Chapter 5: Bound States: Simple Cases

For the harmonic oscillator potential energy, $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$, the ground-state wave function is $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{(}\sqrt{\mathbf{m}\mathbf{k}}\mathbf{/}\mathbf{2}\mathbf{\hslash }\mathbf{)}{\mathbf{x}}^{\mathbf{2}}}$, and its energy is $\frac{\mathbf{1}}{\mathbf{2}}\mathit{\hslash }\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}$.

(a) Find the classical turning points for a particle with this energy.

(b) The Schrödinger equation says that $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and its second derivative should be of the opposite sign when E > Uand of the same sign when E < U . These two regions are divided by the classical turning points. Verify the relationship between $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and its second derivative for the ground-state oscillator wave function.

(Hint:Look for the inflection points.)

Explain to your friend, who is skeptical about energy quantization, the simple evidence provided by distinct colors you see when you hold a CD (serving as grating) near a fluorescent light. It may be helpful to contrast this evidence with the spectrum produced by an incandescent light, which relies on heating to produce a rather nonspecific blackbody spectrum.

A 2kg block oscillates with an amplitude of 10cm on a spring of force constant 120 N/m .

(a) In which quantum state is the block?

(b) The block has a slight electric charge and drops to a lower energy level by generating a photon. What is the minimum energy decrease possible, and what would be the corresponding fractional change in energy?

Air is mostly N₂, diatomic nitrogen, with an effective spring constant of 2.3 x 10³N/m, and an effective oscillating mass of half the atomic mass. For roughly what temperatures should vibration contribute to its heat capacity?

To a good approximation. the hydrogen chloride molecule, HCI, behaves vibrationally as a quantum harmonic ascillator of spring constant 480N/mand with effective osciltating mass just that of the lighter atom, hydrogen If it were in its ground vibtational state, what wave. Iength photon would be just right to bump this molecule. up to its next-higher vibrational energy state?.

In Section 5.5, it was shown that the infinite well energies follow simply from$\mathit{\lambda }\mathbf{=}\raisebox{1ex}{$\mathbf{h}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{p}$}\right.$ the formula for kinetic energy, p²/2m; and a famous standing-wave condition, $\lambda =2L/N$. The arguments are perfectly valid when the potential energy is 0(inside the well) and L is strictly constant, but they can also be useful in other cases. The length L allowed the wave should be roughly the distance between the classical turning points, where there is no kinetic energy left. Apply these arguments to the oscillator potential energy, $\mathit{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$.Find the location x of the classical turning point in terms of E; use twice this distance for L; then insert this into the infinite well energy formula, so that appears on both sides. Thus far, the procedure really only deals with kinetic energy. Assume, as is true for a classical oscillator, that there is as much potential energy, on average, as kinetic energy. What do you obtain for the quantized energies?

The potential energy shared by two atoms in a diatomic molecule, depicted in Figure 17, is often approximated by the fairly simple function $\mathit{U}\left(x\right)\mathbf{=}\left(\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$x^{12}$}\right.\right)\mathbf{-}\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$x^6$}\right.\right)$where constants a and b depend on the atoms involved. In Section 7, it is said that near its minimum value, it can be approximated by an even simpler function—it should “look like” a parabola. (a) In terms ofa and b, find the minimum potential energy U (x₀) and the separation x₀ at which it occurs. (b) The parabolic approximation ${\mathit{U}}_{\mathbf{P}}\left(x\right)\mathbf{=}\mathit{U}\left(x_o\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{\kappa }{\left(x-x_o\right)}^{\mathbf{2}}$has the same minimum value at x₀ and the same first derivative there (i.e., 0). Its second derivative is k , the spring constant of this Hooke’s law potential energy. In terms of a and b, what is the spring constant of U (x)?

To determine the classical expectation value of the position of a particle in a box is $\frac{\mathbf{L}}{\mathbf{2}}$ , the expectation value of the square of the position of a particle in a box isrole="math" localid="1658324625272" $\frac{{\mathbf{L}}^{\mathbf{2}}}{\mathbf{3}}$ , and the uncertainty in the position of a particle in a box is$\frac{\mathbf{L}}{\sqrt{\mathbf{12}}}$ .

Show that the uncertainty in a particle’s position in an infinite well in the general case of arbitrary $\mathit{n}$is given by

$\mathbf{L}\sqrt{\frac{\mathbf{1}}{\mathbf{12}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{n}}^{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}}}$

Discuss the dependence. In what circumstance does it agree with the classical uncertainty of discussed in Exercise 55?

The uncertainty in a particle's momentum in an infinite well in the general case of arbitrary $\mathit{n}$is given by$\frac{\mathbf{n}\mathbf{\pi }\mathbf{h}}{\mathbf{L}}$ .