I've come across this a lot, but I've never understood it. I do know basic Group Theory including Lie Groups. In Introduction to Quantum Mechanics, Griffiths ends the chapter on spin with the remark

"In a mathematical sense this is all group theory- what we are talking about is the decomposition of the direct product of two irreducible representations of the rotation group into a direct sum of the irreducible representations (you can quote that, to impress your friends)."

Could someone explain this translation of the mechanics of spin into the language of group theory?

This is the quantum mechanics part as I understand it:

  1. The operators (that act on states) form a group and have a group algebra.

  2. Raising and lowering operators are constructed. There is a $j=1$ and $j=1/2$ representation of the rotation group.

  3. Addition of spin operators is defined and splits the states into a triplet (eigenvector of $S^2$ with eigenvalue $1$) and a singlet (eigenvector of $S^2$ with eigenvalue $0$).

  4. Spin $1/2$ + Spin $1/2$ = $1$ or $0$.

  5. If you similarly "add" any spin states, then you can get another state with any spin ($s_1 +s_2$) down to ($s_1 −s_2 $) or ($s_2 −s_1 $). The state (say [(3,0)> is expressed as a sum of those possible states ([(2,1)>[(1,-1)> ,[(2,0)>[(1,0)> , [(2,-1)>[(1,1)> ) and the coefficients of those states are the Clebsch-Gordon Coefficients.

  • 4
    $\begingroup$ They talk about Lie groups and Lie algebras. I found my self long time ago with the same problem, not only in QM but also when I was reading about mechanics with the book from Goldstein. The only way you could have this clear is if you read about Lie groups and Lie algebras... I think you'll find this very helpful physics.stackexchange.com/q/148116 $\endgroup$
    – raul
    Oct 26, 2015 at 20:49
  • $\begingroup$ Yes, addition of angular momenta is a straightforward application of the so-called Clebsch-Gordan formula from group theory - case of group SU(2). $\endgroup$
    – DanielC
    Oct 26, 2015 at 22:33
  • 5
    $\begingroup$ To be clear, group theory tells about angular momentum addition, not the other way around. $\endgroup$
    – Prahar
    Oct 27, 2015 at 1:09

1 Answer 1


It is actually all elementary group representation theory, but Dirac, a math major in college, spoiled 3 generations of us physicists by making it too easy and cookbook in his book, to the extent that Griffiths' text describes this basic mathematical structure as a side formality...

Wikipedia says it best: The spin/angular momentum operators act on states and rotate them to each other in ways that keep essential features of the theory invariant, as the Hamiltonian is symmetric with respect to them---it commutes with them. They obey 3 commutation relations, which may be rewritten in more practical, "ladder" bases.

A particle of spin $s_1$ is operated upon by such operators, in an $(2s_1+1)$-dimensional space. Another particle of spin $s_2$ is operated upon by such operators, in an $(2s_2+1)$-dimensional one. But the differently sized matrices acting on these spaces obey the same commutation relations, that is, they are different representations of the same group (here, SO(3), but don't worry about it).

When two such components are put together ("added"), they live in a "tensor product space" of dimension $(2s_1+1)\times (2s_2+1)$, and the sublime mathematical fact underlying this composition is that the composite operators acting on both spaces obey the very same angular momentum commutation relations (their coproduct obeys the same Lie algebra, in mathematese).

As a result, the "total angular momentum operators" are in an $(2s_1+1)(2s_2+1)$-dimensional representation, that is, they are $(2s_1+1) (2s_2+1)\times (2s_1+1)(2s_2+1)$ matrices. These matrices, however, are "reducible", that is, suitably transformed by similarity transformations, they break up into disjoint blocks which do not "talk" to each other... they never get to connect states one block acts on to states of another block. Each such block is thus disjoint, and we say the reducible composite representation has broken up into a direct sum of irreducible ones (the blocks)... this is the Clebsch-Gordan series.

To illustrate this with your example, $s_1=s_2=1/2$, when you "add" the two, (Kronecker multiply the two doublets, in mathematese, ${\bf 2}\otimes{\bf 2}={\bf 4}$ : these guys prefer to write down the dimensionality of the vector spaces in boldface). But, as you've learned from your C-G expansion, the triplet doesn't talk to the singlet--- they live on different alternative worlds, as it were (in mathematese, the quartet is reducible, ${\bf 4}={\bf 3}\oplus {\bf 1}$). In this case, the effective s of the reduced multiplets is the sum or the difference of the two ss of the input multiplets.

In general, it can be a lot of complicated things... E.g., as per Wikipedia, three spin 1/2s add to a spin 3/2 and two spin 1/2s, ${\bf 2}\otimes{\bf 2}\otimes{\bf 2}={\bf 4} \oplus{\bf 2}\oplus{\bf 2}$; or else, a spin 2 composed with a spin 1, yields spins 3, 2, and one, ( ${\bf 5}\otimes{\bf 3}={\bf 7} \oplus{\bf 5}\oplus{\bf 3}$ in mathematese, comporting with the QM textbook rules.)

If you are really curious about how such reductions work in the matrix representation, consider working through Problem 4 of my Some notes of mine. Chapter 16 of the classic text of Mathews & Walker, purportedly written by a young Sidney Coleman, has 95% of the Lie group theory a physicist would need.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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