[I am working with Griffiths Introduction to Quantum Mechanics, 3rd Edition. My problem is general but if you want to look I am reading from ch 4.1 in which the weak-field Zeeman Effect is being calculated when I got stuck.]

We want to calculate $E _{z} = e/2m * \vec B_{ext} \cdot < \vec J + \vec S>$

we work it out so that all we need to find is $<\vec J>$.

I know that $\vec J = \vec L +\vec S$, and thus $ |\vec J|^2= |\vec L|^2+|\vec S|^2+\vec L \cdot \vec S$ where $|\vec L|^2 = \hbar l(l+1)$ and $|\vec s|^2 = \hbar s(s+1)$ in the eigenstates of the Hydrogen atom but griffiths does not appear to use any of these facts and (after stating that the $z$-axis will ie along $\vec B _{ext}$ states

$ \vec B \cdot<\vec J> = \hbar m _{j}$

Maybe I'm just confused about what J is, but how do we goes from one to the other.

  • $\begingroup$ Ah wait, it is because the system was already in a state that was an eigenvalue of the total angular momentum (because when working on fine splitting, the degeneracy of both S and L are broken wheras J still commutes with the hamiltonian). Please correct me if I am wrong here! $\endgroup$
    – zephyrus
    Jun 4, 2015 at 5:04
  • 1
    $\begingroup$ How does what go from one to the other? The first line to the last? Something in between? $\endgroup$
    – Kyle Kanos
    Jun 4, 2015 at 17:08

1 Answer 1


I'm not quite sure what your specific question is, so I'll try to better explain what Griffiths is doing in his book.

In first-order perturbation theory, the Zeeman correction to the energy is:

$$ \begin{align*} E_{Z}^{1} & = \langle n \; l \; j \; m_{j} \vert H_{Z}^{\prime} \vert n \; l \; j \; m_{j} \rangle \\ & = \langle n \; l \; j \; m_{j} \vert \frac{e}{2m}\left( L + 2S \right) \cdot B_{\text{ext}} \vert n \; l \; j \; m_{j} \rangle \\ & = \frac{e}{2m}B_{\text{ext}} \cdot \langle n \; l \; j \; m_{j} \vert \left(L + 2S \right) \vert n \; l \; j \; m_{j} \rangle \\ & = \frac{e}{2m}B_{\text{ext}} \cdot \langle L + 2S \rangle \end{align*} $$

But since $J = L + S$, then $L + 2S$ can be written as $L = J + S$. Since the total angular momentum, $J$, is constant and $L$ and $S$ precess around $J$, we can work out the time average value of $S$ by calculating its projection on $J$:

$$ S_{\text{ave}} = \frac{\left( S \cdot J \right)}{J^{2}}J $$

So now we need to find out what $S \cdot J$ is, which is not immediately obvious. But consider the following:

$$ \begin{align*} L^{2} & = \left( J - S \right) \left( J - S \right) = J \cdot J - 2J \cdot S + S \cdot S \\ & = J^{2} + S^{2} - 2J \cdot S \end{align*} $$

And so if we re-arrange this, we obtain an expression for $S \cdot J$:

$$ S \cdot J = \frac{1}{2}\left(J^{2} + S^{2} - L^{2} \right) $$

But we know that $J^{2} = j\left(j + 1\right)h^{2}$, and similarly with $S^{2}$ and $L^{2}$; so our expression becomes:

$$ \begin{align*} S \cdot J & = \frac{1}{2}\left[j\left(j + 1\right)\hbar^{2} + s\left(s + 1\right)\hbar^{2} - l\left(l + 1\right)\hbar^{2} \right] \\ & = \frac{\hbar^{2}}{2}\left[j\left(j + 1\right) + s\left(s + 1\right) - l\left(l + 1\right) \right] \end{align*} $$

And so it follows that:

$$ \begin{align*} \langle L + 2S \rangle & = \langle J + S \rangle \\ & = \langle \left( 1 + \frac{S \cdot J}{J^{2}} \right)J \rangle \\ & = \left[ 1 + \frac{\frac{\hbar^{2}}{2}\left[j\left(j + 1 \right) + s\left(s + 1\right) - l\left(l + 1\right) \right]}{j\left(j + 1\right)\hbar^{2}}\right]\langle J \rangle \\ & = \left[ 1 + \frac{\left[j\left(j + 1 \right) + s\left(s + 1\right) - l\left(l + 1\right) \right]}{2j\left(j + 1\right)}\right]\langle J \rangle \\ & = g_{J}\langle J \rangle \end{align*} $$

where $g_{J}$ is the Landé g-factor.

Recall our expression for the first-order correction to the energy:

$$ E_{Z}^{1} = \frac{e}{2m}B_{\text{ext}}\cdot \langle L + 2S \rangle $$

We just showed that $\langle L + 2S \rangle = g_{J} \langle J \rangle$, so we have:

$$ E_{Z}^{1} = \frac{e}{2m}B_{\text{ext}} \cdot g_{J} \langle J \rangle $$

At this point, we can choose the z-axis to lie along the direction of $B_{\text{ext}}$. In this case, $B_{\text{ext}} \cdot \langle J \rangle = B_{\text{ext}} \langle J_{z} \rangle$. Of course, the expectation value $\langle J_{z} \rangle = \hbar m_{j}$, and so we have:

$$ \begin{align*} E_{Z}^{1} & = \frac{\hbar e}{2m}B_{\text{ext}}g_{J}m_{j} \\ & = \mu_{B} B_{\text{ext}}g_{J}m_{j} \end{align*} $$

where $\mu_{B}$ is the Bohr magneton.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.