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 differentiable in the usual way (at least not for the infinite well, for the finite well you might use Dirac deltas).

Edit: typo

 

[hokhani]

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. what is $[V(x),P]\psi(x)$ when you expand the commutator?

Very nice separation of the Hamiltonian. If we take $\psi(x)=\sum_m a_m \psi_m$ where $\psi_m$ are the eigenfunction of the Hamiltonian, then it seems that we have $[V(x),P]\psi(x)=0$ since inside the box ($|x|<L$) the potential is zero $V(x)=0$ and for $|x| \geq L$ we have $\psi_m=0$!

 

[hokhani]

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...

For infinite walls I did that in the post #8.

 

[Nugatory]

If we take $\psi(x)=\sum_m a_m \psi_m$ where $\psi_m$ are the eigenfunction of the Hamiltonian, then it seems that we have $[V(x),P]\psi(x)=0$ since inside the box ($|x|<L$) the potential is zero $V(x)=0$ and for $|x| \geq L$ we have $\psi_m=0$!

Take $\psi(x)$ to be an arbitrary function and expand the commutator. You will end up with a term proportional to $\frac{dV}{dx}$, which works just fine when $V$ is continuous and is in general non-zero (but is zero where $V$ is constant, which creates the "at first glance" illusion in your original post).

But $V$ is discontinuous at the walls so that derivative is not defined there.

 

[JimWhoKnew]

which works just fine when $V$ is continuous

The usual textbook demonstration for $~[V(X),P]=iV'(X)~$ assumes that $~V(x),~V'(x)~$ can be expanded as power series. Can you generalize it to other differentiable (let alone continuous) functions?

 

[JimWhoKnew]

Very nice separation of the Hamiltonian. If we take $\psi(x)=\sum_m a_m \psi_m$ where $\psi_m$ are the eigenfunction of the Hamiltonian, then it seems that we have $[V(x),P]\psi(x)=0$ since inside the box ($|x|<L$) the potential is zero $V(x)=0$ and for $|x| \geq L$ we have $\psi_m=0$!

So you assume $~\infty\cdot0=0~$ ?

 

[PeterDonis]

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).

What does "derivable" mean? If you mean sometimes you can't derive its form from first principles, so what? That's true of lots of physics problems; you just have to guess the form of a function.

If you mean its derivative isn't always well-defined, yes, that's true--I addressed that case in post #6.

 

[pines-demon]

What does "derivable" mean? If you mean sometimes you can't derive its form from first principles, so what? That's true of lots of physics problems; you just have to guess the form of a function.

If you mean its derivative isn't always well-defined, yes, that's true--I addressed that case in post #6.

My bad, I meant differentiable.