# Angular momentum coupling

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

## 2 Answers

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

• 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" – Cryo Feb 22 '19 at 22:11

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 $$\left[L^2,S^2\right] = 0$$.

• 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: – Jung Feb 23 '19 at 13:18
• 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? – Jung Feb 23 '19 at 13:19
• 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 $-\frac { 1 } { 2 } \hbar$ – Jung Feb 23 '19 at 13:20
• 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. – Jung Feb 23 '19 at 13:21
• @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. – Thomas Fritsch Feb 23 '19 at 13:34