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Section 11.8: Asymmetric Infinite and Finite Wells




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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) = +V0 for 0 < x < +1 (Region II),     and     V(x) = +∞ for +1 < x.

What happens to the energy eigenfunction as we increase the step height V0?  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 < V0, we have that after applying the boundary conditions at −1 and 1,

ψI(x) = A sin(k[x + 1]) and  ψII(x) = C sinh(κ[x − 1]),

where k ≡ (2mE/ħ2)1/2 and κ ≡ [2m(V0 − E)/ħ2]1/2.  Matching the two energy eigenfunctions at x = 0 (ψI(0) = ψII(0) and ψ'I(0) = ψ'II(0) ) we find: κ tan(ka) = −k tanh(κb) which is the energy-eigenvalue equation for E < V0.

Now for the E > V0 case, and applying the boundary conditions at −1 and 1, we find that

ψI(x) = A sin(k[x + 1]) and  ψII(x) = C sinh(q[x − 1]),

where k ≡ (2mE/ħ2) and q ≡ (2m(E − V0)/ħ2). Matching the two energy eigenfunctions at x = 0, we find: q tan(ka) = −k tan(qb) which is the energy-eigenvalue equation for E > V0.

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?


6For 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, submitted to Eur. J. Phys..
7See 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).

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