I have a Hamiltonian for 3 particles of spin 1 that I boiled down to: \begin{equation} k(\textbf{S}^2+\cdots), \end{equation} where: \begin{equation} \textbf{S}=\textbf{S}_1+\textbf{S}_2+\textbf{S}_3. \end{equation} I read somewhere on the internet that the Clebsch-Gordan coefficients were the thing to use to figure out the value of $\textbf{S}^2$, but I'm not seeing it. Could someone explain to me how to use the Clebsch-Gordan coefficients for this case? - In particular, I understand (more or less) Clebsch-Gordan coefficients for adding two angular momenta, but here, I have 3 particles. Is there an equivalent of Clebsch-Gordan coefficients for three angular momenta $j_1,j_2,j_3$?

  • 1
    $\begingroup$ The Clebsch Gordan coefficients will help you relate the total |SM> basis to the direct product basis |s1m2>|s2m2>|s2m2>. $\endgroup$
    – hft
    May 10, 2015 at 4:05
  • $\begingroup$ Due diligence: Racah coeffs . $\endgroup$ Apr 30, 2016 at 0:57
  • $\begingroup$ Related: Adding 3 electron spins. $\endgroup$ Dec 4, 2016 at 4:36

2 Answers 2


Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system.

For 2 particles with total angular momentum eigenvalues $ l_1,l_2 $, such that for example $ m_1=-l_1,-l_1+1,...,+l_1 $ the composite system is -

$$ (l_1) \otimes (l_2) \sim (l_1+l_2) \oplus ...\oplus(|l_1-l_2|) $$

2 spin half particles reduce to - $$ (\frac{1}{2})\otimes(\frac{1}{2})\sim (1)\oplus(0) $$

the left size decribes 2 particles, each has spin 1/2 ,and you can decribe the sany system configuration by combination of those spins -

$$ |++\rangle , |+-\rangle, |-+\rangle, |--\rangle $$

the right side describes the system as composite system, which can be described by it's total angular momentum and it's total $z$ component -

$$ |S_{tot}=1,m_s=1\rangle ,|S_{tot}=1,m_s=0\rangle , |S_{tot}=1,m_s=-1\rangle , |S_{tot}=0,m_s=0\rangle ,$$

As I said, the Clebsch-Gordan coefficients are the basis transformation matrix elements, for example, one of them is - $$ \langle++|S_{tot}=1,m_s=1\rangle $$

The way to add 3 spins is first to add 2 of the and then to add the third one to the composite system -

$$ [(\frac{1}{2})\otimes (\frac{1}{2})]\otimes (\frac{1}{2})\sim [(1)\oplus(0)]\otimes(\frac{1}{2})\sim [(1)\otimes(\frac{1}{2})]\oplus[(0)\otimes(\frac{1}{2})]\sim (\frac{3}{2})\oplus (\frac{1}{2})\oplus(\frac{1}{2})$$

(notice the first addition. the later one is "conditional" - if S1+S2=1 then ... if S1+S2=0 then)

so the new basis, in order of $ |S_{tot},S_{1+2},m_{tot}\rangle $ is

$ |\frac{3}{2},1,-\frac{3}{2}\rangle,|\frac{3}{2},1,-\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{3}{2}\rangle,|\frac{1}{2},1,\frac{1}{2}\rangle,|\frac{1}{2},1,-\frac{1}{2}\rangle,|\frac{1}{2},0,-\frac{1}{2}\rangle,|\frac{1}{2},0,\frac{1}{2}\rangle $

this is enough for most applications (finding energy spectrum and such)

  • $\begingroup$ Thanks Alexander! So in the case of 3 spins, I need to use the CG coefficients twice, for adding 2 spins, then again to add them to the third spin? $\endgroup$
    – Frank
    May 10, 2015 at 15:13
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    $\begingroup$ Yes, this is the general method to add the spins/angular momentum of any number of particles. Notice that the coefficients are required only if you need an explicit representation of the new states in terms of the old ones. Many questions might be answered even without knowing the specific Clebsch-Gordan coefficients, but only with the final quantum numbers (that determine the configuration of the system) (questions that require to know the eigenvalues of the composite system, such as energy spectrum, but not the eigenvectors) $\endgroup$
    – Alexander
    May 10, 2015 at 16:31
  • $\begingroup$ Thanks - yes, I realized that, I can get energies and degeneracies just with the sum decomposition. Very neat! $\endgroup$
    – Frank
    May 10, 2015 at 16:40

Be aware that the order in which you do the couplings matters, in the sense that $(j_1\otimes j_2)\to j_{12}$ followed by $j_{12}\otimes j_3\to J$, usually written as $(j_1j_2)j_{12}j_3 J$, will not produce the same basis states as coupling $j_2\otimes j_3\to j_{23}$ first, and then coupling $j_1\otimes j_{23}\to J$, or $j_1(j_2j_3)j_{23}J$. The two sets of basis states are related by a unitary transformation that is used to define Racah coefficients as overlaps between the two bases.


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