The Wave Nature of Matter: De Broglie Hypothesis, Uncertainty Principle & Quantum Mechanics - HRK Physics

House of Physics October 02, 2025
The Wave Nature of Matter: Complete Quantum Physics Guide | HRK Chapter 50

The Wave Nature of Matter: Complete Quantum Physics Guide

Mastering De Broglie Hypothesis, Davisson-Germer Experiment, Heisenberg Uncertainty Principle, and Schrodinger Equation
Wave Nature of Matter De Broglie Hypothesis Heisenberg Uncertainty Principle Schrodinger Equation Quantum Mechanics Reading Time: 30 min

📋 Table of Contents

📜 Historical Background

The development of the wave nature of matter concept revolutionized physics in the early 20th century:

  • Louis de Broglie (1924): Proposed that all matter has wave-like properties
  • Clinton Davisson & Lester Germer (1927): Experimentally confirmed electron diffraction
  • George Paget Thomson (1927): Independently confirmed electron wave nature
  • Werner Heisenberg (1927): Formulated the uncertainty principle
  • Erwin Schrödinger (1926): Developed the wave equation for quantum systems

These discoveries established the foundation of quantum mechanics, fundamentally changing our understanding of the microscopic world.

Introduction to Wave Nature of Matter

🔬 What is the Wave Nature of Matter?

This chapter presents experimental evidence supporting the claim that matter, long regarded as made up of particles, has an equally convincing wave aspect. Heisenberg's uncertainty principle shows us the limit to which we can extend the concept of "particle" into quantum mechanics.

We then introduce Schrödinger's equation, the fundamental equation of quantum mechanics, which deals with the wave behavior of particles.

💡 Key Insight

Just as light exhibits both wave-like (interference, diffraction) and particle-like (photoelectric effect) properties, matter also exhibits this wave-particle duality. This dual nature is fundamental to understanding quantum mechanics.

Dual Nature of Radiation

🌊 Wave-Particle Duality of Light

Electromagnetic radiations like light, X-rays, etc., can produce the phenomena of interference, diffraction, and polarization due to their wave nature. But under certain circumstances, they can produce photoelectric effect and Compton Effect, which is evidence of their particle nature.

This means that electromagnetic radiations have dual nature; wave as well as particle nature.

Wave Properties Particle Properties
Interference patterns Photoelectric effect
Diffraction phenomena Compton scattering
Polarization effects Blackbody radiation

Dual Nature of Matter

🔍 Matter Waves

In a similar way, particles like electrons, neutrons, and protons must have dual nature. If a beam of electrons accelerated through a known potential difference \( V \) is made to fall on a double slit and after passing through the double slit, they are allowed to strike a fluorescent screen, the pattern obtained on the screen is similar to the interference pattern of light.

Double Slit Experiment with Electrons

⚡ Electron Double-Slit Experiment

Electron Source
↓
DOUBLE SLIT
↓
Fluorescent Screen
Interference Pattern

Experimental Setup: A filament produces a spray of electrons which are accelerated through a potential difference of about 50 kV. After passing through the double slit, the electrons produce a visible interference pattern on a fluorescent screen, which can be photographed.

Significance: This experiment demonstrates that electrons, traditionally considered particles, exhibit wave-like behavior by producing interference patterns similar to those created by light waves.

De Broglie Hypothesis

🌊 De Broglie's Revolutionary Idea

A wave is associated with every moving particle. Such a wave is called a matter wave whose wavelength \( \lambda \) can be found by the expression:

\[ \lambda = \frac{h}{mv} \]

Where \( h \) is Planck's constant, \( m \) is mass, and \( v \) is velocity of the moving object.

📏 Scale Matters

As the value of Planck's constant \( h \) is very small (\( \approx 10^{-34} \, \text{J·s} \)), the wavelength associated with ordinary objects (e.g., a moving tennis ball) is so small that it is difficult to observe. But for small objects like electrons and neutrons, the wave behavior of particles is dominant.

Sample Problem: De Broglie Wavelength Calculation

Calculate the de Broglie wavelength of (a) a virus particle of mass \( 1.0 \times 10^{-15} \) kg moving at the speed of 2 mm/s, and (b) an electron whose kinetic energy is 120 eV.

(a) For virus particle:
\[ m = 1.0 \times 10^{-15} \, \text{kg} \]
\[ v = 2 \, \text{mm/s} = 2 \times 10^{-3} \, \text{m/s} \]
\[ \lambda = \frac{h}{mv} \]
\[ = \frac{6.63 \times 10^{-34}}{(10^{-15} \times 2 \times 10^{-3})} \]
\[ = 3.3 \times 10^{-16} \, \text{m} \]
(b) For electron:
\[ K.E = 120 \, \text{eV} = 120 \times 1.6 \times 10^{-19} \, \text{J} = 192 \times 10^{-19} \, \text{J} \]
\[ \frac{1}{2}mv^2 = 192 \times 10^{-19} \, \text{J} \]
\[ v^2 = \frac{2 \times 192 \times 10^{-19}}{9.1 \times 10^{-31}} \]
\[ v = \sqrt{\frac{2 \times 192 \times 10^{-19}}{9.1 \times 10^{-31}}} \]
\[ = 6.5 \times 10^6 \, \text{m/s} \]
\[ \lambda = \frac{h}{mv} \]
\[ = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31} \times 6.5 \times 10^6)} \]
\[ = 1.12 \times 10^{-10} \, \text{m} \]

Davisson-Germer Experiment

🔬 Experimental Confirmation

The Davisson-Germer experiment (1927) provided the first experimental evidence for the wave nature of electrons. In this experiment, electrons accelerated through a potential difference \( V \) were made to fall on a nickel crystal.

The scattered electrons showed strong scattering at particular angles, similar to X-ray diffraction patterns, confirming the wave nature of electrons.

⚡ Davisson-Germer Experimental Setup

Electron Gun
↓ Electron Beam
Nickel Crystal
↗ Scattered Electrons
Detector

Experimental Details:

  • Electrons are accelerated through a potential difference of 54 V
  • They strike a nickel crystal at normal incidence
  • Scattered electrons are detected at various angles
  • A sharp peak is observed at scattering angle of 50°

Results: The observed diffraction pattern matched the predictions of de Broglie's hypothesis, providing direct experimental evidence for matter waves.

Sample Problem: Davisson-Germer Analysis

In Davisson-Germer experiment, electrons are accelerated through 54 V and are scattered from a nickel crystal. The first diffraction maximum occurs at scattering angle of 50°. Find the interplanar spacing of the crystal.

Given:
\[ V = 54 \, \text{V} \]
\[ \theta = 50^\circ \]
De Broglie wavelength:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]
\[ = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 54}} \]
\[ = 1.67 \times 10^{-10} \, \text{m} \]
From Bragg's law:
\[ 2d \sin\theta = n\lambda \]
\[ 2d \sin(50^\circ) = 1 \times 1.67 \times 10^{-10} \]
\[ d = \frac{1.67 \times 10^{-10}}{2 \times 0.766} \]
\[ = 1.09 \times 10^{-10} \, \text{m} \]

G.P. Thomson Experiment

🔬 Complementary Evidence

G.P. Thomson (son of J.J. Thomson) independently confirmed the wave nature of electrons by passing high-energy electrons through thin metal foils and observing diffraction rings on a photographic plate.

Interestingly, J.J. Thomson won the Nobel Prize for showing that electrons are particles, while his son G.P. Thomson won the Nobel Prize for showing that electrons are waves.

Waves and Particles

🌊 Wave-Particle Duality

Both radiation and matter have dual nature. They behave as waves in some experiments and as particles in others. The wave nature is described by wavelength \( \lambda \) and frequency \( \nu \), while the particle nature is described by energy \( E \) and momentum \( p \).

These are connected by:

\[ E = h\nu \]
and
\[ p = \frac{h}{\lambda} \]

Localizing Waves in Space and Time

📏 Wave Packets

A pure sinusoidal wave extends throughout space and cannot represent a localized particle. To localize a wave in space, we must form a wave packet by superposing waves of different wavelengths.

The more localized the wave packet, the broader the range of wavelengths needed. This leads to the uncertainty principle.

🧮 Wave Packet Mathematics

Step 1: Superposition of Waves

Consider a wave packet formed by superposing waves with wave numbers in the range \( k_0 \pm \Delta k \):

\[ \Psi(x,t) = \int_{k_0 - \Delta k}^{k_0 + \Delta k} A(k) e^{i(kx - \omega t)} dk \]

Step 2: Spatial Localization

The spatial extent \( \Delta x \) of the wave packet is related to the spread in wave numbers:

\[ \Delta x \cdot \Delta k \approx 1 \]

Step 3: Momentum Relation

Since \( p = \hbar k \), where \( \hbar = h/2\pi \):

\[ \Delta p = \hbar \Delta k \]
\[ \Delta x \cdot \Delta p \approx \hbar \]

Heisenberg's Uncertainty Principle

⚖️ Fundamental Quantum Limit

Heisenberg's uncertainty principle states that it is impossible to simultaneously measure both the position and momentum of a particle with arbitrary precision. The product of the uncertainties has a lower limit:

\[ \Delta x \cdot \Delta p_x \geq \frac{\hbar}{2} \]

where \( \hbar = h/2\pi \) is the reduced Planck's constant.

Derivation from Wave Localization

🧮 Uncertainty Principle Derivation

Step 1: Wave Packet Formation

A localized particle is represented by a wave packet formed by superposing waves with different wavelengths. The spatial extent \( \Delta x \) is related to the spread in wave numbers \( \Delta k \):

\[ \Delta x \cdot \Delta k \approx 1 \]

Step 2: Momentum Relation

Using de Broglie relation \( p = \hbar k \), where \( \hbar = h/2\pi \):

\[ \Delta p = \hbar \Delta k \]

Step 3: Uncertainty Relation

\[ \Delta x \cdot \Delta p \approx \hbar \]
\[ \text{More precisely: } \Delta x \cdot \Delta p_x \geq \frac{\hbar}{2} \]

Single-Slit Diffraction Derivation

🧮 Single-Slit Analysis

Step 1: Diffraction Setup

Consider electrons passing through a single slit of width \( a \). The uncertainty in position is:

\[ \Delta x \approx a \]

Step 2: Momentum Uncertainty

The diffraction pattern has its first minimum at angle \( \theta \) where:

\[ a \sin\theta = \lambda \]
\[ \text{The spread in momentum: } \Delta p_x \approx p \sin\theta \]
\[ = p \cdot \frac{\lambda}{a} \]

Step 3: Using de Broglie Relation

\[ p = \frac{h}{\lambda} \]
\[ \Delta p_x \approx \frac{h}{\lambda} \cdot \frac{\lambda}{a} = \frac{h}{a} \]
\[ \Delta x \cdot \Delta p_x \approx a \cdot \frac{h}{a} = h \]
\[ \text{More precisely: } \Delta x \cdot \Delta p_x \geq \frac{h}{4\pi} = \frac{\hbar}{2} \]

Energy-Time Uncertainty

⏱️ Temporal Uncertainty

There is also an uncertainty relation between energy and time:

\[ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \]

This means that a quantum state with a finite lifetime \( \Delta t \) has an uncertainty in its energy of at least \( \hbar/(2\Delta t) \).

Sample Problem: Uncertainty Principle Application

An electron is confined to a region of width 1 Å (approximately the size of an atom). Estimate the minimum uncertainty in its velocity.

Given:
\[ \Delta x = 1 \, \text{Å} = 10^{-10} \, \text{m} \]
\[ m_e = 9.1 \times 10^{-31} \, \text{kg} \]
\[ \hbar = 1.055 \times 10^{-34} \, \text{J·s} \]
From uncertainty principle:
\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \]
\[ \Delta p \geq \frac{\hbar}{2\Delta x} \]
\[ = \frac{1.055 \times 10^{-34}}{2 \times 10^{-10}} \]
\[ = 5.275 \times 10^{-25} \, \text{kg·m/s} \]
Uncertainty in velocity:
\[ \Delta v = \frac{\Delta p}{m} \]
\[ = \frac{5.275 \times 10^{-25}}{9.1 \times 10^{-31}} \]
\[ = 5.8 \times 10^5 \, \text{m/s} \]

The Wave Function

🌊 Quantum State Description

The state of a quantum system is described by a wave function \( \Psi(x,t) \). According to the Born interpretation, the probability of finding the particle between \( x \) and \( x+dx \) at time \( t \) is:

\[ P(x,t) dx = |\Psi(x,t)|^2 dx \]

The wave function must be normalized so that the total probability is 1:

\[ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx = 1 \]

Schrodinger's Equation

📐 Fundamental Quantum Equation

The time-dependent Schrödinger equation describes how the wave function evolves with time:

\[ i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V(x,t)\Psi \]

For stationary states with definite energy \( E \), the wave function can be separated as \( \Psi(x,t) = \psi(x)e^{-iEt/\hbar} \), leading to the time-independent Schrödinger equation:

\[ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + V(x)\psi = E\psi \]

Free Particle Solution

🧮 Free Particle Wave Function

Step 1: Time-Independent Equation

For a free particle (\( V(x) = 0 \)):

\[ -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = E\psi \]

Step 2: General Solution

\[ \frac{d^2\psi}{dx^2} + k^2\psi = 0 \]
\[ \text{where } k = \frac{\sqrt{2mE}}{\hbar} \]
\[ \psi(x) = Ae^{ikx} + Be^{-ikx} \]

Step 3: Complete Wave Function

\[ \Psi(x,t) = \psi(x)e^{-iEt/\hbar} \]
\[ = Ae^{i(kx - \omega t)} + Be^{-i(kx + \omega t)} \]
\[ \text{where } \omega = \frac{E}{\hbar} \]

Particle in a Box

🧮 Infinite Square Well

Step 1: Potential Definition

Consider a particle confined in a one-dimensional box of length \( L \):

\[ V(x) = \begin{cases} 0 & \text{for } 0 < x < L \\ \infty & \text{otherwise} \end{cases} \]

Step 2: Boundary Conditions

\[ \psi(0) = 0 \quad \text{and} \quad \psi(L) = 0 \]

Step 3: General Solution

\[ \psi(x) = A\sin(kx) + B\cos(kx) \]
\[ \text{Applying } \psi(0) = 0 \Rightarrow B = 0 \]
\[ \psi(x) = A\sin(kx) \]

Step 4: Quantization Condition

\[ \psi(L) = A\sin(kL) = 0 \]
\[ kL = n\pi \quad \text{where } n = 1, 2, 3, \ldots \]
\[ k_n = \frac{n\pi}{L} \]

Step 5: Energy Quantization

\[ E = \frac{\hbar^2 k^2}{2m} \]
\[ E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \]
\[ = \frac{n^2 h^2}{8mL^2} \]

Step 6: Normalized Wave Functions

\[ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \]
Sample Problem: Particle in a Box

An electron is confined in a one-dimensional box of length 2 Å. Calculate the energy of the ground state and the first excited state.

Given:
\[ L = 2 \, \text{Å} = 2 \times 10^{-10} \, \text{m} \]
\[ m_e = 9.1 \times 10^{-31} \, \text{kg} \]
\[ h = 6.626 \times 10^{-34} \, \text{J·s} \]
Energy formula:
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
Ground state (n=1):
\[ E_1 = \frac{(1)^2 (6.626 \times 10^{-34})^2}{8 \times 9.1 \times 10^{-31} \times (2 \times 10^{-10})^2} \]
\[ = \frac{4.39 \times 10^{-67}}{2.912 \times 10^{-49}} \]
\[ = 1.51 \times 10^{-18} \, \text{J} \]
\[ = \frac{1.51 \times 10^{-18}}{1.6 \times 10^{-19}} \, \text{eV} \]
\[ = 9.44 \, \text{eV} \]
First excited state (n=2):
\[ E_2 = 4 \times E_1 = 37.76 \, \text{eV} \]

Potential Step and Barrier Tunneling

🚧 Quantum Tunneling

When a particle encounters a potential barrier of height \( V_0 \) greater than its energy \( E \), classical physics predicts it will be reflected. However, quantum mechanics shows there is a finite probability for the particle to "tunnel" through the barrier.

This phenomenon is crucial for understanding alpha decay, scanning tunneling microscopes, and many semiconductor devices.

🧮 Tunneling Probability

Step 1: Wave Function in Barrier Region

For \( E < V_0 \) and barrier width \( a \), inside the barrier:

\[ \frac{d^2\psi}{dx^2} = \kappa^2 \psi \]
\[ \text{where } \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \]
\[ \psi(x) = Ce^{-\kappa x} + De^{\kappa x} \]

Step 2: Transmission Probability

The probability of tunneling through the barrier is approximately:

\[ T \approx e^{-2\kappa a} \]
\[ = \exp\left[-\frac{2a}{\hbar}\sqrt{2m(V_0 - E)}\right] \]

Practice Problems with Solutions

Problem 1: De Broglie Wavelength

Calculate the de Broglie wavelength of a neutron with kinetic energy 0.025 eV (thermal neutron at room temperature).

Given:
\[ K.E = 0.025 \, \text{eV} = 0.025 \times 1.6 \times 10^{-19} \, \text{J} = 4.0 \times 10^{-21} \, \text{J} \]
\[ m_n = 1.675 \times 10^{-27} \, \text{kg} \]
Velocity calculation:
\[ \frac{1}{2}mv^2 = K.E \]
\[ v = \sqrt{\frac{2K.E}{m}} \]
\[ = \sqrt{\frac{2 \times 4.0 \times 10^{-21}}{1.675 \times 10^{-27}}} \]
\[ = 2.19 \times 10^3 \, \text{m/s} \]
De Broglie wavelength:
\[ \lambda = \frac{h}{mv} \]
\[ = \frac{6.626 \times 10^{-34}}{1.675 \times 10^{-27} \times 2.19 \times 10^3} \]
\[ = 1.81 \times 10^{-10} \, \text{m} = 1.81 \, \text{Å} \]
Problem 2: Uncertainty Principle

The position of a 1 keV electron is measured with an accuracy of ±1 Å. What is the uncertainty in its momentum? What percentage of its momentum is this?

Given:
\[ \Delta x = 1 \, \text{Å} = 10^{-10} \, \text{m} \]
\[ K.E = 1 \, \text{keV} = 1000 \, \text{eV} = 1.6 \times 10^{-16} \, \text{J} \]
Momentum uncertainty:
\[ \Delta p \geq \frac{\hbar}{2\Delta x} \]
\[ = \frac{1.055 \times 10^{-34}}{2 \times 10^{-10}} \]
\[ = 5.275 \times 10^{-25} \, \text{kg·m/s} \]
Electron momentum:
\[ p = \sqrt{2mK.E} \]
\[ = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-16}} \]
\[ = 1.71 \times 10^{-23} \, \text{kg·m/s} \]
Percentage uncertainty:
\[ \frac{\Delta p}{p} \times 100\% = \frac{5.275 \times 10^{-25}}{1.71 \times 10^{-23}} \times 100\% \]
\[ = 3.09\% \]

Frequently Asked Questions

❓ Why don't we observe wave properties in everyday objects?

The de Broglie wavelength is given by \( \lambda = h/p \), where Planck's constant \( h \) is extremely small (\( \approx 6.626 \times 10^{-34} \, \text{J·s} \)). For macroscopic objects, the momentum \( p = mv \) is relatively large, making the wavelength incredibly small and undetectable.

For example, a 1 kg ball moving at 1 m/s has a de Broglie wavelength of:

\[ \lambda = \frac{6.626 \times 10^{-34}}{1 \times 1} = 6.626 \times 10^{-34} \, \text{m} \]

This is much smaller than atomic scales (\( \sim 10^{-10} \, \text{m} \)), so wave effects are negligible. Only for very light particles like electrons do we observe significant wave behavior.

❓ What is the physical meaning of the wave function?

The wave function \( \Psi(x,t) \) itself doesn't have a direct physical interpretation. However, according to Max Born's interpretation, the square of its absolute value \( |\Psi(x,t)|^2 \) gives the probability density of finding the particle at position \( x \) at time \( t \).

More precisely, the probability of finding the particle between \( x \) and \( x+dx \) is:

\[ P(x,t) dx = |\Psi(x,t)|^2 dx \]

The wave function must be normalized so that the total probability over all space is 1:

\[ \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx = 1 \]
❓ How does quantum tunneling work?

Quantum tunneling occurs when a particle passes through a potential barrier even though its energy is less than the barrier height. This is impossible in classical physics but allowed in quantum mechanics due to the wave nature of particles.

The key points are:

  • The wave function doesn't abruptly drop to zero at the barrier
  • It decays exponentially inside the barrier as \( e^{-\kappa x} \)
  • If the barrier is thin enough, there's a finite probability for the particle to appear on the other side

The transmission probability is approximately:

\[ T \approx e^{-2\kappa a} \]
\[ \text{where } \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \]

and \( a \) is the barrier width. This exponential dependence means tunneling is significant only for thin barriers.

📚 Master Quantum Mechanics

Understanding the wave nature of matter is fundamental to quantum mechanics, modern physics, and many technological applications like electron microscopes, semiconductor devices, and quantum computing. Continue your journey into the fascinating world of quantum phenomena.

Read More: Physics HRK Notes of Quantum Mechanics

© House of Physics | HRK Physics Chapter 50: The Wave Nature of Matter

Based on Halliday, Resnick, and Krane's "Physics" with additional insights from university physics curriculum

House of Physics | Contact: aliphy2008@gmail.com

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