How does the Gordon Decomposition of Dirac Current give rise to spin angular momentum?

I used the Gordon Decomposition to split the Probability Current of the Dirac Field into its orbital current and its spin current. I multiplied the currents by $mc$ to covert them into momentum and then crossed the momentum with position to obtain the orbital angular moment and the spin angular momentum. However the spin angular momentum was twice as large as the accepted value of spin angular momentum. I am stuck on how to get rid of the extra factor of two.

The Gordon Decomposition splits the probability current into two terms \begin{align} mc\langle \gamma^{0n} \rangle = \langle \gamma^{0} i\hbar \partial^{n} \rangle + \tfrac{\hbar}{2}\partial_{j}\langle i\gamma^{0nj} \rangle_{n\neq j} , \end{align} where $\langle \gamma^{0} i\hbar \partial^{n} \rangle$ looks like a orbital momentum current and $\tfrac{\hbar}{2}\partial_{j}\langle i\gamma^{0nj} \rangle_{n\neq j}$ looks like a spin momentum current?

The angular moments of $mc\langle \gamma^{0n} \rangle$ can be computed by crossing then with position to obtain the following: \begin{align} mc \left( x^m\langle \gamma^{0n} \rangle - x^n\langle \gamma^{0m} \rangle \right) = \langle \gamma^{0} i\hbar (x^m\partial^{n}-x^n\partial^{m}) \rangle + \tfrac{\hbar}{2}\partial_{j} \left( x^m\langle i\gamma^{0nj} \rangle_{n\neq j} - x^m\langle i\gamma^{0nj} \rangle_{n\neq j} \right) , \end{align}


1 Answer 1


Be careful: The spin contribution to the number current is proportional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution at "Gordon Decomposition" for more details.


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