# The phases from ladder operators and the spin orbit coupling for p orbitals

In this question, I am concerned with $p$ orbitals ($l = 1$) and so I shall use the notation $|0\rangle$ to refer to the state $l = 1, m = 0$ and so forth.

When defining the ladder operators, the phases are typically chosen to be real, such that $L_+ |0\rangle = \sqrt 2 |+\rangle$, $L_-|0\rangle = \sqrt 2 |-\rangle$. However, this introduces an oddity when working with the spin orbit coupling, $$\lambda \vec L \cdot \vec s = \lambda \left(\frac{L_+ s_- + L_- s_+}{2} + L_z s_z \right),$$ if we define $|x\rangle = \frac{1}{\sqrt 2} (|+\rangle + |-\rangle)$, $|y\rangle = \frac{-i}{\sqrt2} (|+\rangle - |-\rangle)$. It can easily be shown that: $$\langle x| \vec L \cdot \vec s |y \rangle = \frac{-i}{2} (\langle +| + \langle -|)( |+\rangle s_z - (-1) |-\rangle s_z) = -i s_z.$$ However, there is no reason why we should favour $z$ over any other axis, and so the result ought to be invariant under permutation of indices, i. e., $$\langle z| \vec L \cdot \vec s |x \rangle = -i s_y.$$ However, if we blindly insert the definitions from above, we get instead: $$\langle z| \vec L \cdot \vec s |x \rangle = \frac{1}{\sqrt 2} \langle 0|\frac{1}{2}(\sqrt 2 |0\rangle s_- + \sqrt 2 |0\rangle s_+) = (s_- + s_+) = s_x.$$ Clearly, something is wrong here. If we had chosen the phase to be $L_+ |0\rangle = i\sqrt 2 |+\rangle$, $L_-|0\rangle = -i\sqrt 2 |-\rangle$, we would get $i s_- - i s_+ = s_y \neq -i s_y$ instead.

Either the physical intuition of invariance under rotation or my understanding of the ladder operators is wrong.

(As reference, http://journals.aps.org/prb/abstract/10.1103/PhysRevB.84.195430 would agree with the rotational invariance, but they do not show their exact calculations.)

I realise I only posted this question half an hour ago, but I have now found the answer, which may be useful to other people.

Basically, the issue lies in the definition of the $|x\rangle$ and $|y\rangle$ orbitals. If we had chosen to define:

$$|x\rangle = \frac{-1}{\sqrt 2} (|+\rangle - |-\rangle),$$ $$|y\rangle = \frac{i}{\sqrt 2} (|+\rangle + |-\rangle),$$ instead, we still have: $$\langle x| \vec L \cdot \vec s |y \rangle = \frac{-i}{2} (\langle +| - \langle -|)( |+\rangle s_z + (-1) |-\rangle s_z) = -i s_z,$$

but now:

$$\langle z| \vec L \cdot \vec s |x \rangle = \frac{-1}{\sqrt 2} \langle 0|\frac{1}{2}(\sqrt 2 |0\rangle s_+ - \sqrt 2 |0\rangle s_-) = -(s_+ - s_-) = -i (-i s_+ + i s_-) = - i s_y,$$

exactly as desired!

I will wait a while before marking this as answered, in case someone else has any remarks.

• Do your redefined have the correct eigenvalue relationship with $L_x$ and $L_y$? – Emilio Pisanty Feb 3 '17 at 23:04
• $L_x |x\rangle = 0$, $L_y |y\rangle = 0$, as expected. In fact, with the other convention, $\frac{1}{\sqrt 2}(L_+ + L_-)(|+\rangle + |-\rangle) = |L_-+\rangle + |L_+-\rangle = 2 |0\rangle$. Is the fault that in this convention, $L_+|-\rangle = -\sqrt 2 |0\rangle \neq \sqrt 2 |0\rangle$? – dalum Feb 4 '17 at 13:07
• Well, so there you go, then. Those eigenvector relations are what really breaks your previous calculation. – Emilio Pisanty Feb 4 '17 at 19:41