# A puzzle about relativistic spin

I'm suffering from a confusion about relativistic spin. I don't believe my question has been asked before, and I'm sure I've made some silly mistake somewhere, but I can't spot it. So I'm appealing to the good people here for some guidance.

For the sake of definiteness, I'm confining attention to massive irreducible representations of the Poincaré group. My confusion stems from two ways relativistic spin shows up:

1. as a pseudo-vector $$\tilde{S}_\mu = \frac{1}{m}W_\mu$$, where $$W_\mu$$ is the Pauli-Lubanski pseudo-vector;
2. as an anti-symmetric rank-2 tensor $$\hat{S}^{\mu\nu}$$, the "intrinsic" component of the total angular momentum tensor $$M^{\mu\nu}$$.

(I'm using tildes and hats to distinguish the two ideas.) My starting point is already a bit controversial, since I'm assuming that $$\hat{S}^{\mu\nu}$$ is extracted from $$M^{\mu\nu}$$ via the identity

$$M^{\mu\nu} = Q^\mu P^\nu - Q^\nu P^\mu + \hat{S}^{\mu\nu}$$

where $$P_\mu$$ is the four-momentum and $$Q^\mu = (t, \mathbf{Q})$$, where $$t$$ is the usual time parameter and $$\mathbf{Q}$$ is the Newton-Wigner position operator (which exists for all massive representations).

Now, for the particle at rest in our chosen frame, $$M^{\mu\nu} = S^{\mu\nu}$$ (I drop the hat for the particle's rest frame), and I assume we want the following identities to hold (in this frame):

• $$S^{0i} = 0$$;
• $$\frac{1}{2}\epsilon_{ijk}S^{jk} = S_i = \frac{1}{m}W_i$$.

(I have also dropped the tilde. Also $$W_0 = 0$$ in this frame.) I believe these identities for the particle's rest frame can in fact be proven, given the assumptions above, but in any case they strike me as sensible, if not compulsory.

The puzzle is this. If I now Lorentz-boost the particle so that it has 3-momentum $$\mathbf{P}$$ in our chosen frame, the transformed pseudo-vector $$\tilde{S}_\mu = \Lambda_\mu^\nu S_\nu$$ is given by

$$\tilde{S}_0 = \frac{\mathbf{S.P}}{m}\ , \qquad \tilde{\mathbf{S}} = \mathbf{S} + \frac{(\mathbf{S.P})\mathbf{P}}{m(H + m)} = \frac{H}{m}\mathbf{S} - \frac{\mathbf{P}\times(\mathbf{S}\times\mathbf{P})}{m(H + m)}\ ,$$

(where $$H = \sqrt{\mathbf{P}^2 + m^2}$$, and forgive me, I'm setting $$c=1$$). The transformation on the 3-vector part is perhaps more clearly written as:

$$\tilde{\mathbf{S}}_\parallel = \gamma\mathbf{S}_\parallel\ , \qquad \tilde{\mathbf{S}}_\perp = \mathbf{S}_\perp \ ,$$

(where $$\mathbf{S}_\parallel$$ is the component of $$\mathbf{S}$$ parallel to $$\mathbf{P}$$, etc., and $$H = \gamma m$$).

But if you transform $$S^{\mu\nu}$$ under the same Lorentz-boost, to obtain $$\hat{S}^{\mu\nu} = \Lambda^\mu_\kappa \Lambda^\nu_\lambda S^{\kappa\lambda}$$ you obtain:

$$\hat{S}^{0i} = \frac{1}{m}(\mathbf{S}\times\mathbf{P})\ , \qquad \frac{1}{2}\epsilon_{ijk}\hat{S}^{jk} = \mathbf{S} + \frac{\mathbf{P}\times(\mathbf{S}\times\mathbf{P})}{m(H + m)}\ .$$

If I now define $$\hat{S}_i = \frac{1}{2}\epsilon_{ijk}\hat{S}^{jk}$$, then the second equation above implies

$$\hat{\mathbf{S}}_\parallel = \mathbf{S}_\parallel\ , \qquad \hat{\mathbf{S}}_\perp = \gamma\mathbf{S}_\perp\ .$$

So clearly $$\tilde{\mathbf{S}} = \hat{\mathbf{S}}$$ if and only if $$\mathbf{P} = \mathbf{0}$$. If $$\mathbf{P} \neq \mathbf{0}$$, then we appear to have oppositely transforming spin 3-vectors. I find this puzzling.

I would have naively expected $$\tilde{\mathbf{S}} = \hat{\mathbf{S}}$$, but at the very least I would expect some simple relation between the two. I have somewhere (I forget where, but it wasn't a reliable source) seen the claim that

$$\hat{S}^{\mu\nu} = \varepsilon^{\mu\nu\kappa\lambda}U_\kappa \tilde{S}_\lambda$$

(where $$U^\mu = \frac{1}{m}P^\mu$$ is the particle's four-velocity, and $$\tilde{S}_\mu$$ and $$\hat{S}^{\mu\nu}$$ are defined as above). This identity has two things going for it:

1. it's consistent with the results above; and
2. you can show that it implies $$\tilde{S}_\kappa = \frac{1}{2}\varepsilon_{\kappa\lambda\mu\nu}\hat{S}^{\lambda\mu}U^\nu$$, which is exactly what you would hope, given the definition of $$\hat{S}^{\mu\nu}$$ as the "intrinsic" component of $$M^{\mu\nu}$$ and the definition of $$\tilde{S}_\mu$$ in terms of the Pauli-Lubanski pseudo-vector $$W_\mu$$.

So my first question is simply: is all this correct?

This leads me to my second puzzle. If all this is correct, then one derives the following identities:

$$\mathbf{K} = M^{0i} = t\mathbf{P} - \mathbf{Q}H + \frac{1}{m}(\mathbf{S}\times\mathbf{P})\ , \qquad \mathbf{J} = \frac{1}{2}\epsilon_{ijk}M^{jk} = \mathbf{Q}\times\mathbf{P} + \mathbf{S}_\parallel + \gamma\mathbf{S}_\perp\ .$$ But work by Foldy (1956) showed rather that $$\mathbf{K} = M^{0i} = t\mathbf{P} - \mathbf{Q}H + \frac{\mathbf{S}\times\mathbf{P}}{H + m}\ , \qquad \mathbf{J} = \frac{1}{2}\epsilon_{ijk}M^{jk} = \mathbf{Q}\times\mathbf{P} + \mathbf{S}.$$

(There are other sources for these identities too, e.g. Sudarshan and Mukunda, Classical Dynamics: A Modern Perspective (1974, Ch. 20, equation 47). There are trivial sign disagreements between these authors and the identities above, plus a disagreement about the $$t\mathbf{P}$$ term, but these are not substantial.) I don't think the conflict is a simple misunderstanding: $$\mathbf{Q}$$ here for me and for Foldy is the Newton-Wigner position operator, and $$\mathbf{S}$$ for me and for Foldy is the spin 3-vector in the particle's own rest frame.

Edit: I should add that the two rival pairs of identities for $$\mathbf{K}$$ and $$\mathbf{J}$$ yield the same Pauli-Lubanski pseudo-vector. But they can’t both be right!

So my second question is: what has gone wrong here? Many thanks in advance.

• Oh, this can of worms. It has been a while and so I am not sure if your sign issue is due to the spin v.s. helicity/chirality difference, or it might be just down to different authors having different metric signature conventions. In any case, the treatment of spin in QFT is best done by considering 1) Wigner and 2) Weyl, both group theoretical, and then getting to the usual expressions. Then you can map that to the angular momentum tensor; this last beast is very annoying because it actually has 4 indices, two of them actually only being labels and the other two are Lorentz indices. Commented Jul 19 at 14:08
• I do not have time to read again this paper, but I think you should find the solutions of your puzzle: "Simple derivation of the Newton-Wigner position operator" by Thomas F. Jordan, Journal of Mathematical Physics 21, 2028 (1980); doi: 10.1063/1.524712 View online: dx.doi.org/10.1063/1.524712 since it connects all formulations you are using Commented Jul 19 at 16:40
• Alas, no joy with Jordan (1980). Commented Jul 19 at 18:45
• The point stays in the definition of $Q$! Your $Q$ referred to Foldy’s one is $Q+ \frac{S\times P}{m(m+H)}$. (Maybe I am wrong for a sign, I am quite tired…) Commented Jul 19 at 21:27
• In summary: The Newton Wigner operator is Foldy’s operator $Q$ and not yours. I am too tired to try to interpret this fact. Perhaps tomorrow. Commented Jul 19 at 21:38

With all thanks to Valter Moretti, I believe we now have a complete solution to my original puzzle. In the identities for $$\mathbf{K}$$ and $$\mathbf{J}$$ that disagreed with the celebrated ones due to Foldy, I mistakenly assumed that the position operator was the Newton-Wigner position operator. In fact it is what Fleming (1965) calls the "centre of inertia". (So it was a simple misunderstanding after all.) Correcting for the discrepancy between the two, one obtains the Foldy identities.

I'd like to lay out here in some more detail what's going on, partly to make the matter clear in my own head, partly because I think it provides a nifty derivation of the Foldy identities. The name of the game is to derive identities for $$\mathbf{K}$$ and $$\mathbf{J}$$ in an arbitrary Lorentz frame by Lorentz-transforming obvious expressions for them in a frame at which the particle is at rest.

I'll consider the spinless case first, since it leads in nicely to the general case. So for a spinless particle, in a frame at which the particle is at rest we have

$$\mathbf{k} = l^{0i} = -m\mathbf{q}\ , \qquad \mathbf{j} = \frac{1}{2}\epsilon_{ijk}l^{jk} = \mathbf{0} \ ,$$

where $$l^{\mu\nu}$$ is the relativistic angular momentum tensor, $$m$$ is the particle's mass and $$\mathbf{q}$$ is its (unchanging) position in our frame. I am using lowercase letters to denote all quantities for the particle at rest. (I will not be changing frames at all; all transformations will be "active", on the particle considered in the same frame.) The expression for $$\mathbf{k}$$ may be justified by considering the non-relativistic case, and reasoning that the relativistic form ought to coincide with it when the particle is at rest.

By Lorentz-transforming $$l^{\mu\nu} \mapsto L^{\mu\nu} = \Lambda^\mu_\kappa \Lambda^\nu_\lambda l^{\kappa\lambda}$$ so that the particle acquires 3-momentum $$\mathbf{P} = \gamma m\mathbf{v}$$, we obtain

$$\begin{array}{rclrcl} \mathbf{K}_\parallel & =& \mathbf{k}_\parallel = -m\mathbf{q}_\parallel \ , \qquad & \mathbf{J}_\parallel & =& \mathbf{j}_\parallel = \mathbf{0} \ , \\ \mathbf{K}_\perp & =& \gamma(\mathbf{k}_\perp -\mathbf{v}\times\mathbf{j}_\perp) = -H\mathbf{q}_\perp\ , \qquad & \mathbf{J}_\perp & =& \gamma(\mathbf{j}_\perp + \mathbf{v}\times\mathbf{k}_\perp) = \mathbf{q}_\perp\times\mathbf{P} . \end{array}$$

These identities look wrong at first blush, but we must re-express $$\mathbf{q}$$ in terms of quantities more suitable, now that the particle is in motion. The particle's new trajectory is given by

$$\mathbf{q}'_\parallel(t') = \gamma(\mathbf{q}_\parallel + \mathbf{v}t)\ , \qquad \mathbf{q}'_\perp(t') = \mathbf{q}_\perp \ , \qquad t' = \gamma(t + \mathbf{v}.\mathbf{q}_\parallel)\ .$$

The quantity we are interested in is $$\mathbf{Q}(t) = \mathbf{q}'(t)$$; this is the (new) position of the particle in our (same) frame at the same time $$t$$. It may be checked that

$$\mathbf{Q}_\parallel = \mathbf{q}'_\parallel(t') - {\mathbf{v}}(t' - t) = \frac{1}{\gamma}(\mathbf{q}_\parallel + \mathbf{v}t)\ .$$ Thus $$m\mathbf{q}_\parallel = \mathbf{Q}_\parallel H - t\mathbf{P}$$. Also $$\mathbf{Q}_\perp = \mathbf{q}_\perp$$. So we finally obtain

$$\mathbf{K} = L^{0i} = t\mathbf{P} - \mathbf{Q}H\ , \qquad \mathbf{J} = \frac{1}{2}\epsilon_{ijk}L^{jk} = \mathbf{Q}\times\mathbf{P}\ .$$

That concludes the spinless case. To introduce spin, we now set, when the particle is at rest,

$$\mathbf{k} = m^{0i} = -m\mathbf{q}\ , \qquad \mathbf{j} = \frac{1}{2}\epsilon_{ijk}m^{jk} = \mathbf{S} \ .$$

Let us decompose $$m^{\mu\nu} = l^{\mu\nu} + s^{\mu\nu}$$ in the obvious way into orbital and intrinsic components, so that $$l^{\mu\nu}$$ here is as in the spinless case. We now transform $$m^{\mu\nu} \mapsto M^{\mu\nu} = \Lambda^\mu_\kappa \Lambda^\nu_\lambda m^{\kappa\lambda}$$ as before. Since the transform is linear, we can take $$l^{\mu\nu}$$ and $$s^{\mu\nu}$$ separately. $$L^{\mu\nu}$$ is as for the spinless case, and, as I showed in my OP, for $$S^{\mu\nu}$$ you get

$$S^{0i} = \frac{1}{m}(\mathbf{S}\times\mathbf{P})\ , \qquad \frac{1}{2}\epsilon_{ijk}S^{jk} = \mathbf{S}_\parallel + \gamma\mathbf{S}_\perp \ .$$ So, putting it all together, $$\mathbf{K} = M^{0i} = t\mathbf{P} - \mathbf{X}H + \frac{1}{m}(\mathbf{S}\times\mathbf{P})\ , \qquad \mathbf{J} = \frac{1}{2}\epsilon_{ijk}M^{jk} = \mathbf{X}\times\mathbf{P} + \mathbf{S}_\parallel + \gamma\mathbf{S}_\perp\ .$$

I am using '$$\mathbf{X}$$' instead of '$$\mathbf{Q}$$' for the position. The first thing to emphasise is that these identities are correct. The second thing to emphasise is that $$\mathbf{X}$$ here is not Newton-Wigner position, as I mistakenly believed. (In the spinless case above, $$\mathbf{Q}$$ is NW position.) We obtained $$\mathbf{X}$$ essentially from a straightforward Lorentz transformation, so $$\mathbf{X}(t)$$ gives us the spatial location, in our frame, of the worldline defined by the particle's centre of mass in its own frame when the time (in our frame) is $$t$$.

Fleming (1965) calls this the particle's "centre of inertia". It has an obviously natural physical interpretation, but a major drawback is that, if the particle has spin, the components of $$\mathbf{X}$$ don't commute (either as quantum operators, or as classical quantities in terms of the Poisson bracket). Fleming considers two other position operators: the "centre of mass" (I've also seen this called the "centre of energy"); and the "centre of spin", which is the Newton-Wigner position, which is what I’m calling $$\mathbf{Q}$$. Newton-Wigner position is distinguished by the fact that its components do commute, no matter the particle's spin.

As Fleming shows (eq. 6.2, and as Valter Moretti saw), $$\mathbf{X}$$ and $$\mathbf{Q}$$ are related by

$$\mathbf{X} = \mathbf{Q} + \frac{\mathbf{S}\times\mathbf{P}}{m(H + m)}\ .$$

(Edit: it’s worth mentioning that the size of this modification term never exceeds $$\frac{|\mathbf{S}|}{mc}$$, the “Møller disc” radius, which is of the order of the Compton wavelength.) By making this substitution, one finally recovers the Foldy identities:

$$\mathbf{K} = M^{0i} = t\mathbf{P} - \mathbf{Q}H + \frac{\mathbf{S}\times\mathbf{P}}{H + m}\ , \qquad \mathbf{J} = \frac{1}{2}\epsilon_{ijk}M^{jk} = \mathbf{Q}\times\mathbf{P} + \mathbf{S}\ .$$

As well as Fleming's paper, I also found this very recent paper by Schwartz and Giulini (2020) very helpful. They justify very convincingly Fleming's name "centre of spin" for the NW position. (The condition on being a centre of spin is essentially that $$\mathbf{J}$$ takes the Foldy form.)

An important lesson I've drawn from this confusion is that if one decomposes the relativistic angular momentum tensor $$M^{\mu\nu}$$ into orbital and intrinsic components in line with Newton and Wigner, then the components taken separately don't transform straightforwardly, as $$M^{\mu\nu}$$ itself does. This is not to say that they cannot be covariantly characterised -- they can, as I believe Fleming originally showed.

• Ok, my answer is not necessary any more, you wrote everything I was about writing and even more than it. I did not know the references by Fleming and Schwartz-Giulini, thanks. Commented Jul 21 at 7:40
• @ValterMoretti okay, then I have shamelessly accepted my own answer. You were an enormous help, many thanks indeed. Commented Jul 21 at 19:37
• It is ok to accept your answer because it is correct. You’re welcome, NW observable is part of my current research. Commented Jul 21 at 19:40
• Actually I am studying the notion of localisation in quantum relativistic theories and NW observable enters the game in several ways. A paper is this one link.springer.com/article/10.1007/s11005-023-01689-5 a more recent one is this link.springer.com/article/10.1007/s11005-024-01817-9 Commented Jul 21 at 19:50
• Thank you, I will read with great interest — not least because you claim to overcome the issue of Hegerfeldt spreading! Commented Jul 21 at 19:54