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}
