# Why does $[xp_{y},x]$ commute?

I'm looking at a solution in my book that says $[xp_{y},x]$ commutes.
Does bracket notation imply:

$[A,B]=AB-BA$

so that

$[xp_{y},x]=xp_{y}x-xxp_{y}$

Taking the comment from Max Graves and solving a slightly different commutation relation:

\begin{align} -[yp_{x},x]f &= yi\hbar\frac{\partial}{\partial x}(xf)-xyi\hbar \\ &=i\hbar y \bigg( (x\frac{\partial f}{\partial x} -\frac{\partial x}{\partial x}f)-x\frac{\partial f}{\partial x} \bigg) \notag\\ &=yi\hbar \bigg( x\frac{\partial f}{\partial x}+ f-x\frac{\partial f}{\partial x} \bigg) \notag\\ &=yi\hbar f\ \Rightarrow -[yp_{x},x] = yi\hbar \end{align} Does this look correct? Do the first and last terms cancel even though the order is not exactly the same?

• you often need to use a test function to check these commutation relations. Nov 7, 2013 at 19:00
• Yeah, that looks ok. See my edit, I cleaned it up a bit. But yeah once you end up deriving a bunch of these relationships it becomes easier to just not use a test function, but the easiest way to not make a mistake is to insert a simple test function. Nov 7, 2013 at 20:14

$$[xp_y,x]=x[p_y,x]+[x,x]p_y$$
Now $[p_y,x]$ vanishes because of the fundamental commutation relation between $p_i$ and $x_i$ which is $$[p_i, x_j]= -i\hbar \delta_{ij}$$
On the other hand $[x,x]=0$ because anything commmutes with itself.
• Thanks Frederico! So $[p_{y},x]$ vanishes because $i=y$ and $j=x$ and the Kronecker delta is $0$ when $i\neq j$? Nov 7, 2013 at 19:48
• Exactly. In this conventions $i,j=1,2,3$ and $p_1=p_x$, $p_y=p_y$, etc. Nov 7, 2013 at 19:50
• @curiousGeorge119 For interest's sake, the property that Frederico uses is that the Lie bracket is a derivation (something that fulfils Liebnitz's rule). Derivations always form a Lie algebra, and in the finite dimensional case the Lie algebra of derivations $Der(\mathfrak{g})\subset \mathfrak{g}$ of the Lie algebra $\mathfrak{g} = {\rm Lie}(\mathfrak{G})$ of a Lie group $\mathfrak{G}$ is the Lie subgroup of automorphisms ${\rm Aut}(\mathfrak{G})\subset\mathfrak{G}$, i.e. ${\rm Der}(\mathfrak{g}) = {\rm Lie}({\rm Aut}(\mathfrak{G})$; somewhat more obtusely: ... Nov 8, 2013 at 6:23
• @curiousGeorge119 ... ${\rm Der}({\rm Lie}(\mathfrak{G})) = {\rm Lie}({\rm Aut}(\mathfrak{G}))$, since I know you are interested in this stuff. See en.wikipedia.org/wiki/Derivation_(abstract_algebra) Nov 8, 2013 at 6:27