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

  1. 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?
  2. 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

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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 $$ [ L \otimes I_s , I_o \otimes S] = [ L , I_o ] \otimes [ I_s , S ] = 0 . $$

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    $\begingroup$ Great reply! I just wanted to add, that the mathematical underpinning for this extension of the operators is the "tensor product of Lie algebra representations" $\endgroup$ – Cryo Feb 22 at 22:11
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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 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 $\psi$ acts 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 $\left[L^2,S^2\right] = 0$.

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  • $\begingroup$ Hi, I like your explanation since i haven't studied tensor product. I think $\hat { L } _ { x } = \hbar i \left( y \frac { \partial } { \partial z } - z \frac { \partial } { \partial y } \right)$ this is correct. I know it’s just a typo :) Few questions: $\endgroup$ – Jung Feb 23 at 13:18
  • $\begingroup$ a. To obtain $\widehat { J }$ you would compute $\widehat { J } _ { x }$, $\widehat { J } _ { y }$, and $\widehat { J } _ { z }$ using $\hat { J } _ { x } = \hat { L } _ { x } + \widehat { S } _ { x }$ respectively and then $\widehat { J } = \hat { J } _ { x } + \hat { J } _ { y } + \hat { J } _ { z | }$. Is that correct ? But similar for orbital and spin quantum operator, we dont discuss about $\widehat { J }$ but instead $\hat { J } ^ { 2 }$ right? $\endgroup$ – Jung Feb 23 at 13:19
  • $\begingroup$ b. I wonder why complex numbers are used to describe electron parts since the eigenvalues of $\hat { S } _ { z }$ are $\frac { 1 } { 2 } \hbar$ and $-\frac { 1 } { 2 } \hbar$ $\endgroup$ – Jung Feb 23 at 13:20
  • $\begingroup$ c. I notice that when you apply $\hat { S } _ { x }$ on the electron WF the spin is flipped. What is the physical reason for this? Thanks a lot. $\endgroup$ – Jung Feb 23 at 13:21
  • $\begingroup$ @jung b. The $i$ in the definition of $L_x$ is needed. Without that $\hat{L}_x$ would not be self-adjoint, i.e. its eigenvalues would not be real numbers. $\endgroup$ – Thomas Fritsch Feb 23 at 13:34

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