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In the book quantum field theory by Itzykson and Zuber, page 53, the authors prove that Dirac's equation has spin 1/2 by showing that if $\psi$ is a solution to Dirac's equation, then compute that $\psi$ is an eigenvectors of $P^\mu P^\mu$ with eigenvalue $m^2$ and of $W^\mu W_\mu$ with eigenvalue $-\frac{3}{4}m^2$. Therefore by Wigner's classification of representations of the Poincare group, we see that Dirac's equation has mass $m$ and spin 1/2. A similar, but easier analysis can be done for the Klein-Gordon equation. (Here $P^\mu$ is the 4-momentum and $W^\mu$ is the Pauli-Lubanski vector.)

Is there a similar analysis to show that Maxwell's equation has helicity $1$ and the Proca equation has spin 1? For example, can we show $W^\mu=P^\mu$ for Maxwell's equation?

$\textbf{Edit:}$After a literature search, I think W.K.Tung's "Group theory in physics" is the best reference on such things.}

$\textbf{Edit:}$ Also, I realized later (after some frustrating failed computation) that this $W^\mu=P^\mu$ business simply doesn't work for the Maxwell's equation, as Weinberg Volume I bottom of page 254 points out, it's impossible to construct a massless field for particle of helicity $\pm 1$.

$\textbf{Edit:}$ I haven't checked yet, but from the first of paragraph of Weinberg Volume I page 255, my guess is that to see why photon has helicity $\pm1$, one has to look at the full antisymmetric Maxwell's tensor.

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    $\begingroup$ I believe the following paper discusses the Pauli Lubanski vector at the level of solutions to the wave equation in quite great detail: arxiv.org/abs/1510.05164. $\endgroup$
    – Gold
    Aug 13, 2022 at 0:12

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I will attempt to give an answer to your question and if there are any points for which something is not clear, please let me know in the comments...

First of all, we shall note that under infinitesimal Lorentz transformations a vector field transforms as $$V^{\alpha}(x)\rightarrow \Big(\delta^{\alpha}_{\beta}- \frac{i}{2}\omega_{\mu\nu}(\mathcal{J}^{\mu\nu})^{\alpha}_{\beta}\Big) V^{\beta}(\Lambda^{-1}x)$$ where the greek indices $\alpha,\beta,...$ are used for the dimensionality of the vector field (which needs not be the same as the spacetime) and $\mu,\nu,...$ are used to denote the spacetime coordinates. The vector on the RHS can be expanded further, given that infinitesimal Lorentz transformations on the spacetime coordinates act as $x^{\mu}\rightarrow \Lambda^{\mu}_{\nu}x^{\nu}$, and hence a (vector valued) function of the spacetime coordinates, upon expanded in its argument, yields the orbital angular momentum operator as the infinitesimal generator, associated to the extrinsic degrees of freedom. I will neglect deriving an expression for the latter, as it has no contribution to the Pauli-Lubanski vector whatsoever (you have probably understood that while studying for the spinor field case).

The generator associated to the intrinsic degrees of freedom must satisfy the commutation relations of the Lorentz algebra. Peskin uses a particular representation for $(\mathcal{J}^{\mu\nu})^{\alpha}_{\beta}$ that satisfies the aforesaid algebra, given by the form $$(\mathcal{J}^{\mu\nu})^{\alpha}_{\beta}= i(g^{\mu\alpha}\delta^{\nu}_{\beta}-g^{\nu\alpha}\delta^{\mu}_{\beta})$$

Massive case: So, the only thing we have to do is simply to derive an expression for the Pauli-Lubanski operator. For convenience, we let it act on the vector field $$w_{\mu}A^{\alpha}= \frac{1}{2}\epsilon_{\mu\nu\rho\sigma}\partial^{\nu} (\mathcal{J}^{\rho\sigma})^{\alpha}_{\beta}A^{\beta}= \frac{1}{2}\epsilon_{\mu\nu\rho\sigma}\partial^{\nu} (g^{\rho\alpha}\delta^{\sigma}_{\beta}-g^{\sigma\alpha}\delta^{\rho}_{\beta}) A^{\beta}= -g^{\alpha\nu}\epsilon_{\mu\nu\rho\sigma}\partial^{\rho}A^{\sigma}$$ where $\partial^{\rho}A^{\sigma}=\frac{1}{2}(\partial^{\rho}A^{\sigma}-\partial^{\sigma}A^{\rho})+\frac{1}{2}(\partial^{\rho}A^{\sigma}+\partial^{\sigma}A^{\rho})$ can be written in terms of the field strength $F^{\rho\sigma}=\partial^{\rho}A^{\sigma}-\partial^{\sigma}A^{\rho}$ and the dual tensor $G^{\rho\sigma}=-\frac{1}{2}\epsilon^{\rho\sigma\mu\nu}F_{\mu\nu}$. Then, the Pauli-Lubanski operator acting on the vector field reduces to $$w_{\mu}A^{\alpha}=g_{\mu\nu}G^{\nu\alpha}$$ and as a last act, acting with another Pauli-Lubanski operator should give us back the vector field multiplied by the eigenvalue $-m^2s(s+1)$ such that we can determine $s$. Indeed, one can show that $$w^{\mu}w_{\mu}A^{\alpha}=-2m^2A^{\alpha}$$ and hence $2=s(s+1)\rightarrow s=1$! This is the desired result for the case of the Proca field (this little handy derivation can be found in https://arxiv.org/abs/1510.05164, as @Gold aptly noted in the comments).

Massless case: In the massless case, according to the previous equation, the square of the Pauli-Lubanski vector vanishes. $$w^{\mu}w_{\mu}=-2m^2 \xrightarrow[\text{}]{m=0} w^{\mu}w_{\mu}=0$$ However, the vanishing of the Pauli-Lubanski vector is not associated in any way with the spin degrees of freedom. So, nothing can be inferred about the spin from the square of the Pauli-Lubanski vector being zero. This does not stop us, however, from deducing the spin of the vector field from the $w^{\mu}w_{\mu}=-2m^2$ equation, as the massless case is a special case of the latter equation and moreover, the vanishing of the mass does not, in any way, interfere with the internal angular momentum degrees of freedom. So, it is safe to say that even massless vector fields have spin 1 as well.

Now, there is another thing I would like to share for the massless case: since the square of the Pauli-Lubanski vector is zero, one can always choose to work in a reference frame in which $p^{\mu}=(p,0,0,p)$ such that $w_{\mu}p^{\mu}=0$ and $w_3=w_0, w_1=w_2=0$ and hence $$w_{\mu}=\lambda p_{\mu}$$ with $\lambda$ being a constant called helicity for the following reason. If we define a unit vector $\eta^{\mu}=(1,0,0,0)$ and dot both sides of the equation above with that unit vector, we can find $$\lambda=\frac{w_{\mu}\eta^{\mu}}{p_{\mu}\eta^{\mu}}=\frac{\vec{J}\cdot\vec{p}}{p}$$ which is the usual definition of helicity and $p$ is the energy carried by the massless vector field.

If there are any questions or if anything is not covered properly, please let me know.

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  • $\begingroup$ Does it also work for the massless case? $\endgroup$ Aug 13, 2022 at 7:05
  • $\begingroup$ I will edit my post now such that it also includes some comments on the massless case! $\endgroup$
    – schris38
    Aug 13, 2022 at 8:57

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