What is the commutator of position and momentum squared in quantum mechanics?

[fluidistic]

Homework Statement

Calculate [tex][\hat X, \hat P^2][/tex].

Homework Equations

[tex][\hat A, \hat B] \Psi =[\hat A \hat B - \hat B \hat A ] \Psi[/tex].

The Attempt at a Solution

I am confused by [tex]P^2[/tex].
P is worth [tex]-i \hbar \frac{\partial}{\partial x}[/tex].
So I believe [tex]P^2= \hbar ^2 \left ( \frac{\partial}{\partial x} \right ) ^2[/tex].
If so, I get that [tex][\hat X, \hat P^2]= \hbar ^2 \left [ x \left ( \frac{\partial \Psi}{\partial x} \right ) ^2 -\Psi ^2 - 2x \Psi \frac {\partial \Psi}{\partial x} - x^2 \left ( \frac{\partial \Psi}{\partial x} \right )^2 \right ][/tex].
However if by [tex]\hat P ^2[/tex] they mean [tex]\hbar ^2 \frac{\partial ^2}{\partial x^2}[/tex], then I get [tex][\hat X, \hat P^2]=-2 \hbar ^2 \frac{\partial \Psi}{\partial x}[/tex].
I think my first approach was correct, but the answer I get seems way too complicated. I would like a feedback.

 

[Mindscrape]

It should the the latter.

[tex]
\hbar ^2 \frac{\partial ^2}{\partial x^2}
[/tex]

The Hamiltonian is often defined with the "P^2" operator.

While not all that physically meaningful, it's interesting to see what P^3 gives... (and so on)

 

[fluidistic]

Oh, thank you. I wasn't aware of that.
I'd like to know if I got a logical result, if it's not too much asked.
In all cases I'll redo the algebra tomorrow.

 

[Mindscrape]

Oh yeah, the result is correct, and I guess see my edit that just made if you are curious.

 

[dingo_d]

Or you could change to impulse representation where the momentum operator acts on [tex]\psi (k)[/tex] the same as the position operator acts on [tex]\psi (x)[/tex] in coordinate representation :) That way you need not mess with derivatives.

But in the end you get the same result ;) So I guess it's the matter of which way is easier to you...