2
$\begingroup$

I have a question about a step in calculating in CG-coefficients for product state

$$\big|l,s, j, m_j \big\rangle=\sum _{m_j, m_s} C^{m_j, m_s}\big|l,m_l\rangle|s,m_s\rangle$$

for given quantum numbers $l=1$ and $s=1/2$.

Obviously $\big|1,1/2, 3/2, 3/2 \big\rangle=\big|1,m_l=1\rangle|1/2,m_s= 1/2\rangle$. Applying ladder operator $j_{-} = l_{-} + s_{-} $ it's easy to see that $$\big|1,1/2, 3/2, 1/2 \big\rangle= j_{-}\big|1,1/2, 3/2, 3/2 \big\rangle= \sqrt{2/3}\big|1,0\rangle|1/2,1/2\rangle + \sqrt{1/3}\big|1,1\rangle|1/2,-1/2\rangle$$

Similar

$$j_{+}\big|1,1/2, 3/2, -3/2 \big\rangle= \sqrt{2/3}\big|1,0\rangle|1/2,-1/2\rangle + \sqrt{1/3}\big|1,-1\rangle|1/2,1/2\rangle$$

My point of interest is to calculate $\big|1,1/2, 1/2, 1/2 \big\rangle$ and $\big|1,1/2, 1/2, -1/2 \big\rangle$.

By ortogonality condition they must be orthogonal to $\big|1,1/2, 3/2, \pm 1/2 \big\rangle$.

Using this I get for example only $$\big|1,1/2, 1/2, 1/2 \big\rangle = \mp \sqrt{1/3}\big|1,0\rangle|1/2,1/2\rangle + \pm \sqrt{2/3}\big|1,1\rangle|1/2,-1/2\rangle$$

so I can calculate the CG's up to the sign $\pm$. By CG are unique determined. How can calculate them?

$\endgroup$
2
$\begingroup$

The CG coefficients are usually determined by first constructing the highest state in each irrep, i.e. the state for which $M=J$. This state is determined by the requirement that it must be killed by $L_+$, i.e. $$ L_+\vert \ell ,s,j,j\rangle=0 $$ From this one can obtain a recursion relation for all the CGs needed for the highest state $\vert \ell ,s,j,j\rangle$, and this recursion can be fully determined in terms of $C^{\ell,m_s}$ with $m_s=j-\ell$. This is enough to fix the relative phases of CGs for the highest state, but not the overall phase. In the well-known example of $\ell=1/2$ and $s=1/2$, the states $$ \vert 00\rangle_\pm=\pm \frac{1}{\sqrt{2}}\left( \vert \textstyle\frac{1}{2}, \frac{1}{2}\rangle \vert \frac{1}{2} ,-\frac{1}{2}\rangle -\vert \frac{1}{2} ,-\frac{1}{2}\rangle \vert \frac{1}{2}, \frac{1}{2}\rangle \right)\, . \tag{1} $$ are both killed by $L_+=L_+^{(1)}+L_+^{(2)}$.

The coefficient $C^{\ell,m_s}$ is often evaluated using a normalization condition, so its sign is arbitrary. The usual Condon-Shortley phase convention is to choose as positive the coefficient $C^{\ell,m_s}$ of the state $\vert \ell,\ell\rangle \vert s,j-\ell\rangle$, i.e. keep the $\vert 00\rangle_+$ state in the example of (1). This phase convention is by far the most commonly used.

Once the relative phases of this highest state are established, the phase of the other coefficients with $m_j\ne j$ follow from the action of the lowering operator.

There is nothing so widely accepted as the Condon-Shortly convention for the phase of CGs for representations of other (Lie) algebras, v.g. $su(3)$. Indeed, there is not even agreement on the signs of the generator matrix elements, although the Gelfan'd Zeitlin scheme as associated matrix elements are often adopted (this scheme can become cumbersome for algebraic manipulations).

The general procedure remains the same: find the highest state by requiring it be killed by all raising operators, fix the sign of one seed coefficient of the recursion for the highest state, and use the lowering operators to get the remaining coefficients.

Beyond Mathematica (which has a built in function ClesbshGordan for the $su(2)$ case), there is a web based interface to numerically compute $su(N)$ CGs for any $N$ using the Gelfan'd Zeitlin basis.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

There is a degree of ambiguity here, because the states $\left|1,\tfrac12, \tfrac12,\tfrac12\right>$ and $\left|1,\tfrac12,\tfrac32,\tfrac12\right>$ live in different representations, and if you applied a global phase to the whole of the $j=1/2$ representation the consequences would be quite limited. Nevertheless, since we do want to have a single unambiguous definition of the Clebsch-Gordan coefficients (as given e.g. in chapter 34 of the DLMF, specifically in equations 34.1.1 and 34.2.4), we do need to fix that phase.

The core criterion for this is given in §3.4 of Edmonds' Angular momentum in quantum mechanics, and the short of it is that you require that

All matrix elements of $J_{1x}$ which are nondiagonal in $j$ are real and non-negative.

This presupposes the condition that if you fix the phase of the $m_J=j$ state using a matrix element to one state in a different $j$ representation, then all possible matrix elements between the two representations will also have that property. This is nontrivial and the proof is in Edmonds but there's no point in repeating it here.

If you want to double-check your work, one handy trick is calculating the coefficient in Mathematica (i.e. as ClebschGordan[{1/2, 1/2}, {1, 0}, {1/2, 1/2}], or more generally using the syntax ClebschGordan[{j1, m1}, {j2, m2}, {j, m}]), or asking Wolfram Alpha to do that calculation for you.

| cite | improve this answer | |
$\endgroup$

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.