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Basic

B-1. Photon Energy Calculation

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Problem:

According to Einstein's light quantum hypothesis, the energy of a photon with frequency \(\nu\) is given by \(E = h\nu\). Using Planck's constant \(h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s}\), find the energy of a single photon for each of the following types of light.

(a) Yellow light from sodium (frequency \(\nu = 5.09 \times 10^{14}\ \mathrm{Hz}\))

(b) Red laser (wavelength \(\lambda = 633\ \mathrm{nm}\), use the speed of light \(c = 3.00 \times 10^{8}\ \mathrm{m/s}\))

(c) Cell phone radio waves (frequency \(\nu = 2.0 \times 10^{9}\ \mathrm{Hz}\))

(a) Yellow light of sodium:

Approach: Substitute directly into \(E = h\nu\).

\[ E = h\nu = (6.63 \times 10^{-34}\ \mathrm{J \cdot s}) \times (5.09 \times 10^{14}\ \mathrm{Hz}) \]
\[ E = 6.63 \times 5.09 \times 10^{-34+14}\ \mathrm{J} = 33.75 \times 10^{-20}\ \mathrm{J} \]
\[ \boxed{E \approx 3.38 \times 10^{-19}\ \mathrm{J}} \]

(b) Red laser (\(\lambda = 633\ \mathrm{nm}\)):

Approach: Convert to frequency using \(\nu = c/\lambda\), then calculate \(E = h\nu = hc/\lambda\).

\[ \lambda = 633\ \mathrm{nm} = 633 \times 10^{-9}\ \mathrm{m} = 6.33 \times 10^{-7}\ \mathrm{m} \]
\[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{6.33 \times 10^{-7}} \]
\[ E = \frac{19.89 \times 10^{-26}}{6.33 \times 10^{-7}} = 3.14 \times 10^{-19}\ \mathrm{J} \]
\[ \boxed{E \approx 3.14 \times 10^{-19}\ \mathrm{J}} \]

(c) Cell phone radio waves (\(\nu = 2.0 \times 10^{9}\ \mathrm{Hz}\)):

\[ E = h\nu = (6.63 \times 10^{-34})(2.0 \times 10^{9}) = 13.26 \times 10^{-25}\ \mathrm{J} \]
\[ \boxed{E \approx 1.33 \times 10^{-24}\ \mathrm{J}} \]

Verification: The order of frequency magnitudes is (a) > (b) > (c), and the energies follow the same order. The energy of visible light is on the order of \(10^{-19}\ \mathrm{J}\), while radio waves are on the order of \(10^{-24}\ \mathrm{J}\), which is physically reasonable.

B-2. Relationship Between Probability Amplitude and Probability

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Problem:

Given a complex probability amplitude \(\phi = 3 + 4i\), calculate the following.

(a) The complex conjugate \(\phi^*\) of \(\phi\)

(b) \(|\phi|^2 = \phi^* \phi\)

(c) When the probability amplitude is written as \(\phi = A e^{i\theta}\) (\(A > 0\), \(\theta\) is real), express \(|\phi|^2\) in terms of \(A\) and \(\theta\).

(a) \(\phi^*\):

The complex conjugate of \(\phi = 3 + 4i\) is obtained by replacing \(i\) with \(-i\):

\[ \boxed{\phi^* = 3 - 4i} \]

(b) \(|\phi|^2 = \phi^* \phi\):

\[ |\phi|^2 = (3 - 4i)(3 + 4i) = 9 + 12i - 12i - 16i^2 = 9 + 16 = 25 \]
\[ \boxed{|\phi|^2 = 25} \]

(c) Express \(|\phi|^2\) in terms of \(A\) and \(\theta\):

When \(\phi = Ae^{i\theta}\), we have \(\phi^* = Ae^{-i\theta}\) (where \(A > 0\) is real), so

\[ |\phi|^2 = \phi^*\phi = Ae^{-i\theta} \cdot Ae^{i\theta} = A^2 e^{0} = A^2 \]
\[ \boxed{|\phi|^2 = A^2} \]

Verification: For (b), \(|\phi| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\), so \(|\phi|^2 = 25\). ✓ For (c), we confirm that the result is independent of the phase \(\theta\), which is consistent with the fundamental property of quantum mechanics that probabilities do not depend on the absolute phase.

B-3. Calculation of Interference Terms

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Problem:

Two probability amplitudes are given by \(\phi_1 = e^{i\alpha}\) and \(\phi_2 = e^{i\beta}\) (where \(\alpha, \beta\) are real numbers).

(a) Expand \(|\phi_1 + \phi_2|^2\) and express it in terms of \(\alpha\) and \(\beta\).

(b) Calculate \(|\phi_1|^2 + |\phi_2|^2\).

(c) Express the interference term \(|\phi_1 + \phi_2|^2 - (|\phi_1|^2 + |\phi_2|^2)\) using the phase difference \(\delta = \alpha - \beta\).

(d) Find \(|\phi_1 + \phi_2|^2\) for \(\delta = 0\) (in phase) and \(\delta = \pi\) (out of phase), respectively.

(a) Expansion of \(|\phi_1 + \phi_2|^2\):

\[ |\phi_1 + \phi_2|^2 = (\phi_1^* + \phi_2^*)(\phi_1 + \phi_2) \]
\[ = |\phi_1|^2 + \phi_1^*\phi_2 + \phi_2^*\phi_1 + |\phi_2|^2 \]

Computing each term. Since \(\phi_1 = e^{i\alpha}\), \(\phi_2 = e^{i\beta}\):

  • \(|\phi_1|^2 = e^{-i\alpha}e^{i\alpha} = 1\)
  • \(|\phi_2|^2 = e^{-i\beta}e^{i\beta} = 1\)
  • \(\phi_1^*\phi_2 = e^{-i\alpha}e^{i\beta} = e^{i(\beta - \alpha)}\)
  • \(\phi_2^*\phi_1 = e^{-i\beta}e^{i\alpha} = e^{i(\alpha - \beta)}\)

Sum of the interference terms:

\[ \phi_1^*\phi_2 + \phi_2^*\phi_1 = e^{i(\beta-\alpha)} + e^{-i(\beta-\alpha)} = 2\cos(\alpha - \beta) \]

Therefore,

\[ \boxed{|\phi_1 + \phi_2|^2 = 2 + 2\cos(\alpha - \beta)} \]

(b) \(|\phi_1|^2 + |\phi_2|^2\):

\[ \boxed{|\phi_1|^2 + |\phi_2|^2 = 1 + 1 = 2} \]

(c) Expressing the interference term in terms of the phase difference \(\delta = \alpha - \beta\):

\[ |\phi_1 + \phi_2|^2 - (|\phi_1|^2 + |\phi_2|^2) = (2 + 2\cos\delta) - 2 \]
\[ \boxed{\text{Interference term} = 2\cos\delta} \]

(d) Special cases:

  • \(\delta = 0\) (in phase): \(|\phi_1 + \phi_2|^2 = 2 + 2\cos 0 = 2 + 2 = \boxed{4}\)
  • \(\delta = \pi\) (out of phase): \(|\phi_1 + \phi_2|^2 = 2 + 2\cos\pi = 2 - 2 = \boxed{0}\)

Verification: When \(\delta = 0\), \(\phi_1 = \phi_2 = e^{i\alpha}\) so \(|\phi_1 + \phi_2|^2 = |2e^{i\alpha}|^2 = 4\). ✓ When \(\delta = \pi\), \(\phi_2 = e^{i(\alpha-\pi)} = -e^{i\alpha} = -\phi_1\) so \(|\phi_1 + \phi_2|^2 = 0\). ✓

B-4. Probability Distributions of Bullets and Electrons

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Problem:

In a double-slit experiment, the probability (for bullets) or the absolute square of the probability amplitude (for electrons) when only slit 1 is open is given at position \(x\) on the screen as follows:

\[ P_1(x) = |\phi_1(x)|^2 = e^{-(x-a)^2}, \quad P_2(x) = |\phi_2(x)|^2 = e^{-(x+a)^2} \]

where \(a > 0\).

(a) Evaluate the combined probability for bullets \(P_{12}^{\text{bullet}}(x) = P_1(x) + P_2(x)\) at \(x = 0\).

(b) For electrons, given that the probability amplitudes are \(\phi_1(x) = e^{-(x-a)^2/2}\) and \(\phi_2(x) = e^{-(x+a)^2/2}\) (here we consider the real-valued case), evaluate \(P_{12}^{\text{electron}}(x) = |\phi_1(x) + \phi_2(x)|^2\) at \(x = 0\).

(c) Compare the results of (a) and (b), and determine which is larger: \(P_{12}^{\text{electron}}(0)\) or \(P_{12}^{\text{bullet}}(0)\).

(a) Bullet case: \(x = 0\):

\[ P_1(0) = e^{-(0-a)^2} = e^{-a^2}, \quad P_2(0) = e^{-(0+a)^2} = e^{-a^2} \]
\[ \boxed{P_{12}^{\text{bullet}}(0) = P_1(0) + P_2(0) = 2e^{-a^2}} \]

(b) Electron case: \(x = 0\):

Calculate the probability amplitudes:

\[ \phi_1(0) = e^{-(0-a)^2/2} = e^{-a^2/2}, \quad \phi_2(0) = e^{-(0+a)^2/2} = e^{-a^2/2} \]
\[ P_{12}^{\text{electron}}(0) = |\phi_1(0) + \phi_2(0)|^2 = (e^{-a^2/2} + e^{-a^2/2})^2 = (2e^{-a^2/2})^2 \]
\[ \boxed{P_{12}^{\text{electron}}(0) = 4e^{-a^2}} \]

(c) Comparison:

\[ P_{12}^{\text{electron}}(0) = 4e^{-a^2}, \quad P_{12}^{\text{bullet}}(0) = 2e^{-a^2} \]
\[ \frac{P_{12}^{\text{electron}}(0)}{P_{12}^{\text{bullet}}(0)} = \frac{4e^{-a^2}}{2e^{-a^2}} = 2 \]
\[ \boxed{P_{12}^{\text{electron}}(0) = 2 \times P_{12}^{\text{bullet}}(0) > P_{12}^{\text{bullet}}(0)} \]

The electron case is larger. This is because at \(x = 0\) (the midpoint between the two slits), the probability amplitudes overlap in phase, resulting in constructive interference.

Verification: \(|\phi_1|^2 = (e^{-a^2/2})^2 = e^{-a^2} = P_1(0)\), which is consistent with the problem setup. ✓

B-5. Polar Form of Complex Numbers and Euler's Formula

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Problem:

Using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\), calculate the following.

(a) \(e^{i\pi}\)

(b) \(e^{i\pi/2}\)

(c) The real and imaginary parts of \(e^{i\pi/4}\)

(d) Find \(|e^{i\theta}|^2\) independent of \(\theta\).

(a) \(e^{i\pi}\):

\[ e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0 = \boxed{-1} \]

(b) \(e^{i\pi/2}\):

\[ e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i = \boxed{i} \]

(c) Real and imaginary parts of \(e^{i\pi/4}\):

\[ e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \]
\[ \boxed{\text{Real part} = \frac{\sqrt{2}}{2}, \quad \text{Imaginary part} = \frac{\sqrt{2}}{2}} \]

(d) \(|e^{i\theta}|^2\):

\[ |e^{i\theta}|^2 = e^{-i\theta} \cdot e^{i\theta} = e^{0} = 1 \]

Alternatively, \(|e^{i\theta}|^2 = \cos^2\theta + \sin^2\theta = 1\).

\[ \boxed{|e^{i\theta}|^2 = 1 \quad (\text{independent of } \theta)} \]

Verification: (a) is the famous Euler identity \(e^{i\pi} + 1 = 0\). (d) corresponds to the fact that a point on the unit circle has absolute value 1. ✓

B-6. Wave Intensity and Interference

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Problem:

Following the notation in the text, suppose the heights of two waves are given by

\[ h_1 = A\cos(\omega t), \quad h_2 = A\cos(\omega t + \delta) \]

where \(A > 0\) and \(\delta\) is the phase difference.

(a) Using the sum-to-product formula, rewrite \(h_1 + h_2\) in product form.

(b) Find the amplitude of \(h_1 + h_2\) when \(\delta = 0\).

(c) Find \(h_1 + h_2\) when \(\delta = \pi\).

(a) Expressing \(h_1 + h_2\) in product form:

Using the sum-to-product formula \(\cos P + \cos Q = 2\cos\!\left(\frac{P+Q}{2}\right)\cos\!\left(\frac{P-Q}{2}\right)\).

Let \(P = \omega t\), \(Q = \omega t + \delta\):

\[ h_1 + h_2 = A(\cos\omega t + \cos(\omega t + \delta)) \]
\[ = 2A\cos\!\left(\frac{\omega t + \omega t + \delta}{2}\right)\cos\!\left(\frac{\omega t - (\omega t + \delta)}{2}\right) \]
\[ \boxed{h_1 + h_2 = 2A\cos\!\left(\frac{\delta}{2}\right)\cos\!\left(\omega t + \frac{\delta}{2}\right)} \]

(b) When \(\delta = 0\):

\[ h_1 + h_2 = 2A\cos(0)\cos(\omega t) = 2A\cos(\omega t) \]

The amplitude is \(\boxed{2A}\)

(c) When \(\delta = \pi\):

\[ h_1 + h_2 = 2A\cos\!\left(\frac{\pi}{2}\right)\cos\!\left(\omega t + \frac{\pi}{2}\right) = 2A \cdot 0 \cdot \cos\!\left(\omega t + \frac{\pi}{2}\right) \]
\[ \boxed{h_1 + h_2 = 0} \]

Verification: (b) In-phase waves interfere constructively, doubling the amplitude. (c) Out-of-phase waves cancel completely. Physically reasonable. ✓

B-7. Discreteness of Energy — The Staircase Analogy with Numbers

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Problem:

As stated in the main text, "the energy of an atom is discrete, like steps on a staircase." The energy levels of the hydrogen atom are given by

\[ E_n = -\frac{13.6\ \mathrm{eV}}{n^2} \quad (n = 1, 2, 3, \ldots) \]

(This will be derived in Ch. 16.) Calculate the following.

(a) Find the energies of the ground state (\(n = 1\)) and the first excited state (\(n = 2\)), respectively.

(b) Find the energy of the photon emitted when a transition occurs from \(n = 2\) to \(n = 1\): \(\Delta E = E_2 - E_1\).

(c) Using \(E = h\nu\), find the frequency \(\nu\) of this photon (use \(1\ \mathrm{eV} = 1.60 \times 10^{-19}\ \mathrm{J}\)).

(d) Does this light fall within the visible light range (frequency \(4.3 \times 10^{14}\ \mathrm{Hz}\) to \(7.5 \times 10^{14}\ \mathrm{Hz}\))?

(a) Energy of the ground state and first excited state:

\[ E_1 = -\frac{13.6\ \mathrm{eV}}{1^2} = \boxed{-13.6\ \mathrm{eV}} \]
\[ E_2 = -\frac{13.6\ \mathrm{eV}}{2^2} = -\frac{13.6}{4}\ \mathrm{eV} = \boxed{-3.4\ \mathrm{eV}} \]

(b) Energy of the emitted photon:

\[ \Delta E = E_2 - E_1 = (-3.4) - (-13.6) = \boxed{10.2\ \mathrm{eV}} \]

(c) Frequency of the photon:

First, convert eV to J:

\[ \Delta E = 10.2\ \mathrm{eV} \times 1.60 \times 10^{-19}\ \mathrm{J/eV} = 16.32 \times 10^{-19}\ \mathrm{J} = 1.632 \times 10^{-18}\ \mathrm{J} \]
\[ \nu = \frac{\Delta E}{h} = \frac{1.632 \times 10^{-18}}{6.63 \times 10^{-34}} = 2.46 \times 10^{15}\ \mathrm{Hz} \]
\[ \boxed{\nu \approx 2.46 \times 10^{15}\ \mathrm{Hz}} \]

(d) Does it fall within the visible light range?:

The frequency range of visible light is \(4.3 \times 10^{14}\ \mathrm{Hz}\) to \(7.5 \times 10^{14}\ \mathrm{Hz}\).

\(\nu \approx 2.46 \times 10^{15}\ \mathrm{Hz}\) is greater than the upper limit of this range, \(7.5 \times 10^{14}\ \mathrm{Hz}\).

\[ \boxed{\text{It does not fall within the visible light range (it corresponds to ultraviolet radiation).}} \]

Verification: It is known that the hydrogen Lyman series (\(n \geq 2 \to n = 1\)) lies in the ultraviolet region, so the result is reasonable. Checking the wavelength: \(\lambda = c/\nu = 3.00 \times 10^8 / 2.46 \times 10^{15} \approx 122\ \mathrm{nm}\), which agrees with the Lyman-\(\alpha\) line (121.6 nm). ✓

B-8. Superposition of Probability Amplitudes — Numerical Example

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Problem:

Consider a 2-state system where the probability amplitude for being in state \(|1\rangle\) is \(\phi_1 = \dfrac{1}{\sqrt{3}}\), and the probability amplitude for being in state \(|2\rangle\) is \(\phi_2 = \sqrt{\dfrac{2}{3}}\, e^{i\pi/3}\).

(a) Find the probability \(P_1 = |\phi_1|^2\) of observing state \(|1\rangle\).

(b) Find the probability \(P_2 = |\phi_2|^2\) of observing state \(|2\rangle\).

(c) Verify that the normalization condition \(P_1 + P_2 = 1\) holds.

(d) Find the real and imaginary parts of \(\phi_2\).

(a) \(P_1 = |\phi_1|^2\):

\[ P_1 = \left|\frac{1}{\sqrt{3}}\right|^2 = \boxed{\frac{1}{3}} \]

(b) \(P_2 = |\phi_2|^2\):

\[ P_2 = \left|\sqrt{\frac{2}{3}}\, e^{i\pi/3}\right|^2 = \frac{2}{3} \cdot |e^{i\pi/3}|^2 = \frac{2}{3} \cdot 1 = \boxed{\frac{2}{3}} \]

(c) Verification of the normalization condition:

\[ P_1 + P_2 = \frac{1}{3} + \frac{2}{3} = 1 \quad \checkmark \]
\[ \boxed{P_1 + P_2 = 1 \text{ is satisfied.}} \]

(d) Real and imaginary parts of \(\phi_2\):

\[ \phi_2 = \sqrt{\frac{2}{3}}\, e^{i\pi/3} = \sqrt{\frac{2}{3}}\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = \sqrt{\frac{2}{3}}\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \]
\[ \boxed{\text{Real part} = \frac{1}{2}\sqrt{\frac{2}{3}} = \frac{1}{\sqrt{6}}, \quad \text{Imaginary part} = \frac{\sqrt{3}}{2}\sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}} \]

Verification: \(|\phi_2|^2 = \left(\frac{1}{\sqrt{6}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}\). ✓

Medium

M-1. Comparison of Classical and Quantum Probability

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Problem:

Based on the Feynman-style comparison of bullets, waves, and electrons described in the main text, answer the following questions.

(a) Explain the physical reason why \(P_{12} = P_1 + P_2\) holds in the double-slit experiment with bullets, based on the fact that "a bullet passes through one slit or the other."

(b) In the double-slit experiment with electrons, each electron is detected as a single particle, yet when many electrons are accumulated, an interference pattern appears. Explain this fact using the mathematical structure of probability amplitudes

\[ P_{12} = |\phi_1 + \phi_2|^2 \]

In particular, clearly identify the reason why \(P_{12} \neq P_1 + P_2\).

(c) Suppose probability amplitudes were restricted to real numbers rather than complex numbers. Would interference terms still appear in that case? Discuss what restrictions arise regarding the sign of the interference term compared to the complex number case.

(a) Why \(P_{12} = P_1 + P_2\) holds for bullets:

Bullets are macroscopic particles that always pass through one slit or the other when traversing a double slit. The event of passing through slit 1 and the event of passing through slit 2 are mutually exclusive events (they cannot occur simultaneously). Therefore, the classical addition rule of probability applies directly, giving

\[ P_{12}(x) = P_1(x) + P_2(x) \]

When a bullet passes through one slit, whether the other slit is open or not does not affect the bullet's trajectory, so the probability distributions from each slit can be independently added together.

(b) Why \(P_{12} \neq P_1 + P_2\) for electrons:

In the case of electrons, probability amplitudes (complex numbers) \(\phi_1(x)\), \(\phi_2(x)\) are assigned to the process of passing through each slit. According to the rules of quantum mechanics, one must first add the probability amplitudes for indistinguishable paths and then take the squared modulus:

\[ P_{12}(x) = |\phi_1(x) + \phi_2(x)|^2 \]

Expanding this gives:

\[ P_{12} = |\phi_1|^2 + |\phi_2|^2 + \underbrace{\phi_1^*\phi_2 + \phi_2^*\phi_1}_{\text{interference term}} \]
\[ = P_1 + P_2 + 2\,\mathrm{Re}(\phi_1^*\phi_2) \]

Since the interference term \(2\,\mathrm{Re}(\phi_1^*\phi_2)\) is generally not zero, \(P_{12} \neq P_1 + P_2\). This interference term can be positive or negative depending on the position \(x\), producing a pattern of bright and dark fringes (interference pattern) on the screen. Although each electron is detected as a single particle, when many electrons are accumulated, this interference pattern emerges statistically.

(c) If probability amplitudes were restricted to real numbers:

Interference terms still appear even when amplitudes are real. When \(\phi_1, \phi_2\) are real:

\[ \phi_1^*\phi_2 + \phi_2^*\phi_1 = 2\phi_1\phi_2 \]

This is not necessarily zero, so interference itself still occurs.

However, a serious limitation arises. With real probability amplitudes, the "phase difference" between two amplitudes can only take 2 values: \(0\) (same sign: \(\phi_1\phi_2 > 0\)) or \(\pi\) (opposite sign: \(\phi_1\phi_2 < 0\)). Consequently, the interference term can only be \(+2|\phi_1||\phi_2|\) (constructive interference) or \(-2|\phi_1||\phi_2|\) (destructive interference), making it impossible to reproduce a smooth interference pattern with continuously varying phase differences.

In the complex case, the phase difference \(\delta\) can vary continuously from \(0\) to \(2\pi\), and the interference term \(2|\phi_1||\phi_2|\cos\delta\) varies smoothly from \(-2|\phi_1||\phi_2|\) to \(+2|\phi_1||\phi_2|\). This allows for a correct description of the continuous interference fringes observed in experiments.

M-2. Light Quantum Hypothesis and the Photoelectric Effect

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Problem:

Based on Einstein's light quantum hypothesis \(E = h\nu\), answer the following questions.

(a) Let the work function of a metal be \(W\). When light of frequency \(\nu\) is incident on the metal, express the maximum kinetic energy \(K_{\max}\) of the ejected electrons in terms of \(h\), \(\nu\), and \(W\).

(b) In the photoelectric effect, if the frequency of light is below a certain value \(\nu_0\), no electrons are ejected regardless of how much the light intensity is increased. Express this threshold frequency \(\nu_0\) in terms of \(W\) and \(h\).

(c) According to classical wave theory (Maxwell's electromagnetism), increasing the intensity of light should be able to provide sufficient energy to the electrons. Nevertheless, the experimental fact that no electrons are ejected below the threshold frequency contradicts a certain assumption of classical theory. Explain what assumption this is.

(d) Light of wavelength \(\lambda = 400\ \mathrm{nm}\) is incident on a metal with work function \(W = 2.3\ \mathrm{eV}\). Find the maximum kinetic energy \(K_{\max}\) of the ejected electrons in units of eV.

(a) Maximum Kinetic Energy \(K_{\max}\):

The energy of a single photon \(h\nu\) is transferred to an electron in the metal. The minimum energy required for an electron to escape from the metal surface is the work function \(W\). From conservation of energy:

\[ h\nu = W + K_{\max} \]
\[ \boxed{K_{\max} = h\nu - W} \]

(b) Threshold Frequency \(\nu_0\):

The condition for an electron to barely escape is \(K_{\max} = 0\), so:

\[ h\nu_0 - W = 0 \]
\[ \boxed{\nu_0 = \frac{W}{h}} \]

(c) Contradiction with Classical Theory:

In classical wave theory, the energy of light is proportional to the square of the amplitude (i.e., the intensity) and does not depend on frequency. Therefore, the classical assumption is:

"By increasing the intensity of light, one can impart an arbitrarily large amount of energy to an electron, regardless of frequency."

However, experiments show that light with \(\nu < \nu_0\) cannot eject electrons no matter how much the intensity is increased. This is because the energy of light is transferred to electrons in discrete units (photons) of \(h\nu\), and if the energy of a single photon \(h\nu\) is less than \(W\), no number of photons can cause a single electron to escape (since each photon interacts with an electron independently). This contradicts the classical assumption that "energy can be accumulated continuously."

(d) Numerical Calculation:

Photon energy:

\[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{400 \times 10^{-9}} = \frac{19.89 \times 10^{-26}}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19}\ \mathrm{J} \]

Converting to eV:

\[ E = \frac{4.97 \times 10^{-19}}{1.60 \times 10^{-19}}\ \mathrm{eV} \approx 3.11\ \mathrm{eV} \]

Maximum kinetic energy:

\[ K_{\max} = E - W = 3.11 - 2.3 = \boxed{0.8\ \mathrm{eV}} \]

Verification: Light with wavelength 400 nm is violet light with energy of approximately 3.1 eV. A work function of 2.3 eV is close to values for metals such as cesium. \(K_{\max} \approx 0.8\ \mathrm{eV}\) is a reasonable value. ✓

M-3. Models in Physics and Falsifiability

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Problem:

Based on the philosophy of science standpoint described in Introduction — Before the Four Journeys at the beginning of this site (models are hypotheses; equations are tools for falsifiability), answer the following questions.

(a) What does it mean to say that Newtonian mechanics is a "hypothesis"? Give one example each of a domain where Newtonian mechanics succeeded and a domain where it broke down, and explain.

(b) The text states that quantum mechanics "has not had a single incorrect prediction in over 100 years." Explain why quantum mechanics is nevertheless called a "hypothesis," referring to its relationship with general relativity.

(c) What is "falsifiability"? Explain why the claim "tomorrow's weather will be sunny, rainy, or cloudy" is not falsifiable.

(a) What it means for Newtonian mechanics to be a "hypothesis":

That Newtonian mechanics is a "hypothesis" means that it is not the "truth" of nature, but merely "the best model that does not currently contradict experiment." If refuted by experiment, it is replaced by a better model.

  • Successful domain: Celestial motion (planetary orbits, lunar motion, etc.). Newton's equations of motion and the law of universal gravitation can predict planetary positions with high precision. The discovery of Neptune was based on predictions from Newtonian mechanics.

  • Domain of failure: Phenomena at the atomic scale. For example, the discreteness of spectral lines of the hydrogen atom and the interference pattern in the double-slit experiment with electrons cannot be explained by Newtonian mechanics. Furthermore, the precise value of Mercury's perihelion precession cannot be reproduced by Newtonian mechanics, requiring general relativity.

(b) Why we call quantum mechanics a "hypothesis":

For over 100 years, the predictions of quantum mechanics have been confirmed in every experiment across atomic physics, solid-state physics, particle physics, and more. Nevertheless, the reasons for still calling it a "hypothesis" are as follows:

Quantum mechanics and general relativity (the model of gravity) are each extremely accurate within their respective domains of applicability, but in situations where both must be applied simultaneously (the center of black holes, the beginning of the universe, and other regimes where gravity is strong at extremely small scales), they contradict each other. A "quantum gravity" model that unifies both has not yet been found. Therefore, quantum mechanics in its current form may be incomplete, and it may eventually be situated as an approximation to a more comprehensive model. No matter how successful it has been, as long as the possibility of being refuted in principle cannot be excluded, it remains a "hypothesis."

(c) What falsifiability is:

Falsifiability is the property that, for a given claim, one can clearly define "what observational result would allow us to judge that claim to be wrong." A scientific hypothesis must be falsifiable.

The statement "tomorrow's weather will be sunny, rainy, or cloudy" covers all conceivable weather conditions, so no observational result can contradict it. Whether it is sunny, rainy, or cloudy, this statement turns out to be "correct." Since there is no way to refute it, this statement is not falsifiable and has no value as a scientific prediction.

M-4. Einstein's Duality — Founder and Critic

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Problem:

Based on the section "Einstein and Quantum Mechanics" in the main text, answer the following questions.

(a) Explain the grounds for calling Einstein "one of the founders" of quantum theory by separately describing the physical content of his 1905 light quantum hypothesis and his 1917 prediction of stimulated emission.

(b) Explain why Einstein became a "critic" of quantum mechanics, taking into account the meaning of his statement "God does not play dice."

(c) Explain the relationship between the EPR paradox (1935) and Bell's theorem (1964) in approximately 200 characters, using ALL of the following keywords: incompleteness, hidden variables, inequality, experiment.

(a) Grounds as a Founder:

The Light Quantum Hypothesis of 1905:

At the time, it was established through Maxwell's electromagnetism that light is a wave. However, in the photoelectric effect (the phenomenon where electrons are ejected when light strikes a metal), the experimental fact that the energy of emitted electrons depends not on the intensity of light but on its frequency could not be explained by wave theory. Einstein proposed that light consists of particles (light quanta, later called photons) possessing energy \(E = h\nu\) proportional to the frequency \(\nu\). This quantitatively explained the photoelectric effect and established the particle nature of light.

The Prediction of Stimulated Emission in 1917:

Through statistical mechanical considerations of atoms and radiation fields in thermal equilibrium, Einstein showed that in addition to absorption and spontaneous emission of light, stimulated emission (a process in which an external photon induces the emission of a photon with the same frequency, same direction, and same phase) must exist for consistency with Planck's blackbody radiation formula. This stimulated emission is the mechanism that creates a state in which all photons are aligned, and it is precisely the operating principle of the laser (LASER: Light Amplification by Stimulated Emission of Radiation), realized in 1960.

(b) Reasons for Becoming a Critic:

The core of quantum mechanics, completed in the 1920s, is a probabilistic, non-deterministic worldview in which "physical quantities do not have definite values until measured" and "only probabilities can be predicted." Einstein fundamentally refused to accept this worldview.

The phrase "God does not play dice (Gott würfelt nicht)" expressed Einstein's conviction that deterministic laws must underlie nature. While Einstein acknowledged that quantum mechanics yields correct experimental predictions, he believed it to be an "incomplete" description, and that a deeper (deterministic) model must exist that determines the values of physical quantities even before measurement. Einstein's position was that probabilities appear only because we are ignoring variables we do not yet know about (hidden variables).

(c) The EPR Paradox and Bell's Theorem:

In the 1935 EPR paper, Einstein argued for the incompleteness of quantum mechanics. In the correlations between two distant particles (quantum entanglement), the fact that a measurement on one particle appears to instantaneously determine the result for the other led him to argue that hidden variables must exist that fix the values before measurement. In 1964, Bell derived Bell's inequality as a statistical constraint following from the assumption of hidden variables. If hidden variables exist, the correlations between measurement results must satisfy this inequality. Subsequent experiments (the Aspect experiment (1982), among others) confirmed that Bell's inequality is violated, thereby ruling out the local hidden variable models that Einstein had envisioned.

M-5. Interference of Probability Amplitudes — Quantitative Analysis

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Problem:

In the double-slit experiment, the probability amplitudes from the two apertures at position \(x\) on the screen are given by

\[ \phi_1(x) = A\, e^{i k r_1(x)}, \quad \phi_2(x) = A\, e^{i k r_2(x)} \]

where \(A > 0\) is a real constant, \(k\) is the wavenumber, and \(r_1(x)\) and \(r_2(x)\) are the distances from aperture 1 and aperture 2 to position \(x\) on the screen, respectively.

(a) Calculate the combined probability \(P_{12}(x) = |\phi_1(x) + \phi_2(x)|^2\) and express it in terms of \(\Delta r(x) = r_1(x) - r_2(x)\) (the path difference).

(b) Express the condition for \(P_{12}(x)\) to take its maximum value in terms of \(\Delta r\) and \(k\). Also, find the value of \(P_{12}\) at that point.

(c) Find the condition for \(P_{12}(x)\) to take its minimum value of \(0\).

(d) Using the wavelength \(\lambda = 2\pi/k\), rewrite the conditions from (b) and (c) in terms of \(\Delta r\) and \(\lambda\).

(a) Combined probability \(P_{12}(x)\):

\[ \phi_1 + \phi_2 = Ae^{ikr_1} + Ae^{ikr_2} = Ae^{ikr_1}\left(1 + e^{ik(r_2 - r_1)}\right) \]

Taking the absolute value squared:

\[ P_{12} = |\phi_1 + \phi_2|^2 = A^2 \left|1 + e^{-ik\Delta r}\right|^2 \]

Here we defined \(\Delta r = r_1 - r_2\), so \(r_2 - r_1 = -\Delta r\).

\[ \left|1 + e^{-ik\Delta r}\right|^2 = (1 + e^{ik\Delta r})(1 + e^{-ik\Delta r}) = 1 + e^{ik\Delta r} + e^{-ik\Delta r} + 1 \]
\[ = 2 + 2\cos(k\Delta r) \]

Therefore:

\[ \boxed{P_{12}(x) = 2A^2\left[1 + \cos(k\Delta r(x))\right] = 4A^2\cos^2\!\left(\frac{k\Delta r(x)}{2}\right)} \]

Alternative derivation (direct expansion):

\[ P_{12} = |\phi_1|^2 + |\phi_2|^2 + \phi_1^*\phi_2 + \phi_2^*\phi_1 \]
\[ = A^2 + A^2 + A^2 e^{-ikr_1}e^{ikr_2} + A^2 e^{-ikr_2}e^{ikr_1} \]
\[ = 2A^2 + A^2\left(e^{ik(r_2-r_1)} + e^{-ik(r_2-r_1)}\right) = 2A^2 + 2A^2\cos(k(r_2-r_1)) \]
\[ = 2A^2(1 + \cos(k\Delta r)) \quad (\text{since } r_2 - r_1 = -\Delta r \text{, we have } \cos(k(r_2-r_1)) = \cos(k\Delta r)) \]

The result is the same.

(b) Condition for maxima:

\(P_{12}\) is maximized when \(\cos(k\Delta r) = 1\), i.e.:

\[ k\Delta r = 2n\pi \quad (n = 0, \pm 1, \pm 2, \ldots) \]
\[ \boxed{\Delta r = \frac{2n\pi}{k} \quad (n = 0, \pm 1, \pm 2, \ldots)} \]

In this case:

\[ \boxed{P_{12}^{\max} = 2A^2(1 + 1) = 4A^2} \]

(c) Condition for minima (zero):

\(P_{12} = 0\) when \(\cos(k\Delta r) = -1\), i.e.:

\[ k\Delta r = (2n+1)\pi \quad (n = 0, \pm 1, \pm 2, \ldots) \]
\[ \boxed{\Delta r = \frac{(2n+1)\pi}{k} \quad (n = 0, \pm 1, \pm 2, \ldots)} \]

(d) Rewriting in terms of \(\lambda = 2\pi/k\):

From \(k = 2\pi/\lambda\), we get \(2\pi/k = \lambda\).

Constructive interference (maximum) condition:

\[ \Delta r = \frac{2n\pi}{k} = n\lambda \]
\[ \boxed{\Delta r = n\lambda \quad (n = 0, \pm 1, \pm 2, \ldots)} \]

Destructive interference (minimum) condition:

\[ \Delta r = \frac{(2n+1)\pi}{k} = \left(n + \frac{1}{2}\right)\lambda \]
\[ \boxed{\Delta r = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, \pm 1, \pm 2, \ldots)} \]

Verification: Constructive interference occurs when the path difference is an integer multiple of the wavelength, and destructive interference occurs when it is a half-integer multiple. This is consistent with the classical wave interference conditions. Also, \(P_{12}^{\max} = 4A^2 = 4 \times |\phi_1|^2\), which is 4 times the probability \(A^2\) from each individual slit. This corresponds to the fact that when the amplitude doubles, the probability quadruples. ✓

Advanced

A-1. Indispensability of Complex Probability Amplitudes

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Problem:

In the main text, it is stated that "probability amplitudes must be complex numbers. Real numbers alone cannot reproduce the predictions of quantum mechanics." Examine this claim by following the steps below.

(a) Consider a two-state system. Let \(\phi_1\), \(\phi_2\) be the probability amplitudes for states \(|1\rangle\) and \(|2\rangle\), satisfying the normalization condition \(|\phi_1|^2 + |\phi_2|^2 = 1\). If \(\phi_1\) and \(\phi_2\) are both real, state the geometric condition satisfied by \((|\phi_1|, |\phi_2|)\).

(b) If \(\phi_1\) and \(\phi_2\) are complex, they can be written as \(\phi_1 = |\phi_1| e^{i\alpha}\), \(\phi_2 = |\phi_2| e^{i\beta}\). How many degrees of freedom (number of independent real parameters) are gained compared to the real case? Take into account that the overall phase (a common phase factor \(e^{i\gamma}\) applied to both \(\phi_1\) and \(\phi_2\)) is physically unobservable.

(c) In the interference pattern of a double-slit experiment, the phase difference \(\delta(x)\) varies continuously with position \(x\) on the screen, and \(P_{12}(x)\) is a smooth function containing \(\cos\delta(x)\). If probability amplitudes were restricted to real numbers, the phase difference could only take the two values \(0\) or \(\pi\). What restrictions would this impose on the interference pattern? Discuss in contrast with the smooth interference fringes observed in experiments.

(d) Based on the above considerations, describe "the physical significance of probability amplitudes being complex numbers in quantum mechanics" in approximately 200 characters (about 3–4 sentences).

(a) Geometric Condition for Real Numbers:

When both \(\phi_1\) and \(\phi_2\) are real, the normalization condition is:

\[ |\phi_1|^2 + |\phi_2|^2 = \phi_1^2 + \phi_2^2 = 1 \]

This represents a point on the unit circle in the \((\phi_1, \phi_2)\) plane. That is, \((|\phi_1|, |\phi_2|)\) satisfies the condition of being a point on the unit circle in the first quadrant (\(|\phi_1|^2 + |\phi_2|^2 = 1\), \(|\phi_1| \geq 0\), \(|\phi_2| \geq 0\)).

However, since \(\phi_1, \phi_2\) themselves can have signs, the degrees of freedom consist of only one angular parameter (which can be written as \(\phi_1 = \cos\theta\), \(\phi_2 = \pm\sin\theta\)) plus the choice of sign (\(+\) or \(-\)) for each amplitude.

(b) Increase of Degrees of Freedom for Complex Numbers:

In the complex case, we can write \(\phi_1 = |\phi_1|e^{i\alpha}\), \(\phi_2 = |\phi_2|e^{i\beta}\). Counting the independent real parameters:

  • \(|\phi_1|\): Due to the normalization condition \(|\phi_1|^2 + |\phi_2|^2 = 1\), once \(|\phi_1|\) is determined, \(|\phi_2|\) is also determined → 1 parameter
  • \(\alpha\): 1 parameter
  • \(\beta\): 1 parameter

Total: 3 parameters. However, since the overall phase \(e^{i\gamma}\) (\(\phi_1 \to \phi_1 e^{i\gamma}\), \(\phi_2 \to \phi_2 e^{i\gamma}\)) is physically unobservable, we subtract 1 parameter.

Therefore, the number of physically independent parameters is \(3 - 1 = 2\).

For real numbers: \(|\phi_1|\) (1 parameter, constrained by normalization) and the choice of signs. Effectively, writing \(\phi_1 = \cos\theta\), \(\phi_2 = \sin\theta\) gives 1 parameter (\(\theta\)). Even including the sign freedom, the phase difference can only take discrete values of \(0\) or \(\pi\), so there is only 1 continuous parameter.

For complex numbers, the physical degrees of freedom are 2 (for example, \(|\phi_1|\) and the relative phase \(\alpha - \beta\)).

\[ \boxed{\text{The degrees of freedom increase by one (the relative phase } \alpha - \beta \text{ is added as a new continuous degree of freedom).}} \]

(c) Restrictions on Interference Patterns for Real Numbers:

For real probability amplitudes, the interference term is:

\[ \phi_1^*\phi_2 + \phi_2^*\phi_1 = 2\phi_1\phi_2 \]

Since \(\phi_1(x)\) and \(\phi_2(x)\) take real values at each position \(x\), the sign of their product \(\phi_1(x)\phi_2(x)\) is positive when both have the same sign and negative when they have opposite signs. Thus the interference term:

\[ 2\phi_1\phi_2 = \pm 2|\phi_1||\phi_2| \]

can only take two values (at each point, it is either constructive or destructive interference—a binary choice).

On the other hand, the interference fringes observed experimentally exhibit a continuous pattern where the probability varies smoothly as a \(\cos\) function depending on position \(x\). With complex probability amplitudes, the phase difference \(\delta(x) = k(r_1(x) - r_2(x))\) varies continuously with position \(x\), and the interference term becomes \(2|\phi_1||\phi_2|\cos\delta(x)\), which varies smoothly from \(-2|\phi_1||\phi_2|\) to \(+2|\phi_1||\phi_2|\).

In the real case, the interference term can only produce discontinuous patterns (or patterns that switch discretely across zero), making it impossible to reproduce the smooth \(\cos\)-type interference fringes observed experimentally.

(d) Summary of Physical Significance:

The physical significance of probability amplitudes being complex numbers in quantum mechanics lies in possessing the degree of freedom of a continuous phase. The phase of a complex number can vary continuously from \(0\) to \(2\pi\), which allows the interference between probability amplitudes of two paths to vary smoothly from constructive to destructive through \(\cos\delta\). This continuous phase degree of freedom enables the smooth interference fringes observed experimentally, phase rotation in time evolution (\(e^{-iEt/\hbar}\)), and non-classical correlations in quantum entanglement. With real numbers, the phase difference is restricted to discrete values of \(0\) or \(\pi\), making it impossible to reproduce the rich predictions of quantum mechanics.

A-2. From the Light Quantum Hypothesis to Stimulated Emission — The Chain of Einstein's Logic

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Problem:

Einstein proposed the light quantum hypothesis \(E = h\nu\) in 1905 and predicted stimulated emission in 1917. These two achievements are not independent but are logically connected. Follow the questions below to trace this chain of reasoning.

(a) Consider a two-level atomic system in thermal equilibrium at temperature \(T\). The ratio of populations of energy levels \(E_1\) (ground state) and \(E_2\) (excited state, \(E_2 > E_1\)) follows the Boltzmann distribution:

\[ \frac{N_2}{N_1} = \exp\!\left(-\frac{E_2 - E_1}{k_B T}\right) \]

where \(k_B\) is the Boltzmann constant. What does \(N_2/N_1\) approach as \(T \to \infty\)? And as \(T \to 0\)? Explain the physical meaning.

(b) When atoms and light (with frequency \(\nu = (E_2 - E_1)/h\)) coexist in thermal equilibrium, atoms repeatedly undergo transitions: absorbing light (\(E_1 \to E_2\)) and emitting light (\(E_2 \to E_1\)). There are two types of emission:

  • Spontaneous emission: An excited atom spontaneously emits a photon. The transition rate per unit time is \(A \cdot N_2\) (\(A\) is a constant).
  • Stimulated emission: Emission of a photon induced by an external photon. The transition rate per unit time is \(B \cdot \rho(\nu) \cdot N_2\) (\(B\) is a constant, \(\rho(\nu)\) is the energy density of light).

The absorption rate is \(B' \cdot \rho(\nu) \cdot N_1\). From the condition that the absorption rate equals the emission rate in thermal equilibrium,

\[ B'\, \rho(\nu)\, N_1 = A\, N_2 + B\, \rho(\nu)\, N_2 \]

express \(\rho(\nu)\) in terms of \(A\), \(B\), \(B'\), \(E_2 - E_1\), and \(k_B T\).

(c) Planck's blackbody radiation formula (studied in detail in Ch. 1) is given by

\[ \rho(\nu) = \frac{8\pi h \nu^3}{c^3} \cdot \frac{1}{e^{h\nu/(k_B T)} - 1} \]

By comparing the result of (b) with this formula, show that \(B' = B\) and express the value of \(A/B\) in terms of \(h\), \(\nu\), and \(c\).

(d) If stimulated emission did not exist (\(B = 0\)), what form would \(\rho(\nu)\) take from the equilibrium condition in (b)? Show that this contradicts Planck's formula, and explain that "stimulated emission is a logical necessity required by thermal equilibrium."

(a) Limits of the Boltzmann Distribution:

\[ \frac{N_2}{N_1} = \exp\!\left(-\frac{E_2 - E_1}{k_B T}\right) \]

When \(T \to \infty\):

\[ \frac{E_2 - E_1}{k_B T} \to 0 \quad \Longrightarrow \quad \frac{N_2}{N_1} \to e^0 = 1 \]
\[ \boxed{T \to \infty: \quad N_2/N_1 \to 1} \]

Physical meaning: In the high-temperature limit, the thermal energy \(k_B T\) is much larger than the energy difference \(E_2 - E_1\), so the ground state and excited state are occupied nearly equally (equipartition).

When \(T \to 0\):

\[ \frac{E_2 - E_1}{k_B T} \to +\infty \quad \Longrightarrow \quad \frac{N_2}{N_1} \to e^{-\infty} = 0 \]
\[ \boxed{T \to 0: \quad N_2/N_1 \to 0} \]

Physical meaning: At absolute zero, all atoms fall into the lowest energy state (ground state), and the population of the excited state becomes zero.

(b) Deriving \(\rho(\nu)\) from the Equilibrium Condition:

Thermal equilibrium condition:

\[ B'\rho(\nu) N_1 = A N_2 + B\rho(\nu) N_2 \]

Solving for \(\rho(\nu)\):

\[ B'\rho(\nu) N_1 - B\rho(\nu) N_2 = A N_2 \]
\[ \rho(\nu)(B' N_1 - B N_2) = A N_2 \]
\[ \rho(\nu) = \frac{A N_2}{B' N_1 - B N_2} = \frac{A}{B'(N_1/N_2) - B} \]

Substituting the Boltzmann distribution \(N_1/N_2 = \exp\!\left(\frac{E_2 - E_1}{k_B T}\right)\). Since \(E_2 - E_1 = h\nu\):

\[ \boxed{\rho(\nu) = \frac{A}{B'\, e^{h\nu/(k_B T)} - B}} \]

(c) Comparison with Planck's Formula:

Planck's blackbody radiation formula:

\[ \rho(\nu) = \frac{8\pi h\nu^3}{c^3} \cdot \frac{1}{e^{h\nu/(k_B T)} - 1} \]

Comparing with the result from (b). For both to agree at all temperatures \(T\), the structure of the denominators must match.

The denominator of the result from (b) is \(B' e^{h\nu/(k_BT)} - B\), while the denominator of Planck's formula (rewritten in a form divided by \(A\)) corresponds to \((A \cdot c^3/(8\pi h\nu^3))(e^{h\nu/(k_BT)} - 1)\).

For direct comparison, rewrite (b) as:

\[ \rho(\nu) = \frac{A/B}{(B'/B)\, e^{h\nu/(k_B T)} - 1} \]

Planck's formula:

\[ \rho(\nu) = \frac{8\pi h\nu^3/c^3}{e^{h\nu/(k_B T)} - 1} \]

For both to agree at all \(T\):

  1. The coefficients of \(e^{h\nu/(k_BT)}\) in the denominators must match: \(B'/B = 1\), i.e.,
\[ \boxed{B' = B} \]
  1. The numerators must match:
\[ \frac{A}{B} = \frac{8\pi h\nu^3}{c^3} \]
\[ \boxed{\frac{A}{B} = \frac{8\pi h\nu^3}{c^3}} \]

Verification (the \(T \to \infty\) limit): When \(T \to \infty\), \(e^{h\nu/(k_BT)} \approx 1 + h\nu/(k_BT)\), so:

Result from (b): \(\rho(\nu) \approx \frac{A}{B'(1 + h\nu/(k_BT)) - B} = \frac{A}{(B'-B) + B' h\nu/(k_BT)}\)

With \(B' = B\): \(\rho(\nu) \approx \frac{A}{B \cdot h\nu/(k_BT)} = \frac{A k_B T}{B h\nu}\)

Planck's formula: \(\rho(\nu) \approx \frac{8\pi h\nu^3}{c^3} \cdot \frac{k_BT}{h\nu} = \frac{8\pi\nu^2 k_BT}{c^3}\)

This is the Rayleigh-Jeans law, consistent with \(A/B = 8\pi h\nu^3/c^3\). ✓

(d) Case Without Stimulated Emission (\(B = 0\)):

Setting \(B = 0\), the equilibrium condition becomes:

\[ B'\rho(\nu) N_1 = A N_2 \]
\[ \rho(\nu) = \frac{A N_2}{B' N_1} = \frac{A}{B'} \cdot \frac{N_2}{N_1} = \frac{A}{B'}\, e^{-h\nu/(k_BT)} \]
\[ \rho(\nu) = \frac{A}{B'}\, e^{-h\nu/(k_B T)} \]

This has the form of Wien's radiation law and contradicts Planck's formula:

\[ \rho(\nu) = \frac{8\pi h\nu^3}{c^3} \cdot \frac{1}{e^{h\nu/(k_BT)} - 1} \]

Specifically:

  1. The \(\nu^3\) factor is missing. In the \(B = 0\) result, the frequency dependence of \(\rho(\nu)\) is only \(e^{-h\nu/(k_BT)}\), lacking the \(\nu^3\) prefactor contained in Planck's formula (assuming \(A/B'\) is independent of \(\nu\)).

  2. Disagreement in the low-frequency (high-temperature) limit. When \(h\nu \ll k_BT\), the \(B = 0\) result gives \(\rho(\nu) \approx A/B'\) (a constant), whereas Planck's formula (and the experimentally confirmed Rayleigh-Jeans law) gives \(\rho(\nu) \propto \nu^2 T\).

  3. The structure of the denominator differs. The denominator of Planck's formula is \(e^{h\nu/(k_BT)} - 1\), whereas the \(B = 0\) case yields a simple exponential \(e^{-h\nu/(k_BT)}\), with the \(-1\) term absent.

Therefore, stimulated emission (\(B \neq 0\)) is logically necessary from the requirement of thermal equilibrium. Since Planck's blackbody radiation formula is experimentally correct, thermal equilibrium cannot be achieved with absorption and spontaneous emission alone; without stimulated emission, detailed balance between atoms and the radiation field cannot hold. Einstein deductively derived the existence of stimulated emission as a new physical process solely from the requirement of thermodynamic consistency.

Verification: The \(B = 0\) result \(\rho \propto e^{-h\nu/(k_BT)}\) corresponds to Wien's radiation law, which is known to be a good approximation to Planck's formula only at high frequencies (\(h\nu \gg k_BT\)). It breaks down at low frequencies, so \(B = 0\) cannot hold in general. ✓