Addition of orbital angular momentum and spin

[gu1t4r5]

Homework Statement

Consider an electron with spin $\frac{1}{2}$ and orbital angular momentum l=1. Write down all possible total angular momentum states as a combination of the product states $| l=1 , m_l > | s = \frac{1}{2} , m_s >$

Homework Equations

Lowering operator : $J_- |j, m> = \sqrt{(j + m)(j - m + 1)} |j, m-1>$

The Attempt at a Solution

Since total angular momentum $| l-s | <= j <= (l+s)$
and its z-component $-j <= m_j <= +j$
I know that the possible $|j, m_j >$ states are:

$| \frac{1}{2} , \frac{-1}{2} >$
$| \frac{1}{2} , \frac{1}{2} >$
$| \frac{3}{2} , \frac{-3}{2} >$
$| \frac{3}{2} , \frac{-1}{2} >$
$| \frac{3}{2} , \frac{1}{2} >$
$| \frac{3}{2} , \frac{3}{2} >$

As for finding the product states, I know that:
$| \frac{3}{2} , \frac{3}{2} > = |1, 1> | \frac{1}{2} , \frac{1}{2} >$
as this is the maximal spin state. I can then find $| \frac{3}{2} , \frac{1}{2} >$, $| \frac{3}{2} , \frac{-1}{2} >$ and $| \frac{3}{2} , \frac{-3}{2} >$ using the lowering operator above. I don't know how I can use this information to find $| \frac{1}{2} , \frac{1}{2} >$ and $| \frac{1}{2} , \frac{-1}{2} >$ however.


Thanks.

 

[scoobmx]

Well the j=1/2 states are the ones that had $m_l$ = 0. As for the actual problem it seems to me like they want you to write out all of the states you listed in terms of the clebsch-gordon coefficients times the uncoupled basis states.

 

[vela]

Homework Statement

Consider an electron with spin $\frac{1}{2}$ and orbital angular momentum l=1. Write down all possible total angular momentum states as a combination of the product states $| l=1 , m_l > | s = \frac{1}{2} , m_s >$

Homework Equations

Lowering operator : $J_- |j, m> = \sqrt{(j + m)(j - m + 1)} |j, m-1>$

The Attempt at a Solution

Since total angular momentum $| l-s | <= j <= (l+s)$
and its z-component $-j <= m_j <= +j$
I know that the possible $|j, m_j >$ states are:

$| \frac{1}{2} , \frac{-1}{2} >$
$| \frac{1}{2} , \frac{1}{2} >$
$| \frac{3}{2} , \frac{-3}{2} >$
$| \frac{3}{2} , \frac{-1}{2} >$
$| \frac{3}{2} , \frac{1}{2} >$
$| \frac{3}{2} , \frac{3}{2} >$

As for finding the product states, I know that:
$| \frac{3}{2} , \frac{3}{2} > = |1, 1> | \frac{1}{2} , \frac{1}{2} >$
as this is the maximal spin state. I can then find $| \frac{3}{2} , \frac{1}{2} >$, $| \frac{3}{2} , \frac{-1}{2} >$ and $| \frac{3}{2} , \frac{-3}{2} >$ using the lowering operator above. I don't know how I can use this information to find $| \frac{1}{2} , \frac{1}{2} >$ and $| \frac{1}{2} , \frac{-1}{2} >$ however.


Thanks.

You want to find $\lvert \frac{1}{2}, \frac{1}{2} \rangle$ so that it's orthogonal to $\lvert \frac{3}{2}, \frac{1}{2} \rangle$.

 

[Cesarth]

In the problem statement they do not ask you to calculate the total momentum $| J, m_j >$. I think that you just have to write down a linear equation in the states $|1,m_l> | 1/2, m_s >$ where $m_l$ has three possible values and $m_s$ two.

 

[vela]

That's what gu1t4r5 has described doing, and he or she explained how to find four of the six linear combinations. The question gu1t4r5 asked was about how to find the last two.

 

[Cesarth]

Oh, yes, I misunderstood the problem.
And it is a good idea to find the remaining two with the requirement that they are orthogonal to the other.
You could also use pre-calculated clebsch-gordon coefficients as scoobmx says.