It seems that I made an error and it was $S^{ij}$ that was used after all. I will not delete the question though because even though it is erroneous, the answer given below is rather insightful.

End of edit.


While reading the textbook "A Modern Introduction to Quantum Field Theory" by Maggiore, I found that the conserved current associated with Lorentz transformations for scalar fields is [p.52, eq.(3.49)] $$j^{(\rho\sigma)\mu}=x^\rho T^{\mu\sigma}-x^\sigma T^{\mu\rho}\ \ \ \ \ \ \ \ (1)$$ with $T^{\mu\nu}$ the energy-momentum tensor. For $\rho, \sigma=1, 2, 3$, the Lorentz transformation is a rotation [Tong's notes, p.99], hence we get the total angular momentum. For scalar particles and $\mu=0$, this gives the orbital angular momentum.

Now, in the case of a Dirac field $\Psi$, this current has an extra term: $$j^{(\rho\sigma)\mu}=x^\rho T^{\mu\sigma}-x^\sigma T^{\mu\rho}-i \bar{\Psi}\gamma^\mu S^{\rho\sigma}\Psi\ \ \ \ \ \ \ \ \ \ \ (2)$$ Now, since I expected spin to correspond to rotations, I expected that upon quantising the Dirac field, we would define a spin operator of the form $$-i\int d^3x\ \bar{\Psi}\gamma^0 S^{ij}\Psi\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ in analogy with the case discussed above. What actually happens is that Maggiore instead takes [p.90, eq.(4.40) and p.31, eq.(2.86)]: $$-i\int d^3x\ \bar{\Psi}\gamma^0 S^{i0}\Psi\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ to be the spin operator, with $S^{i0}$ having $\frac{1}{2}\sigma^i$ on its diagonal. But I know that for indices, say, $\rho=i, \sigma=0$, this is a boost, not a rotation.

Is my analysis correct? Equation (3) seems to give the correct spin for a Dirac field when acted on a 1-particle state with zero momentum, but through the above analogy, I expected equation (3) to be the relevant here. It also seems like (4) has to do with boosts and not rotations. What's happening?


It's sort of both. Spin directly corresponds to rotations, but through them also does define the behaviour of boosts. I can only remember a loose verbal description rather than the rigorous maths, but I feel that might be what you really want anyway.

The relevant thing to look up if you can be bothered is "Wigner's classification" of the irreducible representations of the Poincare group. Essentially, the different particles of different spin transform as irreducible representations of the Poincare group (the Lorentz group plus space and time translations - it is to the Lorentz group what the Galilean group is to the Euclidean group). It's a little difficult to explain, but the gist is that you assign to every future-pointing timelike vector of a particular mass - i.e. all the possible momenta of a particle - a finite-dimensional subspace, which form in total of course an infinite-dimensional space. Boosts take states between the subspaces corresponding to different momenta which you would expect to transform into each other from the vector transformations of the momenta. This means that if a Lorentz transformation leaves a particular momentum invariant, then the Lorentz boost acts to transform vectors only within the subspace attached to that momentum.

Now what class of transformations leave a particular momentum invariant? It's the class which correspond to the spatial rotations in the rest frame of that momentum! The finite-dimensional subspace attached to each momentum vector therefore carries a representation of the 3D rotation group. It was one of Wigner's big insights that for the overall representation to be irreducible these representations must themselves be irreducible. Spin classifies the finite-dimensional irreducible representations of the 3D rotation group, but through this construction - Wigner's construction - it also, along with mass, classifies the infinite-dimensional irreducible representations of the Poincare group. (Then there are also some subtleties to do with parity transformations.)

The physical point to remember, which is counterintuitive, is that the separation between boost and rotation is frame-dependent: what to one observer is a pure spatial rotation, to another observer is a bit of a rotation and a bit of a boost. (Though a pure rotation can never be transformed into a pure boost, as a pure spatial separation can never be transformed into a pure timelike separation.) The spin operators you're applying are "rest frame", in a sense, so this is why they work on zero-momentum states: then you're transforming in the rest frame of the momentum. When you have states of different momenta, then under a Lorentz boost the effects of the different momenta transforming into each other and the spin rotating around the momenta combine in intricate, complicated ways.

This might not make much sense to you. The issue is that the maths is fundamentally very complicated. I'd thoroughly recommend "Group Theory in Physics" by Wu-Ki Tung if you have enough time to sink into this to bother with a whole book.

Edit: Some maths:

So say you have two vector spaces. One is the vector space formed of particles with a particular momentum and a particular spin. Label the momentum $p$, $p^2=m$. Label the spin $\alpha = 1,2$. Call the components of a vector in this space $a_\alpha(p)$. The second vector space is the space of bispinor profiles. This is labelled by a position, $x$, and a bispinor index $l=1,2,3,4$. Call components in this space $\psi_l(x)$. Lorentz transformations in this space, $\Lambda$, are generated by the rule

$$ \Lambda: \quad \psi_l(x) \mapsto \sum_{m }S_{lm}(\Lambda) \psi_m (\Lambda^{-1} x) , $$

i.e. as a field. The $a_\alpha(p)$ transform according to a rule which is a little more convoluted to write down. Essentially one defines the $\alpha$ indices refer to its occupation of the space in the rest frame of the momentum. It really is a little too complicated for me to summarise here exactly how it works, the fundamental point is just that it's different to how the field transforms - it's an entirely different representation on a different space. But, it turns out, that there exists an isomorphism between the spin space and a subspace of the bispinor fields, written,

$$ \psi_l(x) = \sum_\alpha \int d^3p \; \varphi_{l;\alpha}(x;p) a_\alpha(p) . $$

which "respects the group transformation properties" - i.e. if you start in the momentum/spin space, then Lorentz transform, then transform to the field space, you get the same answer as if you start in the momentum/spin space, then transform to the field space, then Lorentz transform. This subspace is the space of solutions to the Dirac equation. (Again, dismissing some subtleties about parity.)

So spin corresponds to spatial rotations in a sense, but the S matrices of the bispinor field doesn't really know much about that, and handles all types of Lorentz transformation. It's only in momentum space, not in position space, where the "rotation-only" nature of spin becomes apparent.

  • $\begingroup$ Thanks for the answer. Although I did not understand all of it, the part about boosts and rotations "mixing" (in the sense described in the answer) makes sense to me through the fact that, say, the commutator of two boosts is a rotation (schematically [K, K]~J). As I found, $S^{i0}=diag(i\sigma^i,i\sigma^i)/2$ and $S^{ij}=\epsilon_{ijk}S^k$ with $S^k=diag(\sigma^k,\sigma^k)/2$. So, at the end of the day, equations (3) and (4) of my question contain single Pauli matrices, but differ by an $i$ and I wonder what the physical implications of this are when trying to find the spin of a spin state. $\endgroup$ Nov 28 '20 at 2:39
  • $\begingroup$ All in all, it is true that the physical role of the differentiation between the boosts and rotation in equations (3) and (4) is not clear to me yet, but again thank you for your input. $\endgroup$ Nov 28 '20 at 2:41
  • $\begingroup$ An issue I think is that the S matrices you're using don't directly correspond with "spin" mathematically. Particle states sort of transform in two different ways: as bispinor fields, which are what transform using your S matrices, and as momenta and spins. It's a highly non-trivial fact that the two transformations can be made equivalent: they're originally defined very differently. "Spin" is a characteristic of the momentum-space transformation, which then induces a corresponding bispinor-field transformation in the subspace of bispinor-fields which are Dirac eq. solutions. $\endgroup$
    – J_B_Phys
    Nov 28 '20 at 2:49
  • $\begingroup$ For the physical moral lesson, I guess think of the S matrices and "spin" as conjugate physical quantities, related similarly to position and momentum. The fact that their transformations "fit together" works in a similar way that it turns out that plane waves transform between each other just like vectors, which corresponds to the fact that the relationship between position and momentum is Lorentz invariant. $\endgroup$
    – J_B_Phys
    Nov 28 '20 at 2:51
  • 1
    $\begingroup$ Alright I'll write a few sums, give me a sec. $\endgroup$
    – J_B_Phys
    Nov 28 '20 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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