# Evaluate Commutator with Partial Derivatives

I need to evaluate the following commutator...

$[x(\frac{\partial}{\partial y})-y(\frac{\partial}{\partial x}),y(\frac{\partial}{\partial z})-z(\frac{\partial}{\partial y})]$

i tried applying an arbitrary function $\psi(x)$ but I am unsure of how to evaluate the whole thing. My first step was:

$=(x(\frac{\partial}{\partial y})-y(\frac{\partial}{\partial x}))*(y(\frac{\partial}{\partial z})-z(\frac{\partial}{\partial y}))\psi(x)-(y(\frac{\partial}{\partial z})-z(\frac{\partial}{\partial y}))*(x(\frac{\partial}{\partial y})-y(\frac{\partial}{\partial x}))\psi(x)$

This is where i am lost, I am unsure of where to go from here. Maybe there is something else I can do as my first step to simplify the question? please let me know what I can do. Thanks.

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Have you worked with partial derivatives before? I would say just keep expanding out the expression you have. – DJBunk Mar 3 '13 at 16:41
Note your wavefunction should be of the form $\psi(x,y,z)$. – Emilio Pisanty Mar 4 '13 at 1:28

## 1 Answer

You can simplify your work using the identities

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

and

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

These are easy to prove by writing out the definition of a commutator.

This seems like it will give a lot of different terms, but most of them are zero. The $x,y,z$ all commute with each other, as do the derivatives. The only non-zero commutators are $[x,\frac{\partial}{\partial x}]$ and likewise for $y$ and $z$.

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Im an unclear of how to apply these two identities. Isn't my questions in a different form? something like... $[aA-bB,cC-dD]$ ??? – quantum savant Mar 3 '13 at 17:00
$A$, $B$, and $C$ in these identities represent arbitrary operators, so they can be $x$ or $\frac{\partial }{\partial x}$, for example. – Mark Eichenlaub Mar 3 '13 at 17:06