1
$\begingroup$

I was reading angular momenta coupling when I came across these CG coefficients, there is a table in Griffith's but doesn't help much.

$\endgroup$
2
$\begingroup$

Is there a simple way to calculate Clebsch-Gordan coefficients?

No, or at least nothing that any working human would qualify as "simple" and that will work for any general Clebsch-Gordan coefficient.

The closest you can get in the general case is given in this Wikipedia page, which puts them as \begin{aligned} \langle j_{1},j_{2};m_{1},m_{2}|j_{1},j_{2};J,M\rangle =\ &\delta _{M,m_{1}+m_{2}}{\sqrt {\frac {(2J+1)(J+j_{1}-j_{2})!(J-j_{1}+j_{2})!(j_{1}+j_{2}-J)!}{(j_{1}+j_{2}+J+1)!}}}\ \times \\&{\sqrt {(J+M)!(J-M)!(j_{1}-m_{1})!(j_{1}+m_{1})!(j_{2}-m_{2})!(j_{2}+m_{2})!}}\ \times \\&\sum _{k}{\frac {(-1)^{k}}{k!(j_{1}+j_{2}-J-k)!(j_{1}-m_{1}-k)!(j_{2}+m_{2}-k)!(J-j_{2}+m_{1}+k)!(J-j_{1}-m_{2}+k)!}}. \end{aligned} That's an explicit expression (so, hoorah), but it's not something that's usable by any human metric.

If you actually want to calculate them on your own from scratch, then let the DLMF lay out the efficient and stable methods of computation:

Methods of computation for 3j and 6j symbols include recursion relations, see [references]; summation of single-sum expressions for these symbols, see [references]; evaluation of the generalized hypergeometric functions of unit argument that represent these symbols, see [references].

But, really: no one calculates CG coefficients, or at least no one who isn't intentionally doing more calculations than they need to or actively developing scientific code on a new platform. Instead, if we want explicit analytical values, we go to the existing tables (like the one in Wikipedia or the one in Griffiths or Edmonds or any other angular-momentum-in-QM book. If we want a software implementation, we just take one of the many existing robust implementations: this not only solves the trivial problem that the task of writing is much simpler, but it also covers the much more important aspect that bugs are minimized and testing of the code is no longer necessary.

For a birds-eye view of good sources for code, see this list of papers, or the DLMF software index, which includes a bunch of books and open-source and proprietary software suites that can calculate them:

enter image description here

$\endgroup$
2
$\begingroup$

It might depend on your definition of "simple".

For easy cases (low numbers, direct steps), yes, it is simple. However, it gets complicated too fast. What we do is: calculate only the easy ones, and let the rest for computers. I really encourage you to do this.

The trick is: at the top of the ladder, there is only one possibility, except for a global phase factor. For example:

$| j_1 \ j_2 ; m_1 m_2 \rangle = |1\ ½; \ 1 \ ½\rangle $ can only be one possibility: $|J \ M\rangle= |3/2\quad 3/2\rangle $

BEcause it is the maximum value of both $j_1$ and $j_2$. They can only yield the maximum $J,M$.

So we know one equivalence:

$|1\ ½; \ 1 \ ½\rangle \equiv|3/2\quad 3/2\rangle $

or, if you want,

$|1\ ½; \ 1 \ ½\rangle \equiv 1\cdot|3/2\quad 3/2\rangle $

I know, I know, except for a phase factor, but let's ignore that.


So, what do we do now? We apply $J_-$ at both sides of the equation.

On the one hand, $J_-=J_{1-}+J_{2-}$, so we know what it does:

$J_- |1\ ½; \ 1 \ ½\rangle = J_{1-}|1\ 1\rangle +J_{2-}|½\ ½\rangle $ (carry on yourself).

On the other hand, the result is also an angular momentum, so:

$J_- | 3/2 \quad 3/2\rangle = \hbar\sqrt{3/2\cdot(3/2+1)-3/2\cdot(3/2-1)} \ \ |3/2 \quad ½\rangle $

So, equating both sides, you can get

$|3/2 \quad ½\rangle $ as a function of $|1\ ½; \ 1 \ -½\rangle$ and $|1\ ½; \ 0 \ ½\rangle$.

You can get an orthogonal vector to that one, so you complete the sub-space.

Now, you can keep applying $J_{-}$ to both of them. Like that, you get all C-G coefficients.

Of course, you can also start by the bottom of the ladder and go upside with $J_+$.

This is the standard procedure. You do it once with the simplest case (for example, two ½ spins). Then, you make the computer program and breath. By the way, there are already many of them, even online.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.