Total angular momentum operator

Question

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.

ZeroTheHero · Accepted Answer

The trick is to expand one basis (say the uncoupled one with elements ${vert LM_Lrangle vert SM_srangle:= vert L M_L;SM_Srangle }$) in terms of another (say the coupled one with elements ${vert JM_Jrangle}$.)
The assumption is that the ${vert JM_Jrangle}$ form a complete set in the sense that the identity 
$$
hat 1=sum_{JM_J}vert JM_Jrangle langle J M_Jvert, .
$$
Hence:
begin{align}
vert LM_L;  S M_Srangle=
sum_{J(M_J)}vert JM_Jrangle langle J M_Jvert LM_L;SM_Srangle, .  tag{1}
end{align}
The overlap coefficients $langle J M_Jvert L M_L;SM_Srangle$ 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 $L+S, L+S-1, L+S-2, ldots, vert L-Svert$, often written more compactly as $L+Sle Jle vert L-Svert$.

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
$$
sum_{J } vert langle JM_Jvert LM_L;S M_Srangle vert^2=1, .
$$
There are additional such formulae.  Starting from $langle JM_Jvert J’M’_{J}rangle=delta_{JJ’}delta_{M_J M’_J}$ and inserting
$$
hat 1=sum_{M_LM_S}vert LM_L;SM_Srangle langle LM_L;SM_Svert
$$
one gets
$$
sum_{M_LM_S}langle JM_Jvert L M_L;SM_Srangle langle LM_L;SM_svert J’M’_Jrangle=delta_{JJ’}delta_{M_JM’_J}
$$
etc.

hft · Answer

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
$$
|1,1rangle=|1/2,1/2rangleotimes|1/2,1/2rangle;,
$$
which is the "stretch-state". The other states can be obtained by applying the lowering operator
$$
J_- = L_{-}otimes 1 +1otimes S_{-};,
$$
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)
$$
sqrt{2}|1,0rangle=|1/2,-1/2rangleotimes|1/2,1/2rangle+|1/2,1/2rangleotimes|1/2,-1/2rangle
$$
and
$$
|1,-1rangle=|1/2,-1/2rangleotimes|1/2,-1/2rangle;.
$$
And the $|0,0rangle$ state is generated by creating a state with $J_z=0$ that is orthogonal to the $|1,0rangle$ state. It is
$$
sqrt{2}|0,0rangle=|1/2,-1/2rangleotimes|1/2,1/2rangle-|1/2,1/2rangleotimes|1/2,-1/2rangle;.
$$
So, you see that two $(|1,1rangle$ and $|1,-1rangle)$ of the eigenfunctions of J in this case are simple tensor-products of the eigenfunctions of L and S, and the other two $(|1,0rangle$ and $|0,0rangle)$ are linear combinations of more than one tensor-product of the eigenfunctions of L and S.

Total angular momentum operator

2 Answers

Add your own answers!

Ask a Question