# 2.35: Addition of Spin Angular Momentum- A Tensor Algebra Approach

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In this entry I work through section 4.4.3 of David Griffithsʹ Introduction to Quantum Mechanics (2nd ed.) in which he treats the addition of angular momentum for two identical spin‐1/2 particles. The tensor algebra approach is illustrated.

The four spin states of two spin‐1/2 particles are written below in the spin‐z basis in tensor format.

$\begin{matrix} | \uparrow \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & | \downarrow \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber$

$\begin{matrix} | \uparrow \uparrow \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} & | \uparrow \downarrow \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} & | \downarrow \uparrow \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} & | \downarrow \downarrow \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber$

The two middle states are not permissible because they distinguish between identical particles. The solution is to form symmetric and anti‐symmetric superpositions of them, but letʹs follow Griffithʹs approach for the time being. The four spin states are labeled as shown below.

$\begin{matrix} \text{a} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} & \text{b} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} & \text{c} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} & \text{d} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber$

The identity and spin operators in units of h/2π are now defined. The final two are the up and down spin ladder operators.

$\begin{matrix} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & S_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & S_y = \frac{1}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} & S_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & S = S_x + S_y + S_z \\ & S_u = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & S_d = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \end{matrix} \nonumber$

Next the total spin operator and total spin operator in the z‐direction are defined using kronecker, Mathcadʹs command for tensor matrix multiplication.

$\begin{matrix} S_{tot} = \text{kronecker(S, I)} + \text{kronecker(I, S)} & Sz_{tot} = \text{kronecker} \left( S_z,~I \right) + \text{kronecker} \left( I,~ S_z \right) \end{matrix} \nonumber$

Calculation of the expectation values for total spin in the z‐direction for spin states a, b, c and d reveals the problem mentioned above. For S = 1 we expect Sz values of ‐1, 0 and 1. The extra value for Sz = 0 indicates an interpretive problem.

$\begin{matrix} \text{a}^T Sz_{tot} \text{a} = 1 & \text{b}^T Sz_{tot} \text{b} = 0 & \text{c}^T Sz_{tot} \text{c} = 0 & \text{d}^T Sz_{tot} \text{d} = -1 \end{matrix} \nonumber$

Griffiths solves the problem by operating with the lowering operator on spin state a, which yields an in‐phase superposition (unnormalized) of spin states b and c.

$\left( \text{kronecker} \left( S_d,~I \right) + \text{kronecker} \left( I,~ S_d \right) \right) \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \nonumber$

Repeating with the result from above yields an unnormalized d spin state.

$\left( \text{kronecker} \left( S_d,~I \right) + \text{kronecker} \left( I,~ S_d \right) \right) \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 2 \end{pmatrix} \nonumber$

Operating on the unnormalized d spin state yield a null vector suggesting that the original spin states might be reformulated as triplet and singlet states.

$\left( \text{kronecker} \left( S_d,~I \right) + \text{kronecker} \left( I,~ S_d \right) \right) \begin{pmatrix} 0 \\ 0 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \nonumber$

Given this hint and the fact that the initial spin states are orthonormal, we preserve this property in the new spin states by constructing an out‐of‐phase superposition of b and c. This give us a revised set of orthonormal spin vectors. In conventional notation these states are |11 >, |10 >, |1‐1 > and |00 >, where the first number is total spin value and the second the spin in the z‐direction. Thus, we have a triplet and a singlet state as will be confirmed below.

$\begin{matrix} \Psi_{1p1} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} & \Psi_{10} = \begin{pmatrix} 0 \\ \frac{1}{ \sqrt{2}} \\ \frac{1}{ \sqrt{2}} \\ 0 \end{pmatrix} & \Psi_{1m1} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} & \Psi_{00} = \begin{pmatrix} 0 \\ \frac{1}{ \sqrt{2}} \\ \frac{-1}{ \sqrt{2}} \\ 0 \end{pmatrix} \end{matrix} \nonumber$

We now establish by calculation of expectation values for S2 (constructed below) and Sz that the first three spin states are members of a triplet state (S = 1) and the final spin state is a singlet (S = 0). The eigenvalues for the S2 operator are S(S + 1) which is 2 for the triplet state and 0 for the singlet state.

$\text{SS = kronecker} \left( S^2,~ I \right) + \text{kronecker} \left( I,~S^2 \right) + 2 \text{kronecker} \left( S_x,~ S_x \right) + \text{kronecker} \left( S_y,~S_y \right) + \text{kronecker} \left( S_z,~S_z \right) \nonumber$

$\begin{matrix} \left( \Psi_{1p1} \right)^T \text{SS} \Psi_{1p1} = 2 & \left( \Psi_{10} \right)^T \text{SS} \Psi_{10} = 2 & \left( \Psi_{1m1} \right)^T \text{SS} \Psi_{1m1} = 2 & \left( \Psi_{00} \right)^T \text{SS} \Psi_{00} = 0 \\ \left( \Psi_{1p1} \right)^T Sz_{tot} \Psi_{1p1} = 1 & \left( \Psi_{10} \right)^T Sz_{tot} \Psi_{10} = 0 & \left( \Psi_{1m1} \right)^T Sz_{tot} \Psi_{1m1} = -1 & \left( \Psi_{00} \right)^T Sz_{tot} \Psi_{00} = 0 \end{matrix} \nonumber$

Next, we look at this problem from the energy perspective and use the spin‐spin interaction Hamiltonian to calculate the energy eigenvalues and eigenvectors for the two spin‐1/2 particle system

$H_{SpinSpin} = \text{kronecker} \left( S_x,~ S_x \right) + \text{kronecker} \left( S_y,~S_y \right) + \text{kronecker} \left( S_z,~ S_z \right) \nonumber$

$H_{SpinSpin} = \begin{pmatrix} 0.25 & 0 & 0 & 0 \\ 0 & -0.25 & 0.5 & 0 \\ 0 & 0.5 & -0.25 & 0 \\ 0 & 0 & 0 & 0.25 \end{pmatrix} \nonumber$

We now ask Mathcad to calculate the eigenvalues and eigenvectors of the spin‐spin operator. These results are displayed by constructing a matrix which contains the eigenvalues in the top row, and their eigenvalues in the columns below the eigenvalues.

$\begin{matrix} i = 1..4 & \text{E = eigenvals} \left( H_{SpinSpin} \right) & \text{EigenvalsEigenvec = rsort} \left( \text{stack} \left( E^T,~ \text{eigenvecs} \left( H_{SpinSpin} \right) \right),~1 \right) \end{matrix} \nonumber$

$\text{EigenvalEigenvec} = \begin{pmatrix} -0.75 & 0.25 & 0.25 & 0.25 \\ 0 & 1 & 0 & 0 \\ 0.707 & 0 & 0 & 0.707 \\ -0.707 & 0 & 0 & 0.707 \\ 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$

These calculations are consistent with those that preceded and with the results presented on page 284 of Griffithsʹ text.

This page titled 2.35: Addition of Spin Angular Momentum- A Tensor Algebra Approach is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Frank Rioux via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.