# Quantum Field Theory (Weinberg): lifting the Weyl equation for a massless particle

I am trying to construct the massless quantum field (Weinberg style) for the $(0,\frac{1}{2})$ representation. So I want to right moving Weyl spinor. My massless quantum field in the general case is the following ($a^{\dagger}(p)$ creating a particle and annihilating an antiparticle of momentum $p$ and the opposite for $b^{\dagger}(p)$ ):

$\psi(x) = \displaystyle \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} \sum_{\sigma} ( u_{\sigma}(p) a_{\sigma}(p) e^{-ip_{\mu} x^{\mu}} + v_{\sigma}(p) b_{\sigma}^{\dagger}(p) e^{ip_{\mu} x^{\mu}})$

Now I proved that in the $(A,B)$ representation of $SL(2,\mathbb{C})$ we must have the helicity (massless case) which verify : $\sigma = B-A$ for the intertwiner $u$ and the opposite for $v$ (which corresponds to the antiparticle).

Thus in my $(0,\frac{1}{2})$ representation (the sum on $\sigma$ vanishes), I have :

$\psi(x) = \displaystyle \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} ( u_{\frac{1}{2}}(p) a_{\frac{1}{2}}(p) e^{-ip_{\mu} x^{\mu}} + v_{-\frac{1}{2}}(p) b_{-\frac{1}{2}}^{\dagger}(p) e^{ip_{\mu} x^{\mu}})$

The intertwiners $u_{\frac{1}{2}} = u$ and $v_{-\frac{1}{2}} = v$ are $2 * 1$ vectors (contrary to the Dirac case in the $(\frac{1}{2},\frac{1}{2})$ where they were $4*2$ matrices.

I now want to derive the "Dirac equation" and that is where I am stuck. My goal (I guess) is to prove that in that case we have :

$\sigma^{\mu} \partial_{\mu} \psi(x) = 0$

where $\sigma_{\mu}$ are the Pauli matrices. In order to do that I thus want to prove that the intertwiners verify:

$\sigma^{\mu} p_{\mu} u(p) = 0$

and the same for $v$. I've managed to prove that for a fiducial massless $k = (1,0,0,1)$ we have the following form for the intertwiners ($c^+$ and $c^-$ being two constants):

$u(k) = c^+ \left(\begin{smallmatrix}1 \\ 0\end{smallmatrix}\right)$ and $v(k) = c^- \left(\begin{smallmatrix}1 \\ 0\end{smallmatrix}\right)$

and following from that :

$\sigma^{\mu} k_{\mu} u(k) = 0$

and the same for $v(k)$. The one small (?) piece which remains is to now lift the equation to any $p$ and I cannot find a way to do that. I do not know the form of the fiducial Lorentz transformation $L(p)$ (which lift $k$ to $p$) in the massless case and it seems painful to construct it as well as its $(0,\frac{1}{2})$ representation (because we also need to lift $u(k)$ to $u(p)$).

Hope it is not too long but I wanted to be precise on the matter as it is quite technical. Thanks for your help and time.

• Perhaps I am not understanding: but isn't just a rotation $R$ that sends $k$ to $p$, and hence the (0,1./2) representation is just $exp(i\sigma^i \omega^i/2)$ where $\omega^i$ is the vector associated to the axis of the rotation with an angle $|\omega|$? – TwoBs Nov 10 '17 at 10:59
• I do not think so. If you want to send $k$ to $p$ then you first need to get the first component $p_0$ right: a boost $B$ in the $z$-axis (for the fiducial $k = (1,0,0,1)$) would give you that. Then I guess you just need a rotation $R$ to get the remaining components right. But that does not easily give you the Weyl relation for any $p$ taking it for granted for the fiducial $k$. – Plussoyeur Nov 10 '17 at 11:15
• yes, that's right yuo need a boost and then a rotation. Therefore the polarization would be $exp(i\sigma^i \omega_i/2)exp(\sigma^i \beta^i/2)\,u(k)$, where $\beta$ is the rapidity of the boost. Say the boost is done in the z-direction for convenience (and the rest is taken care by the rotation), then the $exp(\sigma^i \beta^i/2)=exp(\sigma^3 \beta/2)$ which acts diagonally on the component of the spinor $u(k)$. – TwoBs Nov 10 '17 at 22:57
• The explicit expression of these matrices isn't too hard to get. It is worked out explicitly in an almost equivalent case in Peskin Schroeder around eq. 3.49, to see how it is done – TwoBs Nov 10 '17 at 23:03
• Thanks for your help. So would you write down explicitly the matrices as well as the transformation of k and evaluate directly $\sigma_{\mu} p^{\mu} u(p)$ with the hope we get a $\sigma_{\mu} k^{\mu} u(k)$ somewhere which I know vanishes? I tried to derive it as $\sigma_{\mu} p^{\mu} u(p) = \sigma_{\mu} L(p)^{\mu}_{\nu} k^{\nu} D(L(p)) u(k)$ with the expression of $D(L(p))$ we mentioned. However I cannot obtain something close to $\sigma_{\mu} k^{\mu} u(k)$ to show it vanishes. – Plussoyeur Nov 11 '17 at 0:10

As $\gamma$ matrices transform under Lorentz transformation in the following way: $$\Lambda_s^{-1} \gamma^{\mu} \Lambda_s = (\Lambda_v)^{\mu}_{\text{ }\nu} \gamma^{\nu}$$ it is natural to assume $\sigma$ matrices transform in the same way (because one can think of $\bar{\psi_L} \sigma^{\mu} \psi_R$ as a $4$-vector). Thus we have (in the $(0,\frac{1}{2})$ representation): $$\Lambda_s^{-1} \sigma^{\mu} \Lambda_s = (\Lambda_v)^{\mu}_{\text{ }\nu} \sigma^{\nu}$$
Now using the fiducial representation ($\Lambda_v = L(p)$ in the $4$-vector representation): \begin{align} \sigma^{\mu} p_{\mu} u(p) & = k_{\nu} L(p)_{\mu}^{\text{ }\nu} \sigma^{\mu} \Lambda_s(p) u(k) \\ & = k_{\nu} (L(p)^{-1})_{\text{ }\mu}^{\nu} \sigma^{\mu} \Lambda_s(p) u(k) \\ & = \Lambda_s(p) k_{\nu} \sigma^{\nu} u(k) \\ & = 0 \end{align}
Note: I suspected that hardly any explicit calculation was required to lift the result from $k$ to $p$ that is why I am providing this answer.