regards, Hans

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strangerep
Does anyone know how to prove that $W^2 = 0$ for massless particles, where W is the Pauli-Lubanski vector

$$W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\rho \sigma} P_\nu$$
??

The books I have either give no proof, or they rely on the expression obtained in the massive case,

$$W^2 = - m^2 \vec{J}^2$$

However, the derivation of this formula relies entirely on working in the rest frame of the
particle, so it makes no sense to infer anything about the massless limit from this result.

Which books? It sounds wrong to me, for the reasons you gave, (but don't
take that as gospel :-)

Analysis of massless Poincare irreps is very different from the massive case,
involving the non-compact E(2) little group instead of SO(3), among other things.

Anyway, in this case one should probably be studying the Casimirs and irreps of
the conformal group instead of Poincare.

Homework Helper
Gold Member
There is only a trivial reason: (edit: the reason you gave I see)
The definition of $P_\mu P^\mu=m^2$ as the first Casimir invariant of the
Poincaré group in the massive case.

The Pauli Lubanski vector transforms as a spin vector and the square
of a spin vector is in principle never zero. In the rest frame we have
per definition $W_0=0$ and $W^2=W_x^2+W_y^2+W_z^2$.

Regards, Hans
Thanks Hans for the feedback.

I am trying to work out the massless case, so I can't go to the rest frame.
I am trying to prove that W^2 =0 in that case.

Are you sure that $$W^2 = 0$$ in the massless case? Massless particles can have spins (actually, helicities). This seems to suggest that $$W^2$$ could be non-zero in this case.

I know how to prove $$W^2= m^2 j^2$$ for arbitrary mass m, where $$\mathbf{j}$$ is the spin operator. However $$\mathbf{j}$$ diverges as $$1/m$$, when mass goes to zero. So, this does not prove that $$W^2 = 0$$.

Hans de Vries
Gold Member
Thanks Hans for the feedback.

I am trying to work out the massless case,

Ok, but mass-less particles can have spin, photons, gluons, mass-less neutrinos,
In general the individual chiral currents.

$$j_R ~=~\psi^\dag_R\sigma^\mu\psi_R, ~~~~~~ j_L ~=~ \psi^\dag_L\tilde{\sigma}^\mu\psi_L$$

transform also light-like but they do have spin.

so I can't go to the rest frame.
I am trying to prove that W^2 =0 in that case.

True but W^2 is reference frame independent. Otherwise, the "official" response
would be strangerep's: look at the reduced group. (E.Wigner, page 197)

http://www.pages.drexel.edu/~gln22/On%20Unitary%20Representations%20of%20the%20Inhomogeneous%20Lorentz%20Group.pdf [Broken]

Regards, Hans

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Homework Helper
Gold Member
Ok, but mass-less particles can have spin, photons, gluons, mass-less neutrinos,
In general the individual chiral currents.

$$j_R ~=~\psi^\dag_R\sigma^\mu\psi_R, ~~~~~~ j_L ~=~ \psi^\dag_L\tilde{\sigma}^\mu\psi_L$$

transform also light-like but they do have spin.

True but W^2 is reference frame independent. Otherwise, the "official" response
would be strangerep's: look at the reduced group. (E.Wigner, page 197)

http://www.pages.drexel.edu/~gln22/On%20Unitary%20Representations%20of%20the%20Inhomogeneous%20Lorentz%20Group.pdf [Broken]

Regards, Hans

All the references I have read say that W^2 is zero for massless particle. Do you agree?
I am not saying that $W^\mu$ is zero, only that W^2 = 0. For massless particles, one then have that

$$W^\mu = \lambda P^\mu$$

where $\lambda$ is the helicity. So I am not saying that massless particles have no spin, simply that W^2 =0.

Once I accept that W^2, I can get all the other results, it's just that step which is unclear to me.

I guess there are two possible answers: either there is a trick that I am missing to show that $W^2=0$ or $W^\mu$ has to be defined differently when the particles are massless.

Regards,

Patrick

EDIT: Thanks for the reference by the way, it's neat to see the original paper!

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strangerep
[...] So I am not saying that massless particles have no spin, simply that W^2 =0.

Once I accept that W^2, I can get all the other results, it's just that step which is unclear to me.

I guess there are two possible answers: either there is a trick that I am missing to show that $W^2=0$ or $W^\mu$ has to be defined differently when the particles are massless.
To classify elementary particles under the Poincare group, one must start from the Casimirs,
i.e., P^2 and W^2, which we call mass squared and total angular momentum
squared (or spin squared), respectively. You can't "define $W^\mu$ differently",
because what matters here is the Casimir W^2. If you insist that W^2=0, all you're doing
is restricting your considerations to spinless particles. (Could that be perhaps what those
unnamed books of yours are doing?)

Homework Helper
Gold Member
Are you sure that $$W^2 = 0$$ in the massless case? Massless particles can have spins (actually, helicities). This seems to suggest that $$W^2$$ could be non-zero in this case.

I know how to prove $$W^2= m^2 j^2$$ for arbitrary mass m, where $$\mathbf{j}$$ is the spin operator. However $$\mathbf{j}$$ diverges as $$1/m$$, when mass goes to zero. So, this does not prove that $$W^2 = 0$$.

Bailin and Love in Supersymmetric Gauge Theory and String Theory say (page 16):

For massless particles $W^2 = P^2 = 0$

Ryder (Quantum Field Theory) says W^2 |k> = 0 .

Baer and Tata (Weak Scale Supersymmetry), page 45, says W^2 =0

Actually, I just found out that there is a detailed discussion in Maggiore (A Modern Introduction to Quantum Field Theory) who proves W^2 =0 in the massless case in a careful manner, so I will check it out.

Homework Helper
Gold Member
To classify elementary particles under the Poincare group, one must start from the Casimirs,
i.e., P^2 and W^2, which we call mass squared and total angular momentum
squared (or spin squared), respectively. You can't "define $W^\mu$ differently",
because what matters here is the Casimir W^2. If you insist that W^2=0, all you're doing
is restricting your considerations to spinless particles. (Could that be perhaps what those
unnamed books of yours are doing?)

No, because the conclusion is that we can write
$$M^\mu = \lambda P^\mu$$ where lambda is the helicity. Again, I never implied that $W^\mu$ is zero, just that $W^2$ =0 .

Hans de Vries
Gold Member
Actually, I just found out that there is a detailed discussion in Maggiore (A Modern Introduction to Quantum Field Theory) who proves W^2 =0 in the massless case in a careful manner, so I will check it out.

You're right, I'm looking at Maggiore right now. So the spin is
not zero but W^2 is.

Regards, Hans

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Hans de Vries
Gold Member
OK. Trying to derive things simpler I think this makes sense:

Associating W and P with spin and momentum we have for
mass less particles:

$$P^0=|\vec{P}|, ~~~~~W^0=|\vec{W}|$$

The general rule $W_\mu P^\mu=0$ now forces W and P to be either parallel
or anti-parallel, which gives you the two helicity states.

Regards, Hans

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strangerep
Associating W and P with spin and momentum we have for
mass less particles:

$$P^0=|\vec{P}|, ~~~~~W^0=|\vec{W}|$$

Hi Hans,

I don't have a copy of Maggiore.
Where did your "$W^0=|\vec{W}|$" come from?

Also, Patrick, when you mentioned Ryder, were you referring to his
eq(2.214) on p64? (If so, that seems to rely on eq(2.210) which is
for the massive case, and therefore subject to the same objections
you raised in your first post.)

EDIT: I was able to read enough of Maggiore on Amazon...
What's happening is this: the little group for massless particles
is E(2), which preserves the momentum vector (k,0,0,k).
The two E(2) "translation" generators $N_1:=K_1- J_2$
and $N_2:=K_2- J_1$ are called A,B by Maggiore.
For the massless case, one can show that

$$W^2 ~=~ -k^2 (A^2 + B^2)$$

Then Magiorre gives the standard argument (also given in
Weinberg vol1) that if we had particle eigenstates of A,B,
(A,B commute, btw), then acting on them with $J_3$
produces a whole continuum of eigenvalues -- unless the
eigenvalues of A,B are both 0. Then, "because we don't observe
such particles
" we restrict the Hilbert space just to those
particle types. This of course also implies that the eigenvalue
of $W^2$ for such states is also 0, but this is a rather
different statement than saying that $W^2=0$ as a more
general algebraic result (which does not hold).

I.e., the restriction to states whose eigenvalues for A,B are 0
is imposed by fiat. It is not "derived" from the general representation
theory of the Poincare group. It is an arbitrary restriction.

Other authors (e.g., Kim, Han, et al) see things differently, and show that
the A,B generators can instead be interpreted as generators of gauge
transformations. Compounding this puzzle, when one tries to construct
a 4-vector massless spin-1 field, one finds it to be impossible. Weinberg
covers this in more detail.

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Hans de Vries
Gold Member
Hi Hans,
I don't have a copy of Maggiore.
Where did your "$W^0=|\vec{W}|$" come from?

Just from how it (spin) transforms in the limit case of beta=1.

\begin{aligned} &\{~P^o~,~ \vec{P}~\} ~~~~&\mbox{transforms like}~~~~&\{~\gamma~,~\vec{\beta}\gamma}~\} \\ &\{~W^o,\vec{W}~\} ~~~~&\mbox{transforms like}~~~~&\{~\beta\gamma~,~\gamma}~\} \end{aligned}

Regards, Hans

Hans de Vries
Gold Member
$$\{~W^o,\vec{W}~\} ~~~~\mbox{transforms like}~~~~\{~\beta\gamma~,~\gamma}~\}$$

One remark, it's the component of W along the velocity which transforms like this.
In general we have of course.

$$\vec{W} ~\implies~ \vec{W}_\bot+\gamma \vec{W}_\|$$

Regards, Hans

Hans de Vries
Gold Member
The fields corresponding to the Poincaré generators:

$$\begin{array}{|c|c|c|c|c|c|} \hline &&&&& \\ & \mbox{transl.} & \mbox{Lorentz} & \mbox{rotation} & \mbox{boost} & \mbox{Pauli Lubanski vect.} \\ \hline &&&&& \\ \mbox{Poincar\'{e}:} & P^\mu & J^{\mu\nu} & J^i & K^i & W^\mu=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,J_{\alpha\beta}\,P_\nu \\ &&&&& \\ \hline &&&&& \\ \mbox{EM field:} & A^\mu & F^{\mu\nu} & B^i & E^i & \,C^\mu\,=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}\,A_\nu \\ &&&&& \\ \hline &&&&& \\ \mbox{Spin \tfrac12 field} & j_V^\mu & M^{\mu\nu} & M^i & P^i & j_A^\mu \\ \mbox{bilinear} & = & = & = & = & = \\ \mbox{observables:} & \bar{\varphi}\, \gamma^\mu \varphi & \bar{\varphi}\, \sigma^{\mu\nu} \varphi & \bar{\varphi}\, \Sigma^i \varphi & \bar{\varphi}\, \Pi^i \varphi & \bar{\varphi}\, \gamma^\mu\gamma^5 \varphi \\ &&&&& \\ \hline \end{array}$$

The field $C^\mu$ is the important Chern Simons current, the spin of the electro-
magnetic field. I wrote a chapter about it in a classical EM context here:

http://physics-quest.org/Book_Chapter_EM2_ChernSimonsSpin.pdf

$M^{\mu\nu}$ is the magnetization $(M^i)$ and polarization $(P^i)$ tensor of the electron field $\varphi$

$\Sigma^i$ and $\Pi^i$ are the spinor rotation and boost generators.

Regards, Hans

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