Dimensionality of total angular momentum space

[xago]

Homework Statement

There are 2 electrons, one with n=1, l=0 and the other with n=2, l=1. The question asks what is the dimensionality of total angular momentum space.

Homework Equations

(2$j_{1}$+1)(2$j_{2}$+1)

The Attempt at a Solution

I know for 2 electrons (spin 1/2 each) the possible values of total spin are s=0 or s=1.
the total angular momentum is l=0 + l=1 = 1 (right?)
So does this mean that j1 = 1+0 and j2 = 1+1
which gives a dimensionality of (2(1) +1)(2(2) +1) = 15??
The number seems a little off to me, perhaps the equation I have for dimensionality is incorrect?

 

[vela]

Take the s=0, l=1 combination. Your formula says there are (2s+1)(2l+1) = 3 total states. Those states are
\begin{align*}
&\vert s=0,\ m_s=0;\ l=1,\ m_l=1\rangle \\
&\vert s=0,\ m_s=0;\ l=1,\ m_l=0\rangle \\
&\vert s=0,\ m_s=0;\ l=1,\ m_l=-1\rangle.
\end{align*}The other way to look at it is to sum the angular momenta together, $\vec{J} = \vec{S} + \vec{L}$. According to the rules of addition of angular momenta, there is only a single allowed value for j, namely j=1, so there are 2j+1 = 3 states, corresponding to m_j = 1, 0, and -1. Either way you get three states.

Now you do the s=1, l=1 combination. What are the allowed values of j?

 

[xago]

Take the s=0, l=1 combination. Your formula says there are (2s+1)(2l+1) = 3 total states. Those states are
\begin{align*}
&\vert s=0,\ m_s=0;\ l=1,\ m_l=1\rangle \\
&\vert s=0,\ m_s=0;\ l=1,\ m_l=0\rangle \\
&\vert s=0,\ m_s=0;\ l=1,\ m_l=-1\rangle.
\end{align*}The other way to look at it is to sum the angular momenta together, $\vec{J} = \vec{S} + \vec{L}$. According to the rules of addition of angular momenta, there is only a single allowed value for j, namely j=1, so there are 2j+1 = 3 states, corresponding to m_j = 1, 0, and -1. Either way you get three states.

Now you do the s=1, l=1 combination. What are the allowed values of j?

So if I understand correctly, j= l+s = 2 for l=1, s=1 which means that the possible values of $m_{j}$ are -2,-1,0,1,2 which gives the possible states of:
\begin{align*}
&\vert s=1,\ m_s=-1;\ l=1,\ m_l=1\rangle \\
&\vert s=1,\ m_s=-1;\ l=1,\ m_l=0\rangle \\
&\vert s=1,\ m_s=-1;\ l=1,\ m_l=-1\rangle \\
&\vert s=1,\ m_s=0;\ l=1,\ m_l=1\rangle \\
&\vert s=1,\ m_s=0;\ l=1,\ m_l=0\rangle \\
&\vert s=1,\ m_s=0;\ l=1,\ m_l=-1\rangle \\
&\vert s=1,\ m_s=1;\ l=1,\ m_l=1\rangle \\
&\vert s=1,\ m_s=1;\ l=1,\ m_l=0\rangle \\
&\vert s=1,\ m_s=1;\ l=1,\ m_l=-1\rangle.
\end{align*}

which gives 9 states total... (including the 3 states in the middle there which are the same as the ones given by s=0)

 

[vela]

No, that's not correct. I think you need to go back and study the addition of angular momenta as you seem to have some basic misunderstandings about angular momentum in quantum mechanics.

 

[xago]

I just re-edited my 2nd post there, i confused $m_{j}$ with $m_{s}$, but I know that for s=1 $m_{s}$ is -1,0,1 which combined with $m_{l}$ =-1,0,1 gives 9 states in total including the 3 given by s=0 right?

 

[vela]

Yes, there are nine states. (You meant m_s and m_l, not m_j.)

But think about this. If j=2, then m_j can be -2, -1, 0, 1, 2. That's five states. What are the other four in the |j m_j> basis?

 

[xago]

Would those be the case where only one spin is taken into account aka s=1/2, -1/2
which gives $m_{j}$= -3/2, -1/2, 1/2, 3/2

also, just to clarify, the only possible values for j are 1 and 2 right?

 

[vela]

Would those be the case where only one spin is taken into account aka s=1/2, -1/2
which gives $m_{j}$= -3/2, -1/2, 1/2, 3/2

No.

also, just to clarify, the only possible values for j are 1 and 2 right?

No.