# Commutator and kinetic energy

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1. Jul 23, 2015

### fricke

For particle in the box wave function, it is the eigenfunction of kinetic energy operator but not the eigenfunction of momentum operator. So, do these two operators commute? (or it has nothing to do with commutator stuff?)

How about for free particle? For free particle, the wave function is eigenfunction of both kinetic energy operator and momentum operator. So, does it mean these two operators do not commute?

2. Jul 23, 2015

### Dr. Courtney

3. Jul 23, 2015

### vanhees71

There's no momentum operator for a particle in a infinite square well. So it doesn't even make sense to ask the question whether momentum operator and Hamiltonian commute or not :-).

For the free particle, the momentum operators all commute with the Hamiltonian since the momentum of a free particle is conserved.

4. Jul 23, 2015

### cpsinkule

There is. You couldn't define the hamiltonian without a momentum operator and you wouldn't have an uncertainty from [X,P]=ih1.

5. Jul 23, 2015

### strangerep

Careful! This is a harder issue than it seems. The difficulty is defining a momentum operator for the inf-sq-well case in such a way that it is well-defined on the entire Hilbert space of that problem. This has been discussed at length in older threads. Hmm, now I'll have to go hunting to find one of them...

6. Jul 23, 2015

### Staff: Mentor

Indeed - its this Rigged Hilbert Space stuff which requires care.

The following examines it in the case of the square well:
http://arxiv.org/pdf/quant-ph/0110165v1.pdf

As can be seen its not exactly trivial.

Thanks
Bill

7. Jul 24, 2015

### rubi

Unbounded operators are never defined on the entire Hilbert space. It's true that the momentum operator on $L^2([a,b])$ works differently, though. One starts with a dense domain such that $p=-\mathrm i\partial_x$ is symmetric. Such a domain can be found easily by restricting to continuously differentiable functions that vanish at the boundary. One can then study self-adjoint extensions using the von Neumann deficiency indices method. One finds that there are infinitely many self-adjoint extensions $p_U$, labeled by $U(1)$ matrices (the deficiency spaces are 1-dimensional). They correspond to all possible choices of boundary conditions that keep the operator symmetric. Moreover, $[x,p_U]=\mathrm i$, so each $p_U$ consitutes a possible momentum operator. Depending on the physical situation, one of these $p_U$ will be appropriate. Often, we choose Dirichlet boundary conditions or periodic boundary conditions for example.

However, there is a physically relevant example, where no self-adjoint extensions exist: $L^2([0,\infty))$. So there is no momentum operator conjugate to the radial coordinate $r$.

Last edited: Jul 24, 2015