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In the hydrogen atom, for $n=2$, I want to find all possible states.

For $l=0$ this is easy, and so it is with most of $l=1$. There are two states, however, that I don't understand. They are: $$\Psi_1 =|n=2;l=1;j=1/2;m_j=1/2>$$ $$\Psi_1 =|n=2;l=1;j=1/2;m_j=-1/2>$$ I find this incredibly confusing, because if $j=l+s$, then I get $1+1/2=1/2$, which doesn't make any sense. I know from the Clebsch-Gordon tables and from Griffith's book (Chapter intermediate Zeeman effect) that this state exists, but I can't wrap my head around it. I would appreciate any explanation on how to understand this state

Qmechanic
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Nick Heumann
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1 Answers1

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if $j=l+s$, then I get $1+1/2=1/2$, which doesn't make any sense.

Yes, worded this way it doesn't make sense. It seems you have misunderstood something.

When you have two angular momentum momenta (with $l$ and $s$), then these can couple to give states with several different total angular momenta ($j$):

  • The largest possible value is $j_\text{max}=l+s$,
  • the smallest possible value is $j_\text{min}=|l-s|$.
  • and also the $j$ values in between (with step size $1$) are possible.

Summarizing this you have $$j = l+s,\ l+s-1,\ l+s-2,\ ...,\ |l-s|.$$

For your example ($l=1$ and $s=\frac{1}{2}$) this boils down to just two possible values of $j$: $$j = \frac{3}{2},\ \frac{1}{2}.$$

You can visualize these two groups of states like this. When $\vec{L}$ and $\vec{S}$ are nearly parallel then you get the longest $\vec{J}$ ($j=\frac{3}{2}$). When they are nearly antiparallel then you get the shortest $\vec{J}$ ($j=\frac{1}{2}$). And for every $j$ there are $2j+1$ different states, from $m_j=+j$ (i.e. $\vec{J}$ pointing up) to $m_j=-j$ (i.e. $\vec{J}$ pointing down).

enter image description here
(image from Hyperphysics - Russell-Saunders or L-S coupling)

The $2$ states with $j=\frac{1}{2}$ are (taken from Angular Momentum in the Hydrogen Atom where this is worked out in detail):

$$\begin{align} \left|n=2;l=1;j=\frac{1}{2};m_j=+\frac{1}{2}\right> &=\sqrt{\frac{1}{3}}\left|n=2;l=1;m_l=0;m_s=+\frac{1}{2}\right> \\ &-\sqrt{\frac{2}{3}}\left|n=2;l=1;m_l=+1;m_s=-\frac{1}{2}\right> \\ \left|n=2;l=1;j=\frac{1}{2};m_j=-\frac{1}{2}\right> &=\sqrt{\frac{2}{3}}\left|n=2;l=1;m_l=-1;m_s=+\frac{1}{2}\right> \\ &-\sqrt{\frac{1}{3}}\left|n=2;l=1;m_l=0;m_s=-\frac{1}{2}\right> \end{align}$$

Thomas Fritsch
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  • Why $\vec{L}$ and $\vec{S}$ can't be parallel? – Antonios Sarikas Feb 11 '23 at 19:21
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    @AntoniosSarikas They could be parallel if their lengths would be $|\vec{J}|=j\hbar$, $|\vec{L}|=l\hbar$, $|\vec{S}|=s\hbar$. But actually their lengths are $|\vec{J}|=\sqrt{j(j+1)}\hbar$, $|\vec{L}|=\sqrt{l(l+1)}\hbar$, $|\vec{S}|=\sqrt{s(s+1)}\hbar$. Hence it cannot be $|\vec{J}|=|\vec{L}|+|\vec{S}|$. – Thomas Fritsch Feb 12 '23 at 01:29