# Total angular momentum operator

How do the eigenfunctions of the total angular momentum operator analytically look like?

I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can we explicitely say what they are?

I should add that I am particularly interested in $L$ to be orbital angular momentum operator and $S$ the spin-operator for electrons.

• They're spherical harmonics and Pauli matrices eigenvectors. I don't think they can be represented together, since they are elements of very different vector spaces. – QuantumBrick Mar 22 '15 at 3:28
• for the example of $S=1/2$ take a look at Sakurai equation 3.8.64 – Ali Moh Mar 22 '15 at 3:38
• @QuantumBrick, they're only Pauli matrix eigenvectors for spin 1/2, not in general. – hft Mar 22 '15 at 3:49
• @hft I believe people have the tendency of calling them Pauli matrices for any dimension, as long as they form the desired representation. Sometimes I hear people calling them "generalized Pauli matrices", then they're not talking about SU(2). – QuantumBrick Mar 22 '15 at 4:01
• @QuantumBrick, you should kindly inform those people who have that tendency that they are quite wrong. – hft Mar 22 '15 at 4:06

The trick is to expand one basis (say the uncoupled one with elements $\{\vert LM_L\rangle \vert SM_s\rangle:= \vert L M_L;SM_S\rangle \}$) in terms of another (say the coupled one with elements $\{\vert JM_J\rangle\}$.) The assumption is that the $\{\vert JM_J\rangle\}$ form a complete set in the sense that the identity $$\hat 1=\sum_{JM_J}\vert JM_J\rangle \langle J M_J\vert\, .$$ Hence: \begin{align} \vert LM_L; S M_S\rangle= \sum_{J(M_J)}\vert JM_J\rangle \langle J M_J\vert LM_L;SM_S\rangle\, . \tag{1} \end{align} The overlap coefficients $\langle J M_J\vert L M_L;SM_S\rangle$ are known as Clebsch-Gordan coefficients, sometimes also written as $C^{JM_J}_{LM_L;SM_S}$ or variations on that theme. The coefficients are easiest to calculate from recursion relations but the recursion has been solved and the coefficients have been reduced to summation form ; the simplest cases are often tabulated.

The possible values of $J$ in the sum of Eq.(1) are in the range $L+S, L+S-1, L+S-2, \ldots, \vert L-S\vert$, often written more compactly as $L+S\le J\le \vert L-S\vert$.

In addition, since the total projection $\hat J_z=\hat L_z+\hat S_z$, the eigenvalue $M_J=M_L+M_S$, further restricting the summation in (1). This restricted sum is indicated with the parenthesis around $(M_J)$.

Because they are transition coefficients from one orthonormal basis to another, the CG coefficients satisfy a number of orthonormality conditions, such as $$\sum_{J } \vert \langle JM_J\vert LM_L;S M_S\rangle \vert^2=1\, .$$ There are additional such formulae. Starting from $\langle JM_J\vert J’M’_{J}\rangle=\delta_{JJ’}\delta_{M_J M’_J}$ and inserting $$\hat 1=\sum_{M_LM_S}\vert LM_L;SM_S\rangle \langle LM_L;SM_S\vert$$ one gets $$\sum_{M_LM_S}\langle JM_J\vert L M_L;SM_S\rangle \langle LM_L;SM_s\vert J’M’_J\rangle=\delta_{JJ’}\delta_{M_JM’_J}$$ etc.

How do the eigenfunctions of the total angular momentum operator analytically look like?

I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can we explicitely say what they are?

The eigenfunctions of $J$ are going to be made up of linear combinations of tensor-product states of the eigenfunctions of $L$ and the eigenfunctions of $S$. In general, the linear combinations will involved more than just one tensor-product of the eigenfunctions of $L$ and $S$, however the "stretch-state" is the exception and is the starting point for constructing the others.

For example, if $L$ and $S$ are both spin 1/2 (yes, I know that $L$ usually stands for "orbital" angular momentum, but in this example $L$ and $S$ are both spin 1/2) then total $J$ can be either spin 1=|1/2+1/2| or spin 0=|1/2-1/2|. One of the eigenfunctions of $J=1$ is given by a tensor-product state $$|1,1>=|1/2,1/2>\otimes|1/2,1/2>\;,$$ which is the "stretch-state". The other states can be obtained by applying the lowering operator $$J_- = L_{-}\otimes 1 +1\otimes S_{-}\;,$$ to the stretch-state and normalizing. E.g., we find (I'm putting in the square root of two on the LHS by hand to indicate that the RHS is not generated as a normalized state) $$\sqrt{2}|1,0>=|1/2,-1/2>\otimes|1/2,1/2>+|1/2,1/2>\otimes|1/2,-1/2>$$ and $$|1,-1>=|1/2,-1/2>\otimes|1/2,-1/2>\;.$$ And the $|0,0>$ state is generated by creating a state with $J_z=0$ that is orthogonal to the $|1,0>$ state. It is $$\sqrt{2}|0,0>=|1/2,-1/2>\otimes|1/2,1/2>-|1/2,1/2>\otimes|1/2,-1/2>\;.$$

So, you see that two (|1,1> and |1,-1>) of the eigenfunctions of J in this case are simple tensor-products of the eigenfunctions of L and S, and the other two (|1,0> and |0,0>) are linear combinations of more than one tensor-product of the eigenfunctions of L and S.