Spin-orbit coupling and the Zeemann effect

[Markus Kahn]

Homework Statement

Consider an electron in a hydrogen atom in the presence of a constant magnetic field $B$, which we take to be parallel to the $z$-axis. Without the magnetic field and ignoring the spin-orbit coupling, the eigenfunctions are labelled by $\vert n, l, m, m_s \rangle$, where m s denotes the spin quantum number $m_s = \pm 1/2$. The additional effects lead to $H = H_0 + \Delta H$ with
$$\Delta H = H _ { \mathrm { SO } } + H _ { \mathrm { Z } } = \frac { 1 } { 2 m ^ { 2 } c ^ { 2 } } \frac { 1 } { r } \frac { d V } { d r } \vec { L } \cdot \vec { S } + \mu _ { \mathrm { B } } B _ { z } \left( L _ { z } + 2 S _ { z } \right),$$
where $V (r) = −e^2 /r$ and $\mu_B =e/2mc$ is the Bohr magneton.

1. determine the first order perturbation of the energy spectrum, treating only the spin-orbit coupling as a perturbation.
2. determine the energy spectrum to first order perturbation theory, treating both terms (i.e. $\Delta H$) as a perturbation.
   Hint: You need to diagonalise the $2(2l + 1) \times 2(2l+ 1)$ matrix where $m = −l, ..., l$
   and $m_s = \pm 1/2$ . Since $[L_z + S_z , \Delta H] = 0$, this matrix is block-diagonal.
3. For $B_z = 0$, reproduce the result for the spin orbit coupling that was derived in the lectures. Show that the first order correction in $B_z$ of this result is given by $$\Delta E _ { n \ell j m _ { j } } = g _ { \ell j } \mu _ { B } B _ { z } m _ { j } , \quad m _ { j } = - j , \ldots , j$$
   where $j$ and $m_j$ refer to the diagonal representation of the rotation group corresponding to $\vec{J}=\vec{L}+\vec{S}$ and $$g _ { \ell , \ell + 1 / 2 } = \frac { 2 + 2 \ell } { 1 + 2 \ell } , \quad g _ { \ell , \ell - 1 / 2 } = \frac { 2 \ell } { 1 + 2 \ell }.$$

Homework Equations

All given in the exercise above.

The Attempt at a Solution

1. This boils down to calculating $\langle n, l',m',m_s'\vert H_{\rm SO} \vert n,l,m, m_s\rangle$. If I did the math correctly this should result in $$\langle n, l',m',m_s'\vert H_{\rm SO} \vert n,l,m, m_s\rangle = \frac{e^2}{2m^2c^2}\frac{1}{a_0^3n^3}\frac{1}{(l+1)(2l+1)}\delta_l^{l'}\delta^{m'}_m\delta_{m_s}^{m_s'},$$ where $a_0:=\hbar^2/(me^2)$. This indicates that the matrix of $H_{\rm SO}$ is diagonal and therefore we find the correction of the energy in first order to be $$E_1^n = \frac{e^2}{2m^2c^2}\frac{1}{a_0^3n^3}\frac{1}{(l+1)(2l+1)}.$$
2. Here the problems start. What I've done so far $$\begin{align*}\langle n, l',m',m_s'\vert \Delta H \vert n,l,m, m_s\rangle & = \langle n, l',m',m_s'\vert H_{\rm SO} + H_Z \vert n,l,m, m_s\rangle \\ &= \langle n, l',m',m_s'\vert H_{\rm SO} \vert n,l,m, m_s\rangle + \langle n, l',m',m_s'\vert H_Z \vert n,l,m, m_s\rangle \\ &= \left(E_1^n + \mu B_z (m+2m_s) \right)\delta^{m'}_m\delta_l^{l'}\delta_{m_s}^{m_s'}.\end{align*}$$ Since this would already imply a diagonal-matrix and the hint says I should diganolize it, I'm suspecting that I did something wrong... Could someone explain to me what went wrong?
3. I'm completely out of ideas, especially for the second part.

 

[vela]

Show us your work for part 1. I don't recall the matrix being diagonal in that basis.

 

[Markus Kahn]

Show us your work for part 1. I don't recall the matrix being diagonal in that basis.

$$\begin{align*}
\langle n, l',m',m_s'\vert H_{\rm SO} \vert n,l,m, m_s\rangle &= \langle n, l',m',m_s'\vert \frac{1}{2m^2c^2}\frac{1}{r}\frac{dV}{dr} \vec{L}\vec{S} \vert n,l,m, m_s\rangle \\
&=\frac{1}{2m^2c^2} \langle n, l',m',m_s'\vert \frac{1}{r^3} \frac{1}{2}(J^2-L^2-S^2) \vert n,l,m, m_s\rangle\\
&=\frac{1}{4m^2c^2} \left(\left(l+\frac{1}{2}\right)\left(l+\frac{3}{2}\right)-l\left(l+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right) \langle n, l',m',m_s'\vert \frac{1}{r^3} \vert n,l,m, m_s\rangle\\
&= \frac{1}{4m^2c^2} \left(\left(l+\frac{1}{2}\right)\left(l+\frac{3}{2}\right)-l\left(l+1\right)-\frac{1}{2}\left(\frac{1}{2}+1\right)\right) \frac{1}{a_0^3n^3}\frac{\delta_{l'}^l\delta_{m'}^m\delta_{m_s'}^{m_s}}{l(l+1/2)(l+1)} \\
&= \frac{e^2}{2m^2c^2}\frac{1}{a_0^3n^3}\frac{1}{(l+1)(2l+1)}\delta_l^{l'}\delta^{m'}_m\delta_{m_s}^{m_s'},
\end{align*}$$
where a hint was given that $$\left\langle\frac{1}{r^3}\right\rangle_{nl} = \frac{1}{a_0^3n^3}\frac{1}{l(l+1/2)(l+1)}.$$
Now $n$ is already the same in the last equation of my calculation, but I'd need to explain why $l'$ has to be equal to $l$ to use the hint. The idea was here to say that
$$\langle n,l',m',m_s'\vert r^{-3}\vert n,l,m,m_s\rangle = \int_0^\infty u_n(r)^*u_n(r) r^2 r^{-3} d r \int_0^{2\pi}d\phi\int_0^\pi Y_{l',m'}^* Y_{l,m}\sin\theta d\theta,$$
and since $Y_{l,m}$ is orthogonal to $Y_{l',m'}$ if $l'\neq l$, we know that $l'=l$ or the matrix element is zero.

 

[DrClaude]

$| n,l,m, m_s\rangle$ is not an eigenstate of $\vec{L} \cdot \vec{S}$.

 

[Markus Kahn]

Thank you for the comment. I didn't think about this carefully enough as it seems. I'm quite sure that one can rewrite $\vec{L}\cdot\vec{S}=\frac{1}{2}(J^2-L^2-S^2)$ where $\vec{J}=\vec{L}+\vec{S}$. I also think that $\vert n,l,m,m_s\rangle$ is an eigenstate of $L^2,S^2,L_z$ and $S_z$. Is this correct?

For $L^2$ one would probably need to use the fact that $D_l\otimes D_{1/2}=D_{l-1/2}\oplus D_{l+1/2}$. Let me first introduce $l_s=1/2$, to make sure I include all the relevant quantum number. We can now write the Clebsch-Gordan series as
$$\vert n, l,l_s, j,m\rangle = \sum_{m=m+m_s} \langle n,l,l_s,m,m_s\vert n, l,l_s,j,m\rangle \vert n,l, l_s,m,m_s\rangle, $$
and if I'm not mistaken we know that $L^2\vert n,l,l_s,j,m\rangle = j(j+1)\vert n,l,l_s,j,m\rangle$. Is this the right approach to the problem, or am I making some kind of mistake?

 

[DrClaude]

Thank you for the comment. I didn't think about this carefully enough as it seems. I'm quite sure that one can rewrite $\vec{L}\cdot\vec{S}=\frac{1}{2}(J^2-L^2-S^2)$ where $\vec{J}=\vec{L}+\vec{J}$. I also think that $\vert n,l,m,m_s\rangle$ is an eigenstate of $L^2,S^2,L_z$ and $S_z$. Is this correct?

Yes, although you made a typo: $\vec{J}=\vec{L}+\vec{S}$.

For $L^2$ one would probably need to use the fact that $D_l\otimes D_{1/2}=D_{l-1/2}\oplus D_{l+1/2}$. Let me first introduce $l_s=1/2$, to make sure I include all the relevant quantum number. We can now write the Clebsch-Gordan series as
$$\vert n, l,l_s, j,m\rangle = \sum_{m=m+m_s} \langle n,l,l_s,m,m_s\vert n, l,l_s,j,m\rangle \vert n,l, l_s,m,m_s\rangle, $$
and if I'm not mistaken we know that $L^2\vert n,l,l_s,j,m\rangle = j(j+1)\vert n,l,l_s,j,m\rangle$. Is this the right approach to the problem, or am I making some kind of mistake?

That would be $L^2\vert n,l,l_s,j,m\rangle = l(l+1)\vert n,l,l_s,j,m\rangle$.

To keep working in the $| n,l,l_s,j,m\rangle$ [Edit: I meant $\vert n,l,m, m_s\rangle$] basis, note that
$$
\vec{L} \cdot \vec{S} = L_x S_x + L_y S_y = L_z S_z
$$
and
$$
\begin{align*}
L_+ S_- + L_- S_+ &= \left( L_x + i L_y \right) \left( S_x - i S_y \right) + \left( L_x - i L_y \right) \left( S_x + i S_y \right) \\
&= \left( L_x S_x -i L_x S_y + i L_y S_x + L_y S_y \right) + \left( L_x S_x + i L_x S_y - i L_y S_x + L_y S_y \right) \\
&= 2 \left( L_x S_x + L_y S_y \right)
\end{align*}
$$

 

[Markus Kahn]

I'm sorry, but now I'm completely confused. If I understood what you try to suggest with the post above, then I should rewrite $\vec{L}\cdot \vec{S} = \frac{1}{2}(L_+S_-+L_-S_+) + L_zS_z$. But why would I want to work in the $\vert n,l,l_s,j,m\rangle$ basis in this case? I don't really know how, for example, $L_+$ acts on $\vert n,l,l_s,j,m\rangle$. Couldn't I just continue to use the $\vert n,l,l_s,m,m_s\rangle$ basis in this case, where I'm familiar with what $L_\pm,S_\pm$ do?

EDIT: Since I had some time I calculated the matrix elements with this method:
$$
\begin{align*}
\langle n,l',l_s',m',m_s'\vert H_{\rm SO}\vert n,l,l_s,m,m_s \rangle &= \frac{1}{2m^2c^2 a_0^3n^3}\frac{\delta_{l'}^l}{l(2l+1)(l+1)}
\left( \sqrt{l(l+1)-m(m+1)}\right.\\
&\sqrt{3/4-m_s(m_s-1)}\delta_{m'}^{m+1}\delta_{m_s'}^{m_s-1}
\sqrt{l(l+1)-m(m-1)}\\
&\left.\sqrt{3/4-m_s(m_s+1)} \delta_{m'}^{m-1}\delta_{m_s'}^{m_s+1}(m+m_s)\delta_m^{m'}\delta_{m_s}^{m_s'}
\right)
\end{align*}
$$

 

[kuruman]

But why would I want to work in the ##\vert n,l,l_s,j,m\rangle basis in this case?

Sorry for butting in. @Markus Kahn it seems that you are still confused about what we discussed in this earlier thread,
https://www.physicsforums.com/threads/angular-momentum-coupling.961495/#post-6100752
The gist of that was that $\vec L \cdot \vec S$ is diagonal in the "coupled" $\vert L,S,J,M \rangle$ basis but not in the $\vert l,m_l,s,m_s \rangle$ basis and that you don't have to diagonalize the Hamiltonian if you use that basis. Is that also the case here?

I don't really know how, for example, $L_+$ acts on $\vert n,l,l_s,j,m\rangle.$

But you know what happens when $\vec L \cdot \vec S$ acts on $\vert n,l,l_s,j,m\rangle$ as discussed in the aforementioned thread.
For future reference, it is good practice not to use $l_s$ for the spin. Readability is enhanced if you use $s$ instead of $l_s$ and $s_z$ instead of $l_{s_z}$.

 

[Markus Kahn]

Is that also the case here?

I at least don't see any reason why this shouldn't hold here as well and I'll take the suggestions about notation to heart, thank you very much.

So if I understood correctly we know that $\vec{L}\cdot\vec{S}$ is diagonal in $\vert n,l,s,j,m\rangle$ basis. If I use this basis then I would find
$$
\langle n, l',s',j',m'\vert H_{\rm SO} \vert n,l,s,j, m\rangle = \frac{e^2}{2m^2c^2}\frac{1}{a_0^3n^3}\frac{1}{(l+1)(2l+1)}\delta_l^{l'}\delta^{m'}_m\delta_{j}^{j'}\delta_s^{s'},
$$
is this correct? This would therefore be the solution for the first exercise.

 

[DrClaude]

I'm sorry, but now I'm completely confused. If I understood what you try to suggest with the post above, then I should rewrite $\vec{L}\cdot \vec{S} = \frac{1}{2}(L_+S_-+L_-S_+) + L_zS_z$. But why would I want to work in the $\vert n,l,l_s,j,m\rangle$ basis in this case? I don't really know how, for example, $L_+$ acts on $\vert n,l,l_s,j,m\rangle$. Couldn't I just continue to use the $\vert n,l,l_s,m,m_s\rangle$ basis in this case, where I'm familiar with what $L_\pm,S_\pm$ do?

I'm really sorry, I copied the wrong ket . I meant that you should keep working in the $\vert n,l,m, m_s\rangle$ basis, and therefore need to rewrite $\vec{L}\cdot \vec{S}$ not in terms of $\vec{J}$ but in terms of raising and lowering operators.

 

[Markus Kahn]

I'm really sorry, I copied the wrong ket . I meant that you should keep working in the |n,l,m,m_s⟩ basis, and therefore need to rewrite LS not in terms of J but in terms of raising and lowering operators.

No problem, stuff like that happens to me all the time.^^

So if I should work in terms of lowering and raising operators the Hamiltionian would be given by
$$H_{\rm SO} = \frac { e^2 } { 2m ^ { 2 } c ^ { 2 } } \frac { 1 } { r^3 } \left(\frac{1}{2}(L_+S_-+L_-S_+) + L_zS_z\right)$$
From this I get directly to this
$$
\begin{align*}
\langle n,l',l_s',m',m_s'\vert H_{\rm SO}\vert n,l,l_s,m,m_s \rangle &= \frac{1}{2m^2c^2 a_0^3n^3}\frac{\delta_{l'}^l}{l(2l+1)(l+1)}
\\ &\left( \sqrt{l(l+1)-m(m+1)}\right.
\sqrt{3/4-m_s(m_s-1)}\delta_{m'}^{m+1}\delta_{m_s'}^{m_s-1}\\ &+ \sqrt{l(l+1)-m(m-1)}
\sqrt{3/4-m_s(m_s+1)} \delta_{m'}^{m-1}\delta_{m_s'}^{m_s+1}\\ &+\left.(m+m_s)\delta_m^{m'}\delta_{m_s}^{m_s'}
\right)
\end{align*}
$$
Now the task is to get the first order energy perturbation, but I'm not really sure how to do that if the matrix isn't diagonal (until now this was somehow always the case)... Could someone maybe help a bit?

 

[Dr Transport]

diagonalize the matrix

 

[DrClaude]

diagonalize the matrix

Indeed. That was even given as a hint in the question.

 

[Markus Kahn]

I didn't expect to have to diagonalize the matrix since the hint was given for the second exercise, so I thought that there was a solution without doing this. Nevertheless, I tried to do so but I currently struggle to get the right blocks of the matrix. As far as I can understand the matrix elements are bscly determined by $m,m_s$. For a matrix we need a ordered basis, so I choose
$$(\dots,\vert m-1,-1/2\rangle, \vert m-1,1/2\rangle, \vert m,-1/2\rangle \vert m,1/2\rangle, \vert m+1,-1/2 \rangle,\vert m+1, 1/2\rangle,\dots)$$
Now one block, which I would need to diagonalize, looks like this according to my calculations:
$$
\begin{pmatrix}
\gamma & 0&\beta &0&0&0\\
0&\gamma & 0&0&0&0\\
0 & \alpha & \gamma & 0&0&\beta\\
0&0&0&\gamma &\beta&0\\
0&0&0&\alpha &\gamma &0\\
0&0&0&0&0&\gamma
\end{pmatrix}$$
This seemed really strange (I expected the matrix to at least be symmetric), so I wanted to check before I continue on with this...

 

[vela]

So if I understood correctly we know that $\vec{L}\cdot\vec{S}$ is diagonal in $\vert n,l,s,j,m\rangle$ basis. If I use this basis then I would find
$$
\langle n, l',s',j',m'\vert H_{\rm SO} \vert n,l,s,j, m\rangle = \frac{e^2}{2m^2c^2}\frac{1}{a_0^3n^3}\frac{1}{(l+1)(2l+1)}\delta_l^{l'}\delta^{m'}_m\delta_{j}^{j'}\delta_s^{s'},
$$
is this correct? This would therefore be the solution for the first exercise.

What happened to the contribution from $\vec{L}\cdot\vec{S}$? Perhaps this is what the answer simplifies down to, but I'd expect $j$ to appear. It really helps if you show your work instead of just posting your final answer and asking if it's right.