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I'm trying to calculate the SO-coupling for a single p-electron ($l=1$, $s=\frac{1}{2}$) in the uncoupled representation.

This comes down to calculating these matrix elements:
$$\left\langle nlm_lsm_s\left|\hat{\vec l}.\hat{\vec s}\right|nlm_l^{'}sm_s^{'}\right\rangle$$

It can be easily found that:

$$\hat{\vec l}.\hat{\vec s} = \frac{1}{2}(\hat{ l_+}\hat{ s_-}+\hat{ l_-}\hat{ s_+})+\hat{l_z}\hat{s_z}$$

Which gives me the following matrix representation of $\hat{\vec l}.\hat{\vec s}$:

\begin{bmatrix} \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2} \end{bmatrix}

I recognize this is a diagonal block matrix of the form

\begin{bmatrix} A & 0 \\ 0 & A \end{bmatrix}

And I've found the eigenvalues this way ($\lambda=-1$ and $\lambda=\frac{1}{2}$)

Now I've tried solving for the eigenvectors but I think I'm doing something wrong because the equations I get are trivially equal to zero (The eigenket is represented by a column vector $(x,y,z)$, the calculation is for the $\lambda = \frac{1}{2}$ eigenvalue):

$(A-\lambda I)\ |\phi\rangle=0$ leads to the following equations:

$$-y+z\frac{1}{\sqrt 2}=0$$ $$\frac{1}{\sqrt 2}y-\frac{1}{2}z=0$$

Which in turn gives the trivial equations:

$$z=\frac{2}{\sqrt 2}y$$ $$y= 0, z=0$$

I don't see what I'm doing wrong, I get similar trivial equations when calculating the other eigenvalue.

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The zero vector is always a (trivial) solution of the eigenvector equation. You are looking for a non-zero solution, in fact a normalized solution, so your equations are $z=\sqrt{2}y$ and $y^2 + z^2 = 1$. I get $y=1/\sqrt{3}$, $z=\sqrt{2/3}$.

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  • $\begingroup$ Ah of course... It's been a while since I had linear algebra classes, forgot the obvious. Thanks! $\endgroup$ – Joshua Dec 21 '16 at 15:37

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