Quantum commutator I'm given this commutator:
$$\left[PXP,P\right]$$
Being $P\psi=-i\hbar\partial_x\psi$, and $X\psi=x\psi$
I've solved it in two ways, the first one is just aplying the commutator to some function $\psi$ and see what I get. My final result is:
$$\left[PXP,P\right]=-i\hbar^3\partial_{xx}$$
The second one is using some commutator properties:
$$\left[PXP,P\right]=-\left[P,PXP\right]=-(P\left[P,XP\right]+\left[P,P\right]XP)$$
$\left[P,P\right]=0$, so the second term goes away. I again expand the first term:
$$-P\left[P,XP\right]=-P(X[P,P]+[P,X]P)=-P[P,X]P=i\hbar P^2=\boxed{-i\hbar^3\partial_{xx}}$$
I again get the same result. When the teacher solved it in class, the final result was:
$$\left[PXP,P\right]=2i\hbar P^2$$
I have no idea where that $2$ comes from. Am I missing something? Am I doing something wrong?
 A: You teacher seems to have made a mistake.  I imagine that he/she did something like this:
\begin{align}
  [PXP, P]
&= P[XP,P]+[PX,P]P \\
&= P(X[P,P]+[X,P]P)+(P[X,P]+[P,P]X)P \\
&= P[X,P]P+P[X,P]P \\
&= 2i\hbar P^2
\end{align}
Notice that the first equality is wrong.  You can't peel operators off to the left and right if there are three operators in the first slot of the commutator!
A: I think you are right. 
Using really simple commutator math.
All you need is this:
$$
[AB,C] = A[B,C] + [A,C]B
$$
Then in your case:
$$
A=P$$
$$B=XP$$
$$C=P$$
$$
[PXP,P] = PX [P,P] + [P,P]XP = PX[P,P] + P[X,P]P + [P,P] XP
$$
As you said, [P,P] is antisymmetric to itself, and therefore we can remove all the [p,p] terms. We then have left only one term:
$$[PXP,P] = P[X,P]P$$ and as you know (we defined the operators so they do this) $$[X,P]= i \hbar$$
So we can write:
$$P(i \hbar) P$$. We can move out the scalar so:
$$[PXP,P] = i \hbar P^2$$
I can't find anything wrong with any of my steps, so I am pretty sure the 2 should not be there
A: Both answers are correct but you can do it without rules though they are basic ones
$[PXP,P] = PXP^2-P^2XP$
$[XP^2-P^2X]P$
$[x\frac{d^2}{dx^2}\psi-{\frac{d^2}{dx^2}(x\psi)}]P$
$[x\frac{d^2}{dx^2}\psi -[\frac{d}{dx}(\psi + \psi'x)]]P$
$[(-ih)^2[x\frac{d^2}{dx^2}\psi -[ (2\psi' + \psi''x  )]]]P$
$[(-ih)^2 (-2\psi')]P$
$[(-ih)(-ih)(-2\psi')]P$
$[2ih(-ih)(\psi')]P$
You must know 
$(-ih)(\psi')= P$
so $[XP^2-P^2X] = 2ihP$
So $[XP^2-P^2X]P = 2ihP^2$
