Sudden Perturbation of Potential Well

[unscientific]

Homework Statement

Part (a): Particle originally sits in well V(x) = 0 for 0 < x < a, V = ∞ elsewhere. The well suddenly doubles in length to 2a. What's the probability of the particle staying in its ground state?

Part (b): What is the duration of time that the change occur, for the particle to most likely remain in ground state?

Homework Equations

The Attempt at a Solution

Part (a)

For a well of length a, eigenfunction is:

$$u = \sqrt{\frac{2}{a}} sin \left(\frac{\pi x}{a}\right)$$

For a well of length 2a, eigenfunction is:

$$v = \frac{1}{\sqrt a} sin \left(\frac{\pi x}{2a}\right)$$

For probability amplitude, overlap these 2:

$$\frac{1}{2} \frac{\sqrt 2}{a} \int_0^{2a} 2 sin (\frac{\pi x}{a}) sin (\frac{\pi x}{2a}) dx$$

$$\frac{1}{2} \frac{\sqrt 2}{a} \int_0^{2a} cos (\frac{3\pi x}{2a}) - cos (\frac{\pi x}{2a}) dx$$

which goes to zero.

Part (b)

Am I supposed to use time evolution: $|\psi_t> = U^{-i\frac{E}{\hbar}t}$ ? Doesn't seem to help..

 

[king vitamin]

$$\frac{1}{2} \frac{\sqrt 2}{a} \int_0^{2a} 2 sin (\frac{\pi x}{a}) sin (\frac{\pi x}{2a}) dx$$

$$\frac{1}{2} \frac{\sqrt 2}{a} \int_0^{2a} cos (\frac{3\pi x}{2a}) - cos (\frac{\pi x}{2a}) dx$$

What is the initial wavefunction in the region $a<x<2a$?

 

[unscientific]

What is the initial wavefunction in the region $a<x<2a$?

It is 0 before the change in the potential, since it is a forbidden region.

 

[king vitamin]

It is 0 before the change in the potential, since it is a forbidden region.

Right. Do you see how that affects your overlap integral? It might help to explicitly sketch your before and after wavefunctions.

 

[unscientific]

Right. Do you see how that affects your overlap integral? It might help to explicitly sketch your before and after wavefunctions.

So for a < x < 2a: the overlap is zero. While for 0 < x < a, the overlap is usual - so the limits are from 0 to a, and not from 0 to 2a.

$$\frac{1}{2} \frac{\sqrt 2}{a} \int_0^{a} cos (\frac{3\pi x}{2a}) - cos (\frac{\pi x}{2a}) dx$$

$$= - \frac{8}{3\pi \sqrt 2}$$

So the probability = $\frac{32}{9\pi^2}$, which is less than 1, reassuringly.But for part (b), I haven't got a clue.

 

[unscientific]

Do I need to use adiabatic approximation for part (b)?