I have seen in the algebra of Angular Momentum, how Clebsch-Gordan coefficients arise. However, I do not understand the physical interpretation of it. When I checked the literature it was all in terms of representation group which I don't understand.

Can somebody give me an explanation of Clebsch-Gordan coefficients without going to group theory?

  • $\begingroup$ In what way is the representation theory not "physical"? Seriously: it is a core part of how quantum mechanics describes the world. Learn it, learn to love it, and learn to see how it encodes the physics. $\endgroup$ Jul 4, 2017 at 21:43
  • $\begingroup$ I have to say that reading through a few worked examples, for example in Squires Problems in QM sorted it out for me, without bringing in anything like group theory. cambridge.org/core/books/problems-in-quantum-mechanics/… $\endgroup$
    – user154420
    Jul 4, 2017 at 22:16

2 Answers 2


When combining two angular momenta, it results from this a number of possible values of $j$'s.

In the semiclassical model, the vectors $\vec j_1$ and $\vec j_2$ can be added to produce values of $\vec j$ beyond the simple addition $\vec j_1+\vec j_2$ because the orientations of $\vec j_1$ and $\vec j_2$ need not be colinear. Quantum mechanically, this means that $\vert j_1m_1\rangle \vert j_2m_2\rangle$ will not in general be eigenstates of the total operator $\hat J^2= (\hat J_1+\hat J_2)^2$ as eigenstates of $\hat J^2$ have $j$ as a "good" quantum number.

The values of the projection on the $\hat z$ axes, however, are scalars and additive, i.e. $m_1+m_2=m$ for any of the possible values of $\vec j$.

The Clebsch-Gordan coefficient $\langle jm\vert j_1m_1;j_2m_2\rangle$ with $m=m_1+m_2$ is just the amplitude of finding the state $\vert jm\rangle$ in the coupled state $\vert j_1m_1\rangle \vert j_2m_2\rangle$, just as $\langle \phi\vert \psi\rangle$ is the amplitude of finding the state $\vert \phi\rangle$ in the state $\vert \psi\rangle$.

In particular, $\vert \langle jm\vert j_1m_1;j_2m_2\rangle\vert^2 $ is the probability of finding $\vert jm\rangle $ in $\vert j_1m_1\rangle\vert j_2m_2\rangle$. This implies, for instance, the normalization condition $$ \sum_j\vert \langle jm\vert j_1m_1;j_2m_2\rangle\vert^2=1 $$

Alternatively, to obtain the state $\vert jm\rangle$ for specificed $j$, one must consider linear combinations of states $\vert j_1m_1\rangle\vert j_2m_2\rangle$ with different $m_1$, $m_2$ but such that $m_1+m_2=m$. The coefficients of the linear combinations are precisely the CGs. Thus $$ \vert jm\rangle =\sum_{m_1m_2} \vert j_1m_1\rangle \vert j_2m_2\rangle \langle j_1m_1;j_2m_2\vert jm\rangle \tag{1} $$ In this interpretation, the probability amplitude of finding $\vert j_1m_1\rangle \vert j_2m_2\rangle$ in the state $\vert jm\rangle$ is the CG, and it thus follows that $$ \sum_{m_1m_2}\vert \langle j_1m_1;j_2m_2\vert jm\rangle\vert^2=1\, . \tag{2} $$ In (1) and (2), the sum over $m_1$ restricted to that $m_1+m_2=m$. Finally, note that since the CGs are real, we have $$ \langle j_1m_1;j_2m_2\vert jm\rangle= \langle jm \vert j_1m_1;j_2m_2\rangle\, . $$


From linear Algebra we know, that we can express vectors in different basis. E.g. $$\vec v = \sum_j c_j \; \vec e_j = \sum_j c_j^\prime \; \vec e_j^\prime$$ Furthermore, we know that by changing the basis, we have to multiply the coefficients $c_j$ by the projections of the old basis vectors onto the new basis vectors.

In qm, instead of calling them vectors $\vec e_j$, we call them states $|j\rangle$. Now, if we change the basis, we obtain $$|J, M\rangle = \sum_{m_1, m_2} | j_1, m_1 ; j_2, m_2\rangle \; \langle j_1, m_1 ; j_2, m_2|J, M\rangle $$ where the the expansion coeff $\langle j_1, m_1 ; j_2, m_2|J, M\rangle$ is called Clebsch-Gordan coeff. It insures, that the projections are correct, by automatically fulfilling certain conditions. E.g. if $j_1 = 1$ and $j_2=2$ we won't be able to couple these two stated and obtain a state with $J = 4$. The same is true for the projections $m_1, m_2$.


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