# How to compute the Clebsch Gordan-coefficients for spin-angular momentum coupling of Hydrogen?

Consider a hydrogen atom with Hamiltonian $$H$$, angular momentum $$L$$ and spin operator $$S$$ and total angular momentum $$J:= L+S$$. Since $$H,L^2, J^2$$ and $$J_z$$ all commute with each other, we can find a joint eigenbasis $$\{|n,l,j,m_j\rangle : 0\le l,j\le n-1\,, m_j=-j,\ldots,j\}$$ with canonical quantum numbers. My lecture notes now claim that

$$|n,l,j=l\pm1/2,m_j\rangle=A_{\pm}|n,l,m_j-1/2\rangle\otimes|+\rangle + B_{\pm}|n,l,m_j+1/2\rangle\otimes|-\rangle$$ for $$A_{\pm},B_{\pm}\in\mathbb{C}$$ the Clebsch-Gordan coefficients.

Question: How does one obtain this? As far as I understand, we have $$|n,l,j,m_j\rangle=\sum_{m_l=-l}^{l}c_{m_l,+}|n,l,m_l\rangle\otimes|+\rangle + \sum_{m_l=-l}^{l}c_{m_l,-}|n,l,m_l\rangle\otimes|-\rangle$$ so how does one introduce the quantum number $$m_j$$ here and show that only two summands are non-zero? I know that for the "classical" CG coefficients of the hydrogen atom, we have that the quantum number of $$J_z$$ must be the sum of the $$m_j$$. Can I maybe apply that here? Furthermore, the classical CG formalism does not use the quantum numbers $$n$$ and $$l$$ (or at least assumes that $$l$$ is fixed) so even the expansion $$|n,l,j,m_j\rangle=\sum_{m_l=-l}^{l}c_{m_l,+}|n,l,m_l\rangle\otimes|+\rangle + \sum_{m_l=-l}^{l}c_{m_l,-}|n,l,m_l\rangle\otimes|-\rangle$$ is a mere guess of mine.

• Yes, this conservation law holds whenever you couple angular momentum operators. The formalism is agnostic to which ones are "orbital" and which ones are "spin". Commented Jul 12, 2021 at 17:01
• @ConnorBehan Thanks for the comment but even then I only come as far as $|n,l,j,m_j\rangle=\sum_{m_l=-l}^{l}c_{m_l,+}|n,l,m_l\rangle\otimes|+\rangle + \sum_{m_l=-l}^{l}c_{m_l,-}|n,l,m_l\rangle\otimes|-\rangle$, no? Commented Jul 13, 2021 at 7:08

Let me go back to the beginning and state the question more clearly. We have a quantum system in which there are two angular momenta $$\vec L$$ and $$\vec S$$ that are coupled, so that $$\vec J = \vec L + \vec S$$ is conserved. Then, $$J^2$$, $$L^2$$, $$S^2$$ mutually commute and can be used as good quantum numbers. In the Hydrogen atom example, $$S = 1/2$$. This makes the problem much simpler, and so I will work with this example, although the method can be generalized to any $$S$$.

In the following, I will assume that $$L$$ and $$S$$ are fixed and I will not write $$L$$ and $$S$$ explicitly. The value of $$J$$ will depend on the relative orientation of $$\vec L$$ and $$\vec S$$. There are two possibilities: $$J = L+1/2$$ and $$J = L - 1/2$$.

There are two ways to label the states of the system. We can use the labels of total angular momentum $$J, m_j$$ or we can use the labels of the z angular momentum for $$L$$ and $$S$$, $$m_\ell, m_s$$. Since $$J$$ is the sum of $$L$$ and $$S$$, $$m_j = m_\ell + m_s$$. Since $$S = 1/2$$, $$m_s = \pm 1/2$$.

The total Hilbert space has dimension $$2 (2 L + 1)$$. But this is divided into 2-dimensional subspaces that do not mix with one another. A typical subspace contains the states $$| J = L+{1\over 2}, m_j\rangle \qquad | J = L - {1\over 2}, m_j\rangle$$ or, in the $$m_\ell, m_s$$ basis, $$| m_j-{1\over 2}, +{1\over 2}\rangle \qquad | m_j + {1\over 2}, -{1\over 2}\rangle \ .$$ These pairs of states give two different orthogonal bases for the same Hilbert space. They are related by a unitary transformation. The Clebsch-Gordan coefficients are defined to be the elements of this $$2\times 2$$ unitary matrix. Clebsch-Gordan coefficients can be written $$c_{m_j,+}$$, etc., as in the question, but it is much more transparent to write them as explicitly indicating this change of basis, $$\langle L S m_\ell m_s|LS J m_j\rangle \quad \mbox{or, simply} \quad \langle m_\ell m_s|J m_j\rangle \ .$$ In the standard conventions, Clebsch-Gordon coefficients are real-valued. Then the first equation of the question can be rewritten $$| J m_j> = \langle (m_j-{1\over 2}) (+{1\over 2})|J m_j\rangle \ |(m_j-{1\over 2})(+{1\over 2})\rangle \\ + \langle (m_j+{1\over 2}) (-{1\over 2})|J m_j\rangle \ |(m_j+{1\over 2}) (-{1\over 2})\rangle \ ,$$ which is much more transparent.

Up to this point, everything is formalism. But we are now in a good position to answer the question of how to compute the Clebsh-Gordan coefficients. The arguments below can be found in many textbooks on quantum mechanics. I recommend especially Gordon Baym's "Lectures on Quantum Mechanics", which has a whole chapter (Chapter 15) devoted to the addition of angular momenta.

The key is that, while most of the subspaces are 2-dimensional, there are two special subspaces that are 1-dimensional. There is only one state with $$m_j = J+1/2$$, $$| J = (L+{1\over 2}), m_j = (L+{1\over 2}) \rangle = | L, (+{1\over 2})\rangle \ ,$$ and there is only one state with $$m_j = -L- 1/2$$, $$| J= (L+{1\over 2}), m_j = -(L+{1\over 2})\rangle = \pm | (-L), (-{1\over 2})\rangle \ ,$$ Then $$\langle (L+{1\over 2})(L+{1\over 2}) | L(+1/2)\rangle = 1 \qquad \langle (-L-{1\over 2}), (-L-{1\over 2}) | (-L)(-1/2)\rangle = \pm 1$$ (The relative sign $$\pm$$ has to be worked out by the process given below.)

Taking the first equation here as a starting point, we apply the operator $$J^- = (J^x - i J^y)$$. $$J^-$$ commutes with $$J^2$$, so the action of this operator preserves $$J$$. On a state $$|Jm\rangle$$, it can be shown $$J^- \ |jm\rangle = [ j(j+1) - m(m-1)]^{1/2} | j (m-1)\rangle \ .$$ This equation appears in the proof of the quantization of angle momentum that can be found in any quantum mechanics text. But here we will use it as a tool. Apply $$J^-$$ to $$|(L+{1\over 2})(L+{1\over 2})\rangle$$, $$J^- |(L+{1\over 2})(L+{1\over 2})\rangle = \sqrt{2L+1} |(L+{1\over 2})(L-{1\over 2})\rangle\ .$$ Now write $$J^- = L^- + S^-$$ and apply this to $$| L, (+{1\over 2})\rangle$$, $$J^- | L, (+{1\over 2})\rangle = \sqrt{2L}\cdot | (L-1), (+ {1\over 2})\rangle + 1 \cdot | L, (-{1\over 2})\rangle$$ Now we have found that $$\sqrt{2L+1} |(L+{1\over 2})(L-{1\over 2})\rangle = \sqrt{2L}\cdot | (L-1), (+ {1\over 2})\rangle + 1 \cdot | L, (-{1\over 2})\rangle$$ and so we can identify $$\langle (L-1)(+{1\over 2})| (L+{1\over 2}) (L-{1\over 2})\rangle =\sqrt{{2L\over 2L+1}} \qquad \langle L(-{1\over 2})| (L+{1\over 2})(L -{1\over 2})\rangle =\sqrt{{1\over 2L+1} }\ .$$

There is only one state in the 2-dimensional subspace that is orthogonal to $$|(L+1/2)(L-1/2)\rangle$$. That is the state with $$J = L-1/2$$ and $$m_j = L-1/2$$. Then that state must be the one orthogonal to the above, $$|(L-{1\over 2})(L-{1\over 2})\rangle = - \sqrt{{1\over 2L+1}}\cdot | (L-1), (+ {1\over 2})\rangle + \sqrt{{2L\over 2L+1}} \cdot | L, (-{1\over 2})\rangle$$ up to an overall sign, which can only be fixed by convention. Then we can identify $$\langle (L-1)(+{1\over 2})| (L+{1\over 2}) (L-{1\over 2})\rangle = - \sqrt{{1\over 2L+1}} \qquad \langle L(-{1\over 2})| (L+{1\over 2})(L -{1\over 2})\rangle =\sqrt{{2L\over 2L+1}}\ .$$

We can continue similarly to the states with $$m_j = L-3/2$$ by applying $$J^-$$ one more time. In a similar way, we can compute all of the Clebsch-Gordan coefficients needed for this problem.

It is a nice exercise to carry out this process explicitly for a fixed value of $$L$$, for example, $$L=1$$, and compare your results to those in a reference table. A nice table can be found in Griffiths' Quantum Mechanics book, Table 4.8, or at the original source: https://pdg.lbl.gov/2020/reviews/rpp2020-rev-clebsch-gordan-coefs.pdf .

$$n$$ does not enter in angular momentum coupling as the operators $$\hat L_x, \hat L_\pm$$ do not change $$n$$. Thus the $$n$$’s are those of the original states you want to couple, and the CG’s are the “standard ones” as they do not depend on $$n$$’s.

Finally note that since $$J_z=L_z+S_z$$, indeed $$m_j=m_\ell+ m_s$$ else the CG is $$0$$. Any half-decent QM text will have this, although I recommend

Khersonskii, V.K., Moskalev, A.N. and Varshalovich, D.A., 1988. Quantum Theory of Angular Momemtum. World Scientific.

as the authoritative source of tables and properties of everything angular momentum. AFAIK, there no English e-version available at this time (pity!), although there is a GoogleBook substitute partially available.