IB Physics A QM Question 16B

This question seems to have generated a certain amount of confusion. To explain how it is solved I shall start from what we should expect the wavefunctions to look like from physical ideas (which is how we are supposed to do 16B) and then move on to give graphs of numerical fits to a potential similar to that in the question. The latter approach makes sure, if nothing else, that I haven't made any mistakes in the former.

The Ground State

We know that the ground state must have even parity, and no nodes. Thus we can without loss of generality take psi to be positive everywhere. First identify the classically allowed and forbidden regions: the particle is allowed in the two wells up to the level E_0, and forbidden everywhere else. The Schroedinger equation tells us that the second derivative of psi with respect to x goes as -(E-V) times psi: in other words, the classically forbidden regions have this second derivative with the same sign as psi itself (i.e. positive) and the classically allowed regions have it with the opposite sign (i.e. negative). The evanescent waves of (roughly) decaying exponentials that we expect in the outer forbidden regions naturally have positive second derivative. Similarly in the central forbidden region the wavefunction will have to decay towards the centre, forming a minimum there, and having positive sevond derivative as required. In order to keep the first derivative continuous, and also be even, it will need to be flat in the middle. It isn't allowed to drop to zero, because this would indicate that there would still be lower energy states to find. In the classically allowed region we are left with maxima, approximately Gaussian as the wells are approximately quadratic. Points of inflexion of psi must occur where E=V, as the second derivative is zero there. This should suffice to sketch the function, but a numerical solution never hurts...

An aside - Modelling the Potential

I've approximated the potential sketched in the question with a quartic, the simplest function with three turning points: V(x)=(x^2-1)(x^2-9)+16 looks about right. We also need (in the arbitrary units implied) a value for h_bar^2/2m, and I've set it to 10, as this puts the ground state roughly where it is on the sketch. Finding the wavefunction means finding a solution to the Schroedinger equation. I used a spreadsheet (here in .gnumeric and .xls form) with a fairly low-quality approximation scheme to turn initial values for psi and its first derivative at zero into a complete graph for psi; the spreadsheet's `goal seek' facility was used to find an energy level that met the boundary condition that psi has to go to zero at large x. (Odd and even solutions are generated by setting psi or its first derivative at zero, respectively, to zero.)

The Ground State Illustrated

The ground state, with E=12.4, is shown. Happily it has all of the features described above.

The first excited state

The first excited state must be odd, but still have points of inflexion where the kinetic energy goes through zero. It turns out to have E=13.5, and looks like this. Note that the two central points of inflexion are not very strongly marked as the gradient is just a little shallower in the central region. Note also that the ground state and first excited state are rather close in energy. This is because the linear combinations of the two of them that represent having a particle mostly in one well or the other are fairly stable; classically the particle must be bound in one well or another, and quantum-mechanically it tunnels back and forth fairly infrequently, so the energy difference that governs the transition rate must be small.

The second excited state

The second excited state, with E=30.1, looks like this. Note that the particle isn't classically stuck in one well, so we have a maximum rather than a minimum in the centre.

The third excited state

The third excited state, with E=40.4, looks like this. Since the second and third excited states are not classically bound in one of the wells, we no longer find that they form pairs of solutions close in energy.

The fourth excited state

The fourth excited state, with E=56.0, looks like this. The features of the classical limit are already becoming evident: an approximately sinusoidal oscillation in the classically allowed region is coupled to evanescent waves outside, and the wavefunction oscillates most rapidly where it has most kinetic energy.

The fifth excited state

The fifth excited state, with E=73.4, looks like this. Another feature of the classical limit is becoming apparent: the wavefunction has greatest amplitude where it is moving most slowly classically - that is, on the lump in the centre, and as it runs towards the edges. This is as expected because, classically, it spends most time in these places.

The tenth excited state

The tenth excited state, with E=183.3, looks like this. The features of the classical limit are well established. There is only a small difference in amplitude and `frequency' of oscillation caused by the lump in the middle, as the particle slows down only slightly when it goes over. The larger aplitude approaching the turning points remains marked, though, as it will diverge classically.

The eighteenth excited state

The eighteenth excited state, with E=414.0, looks like this. It is beginning to look similar to the simple harmonic oscillator, though the increased amplitude in the middle is still just visible.