# 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?

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!

• +1 @MyUserIsThis 's conclusion was obviously correct, but I couldn't see what kind of mistake could have given a $2$. As an extra check for the OP, using $[AB,C] = A[B,C] + [A,C]B$ is apparently identity the teacher tried to use, which applied twice would give $[PXP,P] = P[XP,P] + [P,P]XP = PX[P,P] + P[X,P]P = i\hbar P^2$. – Stan Liou Apr 26 '13 at 22:25

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

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 $[XP^2-P^2X]P = 2ihP^2$