Application of Operator Algebras on Angular Momentum of a particle

Question

On page 13 of "Molecular Quantum Mechanics (Fifth edition)" by Peter Atkins and Ronald Friedman, they mentioned that

The quantity $AB-BA$ is called the commutator of $A$ and $B$ and is denoted $[A, B]$: $[A, B]=AB-BA$

On the same page 13 of the same book, they also mentioned

the linear momentum parallel to x is represented by differentiation with respect to $x$. Explicitly: Position representation: $x\rightarrow x\times$ $,$ $p_x\rightarrow \frac{\hbar}{i}\frac{\partial}{\partial{x}}$

They also explicitly defined the operators, note that $x$, $y$, $z$ are variables:

$x = x\times$, $y = y\times$, $z = z\times$, $p_x = \frac{\hbar}{i}\frac{\partial}{\partial{x}}$, $p_y = \frac{\hbar}{i}\frac{\partial}{\partial{y}}$, $p_z = \frac{\hbar}{i}\frac{\partial}{\partial{z}}$

On page 100 of the same book, it was stated that:

$[yp_z - zp_y, zp_x - xp_z]=[yp_z, zp_x]-[yp_z,xp_z]-[zp_y,zp_x]+[zp_y,xp_z]$

I do not understand how to obtain this relation.

Answer 1

This is what you need to do to prove that:

$[A+B,C]=(A+B)C-C(A+B)=AC+BC-CA-CB=AC-CA+BC-CB=[A,C]+[B,C]$

Answer 2

Let $A$ and $B$ be some operators. The exact definition sare largely unimportant. Important is that we can multiply, add and scale them:

• $AB$ and $BA$ exist and make sense.
• $A+B$, $A-B$ exist and make sense.
• $\alpha A$ for some complex or real number $\alpha$.

The commutator of $A$ and $B$ are defined $[A,B]=AB-BA$ which is well defined. First we need to realise that the commutator is linear in the first variable (using distributive law etc.):

$$[\alpha A+C,B]=(\alpha A + C)B-B(\alpha A + C) = \alpha AB + CB - \alpha BA- BC = \alpha(AB-BA)+(CB-BC)=\alpha[A,B]+[B,C]$$.

For the second variable of the commutator it is the same:

$$[A,\beta B+D]=\beta AB+AD-\beta BA-DA=\beta(AB-BA)+AD-DA=\beta[A,B]+[A,D]$$.

Thus for your case:

$$ \begin{align} [yp_z - zp_y, zp_x - xp_z]&=[yp_z,zp_x-xp_z]-[zp_y,zp_x-xp_z]\\ &=[yp_z,zp_x]-[yp_z,xp_z]-[zp_y,zp_x]+[zp_y,xp_z] \end{align} $$