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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.

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1 Answer 1

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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.

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