Fundamental Commutation Relations in Quantum Mechanics

Question

I am trying to compile a list of fundamental commutation relations involving position, linear momentum, total angular momentum, orbital angular momentum, and spin angular momentum. Here is what I have so far:

$$ [x_i,x_j]=0 \\ [p_i,p_j]=0 \\ [x_i,p_j]=i\hbar\delta_{ij} \\ [J_i,J_j]=i\hbar\epsilon_{ijk}J_k \\ [L_i,L_j]=i\hbar\epsilon_{ijk}L_k \\ [S_i,S_j]=i\hbar\epsilon_{ijk}S_k \\ [J^2,J_i]=[L^2,L_i]=[S^2,S_i]=0, where \ X^2=X_i^2+X_j^2+X_k^2 $$

I also know that $L$ and $S$ commute, but I am unsure why. I've heard that it is simply because they act on different variables, but I don't understand exactly what this means. Is there a way to show this explicitly?

What are the remaining commutation relations between $J$, $J^2$, $L$, $L^2$, $S$, and $S^2$?

Answer 1

I also know that L and S commute, but I am unsure why. I've heard that it is simply because they act on difference variables, but I don't understand exactly what this means. Is there a way to show this explicitly?

Suppose we have two Hilbert spaces $H_1$ and $H_2$, an operator $A_1$ acting on $H_1$, and an operator $A_2$ acting on $H_2$. Let $H = H_1 \otimes H_2$. Then we can define $A_1$ and $A_2$ on $H$ by defining \begin{align} A_1(|a_1\rangle \otimes |a_2\rangle) &= A_1|a_1\rangle \otimes |a_2\rangle \\ A_2(|a_1\rangle \otimes |a_2\rangle) &= |a_1\rangle \otimes A_2|a_2\rangle \end{align} where $|a_1\rangle \in H_1, |a_2\rangle \in H_2$; and extending linearly to all of $H$. Then \begin{align} A_1 \circ A_2(|a_1\rangle \otimes |a_2\rangle) &= A_1(|a_1\rangle \otimes A_2|a_2\rangle) \\ &= A_1|a_1\rangle \otimes A_2|a_2\rangle \\ &= A_2(A_1|a_1\rangle \otimes |a_2\rangle) \\ &= A_2 \circ A_1(|a_1\rangle \otimes |a_2\rangle) \end{align} so the commutator vanishes on all pure tensors, and hence on all of $H$.

This is precisely the situation we have with the operators $L$ and $S$. In general, the wave function of a particle lives in a tensor product space. The spatial part of the wave function lives in one space, that of square-integrable functions on $\mathbb{R}^3$. The spin part, on the other hand, lives in a spinor space, i.e., some representation of $SU(2)$. $L$ acts on the spatial part, whereas $S$ acts on the spin part.

What are the remaining commutation relations between $J$, $J^2$, $L$, $L^2$, $S$, and $S^2$?

You should be able to work these out on your own, using the commutation and anti-commutation relations you already know, and properties of commutators and anti-commutators. For example,

$$[J_i, L_j] = [L_i + S_i, L_j] = [L_i, L_j] + [S_i, L_j] = i\hbar\epsilon_{ijk} L_k$$

Likewise:

\begin{align} [J_i^2, L_j] &= J_i[J_i, L_j] + [J_i, L_j]J_i \\ &= J_i(i\hbar \epsilon_{ijk}L_k) + (i\hbar\epsilon_{ijk}L_k)J_i \\ &= i\hbar\epsilon_{ijk}\{J_i, L_k\} \\ &= i\hbar\epsilon_{ijk}\{L_i + S_i, L_k\} \\ &= i\hbar\epsilon_{ijk}(\{L_i, L_k\} + \{S_i, L_k\}) \\ &= i\hbar\epsilon_{ijk}(2\delta_{ik} I + 2S_i L_k) \\ &= 2i\hbar\epsilon_{ijk} S_i L_k \end{align} where we have used the fact that $L_i$ and $S_j$ commute, the linearity of $[,]$ and $\{,\}$, and the identity

$$[AB, C] = A[B, C] + [A, C]B$$

Answer 2

Don't forget the bosonic creation/annihilation operators (harmonic oscillator operators)

$[a,a^{\dagger}] = 1$

and the fermion counterparts

$\{ c, c^{\dagger} \} \equiv cc^{\dagger}+c^{\dagger} c = 1$