*Please wait for the animation to completely load.*

This animation shows a finite potential energy well in which
a constant potential energy function has been added over the right-hand side of
the well. As you drag the slider to the right, the size of this *bump*
or step gets larger. To see the other bound states simply click-drag in
the energy level diagram on the left to select a level. The selected level will
turn red. Consider Region I to be from *x* = −1 to *x* = 0 and
Region II to be from *x* = 0 to *x* = 1 such that

*V*(*x*) = +∞ for *x* < −1, * V*(*x*)
= 0 for 1< *x* < 0 (Region I), * V*(*x*) = +*V*_{0} for 0 <
*x* < +1 (Region II), and * V*(*x*) = +∞ for
+1 < *x*.

What happens to the energy eigenfunction as we increase the step
height *V*_{0}? We begin to notice that the energy eigenfunction, once
having the same amplitude and curviness over both sides of the well, begins to
lose this symmetry. Given the larger potential energy function in Region
II, the energy eigenfunction there has less
curviness. In
addition, the amplitude of the energy eigenfunction should increase in Region II because
it has a higher probability of being found there. (By simple time spent
arguments: a classical particle would spend more time in Region II due to its
reduced speed there.) In addition, since the added potential energy
function is a constant over the entire region, the change in energy eigenfunction curviness and amplitude must be uniform over Region II.

For this asymmetric infinite square well, mathematically we find that for E < *V*_{0}, we have
that after applying the boundary conditions at −1 and 1,

ψ_{I}(*x*) = A sin(*k*[*x* + 1])
and ψ_{II}(x) = C sinh(κ[*x* − 1]),

where *k*^{ }≡ (2*mE*/*ħ*^{2})^{1/2} and κ ≡
[2*m*(*V*_{0 }− E)/*ħ*^{2}]^{1/2}. Matching the two energy eigenfunctions at *x* = 0 (ψ_{I}(0) = ψ_{II}(0) and ψ'_{I}(0) = ψ'_{II}(0)
) we find: κ tan(*ka*) = −*k* tanh(κ*b*) which is the
energy-eigenvalue equation for E < *V*_{0}.

Now for the *E* > *V*_{0} case, and applying the boundary conditions at
−1 and 1, we find that

ψ_{I}(*x*) = A sin(*k*[*x* + 1])
and ψ_{II}(*x*) = C sinh(*q*[*x* − 1]),

where *k* ≡ (2mE/*ħ*^{2}) and
*q* ≡ (2m(E − *V*_{0})/*ħ*^{2}). Matching the two energy eigenfunctions at *x* = 0, we find:
*q* tan(*ka*) = −*k* tan(*qb*) which is the
energy-eigenvalue equation for E > *V*_{0}.

Note that for certain slider values and certain eigenstates, you may notice the
same amplitude in Region I and Region II, despite the potential
energy difference. This is due to the fact that the energy eigenfunctions happen to
match at a node.^{6}

In Animation 2 we have a
finite asymmetric square well.^{7} The main difference between the infinite and finite well is that there are now
exponential tails in the classically forbidden regions *x* < −1 and *x* > 1.

Animation 3 shows a well that is asymmetric in yet another way. In this case it is the sides of the well that are at different potential energies. Change the slider to see the effect of changing the height of the right side of this finite well. Does it behave in the way you might have expected?

___________________

^{6}For
more mathematical details see: M. Doncheski
and R. Robinett, "Comparing Classical and Quantum Probability Distributions for an Asymmetric Infinite Well," *Eur. J. Phys.* **21**, 217-227 (2000) and
and "More on the Asymmetric
Infinite Square Well: Energy Eigenstates with Zero Curvature," L.P. Gilbert, M.
Belloni, M. A. Doncheski, and R. W. Robinett, to appear in *Eur. J. Phys. 2005*.^{
7}See for example, A. Bonvalet, J. Nagle, V. Berger, A. Migus, J.-L.
Martin, and M. Joffre, "Femtosecond Infrared Emission Resulting from Coherent
Charge Oscillations in Quantum Wells," *Phys. Rev. Lett.* **76**,
4392-4395 (1996).