Spin conservation in the Dirac equation

[chingel]

Here I am considering the one particle free Dirac equation. As is known the spin operator does not commute with the Hamiltonian. However, the solutions to the Dirac equation have a constant spinor term and only an overall phase factor which depends on time. So as the solution evolves in time, surely the spin operator will act on the spinor part the same way at any moment.

How to reconcile this with the fact that the spin operator doesn't commute with the Hamiltonian and it is often said that spin is not conserved for the Dirac particle? I mean the solutions are even named spin up and spin down. I know if p is not zero the solutions aren't eigenvectors of the spin operator, but still the spinor is constant, so why isn't spin constant? In the basis of the spin operator the solution has constant components, except for overall phase.

 

[PeterDonis]

the spin operator does not commute with the Hamiltonian.

Which spin operator are you talking about? And which Hamiltonian?

 

[chingel]

The Hamiltonian I am considering is $$H=\vec{\alpha} \cdot \vec{p} + \beta m,$$ where $\alpha^i = \gamma^0\gamma^i$, $\beta=\gamma^0$, $p_k = i\partial_k$, $\vec{p} = -i\vec{\partial}$, so that $i\partial_t \psi=H\psi$ gives just the Dirac equation $(i\gamma^\mu\partial_\mu -m )\psi=0$ multiplied by $\gamma^0$. For the spin operator $$\vec{\Sigma} =
\begin{bmatrix}
\vec{\sigma} & 0 \\
0 & \vec{\sigma} \\
\end{bmatrix}.$$

An example of the calculation for the commutator $[H, \vec{\Sigma}]$ is on slides 90, 91, 92 of this link: https://www.hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/Handout_2_2011.pdf

However, if I look at the solutions on slide 75 (eigenvectors of H), I don't understand how anything changes over time in the way the spin operator acts on the states, the spinors are constant with just an overall phase and the components of the spinor in the eigenbasis of the spin operator should stay the same, up to an overall phase. Looking directly at the solution, why isn't spin conserved?

 

[Spinnor]

Someone raises the same question,

"However, it leads to the question why the spin should change in a free motion? Before trying to answer this question, consider the velocity of a Dirac particle. ..."

From, http://www.phy.ohiou.edu/~elster/lectures/relqm_11.pdf

page 2 of 5.

From, https://www.google.com/search?safe=...0...0...1c.1.64.psy-ab..0.0.0...0.rKJvqBgj-7U

 

[chingel]

Thank you for the link. There they calculate that the expectation value of the change of the spin or angular momentum is zero. While the expectation value does not change, they still seem to say that Zitterbewegung causes the spin to change in some sense. I still don't quite understand why or how, maybe someone can explain it to me.

To maybe make my question clearer, if I have the spin operator $\vec{\Sigma} = \begin{bmatrix} \vec{\sigma} & 0 \\ 0 & \vec{\sigma} \end{bmatrix}$ and a solution of the Dirac equation $u(p) = e^{-ip_\mu x^\mu} \sqrt{E+m} \begin{bmatrix} 1 \\ 0 \\ \frac{p_z}{E+m} \\ \frac{p_x+ip_y}{E+m} \end{bmatrix}$, then if at all and why does the spin of this solution change over time, and if it is Zitterbewegung then how exactly the Zitterbewegung works here to change the spin?

 

[Spinnor]

You are clever and will figure out if the following is right or wrong.

At some point in the derivation of the time derivative of an expectation value don't we need to assume that ψ goes to zero at infinity? If so you need to work with a localized wave packet which contains both positive and negative energies? If that is the case then it is easier to see how interference between the positive and negative components give rise to conservation of only the total angular momentum, L+S?

Interesting question, I hope you get an answer.