Calc Commutator in Infinitely Potential Well: Is it Possible?

[vanhees71]

That's obviously not true. Take a particle on the $x$ axis in a finite potential. Then the generalized momentum eigenfuncrions are
$$u_p(x)=\exp(ipx).$$
This is also a generalized eigenfunction of $p^2$, but so is also $u_{-p}$ and thus also any linear combination,
$$a u_{p}(x)+b u_{-p}(x).$$
The latter is not an eigenfunction of $p$ but of $p^2$.

 

[DrDu]

@Post #14 by vanhees71: Hi Hendrik, see my comment here: https://www.physicsforums.com/showthread.php?t=741204 below your post in that thread.

Well, if you assume the infinite potential outside the box as some limit of a finite potential, the wavefunction will be continuous and hence 0 at the boundaries.

Interestingly there are self adjoint extensions of the operator -d^2/dx^2 which can't be obtained as limits of adding a potential and which lead to wavefunctions with are discontinuous but have continuous derivatives. This is called δ'- interaction.

Sometimes, I prefer to think of the "particle in an infinite box" as a particle moving between two delta functions of infinite strength, as this operator is also defined on the whole real axis as is the free hamiltonian.
This becomes interesting when there is also an additional finite potential, e.g. a finite potential well.
I think it is even possible to do some perturbation expansion in the inverse strength of the delta functions, e.g., to calculate resonance linewidths for a finite potential.

 

[DrDu]

There is a very good article on the topic which was also published in Amer J Phys:

http://arxiv.org/pdf/quant-ph/0103153.pdf?origin=publication_detail

 

[samalkhaiat]

That's obviously not true. Take a particle on the $x$ axis in a finite potential.

My reasoning was based on the premiss that the particle IS a FREE particle.

This is also a generalized eigenfunction of $p^2$, but so is also $u_{-p}$ and thus also any linear combination,
$$a u_{p}(x)+b u_{-p}(x).$$
The latter is not an eigenfunction of $p$ but of $p^2$.

This linear combination DOES NOT represent a FREE PARTICLE:
1)free particles do not just flip the sign of their momenta.
2) it can be made proportional to $\sin ( p x )$ which is not a free particle wavefunction. So to restrict the solutions of $P^{ 2 } \psi = p^{ 2 } \psi$ to a free particle, you should set $b = 0$ or $a = 0$.

The main issue regarding the particle in the box is the following: Inside the box, the particle is never a free particle.

 

[strangerep]

There is a very good article on the topic which was also published in Amer J Phys:

http://arxiv.org/pdf/quant-ph/0103153.pdf?origin=publication_detail

Thanks for the reminder about that paper (Bonneau, Faraut & Valent). I had only skimmed it in the past but this thread motivated me to study it more carefully.

It's interesting how other physical conditions, such as time-reversal invariance, parity invariance, or positivity of energy can play a role in selecting possible self-adjoint extensions, as this highlights how one must consider the entire dynamical problem when attempting quantization.

Most surprising is their remark at the end of sect 7.4: "This discussion shows that an infinite potential cannot be simply described by the limit of a finite one" (because the desired self-adjoint extension does not exist in that limit).

For other readers: if you're not sure what the terms "self-adjoint extension" and "deficiency index" mean in the context of this thread, then studying that paper will help.

 

[vanhees71]

My reasoning was based on the premiss that the particle IS a FREE particle.



This linear combination DOES NOT represent a FREE PARTICLE:
1)free particles do not just flip the sign of their momenta.
2) it can be made proportional to $\sin ( p x )$ which is not a free particle wavefunction. So to restrict the solutions of $P^{ 2 } \psi = p^{ 2 } \psi$ to a free particle, you should set $b = 0$ or $a = 0$.

The main issue regarding the particle in the box is the following: Inside the box, the particle is never a free particle.

I do not understand what you mean. Also for a free particle (a special case for a particle in a finite potential, namely $V=0$ by the way) the two generalized momentum eigenfunctions are $u_{p}$ and $u_{-p}$ both are eigenfunctions of $\hat{p}^2=-\partial_x^2$ and so is any linear combination. Any non-trivial linear combination is, however, obviously not a generalized eigenfunction of $\hat{p}$, which disproves your claim that any eigenfunction of $\hat{p}^2$ must be also one of $\hat{p}$. The other way is right, i.e., any eigenfunction of $\hat{p}$ is also an eigenfunction of any operator that is a function of $\hat{p}$.