# Rotation of plane wave spinor versus plane wave spinor with rotated momentum

Suppose I have a massless fermion with momentum $$p^\mu = (E,\, E\cos\theta,\, 0,\, E\sin\theta)$$. There are two ways to write the plane wave spinor $$u(p)$$ with respect to the spin $$\xi$$, and I seem to find different answers:

1. Plug the momentum into (e.g. Peskin (3.50)) $$u(p) = \begin{pmatrix}\sqrt{p\cdot\sigma}\,\xi\\ \sqrt{p\cdot\bar\sigma}\,\xi \end{pmatrix}$$ For small $$\theta$$ and $$\xi = (1,0)$$, this gives: $$u(p) = \sqrt{2E} \begin{pmatrix} 0\\ \theta/2\\ 1\\ \theta/2 \end{pmatrix}$$ For $$\theta\to 0$$ this matches the expected result for a massless plane wave spinor along the $$z$$-direction.

2. Alternatively, we may start with a massless plane wave spinor along the $$z$$ direction, $$u(k)$$ with $$k^\mu = (E, \, 0, \, 0, \, E)$$. We can then rotate this spinor with the appropriate representation of the Lorentz transformation (Peskin (3.27) and (3.30)). Taking again the case $$\xi=(1,0)$$: $$u(p) = \Lambda_{\frac 12}(\omega) u(k) \, = e^{-\frac i2 \omega_{\mu\nu}S^{\mu\nu}} \, \begin{pmatrix} 0\\ 0\\ \sqrt{2E}\\ 0 \end{pmatrix} = \sqrt{2E} \begin{pmatrix} 0\\ 0\\ 1\\ \theta/2 \end{pmatrix}$$

I am puzzled why the two approaches give different answers.

For a massless fermion, I expected the $$\xi = (1,0)$$ state to stay in the lower-two components of a Dirac spinor (using Peskin's conventions) because chirality should be preserved. Why should plugging in a momentum $$p^\mu$$ that is slightly rotated from $$(E,0,0,E)$$ lead to an opposite chirality component in the first approach?

Am I perhaps confused about helicity being measured with respect to the $$z$$-direction versus with respect to the direction of motion?

## Intermediate Steps

In the first approach, I use the following expressions: $$\sqrt{p\cdot\sigma} =\sqrt{\frac E2} \begin{pmatrix} 1-c_\theta & -s_\theta\\ -s_\theta & 1+c_\theta \end{pmatrix} = \frac{1}{\sqrt{2}} p\cdot\sigma \\ \sqrt{p\cdot\bar\sigma} =\sqrt{\frac E2} \begin{pmatrix} 1+c_\theta & s_\theta\\ s_\theta & 1-c_\theta \end{pmatrix} = \frac{1}{\sqrt{2}} p\cdot\bar\sigma$$ where $$c_\theta \equiv \cos\theta$$ and $$s_\theta \equiv \sin\theta$$. In the small $$\theta$$ limit this gives: $$\sqrt{p\cdot\sigma} =\sqrt{\frac E2} \begin{pmatrix} 0 & -\theta\\ -\theta & 2 \end{pmatrix} \\ \sqrt{p\cdot\bar\sigma} =\sqrt{\frac E2} \begin{pmatrix} 2 & \theta\\ \theta & 0 \end{pmatrix}$$

For the second approach, the transformation parameter is $$\omega_{13}=-\omega_{31} = - \theta$$ which multiplies the generator $$S^{13} = -\frac 12 \begin{pmatrix} \sigma^2 &\\ & \sigma^2 \end{pmatrix}$$ so that the rotation matrix is $$\Lambda_{\frac 12} = e^{-\frac i2 \omega_{13}S^{13} - \frac i2 \omega_{31}S^{31}} = \exp\left[-\frac{i\theta}{2} \begin{pmatrix} \sigma^2 &\\ &\sigma^2 \end{pmatrix}\right] = \mathbb{1}+\frac{\theta}{2} \begin{pmatrix} 0& -1 && \\ 1 &0&&\\ &&0&-1\\ &&1&0 \end{pmatrix}$$ From these two intermediate steps the puzzle is clear: plugging in $$\xi = (1,0)$$ in the first method gives three non-zero components. In the second method, this corresponds to $$u(k)=(0,0,1,0)$$ and we see that $$\Lambda_{\frac 12} u(k)$$ has only two non-zero components.

The mistake is that in the first method, $$\xi = (1,0)$$ is no longer a state of definite helicity. This is because $$\xi=(1,0)$$ is a state of definite spin in the $$z$$-direction, whereas the helicity is defined to be the spin along the direction of motion: $$h(\mathbf{p})\equiv \frac{\mathbf{p}\cdot\sigma}{|\mathbf{p}|}$$ where $$\mathbf{p}$$ is the 3-momentum, in contrast to the 4-momentum $$p$$.
Let us write $$u_\xi(p) = \begin{pmatrix} \sqrt{p\cdot\sigma}\xi\\ \sqrt{p\cdot\bar\sigma}\xi \end{pmatrix}$$ Rather than $$\xi = (1,0)$$, we should use the eigenvector of $$h(\mathbf{p})$$ that equals $$(1,0)$$ in the limit where $$\mathbf{p} = (0,0,E)$$. This is $$\xi_+ = \frac{1}{\sqrt{2}} \begin{pmatrix} \sqrt{1+c_\theta}\\ \sqrt{1-c_\theta} \end{pmatrix}\ ,$$ where $$c_\theta = \cos\theta$$ and $$\mathbf{p} = (E\sin\theta,\,0\,,E\cos\theta)$$. Upon plugging this in, one finds $$u_{\xi_+}(p) = \begin{pmatrix} \sqrt{p\cdot\sigma}\xi_+\\ \sqrt{p\cdot\bar\sigma}\xi_+ \end{pmatrix} = \sqrt{2E} \begin{pmatrix} 0\\ 0\\ 1\\ \theta/2 \end{pmatrix}$$ which indeed matches the result of the second method.
Note that no adjustment was necessary for the second method: you started with a plane wave spinor moving along the $$z$$-direction so the helicity is aligned with the quantization axis. If you start with a state of definite helicity, applying $$\Lambda_{1/2}$$, one rotates the entire plane wave, but it is still a state of the same definite helicity.