Section 2-9 Summary
Section 2-9 Summary
Figure 2-18 (a) A rectangular-hill potential of height U and width 2a. (b) Fraction of beam transmitted as a function of E/U , where U = h2/2π2ma2.
in the transmission function for E > U , with 100% transmission occurring at intervals in E instead of only in the infinite limit. These come about because of interference between waves reflecting off the front and back edges of the barrier. This is most easily understood by recognizing that 100% transmission corresponds to no reflection, so then B = 0. This occurs when the wave reflecting back from x = −a is of opposite phase to that reflecting back from x = +a, and this happens whenever there is an integral number of de Broglie half-wavelengths between x = −a and a. At energies where this happens, the beam behaves as though the potential barrier is not there.
The variation of reflection from thin films (e.g., soap bubbles) of light of different wavelengths results in the perception of colors and is a familiar example of scattering interference. Less familiar is the variation in reflection of a particle beam, outlined above. However, once we recognize the wave nature of matter, we must expect particles to manifest the same sort of wave properties we associate with light.
2-9 Summary
In this chapter we have discussed the following points:
1. A particle constrained in the classical sense (i.e., lacking the energy to overcome barriers preventing its motion over the entire coordinate range) will have quantized energy levels and a finite zero-point energy. In the mathematical analysis, this arises from requirements on ψ at boundaries.
Chapter 2 Quantum Mechanics of Some Simple Systems2. ψ can be nonsmooth, or cusped, where V is infinite at a point. If V is infinite over a finite range, ψ must be zero there.
3. Nondegenerate eigenfunctions of H must be symmetric or antisymmetric for any operation that leaves H unchanged.
4. |ψ|2 may be regarded as a statistical measure—a summary of many measurements of position on independent, but identically prepared, systems.
5. Quantum-mechanical predictions approach classical predictions in the limits of large E, or large mass, or very high quantum number values.
6. Integrals with antisymmetric integrands must vanish.
7. |ψ|2 does not vanish in regions where V > E if V is finite. This is called “barrier penetration.”
8. One-dimensional motion of a free particle has a continuum of energy levels. Except for E = 0, the states are doubly degenerate. Therefore, any mixture of such a pair of states is still an eigenfunction of H . But only two eigenfunctions (for a given E = 0) are also eigenfunctions for the momentum operator. These are the exponential functions. Since they correspond to different momenta, mixing them produces functions that are not eigenfunctions for the momentum operator.
9. Motion of a particle on a ring has quantum-mechanical solutions very similar to those for free-particle motion in one dimension. In both cases, there is no zero-point energy. Both are doubly degenerate for E > 0 because two directional possibilities are present. Both have a set of exponential solutions that are eigenfunctions for momentum. The main difference is that the particle-in-a-ring energies are quantized, due to head-to-tail “joining conditions” on ψ.
10. Increasing the dimensionality of a particle’s range of motion increases the number of quantum numbers needed to define the wavefunctions. In cases where the hamiltonian operator can be written as a sum of operators for different coordinates (i.e., is “separable”), the problem greatly simplifies; the wavefunctions become products, and the energies become sums.
11. Scattering problems are treated by selecting an energy of interest from the con tinuum of possibilities, removing functions that describe nonphysical processes such as backscatter from the trap, and matching wave values and slopes at region boundaries. Resulting wavefunctions show wave interference effects similar to those observed for light.
2-9.A Problems2-1. Ascertain that the expression (2-12) for energy has the proper dimensions.
2-2. Solve Eq. (2-9) for A.
√
2-3. There is a simple way to show that A in Eq. (2-9) must equal 2/L. It involves sketching ψ2, recognizing that sin2 x + cos2 x = 1, and asking what A must equal in order to make the area under ψ2 equal 1. Show this for n = 1, and argue why it must give the same result for all n.
2-4. Evaluate the probability for finding a particle in the middle third of a one dimensional box in states with n = 1, 2, 3, 104. Compare your answers with the sketches in Fig. 2-5 to see if they are reasonable.
2-5. a) Estimate the probability for finding a particle in the n = 1 state in the line element x centered at the midpoint of a one-dimensional box if x = 0.01L.
How does this compare to the classical probability?
b) Repeat the problem, but with x centered one third of the way from a box edge.
2-6. a) Use common sense to evaluate the following integral for the particle in a one-dimensional box, assuming that ψ is normalized.
L/5
ψ2
5 dx
0
b) How does this value compare to that for the integral over the same range, but using ψ1 instead of ψ5? (Larger, smaller, or equal?) Use a sketch to defend your answer.
2-7. Let S and A be respectively symmetric and antisymmetric functions for the operator R. Evaluate the following, where R operates on every function to its right: (a) RS (b) RA (c) RSS (d) RAA (e) RAS (f) RAASASSA (g) RAASASAA.
Can you think of a simple general rule for telling when a product of symmetric and antisymmetric functions will be antisymmetric?
2-8. Using the concept of odd and even functions, ascertain by inspection of sketches whether the following need be identically zero:
a) π sin θ cos θ dθ
0
b)
π
− sin θ cos θ dθ
π
c)
1
− x cos x dx
1
d) a − cos y sin2 y dy a
e) π sin3 θ cos2 θ dθ
0
f) π sin2 θ cos3 θ dθ
0
g)
1
1
−
x2y dx dy
1 −1
h)
π
− x sin x cos x dx
π
i) π sin x d sin2 x dx
0
dx
j)
π
− sin2 x d sin x dx
π
dx
2-9. Verify Eq. (2-23) for the general case n = m by explicit integration.
2-10. For the potential of Fig. 2-8, when E