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Chapter 1 • Theory & Derivations

Physical Basis & Heisenberg Uncertainty Principle

Failures of classical mechanics, emergence of quantization, de Broglie wave-particle duality, and the uncertainty principle.

§1.1The Crisis of Classical Physics and Empirical Anomalies

At the close of the nineteenth century, Newtonian classical mechanics, combined with Maxwell's unified electrodynamics and Boltzmann-Gibbs statistical thermodynamics, was widely considered the complete and final foundation of the physical sciences. However, when applied to microscopic phenomena on atomic and subatomic scales, the classical framework yielded catastrophic contradictions with experimental reality.

1. Blackbody Radiation and the Ultraviolet Catastrophe

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. In thermodynamic equilibrium at absolute temperature $T$, it emits electromagnetic cavity radiation whose spectral energy density is denoted by $u(\nu, T) d\nu$.

Using classical statistical mechanics, the equipartition theorem assigns an average thermal energy of $\langle E \rangle = k_B T$ to each electromagnetic standing wave mode within the cavity. Calculating the number of spatial cavity modes per unit volume in the frequency interval $[\nu, \nu + d\nu]$:

$$g(\nu) d\nu = \frac{8\pi \nu^2}{c^3} d\nu$$

Multiplying $g(\nu)$ by the classical average energy $\langle E \rangle = k_B T$ yields the classical Rayleigh-Jeans Radiation Formula:

$$u_{\text{RJ}}(\nu, T) d\nu = \frac{8\pi \nu^2}{c^3} k_B T d\nu$$

As the frequency approaches the ultraviolet and beyond ($\nu \to \infty$), $u_{\text{RJ}}(\nu, T) \to \infty$. Consequently, the integrated total energy density radiated by a cavity at any non-zero temperature diverges:

$$U = \int_0^\infty u_{\text{RJ}}(\nu, T) d\nu = \frac{8\pi k_B T}{c^3} \int_0^\infty \nu^2 d\nu = \infty$$

This absurd prediction—that an oven or glowing iron bar would radiate infinite energy at high frequencies—is known as the Ultraviolet Catastrophe.

In December 1900, Max Planck resolved this crisis by introducing a revolutionary quantum postulate: the atomic oscillators in the cavity walls cannot absorb or emit energy continuously; rather, energy exchange occurs exclusively in discrete packets, or quanta, proportional to the oscillation frequency:

$$E_n = n h \nu, \quad n \in \{0, 1, 2, 3, \dots\}$$

where $h = 6.62607015 \times 10^{-34} \text{ J}\cdot\text{s}$ is Planck's constant.

Applying Maxwell-Boltzmann statistics, the thermal average energy of a quantum oscillator is given by:

$$\langle E \rangle = \frac{\sum_{n=0}^\infty n h \nu e^{-n h \nu / k_B T}}{\sum_{n=0}^\infty e^{-n h \nu / k_B T}} = \frac{h\nu}{e^{h\nu / k_B T} - 1}$$

Substituting this quantum average into the density of modes yields Planck's Radiation Law:

$$u(\nu, T) d\nu = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu / k_B T} - 1} d\nu$$

In the low-frequency limit ($h\nu \ll k_B T$), expanding the exponential $e^{h\nu / k_B T} \approx 1 + \frac{h\nu}{k_B T}$ recovers the Rayleigh-Jeans formula. In the high-frequency limit ($h\nu \gg k_B T$), the exponential denominator dominates, cutting off emission and preventing any ultraviolet divergence.

2. The Photoelectric Effect

In 1887, Heinrich Hertz discovered that ultraviolet light incident on metallic surfaces ejects electrons. Classical wave theory predicted that: 1. The kinetic energy of ejected electrons should increase with light intensity (the electric field amplitude). 2. For very low light intensities, there should be a measurable time delay while electrons accumulate sufficient energy to escape the metallic surface.

Experiments conducted by Philipp Lenard (1902) directly refuted both classical predictions: 1. The maximum kinetic energy $K_{\text{max}}$ of ejected photoelectrons is strictly independent of light intensity and depends linearly solely on the frequency $\nu$. 2. Emission occurs quasi-instantaneously (within $10^{-9}$ seconds), even at exceptionally low light intensities. 3. Below a characteristic threshold frequency $\nu_0$ (dependent on the specific metal), no electrons are emitted regardless of light intensity.

In 1905, Albert Einstein extended Planck's concept, proposing that electromagnetic radiation is not merely emitted in quanta, but propagates and interacts as localized particle-like energy packets called photons, each with energy:

$$E = h\nu = \hbar\omega$$

where $\hbar = \frac{h}{2\pi} = 1.0545718 \times 10^{-34} \text{ J}\cdot\text{s}$. An incoming photon transfers its entire energy to a single conduction electron. If this energy exceeds the binding work function $\Phi$ of the metal, the electron escapes with kinetic energy governed by Einstein's Photoelectric Equation:

$$K_{\text{max}} = h\nu - \Phi = h(\nu - \nu_0)$$

3. Compton Scattering

In 1923, Arthur Compton directed monochromatic X-rays of wavelength $\lambda$ at a graphite target and observed that the scattered radiation contained a shifted wavelength component $\lambda' > \lambda$.

Treating the collision between an incident photon (energy $E=h\nu$, relativistic momentum $p=h/\lambda$) and a stationary target electron ($m_e$) as an elastic relativistic two-body collision: * Conservation of relativistic energy: $h\nu + m_e c^2 = h\nu' + \sqrt{p_e^2 c^2 + m_e^2 c^4}$ * Conservation of momentum: $\mathbf{p}_\gamma = \mathbf{p}'_\gamma + \mathbf{p}_e$

Solving the relativistic conservation laws yields the Compton Scattering Formula:

$$\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos\theta) = \lambda_C (1 - \cos\theta)$$

where $\lambda_C = \frac{h}{m_e c} \approx 0.02426 \text{ Å} = 2.426 \times 10^{-12} \text{ m}$ is the Compton wavelength of the electron, and $\theta$ is the scattering angle. This demonstrated that photons carry localized momentum $\mathbf{p} = \hbar \mathbf{k}$.

§1.2The Bohr Atom and the Old Quantum Theory

Rutherford Nuclear Model Instability

Ernest Rutherford's 1911 alpha-scattering experiments proved that an atom consists of a tiny, massive, positively charged nucleus surrounded by electrons. However, classical electrodynamics states that an orbiting electron experiences continuous centripetal acceleration $a = v^2/r$. According to Larmor's radiation formula, an accelerated charge radiates electromagnetic power: $$P = \frac{e^2 a^2}{6\pi \epsilon_0 c^3}$$

As the electron loses mechanical orbital energy, its orbital radius must shrink continuously, causing the electron to spiral into the nucleus within an estimated lifetime of $\tau \approx 10^{-11} \text{ seconds}$. Classical mechanics could not explain why atoms exist as stable structures.

Bohr Postulates (1913)

Niels Bohr resolved atomic instability for the hydrogen atom ($Z=1$) by introducing three non-classical postulates: 1. Stationary States: The electron moves in discrete circular orbits without radiating electromagnetic energy. 2. Quantization of Orbital Angular Momentum: The electron's orbital angular momentum $L = m_e v r$ is restricted to integral multiples of $\hbar$: $$L = m_e v_n r_n = n\hbar, \quad n \in \{1, 2, 3, \dots\}$$ 3. Bohr Frequency Condition: Radiation is emitted or absorbed only when an electron undergoes a discrete transition between stationary orbits $E_i$ and $E_f$: $$\Delta E = E_i - E_f = h\nu = \hbar\omega$$

Derivation of Quantized Radii and Energy Levels

Balancing Coulomb attraction with centripetal force: $$\frac{m_e v_n^2}{r_n} = \frac{e^2}{4\pi \epsilon_0 r_n^2} \implies m_e v_n^2 r_n = \frac{e^2}{4\pi \epsilon_0}$$

Combining this with the quantization condition $v_n = \frac{n\hbar}{m_e r_n}$:

$$r_n = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} n^2 = n^2 a_0$$

where $a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \approx 0.529177 \times 10^{-10} \text{ m} = 0.529 \text{ Å}$ is the Bohr radius.

The total mechanical energy $E_n = K_n + U_n = \frac{1}{2}m_e v_n^2 - \frac{e^2}{4\pi\epsilon_0 r_n} = -\frac{e^2}{8\pi\epsilon_0 r_n}$:

$$E_n = -\frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$$

Transitioning between states $n_2 \to n_1$ yields the Rydberg formula:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \quad R_H = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c} \approx 1.09737 \times 10^7 \text{ m}^{-1}$$

While the Bohr-Sommerfeld model successfully explained the Balmer and Lyman spectral series of hydrogen, it was fundamentally limited: it could not explain the spectra of multi-electron atoms (even Helium), chemical bonding, transition rates, or the Zeeman effect without ad-hoc rules.

§1.3de Broglie Hypothesis and Matter Waves

In 1924, Louis de Broglie hypothesized that nature possesses deep physical symmetry: if electromagnetic waves exhibit particle properties (photons with momentum $p = h/\lambda$), then material particles with mass $m$ and momentum $p$ must simultaneously exhibit wave properties.

The de Broglie Relations

For any physical entity with energy $E$ and momentum $\mathbf{p}$: $$\lambda = \frac{h}{p} = \frac{h}{mv}, \qquad \mathbf{p} = \hbar \mathbf{k}, \qquad E = \hbar \omega$$

where $\mathbf{k}$ is the wave vector ($|\mathbf{k}| = 2\pi/\lambda$) and $\omega$ is the angular frequency.

Bohr's Orbit as a Standing Wave

de Broglie provided an intuitive physical basis for Bohr's angular momentum quantization: an electron orbit is stable because it forms a standing de Broglie wave around the nucleus. Constructive interference requires an integral number of wavelengths around the circular circumference: $$2\pi r_n = n \lambda = n \left( \frac{h}{m_e v_n} \right) \implies m_e v_n r_n = n\hbar$$

Experimental Confirmation: The Davisson-Germer Experiment (1927)

Clinton Davisson and Lester Germer fired low-energy electrons ($54 \text{ eV}$) at a crystalline nickel target and recorded an intense diffraction peak at a scattering angle of $\theta = 50^{\circ}$.

Using Bragg's law for crystal diffraction ($n\lambda = 2d \sin\phi$): * Crystal plane spacing $d = 0.091 \text{ nm}$, glancing angle $\phi = 65^{\circ}$ * Measured Bragg wavelength: $\lambda_{\text{exp}} = 2(0.091 \text{ nm})\sin(65^{\circ}) = 0.165 \text{ nm}$. * Theoretical de Broglie calculation for a $54\text{ eV}$ electron: $$\lambda_{\text{theory}} = \frac{h}{\sqrt{2 m_e E}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 (9.109 \times 10^{-31}) (54 \times 1.602 \times 10^{-19})}} = 0.167 \text{ nm}$$

The theoretical and experimental values matched within experimental error, providing definitive proof of matter waves.

Explore this directly in Simulation 1.1 below, where individual electron hits build an interference pattern on a phosphor screen, demonstrating wave-particle duality.

§1.4The Heisenberg Uncertainty Principle

Mathematical Origin: Fourier Conjugacy

In classical mechanics, a particle has simultaneously well-defined position $x(t)$ and momentum $p(t)$. In quantum mechanics, a localized spatial entity is described by a wave packet formed by superposing monochromatic plane waves: $$\psi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{i k x} dk$$

The spatial wavefunction $\psi(x)$ and the wave-number distribution $\phi(k)$ form a Fourier transform pair. From the fundamental bandwidth theorem of harmonic analysis, any wave packet has a spatial spread $\Delta x$ and a wavenumber spread $\Delta k$ bounded by:

$$\Delta x \cdot \Delta k \ge \frac{1}{2}$$

Substituting the de Broglie relation $p = \hbar k \implies \Delta p = \hbar \Delta k$ yields the Heisenberg Uncertainty Principle:

$$\Delta x \cdot \Delta p_x \ge \frac{\hbar}{2}$$

Physical Meaning and Gaussian Minimum Uncertainty

Here, the uncertainties are formally defined as the statistical standard deviations (root-mean-square variances) of the observables: $$\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}, \qquad \Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$$

The lower bound $\frac{\hbar}{2}$ is saturated exclusively by a Gaussian wave packet:

$$\psi_G(x) = \left( \frac{1}{2\pi \sigma^2} \right)^{1/4} e^{-\frac{x^2}{4\sigma^2}} e^{i k_0 x} \implies \Delta x = \sigma, \quad \Delta p = \frac{\hbar}{2\sigma} \implies \Delta x \cdot \Delta p = \frac{\hbar}{2}$$

Any other wave packet shape has an uncertainty product strictly greater than $\hbar/2$.

Energy-Time Uncertainty Relation

A complementary uncertainty relation connects energy and time: $$\Delta E \cdot \Delta t \ge \frac{\hbar}{2}$$

Where $\Delta t$ represents the characteristic time interval over which an expectation value changes appreciably: $\Delta t = \frac{\Delta A}{|d\langle A \rangle / dt|}$. This explains why unstable excited states with short lifetimes $\tau$ display an intrinsic spectral energy line width:

$$\Gamma \approx \frac{\hbar}{\tau}$$

Test this principle in Simulation 1.2 below: adjust the spatial width $\Delta x$ and observe the conjugate momentum distribution $\Delta p$ widen in response.

§1.2The Bohr Atom and the Old Quantum Theory

Ernest Rutherford's 1911 alpha-scattering experiments proved that an atom consists of a tiny, massive, positively charged nucleus surrounded by electrons. However, classical electrodynamics states that an orbiting electron experiences continuous centripetal acceleration $a = v^2/r$. According to Larmor's radiation formula, an accelerated charge radiates electromagnetic power:

$$P = \frac{e^2 a^2}{6\pi \epsilon_0 c^3}$$

As the electron loses mechanical orbital energy, its orbital radius must shrink continuously, causing the electron to spiral into the nucleus within an estimated lifetime of $\tau \approx 10^{-11} \text{ seconds}$. Classical mechanics could not explain why atoms exist as stable structures.

Niels Bohr resolved atomic instability for the hydrogen atom ($Z=1$) by introducing three non-classical postulates: 1. The electron moves in discrete circular orbits without radiating electromagnetic energy. 2. The electron's orbital angular momentum $L = m_e v r$ is restricted to integral multiples of $\hbar$: $$L = m_e v_n r_n = n\hbar, \quad n \in \{1, 2, 3, \dots\}$$ 3. Radiation is emitted or absorbed only when an electron undergoes a discrete transition between stationary orbits $E_i$ and $E_f$: $$\Delta E = E_i - E_f = h
u = \hbar\omega$$

Balancing Coulomb attraction with centripetal force:

$$\frac{m_e v_n^2}{r_n} = \frac{e^2}{4\pi \epsilon_0 r_n^2} \implies m_e v_n^2 r_n = \frac{e^2}{4\pi \epsilon_0}$$

Combining this with the quantization condition $v_n = \frac{n\hbar}{m_e r_n}$:

$$r_n = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} n^2 = n^2 a_0$$

where $a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \approx 0.529177 \times 10^{-10} \text{ m} = 0.529 \text{ Å}$ is the .

The total mechanical energy $E_n = K_n + U_n = \frac{1}{2}m_e v_n^2 - \frac{e^2}{4\pi\epsilon_0 r_n} = -\frac{e^2}{8\pi\epsilon_0 r_n}$:

$$E_n = -\frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$$

Transitioning between states $n_2 \to n_1$ yields the Rydberg formula:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \quad R_H = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c} \approx 1.09737 \times 10^7 \text{ m}^{-1}$$

While the Bohr-Sommerfeld model successfully explained the Balmer and Lyman spectral series of hydrogen, it was fundamentally limited: it could not explain the spectra of multi-electron atoms (even Helium), chemical bonding, transition rates, or the Zeeman effect without ad-hoc rules.

§1.3de Broglie Hypothesis and Matter Waves

In 1924, Louis de Broglie hypothesized that nature possesses deep physical symmetry: if electromagnetic waves exhibit particle properties (photons with momentum $p = h/\lambda$), then material particles with mass $m$ and momentum $p$ must simultaneously exhibit wave properties.

For any physical entity with energy $E$ and momentum $\mathbf{p}$:

$$\lambda = \frac{h}{p} = \frac{h}{mv}, \qquad \mathbf{p} = \hbar \mathbf{k}, \qquad E = \hbar \omega$$

where $\mathbf{k}$ is the wave vector ($|\mathbf{k}| = 2\pi/\lambda$) and $\omega$ is the angular frequency.

de Broglie provided an intuitive physical basis for Bohr's angular momentum quantization: an electron orbit is stable because it forms a around the nucleus. Constructive interference requires an integral number of wavelengths around the circular circumference:

$$2\pi r_n = n \lambda = n \left( \frac{h}{m_e v_n} \right) \implies m_e v_n r_n = n\hbar$$

Clinton Davisson and Lester Germer fired low-energy electrons ($54 \text{ eV}$) at a crystalline nickel target and recorded an intense diffraction peak at a scattering angle of $\theta = 50^{\circ}$.

Using Bragg's law for crystal diffraction ($n\lambda = 2d \sin\phi$):
* Crystal plane spacing $d = 0.091 \text{ nm}$, glancing angle $\phi = 65^{\circ}$
* Measured Bragg wavelength: $\lambda_{\text{exp}} = 2(0.091 \text{ nm})\sin(65^{\circ}) = 0.165 \text{ nm}$.
* Theoretical de Broglie calculation for a $54\text{ eV}$ electron:
$$\lambda_{\text{theory}} = \frac{h}{\sqrt{2 m_e E}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 (9.109 \times 10^{-31}) (54 \times 1.602 \times 10^{-19})}} = 0.167 \text{ nm}$$

The theoretical and experimental values matched within experimental error, providing definitive proof of matter waves.

Explore this directly in below, where individual electron hits build an interference pattern on a phosphor screen, demonstrating wave-particle duality.

§1.4The Heisenberg Uncertainty Principle

In classical mechanics, a particle has simultaneously well-defined position $x(t)$ and momentum $p(t)$. In quantum mechanics, a localized spatial entity is described by a wave packet formed by superposing monochromatic plane waves:

$$\psi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{i k x} dk$$

The spatial wavefunction $\psi(x)$ and the wave-number distribution $\phi(k)$ form a . From the fundamental bandwidth theorem of harmonic analysis, any wave packet has a spatial spread $\Delta x$ and a wavenumber spread $\Delta k$ bounded by:

$$\Delta x \cdot \Delta k \ge \frac{1}{2}$$

Substituting the de Broglie relation $p = \hbar k \implies \Delta p = \hbar \Delta k$ yields the :

$$\Delta x \cdot \Delta p_x \ge \frac{\hbar}{2}$$

Here, the uncertainties are formally defined as the statistical standard deviations (root-mean-square variances) of the observables:

$$\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}, \qquad \Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$$

The lower bound $\frac{\hbar}{2}$ is saturated exclusively by a :

$$\psi_G(x) = \left( \frac{1}{2\pi \sigma^2} \right)^{1/4} e^{-\frac{x^2}{4\sigma^2}} e^{i k_0 x} \implies \Delta x = \sigma, \quad \Delta p = \frac{\hbar}{2\sigma} \implies \Delta x \cdot \Delta p = \frac{\hbar}{2}$$

Any other wave packet shape has an uncertainty product strictly greater than $\hbar/2$.

A complementary uncertainty relation connects energy and time:

$$\Delta E \cdot \Delta t \ge \frac{\hbar}{2}$$

Where $\Delta t$ represents the characteristic time interval over which an expectation value changes appreciably: $\Delta t = \frac{\Delta A}{|d\langle A \rangle / dt|}$. This explains why unstable excited states with short lifetimes $\tau$ display an intrinsic spectral energy line width:

$$\Gamma \approx \frac{\hbar}{\tau}$$

Test this principle in below: adjust the spatial width $\Delta x$ and observe the conjugate momentum distribution $\Delta p$ widen in response.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
EasyExample 1.1: de Broglie Wavelength of Relativistic vs Non-Relativistic Electrons
An electron is accelerated from rest through an electrostatic potential difference of $V = 150 \text{ V}$.\n(a) Determine its de Broglie wavelength using non-relativistic mechanics.\n(b) At what accelerating potential does the relativistic correction to the de Broglie wavelength exceed $1\%$?
Step 1: Non-relativistic Calculation
$$K = e V = 150 \text{ eV} = 150 \times 1.602 \times 10^{-19} \text{ J} = 2.403 \times 10^{-17} \text{ J} $$ $$p = \sqrt{2 m_e K} = \sqrt{2(9.109 \times 10^{-31} \text{ kg})(2.403 \times 10^{-17} \text{ J})} = 6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s} $$ $$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.001 \times 10^{-10} \text{ m} = 1.001 \text{ Å}$$

For quick calculations, note that $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$. For $V = 150 \text{ V}$, $\lambda = 1.00 \text{ Å}$, which corresponds to typical atomic crystal lattice spacings.

Step 2: Relativistic Condition
$$E^2 = p^2 c^2 + m_0^2 c^4 \implies p = \frac{1}{c}\sqrt{K(K + 2m_0 c^2)} $$ $$\lambda_{\text{rel}} = \frac{h c}{\sqrt{K(K + 2m_0 c^2)}} = \frac{\lambda_{\text{class}}}{\sqrt{1 + \frac{K}{2m_0 c^2}}} \approx \lambda_{\text{class}}\left(1 - \frac{K}{4 m_0 c^2}\right) $$ $$\frac{\Delta \lambda}{\lambda} \approx \frac{K}{4 m_0 c^2} \ge 0.01 \implies K \ge 0.04 m_0 c^2 = 0.04 (511 \text{ keV}) \approx 20.44 \text{ keV}$$

When accelerating potentials exceed roughly $20 \text{ kV}$ (typical in transmission electron microscopes), relativistic momentum corrections become necessary.

MediumExample 1.2: Rigorous Proof of Ground-State Energy of Hydrogen via Uncertainty Principle
Using the Heisenberg uncertainty relation $\Delta x \cdot \Delta p \ge \hbar/2$, derive an order-of-magnitude estimate for the ground-state radius (Bohr radius) and binding energy of the hydrogen atom without solving the Schrödinger equation.
Step 1: Express total energy in terms of uncertainty
$$E = K + U = \frac{p^2}{2m_e} - \frac{e^2}{4\pi \epsilon_0 r} $$ $$\text{Setting } r \approx \Delta x \text{ and } p \approx \Delta p \ge \frac{\hbar}{2r} \implies p \approx \frac{\hbar}{r} $$ $$E(r) = \frac{\hbar^2}{2 m_e r^2} - \frac{e^2}{4\pi \epsilon_0 r}$$

The kinetic energy term scales as $1/r^2$ due to quantum confinement (confinement increases momentum spread), while the attractive Coulomb potential scales as $-1/r$. The competition between these two terms creates a stable ground state.

Step 2: Minimize E(r) with respect to r
$$\frac{dE}{dr} = -\frac{\hbar^2}{m_e r^3} + \frac{e^2}{4\pi \epsilon_0 r^2} = 0 $$ $$\frac{\hbar^2}{m_e r^3} = \frac{e^2}{4\pi \epsilon_0 r^2} \implies r_0 = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} = a_0 \approx 0.529 \text{ Å} $$ $$E(r_0) = \frac{\hbar^2}{2 m_e a_0^2} - \frac{e^2}{4\pi \epsilon_0 a_0} = -\frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} = -13.6 \text{ eV}$$

This shows that atomic stability is a direct consequence of the Heisenberg uncertainty principle: if the electron collapsed into the nucleus ($r \to 0$), its kinetic energy would diverge as $+1/r^2$, overwhelming the Coulomb attraction.

An electron is accelerated from rest through an electrostatic potential difference of $V = 150 \text{ V}$.\n(a) Determine its de Broglie wavelength using non-relativistic mechanics.\n(b) At what accelerating potential does the relativistic correction to the de Broglie wavelength exceed $1\%$?
Step 1: Non-relativistic Calculation
$$K = e V = 150 \text{ eV} = 150 \times 1.602 \times 10^{-19} \text{ J} = 2.403 \times 10^{-17} \text{ J} $$ $$p = \sqrt{2 m_e K} = \sqrt{2(9.109 \times 10^{-31} \text{ kg})(2.403 \times 10^{-17} \text{ J})} = 6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s} $$ $$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.001 \times 10^{-10} \text{ m} = 1.001 \text{ Å}$$
For quick calculations, note that $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$. For $V = 150 \text{ V}$, $\lambda = 1.00 \text{ Å}$, which corresponds to typical atomic crystal lattice spacings.
Step 2: Relativistic Condition
$$E^2 = p^2 c^2 + m_0^2 c^4 \implies p = \frac{1}{c}\sqrt{K(K + 2m_0 c^2)} $$ $$\lambda_{\text{rel}} = \frac{h c}{\sqrt{K(K + 2m_0 c^2)}} = \frac{\lambda_{\text{class}}}{\sqrt{1 + \frac{K}{2m_0 c^2}}} \approx \lambda_{\text{class}}\left(1 - \frac{K}{4 m_0 c^2}\right) $$ $$\frac{\Delta \lambda}{\lambda} \approx \frac{K}{4 m_0 c^2} \ge 0.01 \implies K \ge 0.04 m_0 c^2 = 0.04 (511 \text{ keV}) \approx 20.44 \text{ keV}$$
When accelerating potentials exceed roughly $20 \text{ kV}$ (typical in transmission electron microscopes), relativistic momentum corrections become necessary.
MediumExample 1.2: Rigorous Proof of Ground-State Energy of Hydrogen via Uncertainty Principle
Using the Heisenberg uncertainty relation $\Delta x \cdot \Delta p \ge \hbar/2$, derive an order-of-magnitude estimate for the ground-state radius (Bohr radius) and binding energy of the hydrogen atom without solving the Schrödinger equation.
Step 1: Express total energy in terms of uncertainty
$$E = K + U = \frac{p^2}{2m_e} - \frac{e^2}{4\pi \epsilon_0 r} $$ $$\text{Setting } r \approx \Delta x \text{ and } p \approx \Delta p \ge \frac{\hbar}{2r} \implies p \approx \frac{\hbar}{r} $$ $$E(r) = \frac{\hbar^2}{2 m_e r^2} - \frac{e^2}{4\pi \epsilon_0 r}$$
The kinetic energy term scales as $1/r^2$ due to quantum confinement (confinement increases momentum spread), while the attractive Coulomb potential scales as $-1/r$. The competition between these two terms creates a stable ground state.
Step 2: Minimize E(r) with respect to r
$$\frac{dE}{dr} = -\frac{\hbar^2}{m_e r^3} + \frac{e^2}{4\pi \epsilon_0 r^2} = 0 $$ $$\frac{\hbar^2}{m_e r^3} = \frac{e^2}{4\pi \epsilon_0 r^2} \implies r_0 = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} = a_0 \approx 0.529 \text{ Å} $$ $$E(r_0) = \frac{\hbar^2}{2 m_e a_0^2} - \frac{e^2}{4\pi \epsilon_0 a_0} = -\frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} = -13.6 \text{ eV}$$
This shows that atomic stability is a direct consequence of the Heisenberg uncertainty principle: if the electron collapsed into the nucleus ($r \to 0$), its kinetic energy would diverge as $+1/r^2$, overwhelming the Coulomb attraction.