I am working on the following question:


I've so far done the first part of a, finding $H_{so}$ in terms of J L and S. I'm struggling to see how to relate that to $E_{LSJ}$ as it involves the expectation value. I don't see what to integrate.

Edit: is the expectation value just $J^2-L^2-s^2$? That seems to make sense

  • $\begingroup$ Please give us more information. Your first sentence is "I have done the first part of $a$" OK. But what is $a$ ? What is the problem ? Are there many parts in $a$ ? Where do you want to go ? Later on, you say that you don't know "what to integrate". How can we know ? Why would you integrate something ? What for ? To go where ? $\endgroup$
    – Maurice
    Jan 3 at 15:45

1 Answer 1


The states $|LSJ\rangle$ are eigenstates of the $L^2$, $S^2$, and $J^2$ operators (note: these letters are operators) with the corresponding eigenvalues $L^2$, $S^2$, $J^2$ (these letters are quantum numbers used to label the states).

The long way around is to break the operators down into their Cartesian components. We have that $L^2 = L_x^2 + L_y^2 + L_z^2$ (this is a 3D Pythagoras theorem), and likewise $S^2 = S_x^2 + S_y^2 + S_z^2$, and $J^2 = J_x^2 + J_y^2 + J_z^2$. But also, because $\vec{J} = \vec{L} + \vec{S}$, we have that:

$$\begin{align} J_x = L_x + S_x; \quad J_y = L_y + S_y; \quad J_z = L_z + S_z. \end{align}$$


$$\begin{align} H_\text{so} &= \xi (\vec{L} \cdot \vec{S}) \\ &= \xi (L_xS_x + L_yS_y + L_zS_z) \\ &= \frac{\xi}{2} (2L_xS_x + 2L_yS_y + 2L_zS_z) \\ &= \frac{\xi}{2} [(L_x + S_x)^2 - L_x^2 - S_x^2 \\ &\quad\quad+ (L_y + S_y)^2 - L_y^2 - S_y^2 \\ &\quad\quad+ (L_z + S_z)^2 - L_z^2 - S_z^2] \\ &= \frac{\xi}{2} [J_x^2 + J_y^2 + J_z^2 - (L_x^2 + L_y^2 + L_z^2) - (S_x^2 + S_y^2 + S_z^2)] \\ &= \frac{\xi}{2} (J^2 - L^2 - S^2) \end{align}$$

and so the corresponding expectation value is

$$\langle LSJ | H_\text{so} | LSJ \rangle = \left\langle LSJ \middle| \frac{\xi}{2} (J^2 - L^2 - S^2) \middle| LSJ \right\rangle = \frac{\xi}{2} (J^2 - L^2 - S^2).$$

The 'short' way, which is basically saying the same thing as above but in a much more concise manner, is:

$$\begin{align} J^2 &= |\vec{L} + \vec{S}|^2 \\ &= L^2 + S^2 + 2(\vec{L}\cdot\vec{S}) \\ \vec{L}\cdot\vec{S} &= \frac{1}{2}(J^2 - L^2 - S^2) \\ \xi(\vec{L}\cdot\vec{S}) &= \frac{\xi}{2}(J^2 - L^2 - S^2) \\ \end{align}$$

but to convince yourself of the equality going from the the first and second line you need to expand the vectors out into their Cartesian components.

Addendum: Ian Bush has pointed out in the comments that this also assumes that the components of $\vec{L}$ and $\vec{S}$ commute with one another: otherwise, in principle we have that

$$(L_x + S_x)^2 = L_x^2 + S_x^2 + L_xS_x + S_xL_x$$


$$|\vec{L} + \vec{S}|^2 = L^2 + S^2 + (\vec{L} \cdot \vec{S}) + (\vec{S} \cdot \vec{L})$$

for the 'short' derivation. Thankfully, because $\vec{L}$ and $\vec{S}$ are different sources of angular momentum, they do fully commute, and so the last two terms in both lines above are equal. Thanks to Ian for the reminder!

  • 1
    $\begingroup$ Note in the above you have assumed L commutes with S. I have no idea if this subtly is above the pay grade if the OP $\endgroup$
    – Ian Bush
    Jan 3 at 16:59
  • 1
    $\begingroup$ @IanBush Thanks for pointing that out! I shall edit. To be honest — I am starting to realise I am very rusty. One day not too far from now these questions may be above my pay grade. :-) $\endgroup$ Jan 3 at 19:16
  • 1
    $\begingroup$ to indicate operators, you can use the command \hat{}, eg \hat{H} gives $\hat{H}$ $\endgroup$
    – Andrew
    Jan 3 at 20:40
  • $\begingroup$ @Andrew, thank you! I'm aware. I just wanted to follow the style of the question, and I think the hats are usually dropped in the wild anyway (great if you're used to it... potentially very confusing for beginners). $\endgroup$ Jan 4 at 0:01

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