What is the value of the commutator $[L_x, S_y]$? [closed]

Question

I would like to know how to calculate the commutator $[L_x, S_y]$ in quantum mechanics, if $L_x$ is the x-component of the angular momentum operator and $S_y$ the y-component of the spin operator.

I would like to know this, because I'm trying to determine $[J_x, J_y]$, where $J=L+S$ is the total angular momentum of a particle. With some algebra we can write the commutator as: $$[J_x, J_y]= [L_x, L_y]+ [S_x, L_y] + [L_x, S_y] + [S_x, S_y]$$

I already know the values for $[L_x, L_y]=ihL_z$ and $[S_x, S_y]=ihS_z$ and I suspect that $[S_x, L_y] + [L_x, S_y]=0$, so that $[J_x, J_y]=ihJ_z$ but I don't know how to show this.

Answer 1

As the comments already said, a commutator like $[L_x,S_y]$ is zero for a simple reason. To understand this, take for example a single particle with spin $\frac 12$. This can be described by a $2$-component spinor wavefunction. $$\psi = \begin{pmatrix} \psi_+(x,y,z) \\ \psi_-(x,y,z) \end{pmatrix}$$

Then $L_x$ is a differential operator $$L_x=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)$$ which acts different for different positions $(x,y,z)$, but in the same way on $\psi_+$ and $\psi_-$.
And $S_y$ is a matrix multiplication operator $$S_y=\frac 12 \hbar \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$$ which scrambles $\psi_+$ and $\psi_-$, but does so independent of $(x,y,z)$.

Now you can calculate $L_xS_y\psi$ and $S_yL_x\psi$ (and I recommend you do it on your sheet of paper, to see there is no magic). For both you get the same result

$$\frac 12\hbar^2\begin{pmatrix} -y\frac{\partial\psi_-(x,y,z)}{\partial z}+z\frac{\partial\psi_-(x,y,z)}{\partial y} \\ y\frac{\partial\psi_+(x,y,z)}{\partial z}-z\frac{\partial\psi_+(x,y,z)}{\partial y} \end{pmatrix}$$

So it is $L_xS_y\psi=S_yL_x\psi$ for every $\psi$, and hence $[L_x,S_y]=0$.

Answer 2

The current answer correctly addresses your question from a concrete standpoint, so I will briefly mention a more abstract one. If you have a composite Hilbert space $\mathscr H = \mathscr H_A\otimes \mathscr H_B$, then a generic element of $\mathscr H$ can be written as $$\psi = \sum_n \alpha_n \otimes \beta_n$$

where the $\alpha_n$'s and $\beta_n$'s are elements of $\mathscr H_A$ and $\mathscr H_B$, respectively. If $O_A$ is an operator on $\mathscr H_A$ alone, we can "promote" it to an operator on $\mathscr H$ by letting $O_A \mapsto O_A\otimes \mathbb 1$, which acts on elements of $\mathscr H$ as $$\big(O_A\otimes \mathbb 1\big) \psi = \sum_n \big(O_A \alpha_n\big)\otimes\big(\mathbb 1 \beta_n\big) = \sum_n \big(O_A \alpha_n\big) \otimes \beta_n$$

Similarly, we can promote an operator $O_B$ on $\mathscr H_B$ to an operator on the full composite space by letting $O_B \mapsto \mathbb 1 \otimes O_B$.

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Having made this point, it is trivially easy to show that if $O_A$ and $O_B$ are operators which act on $\mathscr H_A$ and $\mathscr H_B$, then their promoted partners must commute with one another:

$$\big(\mathbb 1 \otimes O_B\big) \psi = \sum_n \alpha_n \otimes \big(O_B \beta_n\big) \qquad \big(O_A \otimes \mathbb 1\big)\big(\mathbb 1 \otimes O_B\big) \psi = \sum_n \big(O_A\alpha_n\big) \otimes \big(O_B \beta_n\big)$$ $$\big(O_A \otimes \mathbb 1\big) \psi = \sum_n\big(O_A \alpha_n\big) \otimes \beta_n \qquad \big(\mathbb 1 \otimes O_B\big)\big(O_A \otimes \mathbb 1\big) \psi = \sum_n \big(O_A\alpha_n\big) \otimes \big(O_B \beta_n\big)$$

$$\implies \big(O_A \otimes \mathbb 1\big)\big(\mathbb 1 \otimes O_B\big)= \big(\mathbb 1 \otimes O_B\big)\big(O_A \otimes \mathbb 1\big) $$ $$\implies \bigg[O_A \otimes \mathbb 1 , \mathbb 1 \otimes O_B \bigg] = 0$$

Your question asks specifically about the commutation of $L_n\otimes \mathbb 1$ and $\mathbb 1 \otimes S_n$ which act on the composite Hilbert space $L^2(\mathbb R^3) \otimes \mathbb C^2$. State vectors in this space take the general form $$\psi = \pmatrix{\phi_1 \\ \phi_2} = \phi_1 \otimes \pmatrix{1\\0} + \phi_2 \otimes \pmatrix{0\\1}$$ where $\phi_1,\phi_2\in L^2(\mathbb R^3)$. The operators $L_n$ on $L^2(\mathbb R^3)$ are the familiar differential operators which correspond to the angular momentum associated to spatial wavefunctions, while the operators $S_n$ on $\mathbb C^2$ are the Pauli operators which correspond to the angular momentum associated to 2-component spinors.

For historical and conventional reasons, the former are referred to as orbital angular momentum operators while the latter are referred to as spin angular momentum operators. The full operators $J_n = L_n\otimes \mathbb 1 + \mathbb 1 \otimes S_n$ which correspond to the angular momentum of a spin-1/2 particle in $d=3$ spatial dimensions are referred to as the "total" angular momentum operators.