Gerald Rav Mlller University of Maryland College Park, MD 20742
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The Particle in ihe One-DimmionaI Champagne Bottle
At the senior-beginning graduate student level, there are relatively few problems suitable for illustrating perturhation theory in the classroom or for assigning as homework. A very useful problem results if one introduces a half-sine wave bottom to the usual one-dimensional particle-in-the-box problem by defining V = A sin axla inside the box. The resulting potential energy diagram is concave-hottomed, resembling the bottom of a champagne bottle (Fig. 1). In order to enable comparisons to he made hetween the amplitude, A, of the perturhation and the zero-order and first-order energies, it is best to specify A. Below, A is defined as h2/32 ma2, which is one-fourth the zero-order ground state energy, though other choices would he satisfactory. Inside the box, h2 sm . rz -h2 d2 H=--+2m dx2 32ma2 o liX H ' = - h2 Ha=--h2 d2 2m Z$ 32ma2sin T +,O=@sin?,
-
Figure 1. The potential energy function
V= A
sin rxla, 0 < x < a: V =
-, x
=Oandx=a.
n = 1,2,3,.. .
n2h2 8mo The first-order correction t o the energy of the ground state, El', is simply E.0 =
El' =
A"
-2
+,0'
hZ
-sin
rx
dx
So'
r x dx 2 sin%32m020 a an integral readily found in the "Handbook of Chemistry and Physics." The first-order correction to the energy is comparable in magnitude to the perturhation (El' = 0.849A) and has the same sign. Students who have previously studied the traditional particle-in-the-box problem can visualize this perturbation, and the results of the calculation agree with their intuitive idea of what should happen if one pushes up the bottom of the box. The calculation of the first-order correction t o the ground state wavefunction is also instructive. The wavefnnction for the ground state, corrected through first-order, is
Figure 2. Comparison of mtt first three j
= odd zero-order
wavefunctions
chosen for our perturhation, one obtains as, = +(1/60d and aai = +(1/1260s). Our ground state wavefunction then becomes 3rx sm - + 0.005305sin a
The ajl integral is zero for all even j since sin2 ax/a is symmetric about the center of the box, x = a12, and the even-j sine functions are antisymmetric about the center of the box. Therefore, the first contribution is from j = 3 and the second is from j = 5. The resulting integrals for as1 and as, are easily evaluated using the trigonometric identities1 for sin 3x and sin 5x and the integral1 for sin" xdx. For the amplitude, A,
' "CRC Handbook of Chemistry and Physics," 47th Ed. Robert C.
Weast, Ed. The Chemical Rubber Co. 1966.
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in unrenormalized form. There are a number of pedagogically useful points to make with these values for ajl. First, of all, the j = even contributions are zero. If they were not, the resulting first-order wavefunction would he unsymmetrical with respect to the center of the box resulting in a greater probability of finding the particle on one side of x = a12 than on the other side. Secondly, the addition of the small amount of &"character (Fig. 2) to the ground-state wavefunction reduces the wavefunction near the center of the box, where the perturhation Volume 56, Number 11, November 1979 1 709
is greatest, and builds up the wavefunction toward the boundaries of the box as students expect should happen if the box has its bottom raised in the middle. Thirdly, the contribution of 6s" is much smaller than that of illustrating the reduced cbntribution to be expected from the higher excited states. The sign of asl is also positive which leads to a further small increase in II. near the boundaries of the box as one would expect. Perhaps unexpected is the slight increase a t x = a12 from the addition of 6s'' --.-~. - character. The contributions from $jQ,j = odd, alternate in sign and decrease in magnitude a t 1: = a12 with increasine -.i. I t is also worth pointing out that reversing the sign of the perturbation reverses the signs of the ajl's. The resulting box ~~~~
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710 1 Journal of ChemicalEducation
in convex-bottomed and the first-order correction to the ground state wavefunction increases the magnitude of J. (and therefore +2) near the center of the box and decreases it near the boundaries. While for many students this result is intuitively obvious, for others it is not and they learn through this example the relationship between the perturbation and response of the wavefunction for a simple system. The use of the ground-state wavefunction, *lo, for the functional form of the perturbation, H', might at first appear to be confusing to the students. However, I have found over the years of using this example that the use of a specific amplitude (such as our h2/32ma2)prevents possible confusion. The use of the same functional form, of course, led to the relatively painless integration.
