Question

Consider the quantum mechanical analog to the classical problem of a ball (mass m) bouncing elastically on the floor.
(a) What is the potential energy, as a function of height x above the floor? (For negative x, the potential is infinite x - the ball can't get there at all.)
(b) Solve the Schrödinger equation for this potential, expressing your answer in terms of the appropriate Airy function (note that Bi(z) blows up for large z, and must therefore be rejected). Don’t bother to normalize $\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
(c) Using $\mathbf{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{80}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}$and $\mathbf{}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{kg}$ , find the first four allowed energies, in joules, correct to three significant digits. Hint: See Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions, Dover, New York (1970), page 478; the notation is defined on page 450.
(d) What is the ground state energy, in ,eV of an electron in this gravitational field? How high off the ground is this electron, on the average? Hint: Use the virial theorem to determine $\mathbf{<}\mathbf{x}\mathbf{>}$ .

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{V}\left(\mathrm{x}\right)=\mathrm{mgx}. \\ \left(\mathrm{b}\right)\mathrm{\psi }\left(\mathrm{x}\right)={\mathrm{c}}_1\mathrm{Ai}\left[\mathrm{\alpha }\right(\mathrm{x}-\mathrm{E}/\mathrm{mg}\left)\right] \\ \left(\mathrm{c}\right){\mathrm{E}}_1=8.805\times {10}^{-23}\mathrm{J}; \\ {\mathrm{E}}_2=1.540\times {10}^{-22}\mathrm{J}; \\ {\mathrm{E}}_3=2.079\times {10}^{-22}\mathrm{J}; \\ {\mathrm{E}}_4=2.556\times {10}^{-22}\mathrm{J} \\ \left(\mathrm{d}\right){\mathrm{E}}_{\mathrm{gs}}\approx 1.149\times {10}^{-13}\mathrm{eV};⟨\mathrm{x}⟩\approx 1.4\mathrm{mm}. \end{gathered}$$

Step-by-step solution

(a) To find potential Energy.

For a free fall under the influence of gravity, the potential energy is

$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{mgx}$,

Wheregis the gravitational acceleration andxis the height above the floor.

Hence,

$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\mathrm{mgx},\\ \infty ,\end{array}\begin{array}{c}\mathbf{x}\mathbf{>}\mathbf{0}\\ \mathbf{x}\mathbf{<}\mathbf{0}\end{array}$

(b) To solve Schrodinger equation by using this potential equation.

Using the potential in part (a) (for $x>0$), the Schrodinger equation reads

$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+mgx\psi =E\psi$

(or)

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{dx}}^2}=\frac{2\mathrm{m}(\mathrm{mgx}-\mathrm{E})}{{\hslash }^2}\mathrm{\psi } \\ \frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{dx}}^2}=\frac{2{\mathrm{m}}^2\mathrm{g}}{{\hslash }^2}\left(\mathrm{x}-\frac{\mathrm{E}}{\mathrm{mg}}\right)\mathrm{\psi } \end{gathered}$$

If we let $\mathrm{u}\equiv \mathrm{x}-\mathrm{E}/\mathrm{mg}$ and$\mathrm{\alpha }\equiv {(2{\mathrm{m}}^2\mathrm{g}/{\hslash }^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$,

We can write the equation as

$\frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{du}}^2}={\mathrm{\alpha }}^3\mathrm{u\psi }$

Thecan be absorbed into the independent variable by defining$z\equiv \alpha u$ , so that

$\frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{dz}}^2}=\mathrm{z\psi }$

This is Airy's equation, and the solutions are expressed in terms of the linearly independentAiry functions,$Ai\left(z\right)$and$Bi\left(z\right)$ :

$\mathrm{\psi }\left(\mathrm{z}\right)={\mathrm{c}}_1\mathrm{Ai}\left(\mathrm{z}\right)+{\mathrm{c}}_2\mathrm{Bi}\left(\mathrm{z}\right)$.

Note down$\mathrm{Bi}\left(\mathrm{z}\right)$ blows up for large z (or x, and therefore we require$c_2=0$ . Thus,

$\psi \left(x\right)=c_{1}Ai\left[\alpha \left(x-\frac{E}{mg}\right)\right]$

Where$c_1$ is a constant fixed by normalization.

(c) To find first four energies.

The potential is infinite for $x<0$, so we require $\psi \left(0\right)=0$, or

$\mathrm{Ai}\left(-\frac{\mathrm{\alpha }}{\mathrm{m}\mathrm{g}}\mathrm{E}\right)=0$

$(c_1\ne 0)$. It follows that,

${\mathrm{E}}_{\mathrm{n}}=-\frac{\mathrm{mg}}{\mathrm{a}}{\mathrm{a}}_{\mathrm{n}}=-{\left(\frac{1}{2}{\mathrm{mg}}^2{\hslash }^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\mathrm{a}}_{\mathrm{n}.......\left(1\right)}$

Where $a_n$are the zeros of Ai .

Using $\mathrm{g}=9.80\mathrm{m}/{\mathrm{s}}^2$and $\mathrm{m}=0.100\mathrm{kg}$

${({\mathrm{mg}}^2{\hslash }^2/2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=3.766\times {10}^{-23}\mathrm{J}$.

Hence, the first four allowed energies (correct to three significant digits) are

$$\begin{gathered} {\mathrm{E}}_1=8.805\times {10}^{-23}\mathrm{J}; \\ {\mathrm{E}}_2=1.540\times {10}^{-22}\mathrm{J}; \\ {\mathrm{E}}_3=2.079\times {10}^{-22}\mathrm{J}; \\ {\mathrm{E}}_4=2.556\times {10}^{-22}\mathrm{J} \end{gathered}$$

(d) To find the electron in the ground state.

For an electron,

${({\mathrm{m}}_{\mathrm{e}}{\mathrm{g}}^2{\hslash }^2/2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=7.865\times {10}^{-33}\mathrm{J}$

It follows from Equation (1) that the ground state energy is

$$\begin{gathered} E_{gs}\approx 1.839\times {10}^{-32}J \\ {E}_{gs}\approx 1.149\times {10}^{-13}eV \end{gathered}$$

The Virial theorem for the case of a stationary state says that

In our case$V\left(x\right)=mgx$ ,

And so

$2⟨T⟩=⟨mgx⟩=⟨V⟩$

But we know that

$H=T+V.$

Thus

(or)

For an electron in the ground state, we have

We see that the electron should be, on the average, about 1.4mm high off the ground.