Dirac equation in a central field (Schiff)

[DrClaude]

Not really a homework problem, but I think it fits better in this section.

Homework Statement

I'm having a problem with eq. (53.12) in the book Quantum Mechanics by Schiff. In the context of the Dirac equation, we have
$$
\hbar^2 k^2 = (\vec{\sigma}' \cdot \vec{L})^2 + 2\hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2 = (\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 + \frac{1}{4} \hbar^2
$$
The first equality is fine, it is the second one that I can't reproduce. The point of the equation is to recover $(\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 = (\vec{L} + \vec{S})^2 = \vec{J}^2$.

Homework Equations

$$
\vec{\sigma}' = ( \sigma_x', \sigma_y', \sigma_z')
$$
where the $\sigma_i'$ are $4\times4$ matrices related to the Pauli matrices $\sigma_i$ through
$$
\sigma_i' \equiv \begin{pmatrix} \sigma_i & 0 \\ 0 & \sigma_i \end{pmatrix}
$$
$\vec{L}$ is the orbital angular momentum (actually an operator, but that's not important in the present context).

The Attempt at a Solution

I start by looking at the square term on the LHS:
$$
\begin{array}{}
(\vec{\sigma}' \cdot \vec{L})^2 &= ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)^2 \\
&= (\sigma_x')^2 L_x^2 + \sigma_x' \sigma_y' L_x L_y + \sigma_x' \sigma_z' L_x L_z \\
&\quad + \sigma_y' \sigma_x' L_y L_x + (\sigma_y')^2 L_y^2 + \sigma_y' \sigma_z' L_y L_z \\
& \quad + \sigma_z' \sigma_x' L_z L_x + \sigma_z' \sigma_y' L_z L_y + (\sigma_z')^2 L_z^2 \\
&= L_x^2 \mathbf{1} + L_y^2 \mathbf{1} + L_z^2 \mathbf{1} = \vec{L}^2
\end{array}
$$
where I have used the properties of the Pauli matrices, namely $\sigma_i^2 = \mathbf{1}$, with $\mathbf{1}$ the identity matrix, and $\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}$.

Now, for the square term on the RHS:
$$
\begin{array}{}
(\vec{L} + \frac{1}{2} \hbar \vec{\sigma}')^2 &= \vec{L} \cdot \vec{L} + \frac{1}{2} \hbar \vec{L} \cdot \vec{\sigma}' + \frac{1}{2} \hbar \vec{\sigma}' \cdot \vec{L} + \frac{1}{4} \hbar^2 \vec{\sigma}' \cdot \vec{\sigma}' \\
&= \vec{L}^2 + \hbar \vec{\sigma}' \cdot \vec{L} + \frac{1}{4} \hbar^2 (\sigma_x^2 + \sigma_y^2 + \sigma_z^2) \\
&= \vec{L}^2 + \hbar \vec{\sigma}' \cdot \vec{L} + \frac{3}{4} \hbar^2
\end{array}
$$

Putting all this together, I can rewrite the original equality as
$$
\vec{L}^2 + 2\hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2 = \vec{L}^2 + \hbar (\vec{\sigma}' \cdot \vec{L}) + \hbar^2
$$
There is a factor 2 missing in the RHS that I can't find.

 

[TSny]

Consider $$ \sigma_x' \sigma_y' L_x L_y + \sigma_y' \sigma_x' L_y L_x $$ Does this reduce to zero or something else?

 

[DrClaude]

Consider $$ \sigma_x' \sigma_y' L_x L_y + \sigma_y' \sigma_x' L_y L_x $$ Does this reduce to zero or something else?

I was so concentrated on the $\sigma'$ that I only saw $\sigma_x' \sigma_y' + \sigma_y' \sigma_x' = 0$. But of course
$$
\begin{array}{}
\sigma_x' \sigma_y' L_x L_y + \sigma_y' \sigma_x' L_y L_x &= i \sigma_z' L_x L_y - i \sigma_z' L_y L_x \\
&= i \sigma_z' (L_x L_y - L_y L_x) \\
&= i \sigma_z' (i \hbar L_z) =-\hbar \sigma_z' L_z
\end{array}
$$
and so on, such that
\begin{array}{}
(\vec{\sigma}' \cdot \vec{L})^2 &= ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)^2 \\
&= (\sigma_x')^2 L_x^2 + \sigma_x' \sigma_y' L_x L_y + \sigma_x' \sigma_z' L_x L_z \\
&\quad + \sigma_y' \sigma_x' L_y L_x + (\sigma_y')^2 L_y^2 + \sigma_y' \sigma_z' L_y L_z \\
& \quad + \sigma_z' \sigma_x' L_z L_x + \sigma_z' \sigma_y' L_z L_y + (\sigma_z')^2 L_z^2 \\
&= (L_x^2 + L_y^2 + L_z^2) \mathbf{1} - \hbar ( \sigma_x' L_x + \sigma_y' L_y + \sigma_z' L_z)\\
&= \vec{L}^2 - \hbar ( \vec{\sigma}' \cdot \vec{L}),
\end{array}
leading to the correct equality.

Thanks a lot for pointing me in the right direction.

 

[TSny]

OK. Good.