Riemann surface
matlab

Quantum phenomena are often negligible in the “macroscopic” world. Show this numerically for the following cases:
(a) The amplitude of the zero-point oscillation for a pendulum of length $l=1 \mathrm{~m}$ and mass $m=1 \mathrm{~kg}$.
(b) The tunneling probability for a marble of mass $m=5 \mathrm{~g}$ moving at a speed of $10 \mathrm{~cm} / \mathrm{sec}$ against a rigid obstacle of height $H=5 \mathrm{~cm}$ and width $\mathrm{w}=1 \mathrm{~cm}$.
(c) The diffraction of a tennis ball of mass $m=0.1 \mathrm{~kg}$ moving at a speed $v=0.5 \mathrm{~m} / \mathrm{sec}$ by a window of size $1 \times 1.5 \mathrm{~m}^2$.
(Wisconsin)

Solution:
(a) The theory of the harmonic oscillator gives the average kinetic energy as $\bar{V}=\frac{1}{2} E$, i.e., $\frac{1}{2} m \omega^2 A^2=\frac{1}{4} \hbar \omega$, where $w=\sqrt{g / l}$ and $A$ is the root-meansquare amplitude of the zero-point oscillation. Hence
$$A=\sqrt{\frac{\hbar}{2 m \omega}} \approx 0.41 \times 10^{-17} \mathrm{~m} .$$
Thus the zero-point oscillation of a macroscopic pendulum is negligible.
(b) If we regard the width and height of the rigid obstacle as the width and height of a gravity potential barrier, the tunneling probability is
\begin{aligned} T & \approx \exp \left[-\frac{2 w}{\hbar} \sqrt{2 m\left(m g H-\frac{1}{2} m v^2\right)}\right] \ & =\exp \left(-\frac{2 m w}{\hbar} \sqrt{2 g H-v^2}\right), \end{aligned}
where
$$\frac{2 m w}{\hbar} \sqrt{2 g H-v^2} \approx 0.9 \times 10^{30}$$
Hence
$$T \sim e^{-0.9 \times 10^{30}} \approx 0$$
That is, the tunneling probability for the marble is essentially zero.
(c) The de Broglie wavelength of the tennis ball is
$$\lambda=\mathrm{h} / \mathrm{p}=h / m v=1.3 \times 10^{-30} \mathrm{~cm} \text {, }$$

and the diffraction angles in the horizontal and the vertical directions are respectively
$$\theta_1 \approx \lambda / D=1.3 \times 10^{-32} \mathrm{rad}, \quad \theta_2 \approx \lambda / L=9 \times 10^{-33} \mathrm{rad} .$$
Thus there is no diffraction in any direction.

Express each of the following quantities in terms of $\hbar, \mathrm{e}, \mathrm{c}, m=$ electron mass, $M=$ proton mass. Also give a rough estimate of numerical size for each.
(a) Bohr radius (cm).
(b) Binding energy of hydrogen (eV).
(c) Bohr magneton (choosing your own unit).
(d) Compton wavelength of an electron (cm).
(e) Classical electron radius (cm).
(f) Electron rest energy (MeV).
(g) Proton rest energy $(\mathrm{MeV})$.
(h) Fine structure constant.
(i) Typical hydrogen fine-structure splitting (eV).

Solution:
(a) $\mathrm{a}=\hbar^2 / m e^2=5.29 \times 10^{-9} \mathrm{~cm}$.
(b) $\mathrm{E}=m e^4 / 2 \hbar^2=13.6 \mathrm{eV}$.
(c) $\mu_B=e \hbar / 2 m c=9.27 \times 10^{-21} \mathrm{erg} \cdot \mathrm{Gs}^{-1}$.
(d) $\lambda=2 \pi \hbar / m c=2.43 \times 10^{-10} \mathrm{~cm}$.
(e) $r_e=e^2 / m c^2=2.82 \times 10^{-13} \mathrm{~cm}$.
(f) $E_e=m c^2=0.511 \mathrm{MeV}$.
(g) $E_p=M c^2=938 \mathrm{MeV}$.
(h) $\alpha=e^2 / \hbar c=7.30 \times 10^{-3} \approx 1 / 137$.
(i) $\mathrm{AE}=e^8 m c^2 / 8 \hbar^2 c^4=\frac{1}{8} \alpha^4 m c^2=1.8 \times 10^{-4} \mathrm{eV}$

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