Calculating Clebsch–Gordan coefficients

[Markus Kahn]

Homework Statement

Prove that the Clebsch-Grodan coefficients (in the notation $\langle j_1j_2m_1m_2|j_1j_2jm\rangle$) for the decomposition of the tensor product of spin $l$ and spin $1/2$ to spin $l+1/2$ are
$$\left\langle l,\frac{1}{2},m\mp \frac{1}{2}, \pm \frac{1}{2} \Bigg\vert l, \frac{1}{2}, l+\frac{1}{2},m\right\rangle = \sqrt{\frac{l\pm m +\frac{1}{2}}{2l+1}}$$

Homework Equations

Recursion relations for this case are
$$\begin{align*}\sqrt{j(j+1)-m(m\mp 1)}&\langle j_1j_2m_1m_2\vert j_1j_2j(m\mp 1)\rangle \\
&= \sqrt{j_1(j_1+1)-m_1(m_1\pm 1)}\langle j_1j_2(m_1\pm 1)m_2\vert j_1j_2jm\rangle\\
&+ \sqrt{j_2(j_2+1)-m_2(m_2\pm 1)}\langle j_1j_2m_1(m_2\pm 1)\vert j_1j_2jm\rangle .\end{align*}$$

The Attempt at a Solution

A hint in the exercise suggests to first focus on the $m_2=1/2$ coefficients. One should then prove that
$$\left\langle l,\frac{1}{2},m- \frac{3}{2}, \frac{1}{2} \Bigg\vert l, \frac{1}{2}, l+\frac{1}{2},m-1\right\rangle = \sqrt{\frac{l+m-\frac{1}{2}}{l+m+\frac{1}{2}}} \left\langle l,\frac{1}{2},m - \frac{1}{2}, \frac{1}{2} \Bigg\vert l, \frac{1}{2}, l+\frac{1}{2},m\right\rangle .$$
This worked out just fine by using the recursion relation with the minus sign on the left-hand-side of the given eq. in 2 but at this point I have already two questions:

• Why should I focus on $m_2=1/2$ first? What indicates that this is the right choice?
• If I now assume to start with the coefficients of $m_2=1/2$, how exactly do I arrive at the fist eq. that I should show? I mean the hint is nice and all, but I don't understand how one should come to this, or to be maybe a bit more precise, to the specific values in the bras/kets.

Now the hint suggests to identify the state of highest spin and use the new relation to find all the $m_2=1/2$ coefficients. Here I'm completely lost. As fare as I understand the state of highest spin for a fixed $m_2$ would be $m_1=l_1$ but I really don't see how that is useful in this context.

 

[stevendaryl]

Look at your recurrence relation:

$\sqrt{j(j+1) - m(m\mp 1)} \langle j_1 j_2 m_1 m_2| j_1 j_2 j (m \mp 1)\rangle = \sqrt{j_1(j_1+1) - m_1(m_1\pm 1)} \langle j_1 j_2 (m_1 \pm 1) m_2| j_1 j_2 j m)\rangle + \sqrt{j_2(j_2+1) - m_2(m_2 \pm 1)} \langle j_1 j_2 m_1 (m_2 \pm 1)| j_1 j_2 j m\rangle$

Now, look at the special case where

• $j_2 = \frac{+1}{2}$
• $m_2 = \frac{+1}{2}$
• $j_1 = l$
• $m_1 = m - \frac{3}{2}$
• $j = l + \frac{1}{2}$
• Choose the minus sign in the $\mp$ on the left side and the corresponding plus sign in the $\pm$ on the right side

$\sqrt{(l+\frac{1}{2})(l + \frac{3}{2}) - m(m - 1)} \langle l \frac{1}{2} (m-\frac{3}{2}) \frac{+1}{2} | l \frac{1}{2} (l + \frac{1}{2}) (m - 1)\rangle = \sqrt{l(l+1) - (m-\frac{3}{2})(m-\frac{1}{2})} \langle l \frac{1}{2} (m - \frac{1}{2}) \frac{+1}{2} |l \frac{1}{2} (l + \frac{1}{2}) m)\rangle + \sqrt{\frac{1}{2}(\frac{3}{2}) - \frac{1}{2}(\frac{3}{2})} \langle l \frac{1}{2} (m-\frac{3}{2}) (\frac{3}{2})|l \frac{1}{2} (l + \frac{1}{2}) m\rangle$

The last term on the right is 0, so this simplifies to:

$\sqrt{(l+\frac{1}{2})(l + \frac{3}{2}) - m(m - 1)} \langle l \frac{1}{2} (m-\frac{3}{2}) \frac{+1}{2} | l \frac{1}{2} (l + \frac{1}{2}) (m - 1)\rangle = \sqrt{l(l+1) - (m-\frac{3}{2})(m-\frac{1}{2})} \langle l \frac{1}{2} (m - \frac{1}{2}) \frac{+1}{2} |l \frac{1}{2} (l + \frac{1}{2}) m)\rangle$

With some algebra, that should be the same as the thing you need to prove in the hint.

 

[Markus Kahn]

First of all, thank you very much for the explanation, but my problem isn't proving the hint but rather understanding it (I'm sorry if that wasn't clear enough from my description). You say that I should look at the special case $j_2=1/2$, etc., but I don't understand how you come to that. What exactly makes you chose this special configuration for plugging in in the recurrence relation?

What I meant in my second question was how one can use the result for $m_2=1/2$ to find the other coefficients. Sorry about the inaccuracy in my question.

 

[stevendaryl]

First of all, thank you very much for the explanation, but my problem isn't proving the hint but rather understanding it (I'm sorry if that wasn't clear enough from my description). You say that I should look at the special case $j_2=1/2$, etc., but I don't understand how you come to that. What exactly makes you chose this special configuration for plugging in in the recurrence relation?

What I meant in my second question was how one can use the result for $m_2=1/2$ to find the other coefficients. Sorry about the inaccuracy in my question.

I'm not exactly sure what needs explaining. The original problem is talking about combining orbital angular momentum with spin angular momentum. That means that $j_1 = l$, $j_2 = \frac{1}{2}$, $m_2 = \pm \frac{1}{2}$. So there are just two possibilities for $m_2$. Just pick one. The results for the other one are analogous.

The simplification comes because the recurrence relation normally involves two terms. But if you pick the parameters carefully, one of the terms vanishes. That's because the second term is multiplied by $\sqrt{j_2 (j_2 +1) - m_2 (m_2 \pm 1)}$. If you pick $m_2 = +\frac{1}{2}$ and pick the plus sign in the $\pm 1$, then that term vanishes. You could also make it vanish by picking $m_2 = -\frac{1}{2}$ and picking the minus sign in the $\pm 1$. The recurrence relation is true for any values of $m_2$ and for any choice of $\pm 1$, so just pick one that makes the equation simpler.

 

[Markus Kahn]

Thank you for the explanation. In the heat of the moment I forgot that $j_1, j_2$ and $m_2$ are fixed. Therefore $j=j_1+j_2$ is also fixed and for finding $m_1$ you just used the relation $m=m_1+m_2$, right?

If you pick m2=+12m2=+12m_2 = +\frac{1}{2} and pick the plus sign in the ±1±1\pm 1, then that term vanishes. You could also make it vanish by picking m2=−12m2=−12m_2 = -\frac{1}{2} and picking the minus sign in the ±1

I understand what you mean by that, but I'm still struggling in finding the other values... Could you maybe explain what is meant by "highest state of spin" and how this is related to the calculations from this point on?

 

[stevendaryl]

Thank you for the explanation. In the heat of the moment I forgot that $j_1, j_2$ and $m_2$ are fixed. Therefore $j=j_1+j_2$ is also fixed and for finding $m_1$ you just used the relation $m=m_1+m_2$, right?

I understand what you mean by that, but I'm still struggling in finding the other values... Could you maybe explain what is meant by "highest state of spin" and how this is related to the calculations from this point on?

Do you understand how the fact referred to in the hint---the one introduced by "One should prove that..."--is derived? It's a special case of the general recurrence relation.

To simplify the notation, let me define

$A_m \equiv \langle l \frac{1}{2} (m - \frac{1}{2}) \frac{+1}{2} |l \frac{1}{2} (l + \frac{1}{2}) m)\rangle$

The fact quoted in the hint tells us that:

$A_{m-1} = \sqrt{\frac{l+m-\frac{1}{2}}{l+m+\frac{1}{2}}} A_m$

The biggest that $m$ can be is $l+\frac{1}{2}$. That is the "highest state of spin".

If you know what $A_{l+\frac{1}{2}}$ is, then you can use the above equation to figure out $A_{l-\frac{1}{2}}$ and $A_{l-\frac{3}{2}}$, etc. So the problem is to figure out what $A_{l+\frac{1}{2}}$ is.

And the answer to that is pretty easy. In general, $|j_1, j_2, j, m\rangle$ is a linear combination of $|j_1, j_2, m_1, m_2\rangle$ with $m_2$ ranging from $-j_2$ to $+j_2$, and $m_1 = m - m_2$. But for the case $j_1 = l, j_2 = \frac{1}{2}, j = l+\frac{1}{2}, m = l+\frac{1}{2}$, there is only one possibility, namely $m_1 = l$ and $m_2 = \frac{+1}{2}$. So it must be the case that

$|l, \frac{1}{2}, l+\frac{1}{2}, l+\frac{1}{2}\rangle = |l, \frac{1}{2}, l, \frac{+1}{2}\rangle$

That implies that $A_{l + \frac{1}{2}} = 1$. Then you can work your way backwards to find

$A_{l - \frac{1}{2}} = \sqrt{\frac{2l}{2l+1}}$
$A_{l - \frac{3}{2}} = \sqrt{\frac{2l}{2l+1}} \sqrt{\frac{2l-1}{2l}}$
$A_{l - \frac{5}{2}} = \sqrt{\frac{2l}{2l+1}} \sqrt{\frac{2l-1}{2l}} \sqrt{\frac{2l-2}{2l-1}}$
etc.

 

[Markus Kahn]

Do you understand how the fact referred to in the hint---the one introduced by "One should prove that..."--is derived? It's a special case of the general recurrence relation.

Yes, I understand that part. By using
$$
\begin{align*}\sqrt{j(j+1)-m(m - 1)}&\langle j_1j_2m_1m_2\vert j_1j_2j(m - 1)\rangle \\
&= \sqrt{j_1(j_1+1)-m_1(m_1 + 1)}\langle j_1j_2(m_1 + 1)m_2\vert j_1j_2jm\rangle\\
&+ \sqrt{j_2(j_2+1)-m_2(m_2 + 1)}\langle j_1j_2m_1(m_2 + 1)\vert j_1j_2jm\rangle,\end{align*}
$$
where $j_1=l, j_2=1/2$, etc. one can find
$$
\left\langle l,\frac{1}{2},m- \frac{3}{2}, \frac{1}{2} \Bigg\vert l, \frac{1}{2}, l+\frac{1}{2},m-1\right\rangle = \sqrt{\frac{l (l+1)-(m-\frac{3}{2})(m-\frac{1}{2})}{(l+\frac{1}{2})(l+\frac{3}{2})-m(m-1)}} \left\langle l,\frac{1}{2},m - \frac{1}{2}, \frac{1}{2} \Bigg\vert l, \frac{1}{2}, l+\frac{1}{2},m\right\rangle .
$$
And in a last step you can then show that
$$
\sqrt{\frac{l (l+1)-(m-\frac{3}{2})(m-\frac{1}{2})}{(l+\frac{1}{2})(l+\frac{3}{2})-m(m-1)}} =
\sqrt{\frac{l+m-\frac{1}{2}}{l+m+\frac{1}{2}}}.
$$
You have therefore proven the eq. given in the hint.

I can follow your explanation about the $A_m$ and how one can find the one for the "highest spin" as well as all the following. If I understand your notation correctly the whole point of this exercise is to prove that $A_m = \sqrt{\frac{l + m-\frac{1}{2}}{2l+1}}$. The formula works just fine with the derived recurrence relation, but I'm wondering how one would find this closed form as suggested in the exercise.

Anyway, Thank you very much for the help and explanation!