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Chapter 5: Q36E (page 188)

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height $U_{0}$ to the kinetic energy and the kinetic energy (through $λ$ ) to n and L. Show that the number of bound states is given roughly by $\sqrt{8 m l^{2} U_{0} / h^{2}}$ (Assume that the number is large.)

Short Answer

Expert verified

Hence, the number of bound states is given by $n_{\max} \approx \sqrt{\frac{8 U_{0} m L^{2}}{h^{2}}}$ is proved

Step by step solution

01

Assumption

We assume that the wave function and energies correspond to the infinite well.

If, $E < U_{0}$ then inside the well, the energy is totally kinetic and is given by,

$\frac{h^{2} k^{2}}{2 m} = \frac{n^{2} π^{2} h^{2}}{2 m L^{2}}$

02

Calculation

The highest bound state exists when $E = U_{0}$ . i.e., the difference $E - U_{0} \to 0$ . For this state,

$\begin{matrix} \frac{n_{\max}^{2} π^{2} h^{2}}{2 m L^{2}} \approx U_{0} \\ n_{\max} \approx \sqrt{\frac{8 U_{0} m L^{2}}{h^{2}}} \end{matrix}$

Hence the number of bounded states $n_{\max} \approx \sqrt{\frac{8 U_{0} m L^{2}}{h^{2}}}$ is proved.

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Most popular questions from this chapter

Verify that solution (5-19) satisfies the Schrodinger equation in form (5.18).

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant $C$ . First, use the continuity/smoothness conditions to eliminate $A$ , $B$ , and $G$ in favor of $C$ in (21). Then make the change of variables $z = x - L / 2$ and use the trigonometric relations

$\sin (a + b) = sinacosb + cosasinb$ and

$\cos (a + b) = cosacosb - sinasinb$ on the

functions in region I, $- L / 2 < z < L / 2$ . The change of variables shifts the problem so that it is symmetric about $z = 0$ , which requires that the probability density be symmetric and thus that $ψ (z)$ be either an odd or even function of $z$ . By comparing the region II and region III functions, argue that this in turn demands that $(α / k) sinkL + coskL$ must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as $\frac{α}{k} = \tan \frac{kL}{2}$ and $\frac{α}{k} = - \cot \frac{kL}{2}$ . Finally, plug these separately back into the region I solutions and show that

$ψ (z) = C \times {\begin{cases} \begin{matrix} e^{α (z + L / 2)} z < L / 2 \\ \frac{coskz}{\cos \frac{kL}{2}} - L / 2 < z < L / 2 \\ e^{- α (z - L / 2)} z > L / 2 \end{matrix} \end{cases}$

or

$ψ (z) = C \times {\begin{matrix} e^{α (z + L / 2)} z < L / 2 \\ - \frac{sinkz}{\sin \frac{kL}{2}} - L / 2 < z < L / 2 \\ e^{- α (z - L / 2)} z > L / 2 \end{matrix}$

Note that $C$ is now a standard multiplicative normalization constant. Setting the integral of ${| ψ (z) |}^{2}$ over all space to 1 would give it in terms of $k$ and $α$ , but because we can’t solve (22) exactly for k(or E), neither can we obtain an exact value for $C$ .

The harmonic oscillator potential energy is proportional to $x^{2}$ , and the energy levels are equally spaced:

$E_{n} \propto (n + \frac{1}{2})$ . The energy levels in the infinite well become farther apart as energy increases: $E_{n} \propto n^{2}$ .Because the function $\lim_{b \to \infty} {| x / L |}^{b}$ is 0 for $| x | < L$ and infinitely large for $| x | > L$ . the infinite well potential energy may be thought of as proportional to ${| x |}^{\infty}$ .

How would you expect energy levels to be spaced in a potential well that is (a) proportional to $| x |^{1}$ and (b) proportional to $- | x |^{- 1}$ ? For the harmonic oscillator and infinite well. the number of bound-state energies is infinite, and arbitrarily large bound-state energies are possible. Are these characteristics shared (c) by the ${| x |}^{1}$ well and (d) by the $- {| x |}^{- 1}$ well? V

A classical particle confined to the positive x-axis experiences a force whose potential energy is-

$U (x) = \frac{1}{x^{2}} - \frac{2}{x} + 1$

a) By finding its minimum value and determining its behaviors at $x = 0$ and role="math" localid="1660119698069" $x = \infty$ , sketch this potential energy.

b) Suppose the particle has energy of $0.5 J$ . Find any turning points. Would the particle be bound?

c) Suppose the particle has the energy of $2.0 J$ . Find any turning points. Would the particle be bound?

Figure 5.15 shows that the allowed wave functions for a finite well whose depth $U_{0}$ was chosen to be $6 π ℏ^{2} / m L^{2}$ .

(a) Insert this value in equation (5-23), then using a calculator or computer, solve for the allowed value of $k L$ , of which there are four.

(b) Using $k = \frac{\sqrt{2 m E}}{ℏ}$ find corresponding values of E. Do they appear to agree with figure 5.15?

(c) Show that the chosen $U_{0}$ implies that $α ≃ \sqrt{\frac{12 π^{2}}{L^{2}} - k^{2}}$ .

(d) Defining $L$ and $C$ to be 1 for convenience, plug your $K L$ and $α$ values into the wave function given in exercise 46, then plot the results. ( Note: Your first and third $K L$ values should correspond to even function of z, thus using the form with $C O S K Z$ , while the second and forth correspond to odd functions. Do the plots also agree with Figure 5.15?

See all solutions

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