/sci/ - Science & Math

No replies... Is this one of those questions with a not so obvious obviously obvious answer that doesn't even need to stated?

>>6380204

Yes, it's the classical particle in a box.

Well, since Hpsi * Vpsi = Epsi, if V is infinite psi is zero.

>>6380213
Since if it wasn't 0, the derivatives in schrödinger's would be infnite, so Ψ would then be infinite which makes it either not continuous or even normalizable? Okay then. Ty.

>>6380166

consider the time-indipendent shroedinger equation for psi, which is the equation for energy eigenstates.

Assume V outside isn't infinite, but a large finite value which you'll send to infty later.

Explicitly solve schrod eq outside the box, ignore boundary conditions at the box, but don't forget those at infty. Check that all solutions have pointwise limit 0 for V going to infty. The actual solutions extended to the box will be a subset of these, and so the property carries.

So all energy eigenfunctions are 0 outside the box in the limit.

Now all wavefunctions must be a linear combination (sum/integral) of these eigenfunctions, and the property carries onto them (check this using linearity of integral, shouldn't be too hard)

And step two is solved.

Step one is easy. Assume psi normalized and with support in the box. You just need to prove that the integral of x*|psi|^2 is a value in the box. There's many ways you could go about this. The smartest I guess would be trivially decomposing psi in an integral of position eigenfunctionals.

>>6380224
I've just started QM so I'm not too sure I understand, though I think I do-ish.
Make V=c outside box. Decompose Ψ, in for example standing waves. Solve for standing waves. As c->+inf, (std. waves)->0. So the linear combination of the standing waves, which is Ψ, also goes to 0.
Is that it-ish?

>>6380255
Classically setting V=c doesn't do anything. Now I'm not so sure... Maybe using a linear growing V instead.
Well, there's matsci answer but yours seems more precise.

Got an "A" in this class last year, but already forgot this shit. lel. gl, OP.

>>6380255
yeah kind-ish

this was as formal as I could make it without puking. But depending on the anality of your course, you might wanna check something more intuitive.

Out of the top of my head: take an energy eigenstate. If it was ("significantly") nonzero outside the box then you'd have a nonzero probability of measuring x outside the box. Since V is a function of x, you have a nonzero prob of measuring infty for V, so V has expectation value infty. However, the e.v. for H is always greater than or equal to that of V, you should have proven this in class somewhere. If you have, then you easily have your contradiction, as the e.v. for H is just the finite energy eigenvalue for that state.

Idk I guess with some clever thinking you could reduce the proof to something really easy but right now I can't do better