1.

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)

Before we start, let’s lay down some basic quantum mechanical formulas that we will be using:

(a) The ground state energy (zero-point energy) of a quantum harmonic oscillator is given by E = ħω/2, where ω is the angular frequency and ħ is the reduced Planck constant.

(b) The tunneling probability for a particle against a potential barrier is given by the formula derived from solving Schrödinger’s equation, which can be approximated for a particle with energy E moving against a barrier of width w and height V as follows: P ~ exp(-2*w*(2m*(V-E))^(1/2)/ħ), where m is the mass of the particle and ħ is the reduced Planck constant. Note that this is a simplified version of the formula and is generally applicable only when E < V.

(c) The diffraction pattern of particles is described by the De Broglie wavelength λ = h/p, where h is the Planck constant and p is the momentum of the particle.

Let’s proceed to answer the questions:

(a) First we have to calculate the angular frequency ω = sqrt(g/l) = sqrt(9.8 m/s^2 / 1m) ≈ 3.13 s^-1. Therefore, the zero-point energy is E = ħω/2 ≈ 1.05e-34 J*s * 3.13 s^-1 / 2 ≈ 1.64e-34 J. This is a very small energy compared to the energy scales we usually consider in the macroscopic world, showing that quantum effects are negligible in this case.

(b) In the given problem, the energy of the marble is E = 1/2 * m * v^2 = 1/2 * 0.005 kg * (0.1 m/s)^2 = 2.5e-5 J, and the height of the barrier in terms of energy is V = m*g*h = 0.005 kg * 9.8 m/s^2 * 0.05 m = 2.45e-3 J. The tunneling probability is therefore P ~ exp(-2*0.01 m*(2*0.005 kg*(2.45e-3 J – 2.5e-5 J))^(1/2)/(1.05e-34 J*s)). This yields an extremely small value, implying that the probability of the marble quantum tunneling through the barrier is essentially zero in our macroscopic world.

(c) The momentum of the tennis ball is p = mv = 0.1 kg * 0.5 m/s = 0.05 kg*m/s. Therefore, the De Broglie wavelength is λ = h/p = 6.626e-34 J*s / 0.05 kg*m/s = 1.325e-34 m. This is much smaller than the size of the window (1 m or 1.5 m), so the diffraction of the tennis ball is not observable in the macroscopic world.

This demonstrates that quantum effects are not typically observable on macroscopic scales, due to the small values of the Planck constant and reduced Planck constant.

Express each of the following quantities in terms of $\hbar$, e, c, $m=$ electron mass, $M=$ proton mass. Also give a rough estimate of numerical size for each.
(a) Bohr radius $(\mathrm{cm})$.
(b) Binding energy of hydrogen (eV).
(c) Bohr magneton (choosing your own unit).
(d) Compton wavelength of an electron $(\mathrm{cm})$.
(e) Classical electron radius $(\mathrm{cm})$.
(f) Electron rest energy $(\mathrm{MeV})$.
(g) Proton rest energy $(\mathrm{MeV})$.
(h) Fine structure constant.
(i) Typical hydrogen fine-structure splitting $(\mathrm{eV})$.
(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) $\mathrm{AE}=e^8 m c^2 / 8 \hbar^2 c^4=\frac{1}{8} \alpha^4 m c^2=1.8 \times 10^{-4} \mathrm{eV}$.

(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 $\mathrm{w}=\sqrt{g / l}$ and $A$ is the root-meansquare amplitude of the zero-point oscillation. Hence
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
$$\frac{2 m w}{\hbar} \sqrt{2 g H-v^2} \approx 0.9 \times 10^{30} \text {. }$$
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
$$\lambda=\mathrm{h} / \mathrm{p}=h / m v=1.3 \times 10^{-30} \mathrm{~cm},$$

Problems and Solutions on Electromagnetism
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.

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