Angular momentum coupling I read about angular momentum coupling on wikipedia and there are a few things i dont understand.


*

*What does this mean "spin and orbital angular momentum of a single object belong to different Hilbert spaces" ? How does the addition of 2 vectors, which belong to different vector spaces work?

*When spin-orbit coupling is taken into consideration, then $$\left[ \hat { L } ^ { 2 } , \hat { H } \right] = 0$$
$$\left[ \hat { S } ^ { 2 } , \hat { H } \right] = 0$$ Can we deduce something about $\left[ \hat { L } ^ { 2 } , \hat { S } ^ { 2 } \right]$ ? I am not sure how a combination of 2 eigenstates of 2 operators($\hat { L } ^ { 2 }$ and $\hat { S } ^ { 2 }$), that do not commute with $\hat {H}$, will end up to be the eigenfunction of $\hat {H}$ $$\left[ \hat { J } ^ { 2 } , \hat { H } \right] = 0$$ where $J = L + S$
Link to Wiki page: https://en.wikipedia.org/wiki/Angular_momentum_coupling
 A: The orbital and spin angular momentum spaces are indeed different Hilbert spaces. The full Hilbert space is a tensor product of the two spaces. Let me denote a state in the orbital angular momentum Hilbert space by $| o \rangle$ and denote the a state in the spin angular momentum Hilbert space by $| s \rangle$. The full state of the particle is a tensor product
$$
|o\rangle \otimes | s \rangle
$$
The addition of operators in two different Hilbert spaces also happens via tensor product.
First, lets look at the operator $L$. This acts on the state $| o \rangle$ but not on $|s\rangle$. We can extend this operator on the full Hilbert space by taking a tensor product with the identity operator in the spin Hilbert space. Thus, what I really mean by $L$ is $L \otimes I_s$. Similarly what I really mean by $S$ is $I_o \otimes S$. Then, what I really mean by $L+S$ is
$$
L \otimes I_s + I_o \otimes S
$$
Its action on the state is similarly determined
\begin{align}
\left( L \otimes I_s + I_o \otimes S \right) |o\rangle \otimes | s \rangle   &=    L  |o\rangle \otimes I_s | s \rangle +  I_o 
 |o\rangle \otimes  S   | s \rangle \\
&=   L  |o\rangle \otimes  | s \rangle +  |o\rangle \otimes  S   | s \rangle
\end{align}
From this the commutator between $L$ and $S$ is obvious. What we really mean by $[L,S]$ is actually
$$
[ L \otimes I_s , I_o \otimes S] = [ L , I_o ] \otimes [ I_s , S ] = 0 . 
$$
A: 
What does this mean "spin and orbital angular momentum of a single object
operate on different Hilbert spaces" ?

To keep the math simple and concrete I choose a single electron as the system to be described,
and use the Schrödinger representation for describing it.
For describing the electron's orbital part only,
we would use a complex-valued function
$\psi(x,y,z)$ on the 3-dimensional $xyz$ space.
Here the Hilbert space is the space of functions mapping from $\mathbb{R}^3$ to $\mathbb{C}$.
For describing the electron's spin part only,
we would use 2 complex numbers: $\begin{pmatrix} \psi_+ \\ \psi_- \end{pmatrix}$.
Here the Hilbert space is just $\mathbb{C}^2$.
Putting the orbital and spin part together, we can describe the electron
by two complex-valued functions $\begin{pmatrix} \psi_+(x,y,z) \\ \psi_-(x,y,z) \end{pmatrix}$
on the 3-dimensional $xyz$ space.
The full Hilbert space now is the space of functions mapping from $\mathbb{R}^3$ to $\mathbb{C}^2$.
Mathematically speaking: the full Hilbert space is the tensor product of the two Hilbert spaces from above.
Now, for example, consider the $x$-component of the orbital angular momentum operator:
$$ \hat{L}_x = \frac{\hbar}{i}(\pmb{r}\times\pmb{\nabla})_x
= \frac{\hbar}{i} \left(y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}\right) $$
When applying this $\hat{L}_x$ operator on $\psi$ we get
$$
\hat{L}_x \begin{pmatrix} \psi_+(x,y,z) \\ \psi_-(x,y,z) \end{pmatrix}
= \frac{\hbar}{i} \begin{pmatrix} 
  y \frac{\partial}{\partial z} \psi_+(x,y,z) - z \frac{\partial}{\partial y} \psi_+(x,y,z)  \\
  y \frac{\partial}{\partial z} \psi_-(x,y,z) - z \frac{\partial}{\partial y} \psi_-(x,y,z)
\end{pmatrix}
$$
Obviously the $\hat{L}_x$ operator acts on the upper and lower $\psi$ components in the same way.
It acts only on the orbital $(x,y,z)$ part, not on the spin $(+,-)$ part.
As a second example, consider the $x$-component of the spin angular momentum operator:
$$ \hat{S}_x = \frac{\hbar}{2}\sigma_x
= \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
When applying this $\hat{S}_x$ operator on $\psi$ we get
$$
\hat{S}_x \begin{pmatrix} \psi_+(x,y,z) \\ \psi_-(x,y,z) \end{pmatrix}
= \frac{\hbar}{2} \begin{pmatrix} \psi_-(x,y,z) \\ \psi_+(x,y,z) \end{pmatrix}
$$
Obviously the $\hat{S}_x$ operator acts on $\psi$ independent of $x$, $y$, $z$.
It acts only on the spin $(+,-)$ part, not on the orbital $(x,y,z)$ part.

How does the addition of 2 vectors operators, which operate on different Hilbert spaces work?

Now you can write down other operators, like for example the
$x$-component of the total angular momentum operator:
$$
\hat{J}_x
= \hat{L}_x + \hat{S}_x
= \frac{\hbar}{i} \left(y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}\right) 
+ \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
$$
When applying this $\hat{J}_x$ operator on $\psi$ we get
$$
\hat{J}_x \begin{pmatrix} \psi_+(x,y,z) \\ \psi_-(x,y,z) \end{pmatrix}
= \hbar \begin{pmatrix} 
  -i y \frac{\partial}{\partial z} \psi_+(x,y,z) + i z \frac{\partial}{\partial y} \psi_+(x,y,z) + \frac{1}{2} \psi_-(x,y,z) \\
  -i y \frac{\partial}{\partial z} \psi_-(x,y,z) + i z \frac{\partial}{\partial y} \psi_-(x,y,z) + \frac{1}{2} \psi_+(x,y,z)
\end{pmatrix}
$$
We see, the $\hat{J}_x$ operator acts in a more complicated way both on the
orbital $(x,y,z)$ and the spin $(+,-)$ part of $\psi$.

Can we deduce something about $\left[L^2,S^2\right]$ ?

With the technique described above it is trivial to show that
$$ L^2 S^2 \psi = S^2 L^2 \psi $$
and hence $\left[L^2,S^2\right] = 0$.
