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Reading Griffiths' Quantum Mechanics.

We have the electronic confirmation of Carbon as

$$(1s)^2 (2s)^2 (2p)^2$$

in the ground state.

He says

There are two electrons with orbital angular momentum quantum number $1$, so the total angular momentum quantum number could be $2, 1$ or $0$.

How? The angular momentum quantum number $l$ is positive. There are two electrons with $l=1$, so I can only see that you can combine them to get $L=2$.

Not sure how the other options come about - only can see $0 = 1-1$, but I didn't think you would take them away to get the total angular momentum

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    $\begingroup$ Are you familiar with the fact that for two spin half particles the states can be rearranged into a spin 1 triplet and a spin 0 singlet? This is the same thing. $\endgroup$ – jacob1729 Apr 16 at 15:33
  • $\begingroup$ I think spin 1 triplet is when the up and down is paired, but I'm not sure what a spin 0 triplet is. I'm also having similar difficulties when he talks about the total spin angular momentum quantum number $S$, in that I can't see how he has combined them, so I would say I'm 'unfamiliar'. $\endgroup$ – PhysicsMathsLove Apr 16 at 15:35
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With two electrons with angular momentum quantum numer $l=1$ there are three possibilities, you can heuristically think of them like this:

$L=0$: The angular momentum vectors of the different electrons are anti-aligned. So the total angular momentum is $1-1=0$.

$L=1$: The angular momentum vector of one of the electrons is pointing along the z-axis, while the others doesn't. So the total angular momentum for this configuration is $1+0=1$

$L=2$: The angular momentum vectors are both aligned, giving $1+1=2$.

You can also look at the $m_l$ quantum number. You always have two electrons with $l=1$, but the different possible combinations of $m_{l_1}$ and $m_{l_2}$ give you different total angular momentum:

($m_{l_1}=1$ and $m_{l_2}=-1$) or ($m_{l_1}=-1$ and $m_{l_2}=1$) gives $L=0$,

($m_{l_1}=1$ and $m_{l_2}=0$) or ($m_{l_1}=0$ and $m_{l_2}=1$) or ($m_{l_1}=-1$ and $m_{l_2}=0$) or ($m_{l_1}=0$ and $m_{l_2}=-1$) gives $L=1$,

($m_{l_1}=1$ and $m_{l_2}=1$) or ($m_{l_1}=-1$ and $m_{l_2}=-1$) gives $L=2$.

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  • $\begingroup$ so while $l$ does not take into account direction, $L$ does? $\endgroup$ – PhysicsMathsLove Apr 16 at 15:39
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    $\begingroup$ This answer isn't entirely correct. The $|1,0\rangle=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt{2}$ state has $L=1$ and the angular momenta of the electrons are anti-aligned. $\endgroup$ – probably_someone Apr 16 at 15:41
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    $\begingroup$ @PhysicsMathsLove $l$ and $L$ both denote the total angular momentum of something. In the case of $l$ it is the total angular momentum on one electron, and for $L$ it is the total angular momentum of both combined. It is the $m_l$ quantum number that tells you about direction. $\endgroup$ – JSorngard Apr 16 at 15:41
  • $\begingroup$ ohhh I see now! $\endgroup$ – PhysicsMathsLove Apr 16 at 15:42
  • $\begingroup$ @probably_someone indeed! This heuristic misses all non obvious cases where the $m_l$s do not sum to the total $L$. $\endgroup$ – JSorngard Apr 16 at 15:50
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Let's review the simpler case of adding two spin 1/2 particles before doing the calculation for two spin 1 particles. The discussion hinges on the ladder operators crucially so maybe review those if you're unfamiliar.

Spin 1/2

The single particle states are spanned by $\{|+\rangle , |-\rangle\}$ so the two particle state has basis: $$ \{ |--\rangle ,|-+\rangle , |+-\rangle , |++\rangle \}$$ note that two of these obey an exchange symmetry but the middle two do not. We can fix this by rewriting the middle two states as combinations of symmetric/anti-symmetric states: $$|-+\rangle = \frac{1}{2}(|-+\rangle +|+-\rangle )+\frac{1}{2}(|-+\rangle -|+-\rangle )$$ and likewise for $|+-\rangle$, thus our two spin system has an exchange symmetry adapted basis: $$ \{ |--\rangle ,\frac{1}{\sqrt{2}}(|-+\rangle +|+-\rangle ) , \frac{1}{\sqrt{2}}(|-+\rangle -|+-\rangle ) , |++\rangle \}$$

We can now ask the question: what is the total spin in each of these states? For this you need to know the total spin operator $J^2=(\vec{J}_1+\vec{J}_2)^2=J_1^2+J_2^2+2\vec{J}_1\cdot\vec{J}_2$ The last term can be written in terms of components as $2\vec{J}_1\cdot\vec{J}_2 = 2J_1^z1J_2^z + 2(J_1^xJ_2^x + J_1^yJ_2^y)$ and the last term can again be rewritten in terms of ladder operators:

$$2(J_1^xJ_2^x + J_1^yJ_2^y) = J_1^+J_2^- + J_1^-J_2^+$$

which finally allows one to write $$J^2 = J_1^2+J_2^2 + 2J_1^zJ_2^z + J_1^+J_2^- + J_1^-J_2^+$$ (this expression is not limited to the $J=1/2$ case).

You can then simply apply this to each state and recover that the three symmetric basis states we wrote down are all $J=1$ states and the anti-symmetric states are $J=0$.

Spin 1

In this case there are still two particles but now three states each. We get an unsymmetrised basis with $3^2=9$ states. These states are not eigenstates of the total angular momentum $J^2$ which should behave well under particle exchange. When all is said and done, we will get eight symmetry adapted states which break up into a set of 5 symmetric states ($J=2$) a set of 3 anti-symmetric states($J=1$) and a singlet symmetric state ($J=0$).

You can think of the $J=0$ states as having spins that are anti-aligned and so cancelling out, but be wary of the fact that these states are in fact (anti)-symmetrised combinations of the single particle states that may make sense to you, and the relative phases play a big role in determining the final angular momentum. However, the intuitive picture does correctly predict that adding two spins $j_1, j_2$ gives a total spin $|j_1-j_2|\leq J \leq |j_1+j_2|$.

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