# How to prove that the L and S (of the total angular momentum commute?

• I
• Heidi
In non-relativstic theory all three spin-component operators commute with ##\hat{\vec{x}}## and ##\hat{\vec{p}}## and thus ##\hat{\vec{S}}^2## and ##\hat{S}_3## can be used as two additional compatible observables in addition to ##\hat{H}##, ##\hat{\vec{L}}^2##, and ##\hat{L}_3##. By the usual construction these common eigenvectors ##|E,L,m,s,\sigma \rangle## span the quantum-mechanical Hilbert space of a particle with spin.

The relativistic case is much more involved. Spin and orbital angular momentum cannot be distinguished. The observables are rather total angular momentum and "polarization states" (Pauli-Lubanski vector).

topsquark and PeroK
If L and S commute one can find a commun basis in which they would be diagonal
Son on this diagonal we would find 2 discrete values of S and the eigenvalues of L.
What is the mistake?

Heidi said:
If L and S commute one can find a commun basis in which they would be diagonal
Son on this diagonal we would find 2 discrete values of S and the eigenvalues of L.
What is the mistake?
L and S act on different Hilbert spaces. If you want to study them combined, then you need to use the tensor product of those Hilbert spaces, as your new Hilbert space on which those operators are appropriately defined.

vanhees71 and topsquark
As I said, a complete basis, spanning the Hilbert space of a single particle with spin ##s## is given by, e.g.,
$$| |\vec{p}|,\ell,m,s,\sigma \rangle$$
with ##|\vec{p}| \in \mathbb{R}_{\geq 0}##, ##\ell \in \mathbb{N}_0##, ##m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}##, ##s \in \mathbb{N}_0/2##, ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}##.

A "wave function" can then be represented wrt. this basis,
$$\psi_\sigma(|\vec{p}|,\ell,m,s)=\langle |\vec{p}|,\ell,m,s,\sigma|\psi \rangle.$$

topsquark
I don't see the statement there, although my French is very limited. Of course, you can as well have another complete set, i.e., with ##\vec{J}=\vec{L}+\vec{S}##, you can use ##\hat{\vec{J}}^2##, ##\hat{\vec{L}}^2##, ##\hat{\vec{S}}^2##, and ##\hat{J}_3## as a complete set of angular-momentum observables. It's just a basis transformation from the complete orthonormal set to this other one, introducing the corresponding Clebsch-Gordon coefficients dealing with the addition of two independent angular momenta (as here orbital and spin angular momentum of a non-relativistic particle).

topsquark
the french sentence is:
il convient de remarquer que le terme
ne commute pas avec
et

it should be noted that the term L.S does not commute with L and S (with arrows)
i agree that we can take the tensor product of the two Hilbert spaces (on which they act separately) to get one Hilbert space on which L ans S act. But how do you compure [L,S] to see if they commute?

Heidi said:
the french sentence is:
il convient de remarquer que le terme
ne commute pas avec
et

it should be noted that the term L.S does not commute with L and S (with arrows)
i agree that we can take the tensor product of the two Hilbert spaces (on which they act separately) to get one Hilbert space on which L ans S act. But how do you compure [L,S] to see if they commute?
Denote the space on which L acts as ##\Lambda## and the space S acts on as ##\Sigma##, with the usual eigen-relations.

Then we can act L and S on the tensor product space ##\Lambda \bigotimes \Sigma##. If I have my notation right:

##[L, S] ( \lambda \otimes \sigma ) \equiv ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle - ( S \bigotimes L ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= \lambda \sigma ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle - \sigma \lambda ( S \bigotimes L ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= ( \lambda \sigma - \sigma \lambda ) ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= 0##

Because the two spaces are independent, the eigenvalue relations will only act on those individual spaces, so commutativity is almost trivial.

-Dan

I leave out the hats. All angular-momentum components are operators in this posting.

We have
$$[L_j,L_k]=\mathrm{i} \epsilon_{jkl} L_l, \quad [S_j,S_k]=\mathrm{i} \epsilon_{jkl} S_l, \quad [L_j,S_k]=0.$$

Of course from this it follows that the ##\vec{L} \cdot \vec{S}## doesn't commute with ##\vec{L}## and also not with the ##\vec{S}## but with the ##\vec{J}##:
$$[L_j,L_k S_k]=[L_j,L_k]S_k+L_k [L_j,S_k]=\mathrm{i} \epsilon_{jkl} L_l S_k=\mathrm{i} (\vec{S} \times \vec{L})_j,$$
$$[S_j,L_k S_k]=[S_j,L_k]S_k + L_k [S_j,S_k] = L_k \mathrm{i} \epsilon_{jkl} S_l = \mathrm{i} (\vec{L} \times \vec{S})_j$$
and from that
$$[S_j,L_k S_k]=0.$$
So you can have the above mentioned new cons of common eigenvectors for ##\vec{J}^2##, ##\vec{L}^2##, ##\vec{S}^2##, and ##J_z##.

strangerep, Heidi and topsquark
topsquark said:
Denote the space on which L acts as ##\Lambda## and the space S acts on as ##\Sigma##, with the usual eigen-relations.

Then we can act L and S on the tensor product space ##\Lambda \bigotimes \Sigma##. If I have my notation right:

##[L, S] ( \lambda \otimes \sigma ) \equiv ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle - ( S \bigotimes L ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= \lambda \sigma ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle - \sigma \lambda ( S \bigotimes L ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= ( \lambda \sigma - \sigma \lambda ) ( L \bigotimes S ) \mid \lambda \rangle \, \otimes \mid \sigma \rangle##

##= 0##

Because the two spaces are independent, the eigenvalue relations will only act on those individual spaces, so commutativity is almost trivial.

-Dan
The point is that ##\vec{L} \cdot \vec{S}## acts on both products of the tensor product. There's no contradiction that neither ##\vec{L}## nor ##\vec{S}## commute with this spin-orbit term, but on the other hand it's a scalar quantity, and thus it must commute under rotations, and rotations are generated by the total angular momentum operators, i.e., ##\vec{J}=\vec{L}+\vec{S}##.

topsquark
Is the french wiki's sentence false?

No, why? It states what I've proved in #10. How do you come to the conclusion that ##\vec{L} \cdot \vec{S}## should commute with the ##\vec{L}## and/or the ##\vec{S}## components? It's a scalar under rotations, and rotations are represented by ##\vec{J}=\vec{L}+\vec{S}##, and indeed it commutes with ##\vec{J}## as it should be for a scalar.

I would like to be able to manipulate the algebra of opérators and the Hilbert space machinary With just the separable l2 global Hilbert space.
there is an hilbertian basis. an inner scalar product ans so on
How do you define L.S operator? What is this dot berween operators? of course it is not the scalar product between vectors. Is it a scalar product in the C* algebra?
And how to calculate [L, L.S] and give its non null value?

Heidi said:
I would like to be able to manipulate the algebra of opérators and the Hilbert space machinary With just the separable l2 global Hilbert space.
Only ##L## is an operator in this space; ##S## is not. So you can't even frame any questions about ##S##, or any expression involving ##S##, if you restrict yourself to this Hilbert space.

Heidi said:
What is this dot berween operators? of course it is not the scalar product between vectors
Yes, it is the scalar product between vectors--vectors of operators. The vector ##L## has components ##L_x##, ##L_y##, and ##L_z##, and the vector ##S## has components ##S_x##, ##S_y##, and ##S_z##. The dot product then works just like you would expect.

Heidi said:
And how to calculate [L, L.S] and give its non null value?
@vanhees71 already showed you how to do that in post #10.

vanhees71 and Heidi
thank you for the answers. I understant now.

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