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[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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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