I Commutation of operators for particle in a box

[hokhani]

TL;DR Summary

How to calculate $[H, P}$ for a particle in a box?

I would like to know how to calculate the $[\hat{H}, \hat{P}]$ for a particle in a 1D box? At the first glance it seems that they commute but they don't get diagonalized in identical basis. How to calculate this commutation?

 

[Nugatory]

At the first glance it seems that they commute

Does it? We have $[H,P]=[P^2/2m,P]+[V(x),P]$; the first term is trivially zero but there’s no reason to expect the second one to be.

You can calculate the in general non-zero second term by applying it to an arbitrary test function: what is $[V(x),P]\psi(x)$ when you expand the commutator?

 

[dextercioby]

Finite walls or infinite walls? Start down by writing the Hilbert space, then the two operators. You may discover that the commutator is not even defined...

 

[pines-demon]

Well usually $[f(x),p]=i\hbar \partial f/\partial x$, for a function $f(x)$, but the potential here is kind of troublesome.

 

[PeterDonis]

Well usually $[f(x),p]=i\hbar \partial f/\partial x$, for a function $f(x)$, but the potential here is kind of troublesome.

I don't see why the potential is "troublesome"; it's a function of $x$ whose derivative won't in general be zero, so it won't commute with $p$.

 

[PeterDonis]

You may discover that the commutator is not even defined...

In which case it's certainly not zero. So either way the OP's expectation of being able to find a common diagonalization for $H$ and $P$ is false.

 

[pines-demon]

I don't see why the potential is "troublesome"; it's a function of $x$ whose derivative won't in general be zero, so it won't commute with $p$.

The potential is not a function that is derivable in the usual way (at least not for the infinite well, for the finite well you might use Dirac deltas).