Constants of the Motion

Constants of the motions are those dynamic variables that commute with the Hamiltonian. For a free particle ($\hbar=c=1$)

$$H = \vec{\alpha}\cdot\vec{p} + \beta m .$$ (5.47)

With the usual definition of the orbital angular momentum

$$\vec{L} = \vec{x}\times\vec{p}$$ (5.48)

we find that none of the components of $\vec{L}$ commute with $H$. Thus the orbital angular momentum is not a constant of the motion. On the other hand we find that

$$\vec{J} = \vec{L} + \frac{1}{2}\vec{\Sigma}$$ (5.49)

with

$$\vec{\Sigma} = \left( \begin{array}{cc} \vec{\sigma} & 0 \\ 0 & \vec{\sigma} \end{array} \right)$$ (5.50)

does commute with $H$ and thus is a constant of the motion.

We now prove this statement. To do this, we simply verify that this result is true for $J_1$. Clearly,

$$[J_1,H]= [J_1,\vec{\alpha}\cdot\vec{p}] = [L_1,\vec{\alpha}\cdot\vec{p}] + \frac{1}{2} [\Sigma_1,\vec{\alpha}\cdot\vec{p}] .$$ (5.51)

For the first term in equation 5.51,

$$[L_1,\vec{\alpha}\cdot\vec{p}]= \alpha_i [x_2 p_3 - x_3 p_2,p... ...i(\alpha_2p_3-\alpha_3p_2) = -i(\vec{p}\times\vec{\alpha})_1 .$$ (5.52)

For the second term in equation 5.51, we make use of the relationship

$$\vec{\Sigma} = \gamma^5 \vec{\alpha} = \vec{\alpha} \gamma^5 ,$$ (5.53)

where $\gamma^5 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$ . Using $\gamma^5$ we have

$$[\Sigma_1,\vec{\alpha}\cdot\vec{p}]= [\gamma^5\alpha_1,\vec{\alpha}\cdot\vec{p}] = \gamma^5[\alpha_1,\alpha_i]p_i .$$ (5.54)

Since

$$[\alpha_i,\alpha_j]= 2i\epsilon_{ijk}\Sigma_k ,$$ (5.55)

we get

$$\frac{1}{2}[\Sigma_1,\vec{\alpha}\cdot\vec{p}] = i\gamma^5\epsilon_{1ij} \Sigma_j p_i = i(\vec{p}\times\vec{\alpha})_1 .$$ (5.56)

Thus, we have shown that

$$[J_1,\vec{\alpha}\cdot\vec{p}]= 0 .$$ (5.57)

By symmetry it now follows that quite generally

$$[\vec{J},\vec{\alpha}\cdot\vec{p}]= 0 .$$ (5.58)

As a consequence we have that $J^2, J_3$ and $H$ can be simultaneously diagonalized.

The bonus embodied in the Dirac equation is the extra twofold degeneracy. This means that there must be another observable which commutes with $H$ and $\hat{p}$, whose eigenvalues can be taken to distinguish the states. On inspection, we see that the operator $\vec{\Sigma}\cdot\hat{p}$ commutes with $H$ and $\hat{p}$; $\hat{p}$ is the unit vector pointing in the direction of the momentum, $\vec{p}/\vert\vec{p}\vert$. The ``spin'' component in the direction of the motion, $\frac{1}{2}\vec{\sigma}\cdot\hat{p}$, is therefore a ``good'' quantum number and can be used to label the solutions. We call this quantum number the helicity of the state.