I Particle in the box eigenfunctions

[DrDu]

Yes, your hamiltonian is not defined on momentum eigenstates. You basically discussed the evolution under the free particle hamiltonian, not the box hamiltonian. You could try to repeat your analysis expressing the delta function in terms of the eigenstates of the box hamiltonian.

 

[fog37]

I see. But isn't the Hamiltonian where the particle exists, inside the well, essentially equal to the free-particle Hamiltonian since $V(x)=0$ in that region? That seems to be what is done to find the energy eigenstates. Setting $V(x)=0$ is how the energy eigenvalue equation arises to find those energy eigenstates...

 

[PeterDonis]

where the particle exists, inside the well

Are you assuming an infinitely high potential outside the well? In other words, $V(x) = \infty$ outside $-\frac{L}{2} \le x \le \frac{L}{2}$?

 

[fog37]

Yes, that I have been thinking/assuming that infinite potential at the edges of the well.

 

[PeterDonis]

I have been thinking/assuming that infinite potential at the edges of the well.

Then that won't affect the energy eigenvalues, but it will affect the relationship between energy eigenstates and other eigenstates. For example, a momentum eigenstate won't just be $e^{ikx}$, since the wave function in any representation has to vanish outside the box, and that function doesn't. I don't think that affects the general arguments in this thread (for example, the general argument for why position eigenstates can't be stationary), but it does affect specific details like the form of the wave functions.

 

[fog37]

Well, if that is the case, even the accepted energy eigenfunctions in the form $$\sqrt \frac {2}{L} sin(\frac {\pi n x}{L})$$ are not mathematically acceptable since a sine is not zero outside the walls of the well. The function should be multiplied by another function that renders it zero (step function)...

 

[DrDu]

Either this or you consider from the very beginning only functions on 0<x<L.

 

[DrDu]

You could express a delta function as $\delta(x-a)=\sum_{n=1}^\infty \psi_n(a)\psi_n(x)$ where $\psi_n(x)=\sqrt{2/L} \sin(\pi n x/L)$. Then it is easy to write down its time dependence.

 

[houlahound]

I can read all the equations in this thread, the last post is unreadable on my screen?

 

[DrDu]

I can read all the equations in this thread, the last post is unreadable on my screen?

Sorry, I did correct the Latex tags.

 

[fog37]

Hello.
The Hamiltonian operator $\hat H= \frac {\hat p^2} {2m}+ \hat V(x)$ is specific to the problem and physical system under study. Its specificity is tied to the potential energy $V(x)$ which varies from problem to problem. The kinetic energy operator $\hat H= \frac {\hat p^2} {2m}$ remains the same for every type of problem.

As far as finding the eigenstates of different operators for a specific problem, I would say that the eigenvalue equation of a specific operator, to find its eigenstates, is the same for all different physical problems. What changes in each problem are the applied boundary conditions. Different problems will have different energy eigenstates or momentum eigenstates or angular momentum eigenstates not because the eigenvalue equations in each problem (they always have the form $\hat A |\Psi> = a |\Psi>$ where $\hat A$ is a any Hermitian operator) are different but because the boundary conditions that are imposed are different, correct?

Only the energy eigenvalue equation $\hat H |\Psi> = E |\Psi>$ seems to naturally arise directly from Schrodinger equation once we set $V(x)=0$ in the regions of space where that is applicable. The applications of BCs will determine which type of energy eigenfunctions will spur out from this eigenvalue equation.

The eigenvalue equations for other operators don't arise the same way and are just considered as the starting point for finding the respective eigenfunctions...