# I Adding types of angular momenta

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1. May 14, 2017

### Jezza

There are two types of angular momentum: orbital and spin. If we define their operators as pseudo-vectors $\vec{L}$ and $\vec{S}$, then we can also define the total angular momentum operator $\vec{J} = \vec{L}+\vec{S}$.

Standard commutation relations will show that we can have simultaneous well defined values for $J^2$ and $J_z$ and etc. for $\vec{L}$ & $\vec{S}$. i.e. We can have well defined total angular momentum and one component of it (usually z) for each type. The eigenvalues of these operators are then $\hbar^2 j(j+1)$ and $\hbar m_j$ respectively when we consider a simultaneous eigenstate $|j,m_j>$ of $J^2$ and $J_z$ only, and etc. for $l$ and $s$

My question is really about how these different types combine.

Using these standard commutation relations:
$$[J_i, J_j] = i \sum_k \epsilon_{ijk} J_k \hspace{10mm} [J_i, L_j] = i \sum_k \epsilon_{ijk} L_k \hspace{10mm} [J_i, S_j] = i \sum_k \epsilon_{ijk} S_k$$
It's very easy to show that we can have a simultaneous eigenstate $|m_j,m_l,m_s>$ of $J_z, L_z, S_z$ respectively, and thus the relation between the eigenvalues is
$$m_j=m_l+m_s$$
We also have the commutation relations:
$$[J^2, L^2] = 0 \hspace{10mm} [J^2, S^2] = 0 \hspace{10mm} [L^2, S^2] = 0$$
So we can have a simultaneous eigenstate $|jls>$ of $J^2, L^2, S^2$. My question is then what is the relationship between $j, l$ and $s$? So far as I can see, it is not straightforwards because:
$$J^2 = L^2+S^2+2\vec{L}\cdot\vec{S} = L^2+S^2+2\sum_iL_iS_i \\ j(j+1) = l(l+1) + s(s+1) + \frac{2}{\hbar^2} \sum_i <jls|L_iS_i|jls>$$

Which presents a problem, since $[L_i, J^2], [S_i,J^2] \neq 0$ so the state $|jls>$ cannot be an eigenstate of any of $S_i, L_i$ and so the relationship between the 3 numbers is not well defined.

How can we have a state in which $j,l,s$ are well defined and yet their relationship is not well defined?

2. May 14, 2017

### blue_leaf77

Depends on the Hamiltonian, if it doesn't contain a coupling term between orbital and spin angular momenta then yes you can form a simultaneous eigenstates of $J_z$, $L_z$, and $S_z$.
It is a well-known relationship that if $\vec J＝ \vec L + \vec S$ then $j = |l+s|, |l+s|-1, \ldots, |l-s|+1, |l-s|$. There is a way to prove it unfortunately I forgot it already now.

3. May 14, 2017

### Staff: Mentor

I don't see how that follows. The bracket $<jls|L_i S_i|jls>$ is well-defined for any state, whether it's an eigenstate or not.

4. May 14, 2017

### vanhees71

In non-relativistic QT, where spin and orbital angular momentum are well-defined observables you have the commutation relations
$$[S_i,S_j]=\mathrm{i} \epsilon_{ijk} S_k, \quad [L_i,L_j]=\mathrm{i} \epsilon_{ijk} L_k, \quad [S_i,L_j]=0$$
from which you also find
$$[\vec{L}^2,L_j]=[\vec{S}^2,S_j]=[\vec{S}^2,L_j]=[\vec{L}^2,S_j]=0.$$
So you can define common eigenvectors of $\vec{L}^2$, $\vec{S}^2$, $L_3$, and $S_3$.

Of course you can also define common eigenvectors of $\vec{J}^2$, $\vec{L}^2$, $\vec{S}^2$, and $J_3$ and express everything in the one or the other basis. The corresponding coefficients are known as Clebsch-Gordan coefficients,
$$C_{L,S,m,\sigma}^{L,S,J,M}=\langle L,s,m,\sigma|L,S,J,M \rangle.$$
As stated in #2 for given $L$ and $S$, the coefficients are different from zero only for
$$J \in \{|L-S|,|L-S|+1,\ldots,L+S \}, \quad M=m+\sigma.$$

If you want in addition that you have also an eigenstate of the Hamiltonian, $H$, it of course depends on the Hamiltonian, which angular-momentum parts are commuting with it. For a closed system $H$ must commute with $\vec{J}$ (the total angular momentum), because of rotational symmetry of Galileo space-time, i.e., the total angular momentum is conserved (Noether's theorem for rotations). In general systems neither $\vec{L}$ nor $\vec{S}$ are conserved for themselves but only $\vec{J}=\vec{L}+\vec{S}$.

5. May 15, 2017

### Jezza

I suppose what I meant by this is one cannot deduce it's value merely from its being in an eigenstate of $J^2, L^2, S^2$. Is that fair to say?

But thank you everyone I think this makes a lot more sense now. I think, then, the short answer is $|l-s| \leq j \leq l+s$.

6. May 15, 2017

### blue_leaf77

Just be careful the the allowed values of $J$ within that interval hop by one unit, not continuous.

7. May 15, 2017

### Jezza

I've never understood Clebsch-Gordan coefficients, but I've never thought of them as simply the coefficients in a basis transformation, so thanks for this. I'll go away and read about them again and hopefully they'll make much more sense to me!

8. May 15, 2017

### vanhees71

It's not an easy topic. It takes some time to understand how to evaluate the CG coefficients. I pretty clear explanation is given in

J. J. Sakurai et al, Modern Quantum Mechanics

9. May 15, 2017

### Jezza

Thanks for the recommendation, our library has it so I'll go and have a look.

10. May 15, 2017

### Jezza

The book does indeed have a very clear explanation. Thank you very much :)

11. May 15, 2017

### Staff: Mentor

No. The term $\Sigma_i <jls|L_i S_i|jls>$ is just the expectation value of the operator $\vec{L} \cdot \vec{S}$ for the state $|jls>$. You can always compute an expectation value of an operator for any state.

12. May 15, 2017

### dextercioby

The last part is not true, for there are states (vectors as pure state representatives) which don't belong to the maximal domain of an operator, hence the expectation value is undefined. Which leads me to the question: what does $|jls\rangle$ stand for?

13. May 15, 2017

### Staff: Mentor

Can you give an example?

As I understand it from the OP, it's a state which is a simultaneous eigenstate of $J^2$, $L^2$, and $S^2$.

14. May 15, 2017

### dextercioby

For a Galilean particle moving freely in R from -infinity to +infinity, the maximal domain for the coordinate x is

$D_x = \{\psi (x) \in \mathbb{L^2 (R)} | \int_R x^2 \psi^2 (x) {} dx < \infty \}$

For $\phi \in \mathbb{L^2 (R)} \setminus D_x$, $x\phi$ is undefined, hence the expectation value as well.

15. May 15, 2017

### dextercioby

Perhaps, I am not seeing it, but if L and S "live" in separate Hilbert spaces, how do their squares share eigenstates? If one then switches to the tensor product space, how's $|jls\rangle$ defined in a tensor product space of two spaces?

16. May 15, 2017

### Staff: Mentor

Can you pose this question in the context of post #4? That post by @vanhees71 looks to me to be the best presentation in the thread of the applicable underlying theory.

17. May 15, 2017

### dextercioby

Well, in post# 4, there is no supposition about the underlying Hilbert space(s), but if L is orbital angular momentum ("living" in L^2 (R^3)), and S is spin angular momentum ("living" in C^{k}, k in N), then, it's clear that their sum is ill-defined, unless one uses tensor products of spaces.

18. May 15, 2017

### blue_leaf77

Yes, the two operators are actually defined in a tensor product space. When both spin and orbital angular momenta are considered, physicists like to use $L^2$ and $S^2$ as shorthand notations for $L^2 \otimes I_S$ and $I_L \otimes S^2$ respectively, where $I_L$ and $I_S$ are identity operators in orbital and spin space. The same also applies to the components of $\vec L$ and $\vec S$. The notation $|jls\rangle$ is actually not a complete specification of the states, usually one more is required which is $m_j$.

Last edited: May 15, 2017
19. May 16, 2017

### vanhees71

Of course in non-relativistic physics, where there is a unique split of total angular momentum in to orbital and spin angular momentum and all spin components commute with all orbitarl-angular-momentum components the Hilbert space representing $\vec{L}$ and $\vec{S}$ is the tensor product $\mathcal{H}_L \otimes \mathcal{H}_S$. The common eigenvectors of $\vec{L}^2$, $L_z$, $\vec{S}^2$, and $S_z$ are given by the product basis
$$|l,m,s,\sigma \rangle=|l,m \rangle \otimes |s,\sigma \rangle.$$