Let $u_\sigma(p)$ be a Dirac spinor. As far as I know, it transforms under changes of reference frame according to $$ u_\sigma(p)=S(\Lambda)u_\sigma(\Lambda p)\tag{1} $$ where the $\sigma$ label doesn't mix. Why is this? shouldn't the polarisations be frame-dependent? After all, the "spin quantisation axis" is frame dependent.

Put it another way: the relation above is equivalent to $$ U(\Lambda)|p,\sigma\rangle=|\Lambda p,\sigma\rangle\tag{2} $$ without $\sigma$ mixing. To me, one-particle states should transform according to $$ U(\Lambda)|p,\sigma\rangle\stackrel?=\sum_{\sigma'}D_{\sigma\sigma'}(\Lambda)|\Lambda p,\sigma'\rangle\tag{3} $$ though this is obviously not the case. If this were true, then we would have $$ u_\sigma(p)\stackrel?=S(\Lambda)\sum_{\sigma'}D_{\sigma\sigma'}(\Lambda)u_{\sigma'}(\Lambda p)\tag{4} $$ instead of $(1)$.

Question: why don't polarisations mix under Lorentz transformations (in neither $(1)$ nor $(2)$)?


As pointed out by Blazej, the spin components $\sigma$ do mix under Lorentz transformation, and the correct law is $$ u_\sigma(p)=\sum_{\sigma'}M_{\sigma\sigma'}u_{\sigma'}(\Lambda p) $$ for some matrix $M$ (which is actually related to a Wigner's little group matrix, but whose form is not that relevant to me; the important part is that the $\sigma$ components mix, and not what is the actual matrix that mixes them).

My concern is that this is not what I find online: for example, see this answer in physics.SE (last equation). Also, see this wikipedia article. Who is right and who is wrong?

  • $\begingroup$ Note that $(3)$ is also not corect transformation rule. It should be $U(\Lambda)|p s \rangle = \sum_{s'}D(W(\Lambda,p))_{s's}|\Lambda p, s' \rangle$, where $W(\Lambda, p)$ is Wigner's rotation. Confront Weinberg, volume 1, chapter 2 for details. $\endgroup$
    – Blazej
    Aug 10 '16 at 14:34
  • $\begingroup$ @Blazej Thank you! I dont really see the difference between your formula and mine :-S you just wrote the $D$ matrix in full detail, but the content of both our formulas are the same, right? (I wrote $D_{\sigma\sigma'}(\Lambda)$ and you wrote $D_{ss'}(W(\Lambda,p))$, but these are the same, right?). $\endgroup$ Aug 10 '16 at 14:39

We can answer this question in more generality: what are the transformation properties for the polarizations associated to a massive field? (I am restricting to massive because the polarizations for massless particles with spin greater or equal than 1 require the discussion of gauge invariance; I'll leave this to another day).

The polarizations can be defined with no reference to the field equation: they are defined as matrix elements for a field between the vacuum and the one-particle states: $$ \langle 0| \psi_\ell(0) | \mathbf{p},\sigma\rangle \propto u^\sigma_\ell(\mathbf{p}) $$ where the proportionality factor is a constant (known as wavefunction renormalization) in the relativistic normalization for the state $|\mathbf{p},\sigma\rangle$. Here the index $\ell$ is the index in the Lorentz representation carried by the field $\psi_\ell$ (for example, $\ell=\mu$ for a 4-vector, $\ell=\alpha$ for a dirac spinor, more generally $\ell=(\alpha,\beta)$ is a pair of indexes in the $(A,B)$ representation of $SO(3,1)\sim SU(2)\times SU(2)$). The polarizations carry also another index, $\sigma$, which represents the spin of the particle. More precisely, it is the little-group index carried by the particle, either the spin or the helicity. This definition immediately tells us how $u^\sigma_\ell$ transforms given that $\psi$ carries a Lorentz representation $$ U(\Lambda)\psi_\ell(x) U^{-1}(\Lambda)=D(\Lambda^{-1})_{\ell\ell^\prime}\psi_{\ell^\prime}(\Lambda x) $$ and the one-particle state transform with respect to the little group with a Wigner rotation $$ U(\Lambda)|\mathbf{p},\sigma\rangle = \mathcal{L}_{\sigma^\prime\sigma}(W(\Lambda),p)|\mathbf{p}_\Lambda, \sigma^\prime\rangle $$ which imply $$ D(\Lambda)_{\ell\ell^\prime}u^\sigma_{\ell^\prime}(\mathbf{p})=u^{\sigma^\prime}_{\ell}(\mathbf{p}_\Lambda)\mathcal{L}_{\sigma^\prime\sigma}(W(\Lambda),p) $$ where $\mathbf{p}_\Lambda$ is 3-vector part of the 4-vector $\Lambda p$. (One way to read this equation is by saying that the polarizations transform on the left under Lorentz and on the right under the little-group transformations: this has to be so such that one can convert the Lorentz indexes of correlations functions of fields into the little group indexes of the scattering matrix $S$ as dictated by the LSZ reduction formula).

This answer your question. But in fact we can say more: those transformation properties are constructive since they allow you to determine explicitly the polarization (and show that they satisfy certain equations, e.g. Dirac for spin-1/2,...) as it was shown long ago in the 60's by Weinberg (see the discussion in his textbook on QFT vol.1 chapter 5). For example, take $k=(m,\mathbf{0})$ (for a massive particle) and apply the canonical Lorentz transformation $L=\Lambda$ that brings it to $p=(E,\mathbf{p})=L\,k$. In this case the Wigner rotation is trivial, $W=1$, and hence $$ u^{\sigma}_{\ell}(\mathbf{p})=D_{\ell\ell^\prime}(L)u^\sigma_{\ell^\prime}(\mathbf{0}) $$ which means that we just need to know them at zero momentum (or, for massless particles with respect to the reference vector used for the little group). Moreover, for an abitrary rotation $\Lambda=R$ we have $W=R$ for any $p$ so that $$ D_{\ell\ell^\prime}(R)u_{\ell^\prime}^\sigma=u^{\sigma^\prime}_\ell(\mathbf{0})\mathcal{L}_{\sigma^\prime\sigma}(R) $$ Taking diagonal the rotations around the $z$ axis, $\mathcal{L}_{\sigma^\prime\sigma}(R_z)=e^{i\sigma\theta}\delta_{\sigma^\prime\sigma}$, the polarization can be extracted by the infinitesimal $z$-rotation $$ D_{\ell\ell^\prime}(J^z) u^\sigma_{\ell^\prime}{\mathbf{0}}=\sigma u^\sigma_{\ell}(\mathbf{0})\,. $$

Let me just give an instructive example: a spin-1 massive state (where $\ell$ is an index in the irrep $(1/2,1/2)$, that is $\ell=\mu=0,1,2,3$ is 4-vector index) has a the 3-dimensional representation of $J^z$ where $D(J^z)_{ij}=-i\epsilon_{3ij}$ and $D(J^z)_{00}=D(J^z)_{i0}=0$ so that the polarizations $$ \epsilon_{\mu}^{\pm}(\mathbf{0})=\frac{1}{\sqrt{2}}(0,\mp 1,-1,0)^T\,,\qquad \epsilon_\mu^0=(0,0,0,1)^T\,. $$ solve the desired equations above. Clearly they satisfy $p^\mu \epsilon_\mu^\sigma=0$, once we boost $(m,\mathbf{0})$ to $p$. Therefore, the matrix element $$ \Psi_\mu(x)\equiv \langle 0| \psi_\mu(x) | \mathbf{p},\sigma\rangle =e^{ipx} \langle 0| \psi_\mu(0) | \mathbf{p},\sigma\rangle \propto e^{ipx} u^\sigma_\mu(\mathbf{p}) $$ satisfies $$ (\square+m^2)\Psi_\mu(x)=0\,,\qquad \partial_\mu\Psi^\mu(x)=0 $$ which are derived rather that used as starting point.

The same can be done for any spin, in particular for the spin-1/2 and see that they solve the Dirac equation. More generally, since the Lorentz group $SO(3,1)\sim SU(2)_A \times SU(2)_B$ the angular momentum is given by $J=J_A+J_B$ which tells us that $$ D(J_A)_{\alpha\alpha^\prime}u^\sigma_{\alpha^\prime\beta}(\mathbf{0})+D(J_B)_{\beta\beta^\prime}u^\sigma_{\alpha\beta^\prime}(\mathbf{0})=\mathcal{L}(J_S)_{\sigma^\prime\sigma} u^{\sigma^\prime}_{\alpha\beta}(\mathbf{0}) $$ where $\ell=(\alpha,\beta)$. In other words, the polarizations are (proportional to) the Clebsch Gordan coefficient for the spin $S$ found inside $A\otimes B$ $$u^\sigma_{\alpha\beta}(\mathbf{0})\propto C^{(S)\sigma}_{(AB)\alpha\beta} $$ Several of the properties of the polarizations come indeed from the unitary condition for these Clebsch Gordan coefficients.

  • $\begingroup$ So, to put it in simple terms, the short answer is that those links at the bottom of my post are simply wrong? Someone should edit the wikipedia entry then, right? $\endgroup$ Oct 29 '16 at 12:14
  • $\begingroup$ @AccidentalFourierTransform No, those links are correct, and so its my answer. You are missing the point that I have tried to stressed carefully: that the polarizations have two indexes, one spinorial which thus transform as in those links, and one little-group index that transforms too, but with the Wigner rotations. There is no contradiction, those links simply discuss the case of objects with a single spinorial index (as e.g. a spin 1/2 field) while polarizations have another, extra, little group index. You can even see that my second equation is in agreement with what claimed in the links $\endgroup$
    – TwoBs
    Oct 29 '16 at 21:10
  • $\begingroup$ @AccidentalFourierTransform sorry, I have looked only a the wiki page which is correct as I was saying in the comment just posted. On the other hand, the old answer on SEhttp://physics.stackexchange.com/questions/87575/unitary-lorentz-transformation-on-quantized-dirac-spinor/228078#228078 that you were linking is just simply wrong. $\endgroup$
    – TwoBs
    Oct 29 '16 at 21:18

For first, note that u(ps)u(ps) are not states in Hilbert space of some quantum theory. Instead they are solution to certain equation, namely $(\gamma^{\mu}p_{\mu}−m)u(ps)=0$(equivalently: $u(ps)e^{−ipx}$ solves Dirac equation). Therefore bracket notation is not really in place (though tempting!) Second remark is that you should think about how is spin even defined? Usual convention says that spin of a moving particle is defined as its spin in the reference frame where it doesn't move. Now let $p_0=(m,0,0,0)$ and define $u(p_0s)$ as solution of Dirac equation describing particle at rest with spin $s$. For any rotation (which is just Lorentz transformation $\Lambda$ such that $\Lambda p_0=p_0$) we have relation familiar from ordinary QM $$ S(\Lambda)u(p_0s) = D(\Lambda)_{s's}u(p_0s') $$ Then for any possible momentum $p$ of this particle we choose some standard boost $\Lambda_0(p)$ which transforms $p_0$ to $p$. Standard choice is just boost in the $\vec p$ direction. Now define $u(ps)=S(\Lambda_0(p))u(p_0s)$. This is solution describing moving particle with spin $s$ in its rest frame. Now let's choose some arbitrary Lorentz transformation $\Lambda$ and calculate its action $$ S(\Lambda) u(ps) = S(\Lambda \Lambda_0(p)) u(p_0 s)=S(\Lambda_0(\Lambda p))S(W(\Lambda,p)) u(p_0s),$$ where $W(\Lambda,p)=\Lambda_0(\Lambda p)^{-1}\Lambda \Lambda_0(p)$. This Lorentz transformation is called Wigner's rotation. It is easy to see that $W(\Lambda,p)p_0=p_0$, so previous formula applies. Therefore $$ S(\Lambda)u(ps) = S(\Lambda_0(\Lambda p)) \sum_{s'}D_{s's}(W(\Lambda,p)) u(p_0s')=\sum_{s'}D_{s's}(W(\Lambda,p)) u(\Lambda p,s').$$ The second equality follows from the definition of $u(ps)$ for $p \neq p_0$.

  • $\begingroup$ So, in simple terms, you are saying that the $s$ components do mix under Lorentz transformations, right? Or put it another way, the right formula is $Su_s=\sum_{s'}D_{ss'}u_{s'}$, instead of $Su_s=u_s$, right? $\endgroup$ Aug 10 '16 at 18:54
  • $\begingroup$ maybe I should have written $Su_s=\sum_{s'}M_{ss'}u_{s'}$ for some matrix $M$. I wanted to write the general structure of the equation, without worrying about the details. To me, the important part is that they mix, but not that the matrix is $D$ or some other matrix. Anyway, I shall edit my post in a minute or two... $\endgroup$ Aug 10 '16 at 18:59
  • $\begingroup$ They do mix but not through $D(\Lambda)$ but rather than through $D(W(\Lambda,p))$. Please note several important facts. First, that this Wigner rotation depends on momentum of the particle explicitly. Second, that $D(\Lambda)$ wouldn't even make sense because Wigner's matrix $D$ is defined for rotations and not for general Lorentz transformations. Second of all, using these formulas you can convince yourself that if you take a particle at rest with some spin configuration, boost it once and then again in other direction you will end up with something with different polatization. $\endgroup$
    – Blazej
    Aug 10 '16 at 18:59

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