Spin Angular Momentum Dirac Equation

[Bob Dylan]

In the Dirac equation, the wave-function is broken into four wave-functions in four entries in a column of a matrix. Since there are four separate versions of the wave-function, does each version have the spin angular momentum of h-bar/2? This seems overly simplistic. How does spin angular momentum work for the Dirac equation?

 

[stevendaryl]

In the Dirac equation, the wave-function is broken into four wave-functions in four entries in a column of a matrix. Since there are four separate versions of the wave-function, does each version have the spin angular momentum of h-bar/2? This seems overly simplistic. How does spin angular momentum work for the Dirac equation?

The spin operator mixes the components of the Dirac wave function, so it's not a property of anyone component, but of all 4. On the other hand, orbital angular momentum does not mix the components, so it makes sense to say that a single component has an orbital angular momentum, but not to say that it has spin angular momentum.

 

[Bob Dylan]

Does this mean all four components share the same phase?

 

[stevendaryl]

Does this mean all four components share the same phase?

If a particle is in a spin eigenstate, then the phases of the components must be related.

It's easiest to see with the two-component nonrelativistic limit of the Dirac equation, the Pauli equation.

With the Pauli equation, the wave function has two components: $\psi = \left( \begin{array} \\ \alpha \\ \beta \end{array} \right)$.

If $\psi$ is spin-up in the z-direction, that means $\left( \begin{array} \\ 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array} \\ \alpha \\ \beta \end{array} \right) = +1 \left( \begin{array} \\ \alpha \\ \beta \end{array} \right)$. That implies $\beta = 0$.

If $\psi$ is spin-up in the x-direction, that means $\left( \begin{array} \\ 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array} \\ \alpha \\ \beta \end{array} \right) = +1 \left( \begin{array} \\ \alpha \\ \beta \end{array} \right)$. That implies $\alpha = \beta$.

 

[Bob Dylan]

So is h-bar (or h-bar divided by 2) a form of spin or orbital angular momentum?