Chapter 39: More About Matter Waves

Matter waves, crucial for quantum applications like spacecraft nanotechnology, deepen our understanding of quantum mechanics. Building on the basics of photons and matter waves (Chapter 38) and relativity (Chapter 37), this chapter explores advanced concepts of matter waves. For JEE Main, JEE Advanced, and NEET students, mastering these topics is essential, as they frequently appear in problems involving wave functions, energy quantization, and tunneling. This chapter, More About Matter Waves, covers wave function and probability, particle in a box, quantum tunneling, and uncertainty principle and applications, providing detailed explanations, derivations, solved examples, and practical applications to ensure conceptual clarity and problem-solving proficiency.

39.1 Wave Function and Probability

The wave function describes the quantum state of a particle, a foundational concept for JEE/NEET.

Wave Function

• The wave function $\psi(x, t)$ describes the quantum state of a particle.
• For a free particle, a simple form is:
  $$ \psi(x, t) = A e^{i (kx - \omega t)} $$
• $A$: Amplitude, $k = \frac{2\pi}{\lambda}$: Wave number, $\omega = 2\pi \nu$: Angular frequency.

Probability Interpretation

• The probability density of finding a particle at position $x$ is given by the Born interpretation:
  $$ P(x) = |\psi(x)|^2 $$
• The probability of finding the particle in a region $dx$ is $|\psi(x)|^2 dx$.
• Normalization condition: The total probability over all space must be 1:
  $$ \int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1 $$

Expectation Values

• The expectation value of position $\langle x \rangle$ (average position) is:
  $$ \langle x \rangle = \int_{-\infty}^{\infty} x |\psi(x)|^2 dx $$
• Similarly, for momentum $\langle p \rangle$:
  $$ \langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x) \left(-i \hbar \frac{\partial}{\partial x}\right) \psi(x) dx $$

Derivation: Probability Density for a Free Particle
Consider a free particle wave function $\psi(x) = A e^{i k x}$ (time-independent part). The probability density is:
$$ |\psi(x)|^2 = \psi^*(x) \psi(x) = (A e^{-i k x}) (A e^{i k x}) = A^2 $$ To normalize over a finite region (e.g., box of length $L$), $\int_0^L A^2 dx = 1$, so $A^2 L = 1$, $A = \frac{1}{\sqrt{L}}$, thus $|\psi(x)|^2 = \frac{1}{L}$, a uniform probability density.

Derivation: Expectation Value in Rocket Nanotechnology
A spacecraft quantum sensor uses a particle with $\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)$ in a box of length $L = 1 , \text{nm}$. Compute $\langle x \rangle$:
$$ \langle x \rangle = \int_0^L x \left(\frac{2}{L}\right) \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{2}{L} \int_0^L x \left(\frac{1 - \cos\left(\frac{2\pi x}{L}\right)}{2}\right) dx = \frac{L}{2} = 0.5 , \text{nm} $$ This symmetry aids precision in nanotechnology (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a normalized wave function $\psi(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right)$ for $0 \leq x \leq a$. Find the probability of finding the particle between $x = 0$ and $x = \frac{a}{2}$.

• Solution:
  Probability $P = \int_0^{a/2} |\psi(x)|^2 dx = \int_0^{a/2} \frac{2}{a} \sin^2\left(\frac{\pi x}{a}\right) dx$. Use identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
  $$ P = \frac{2}{a} \int_0^{a/2} \frac{1 - \cos\left(\frac{2\pi x}{a}\right)}{2} dx = \frac{1}{a} \left[ x - \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) \right]_0^{a/2} = \frac{1}{a} \left( \frac{a}{2} - \frac{a}{2\pi} \sin \pi \right) = \frac{1}{2} $$
  • JEE Tip: Use trigonometric identities for $\sin^2$; confirm normalization. Common error: Forgetting to normalize or incorrect limits.

Solved Example: A NEET problem involves a wave function $\psi(x) = A e^{i k x}$. Find $|\psi(x)|^2$.

• Solution:
  $|\psi(x)|^2 = \psi^*(x) \psi(x) = (A e^{-i k x}) (A e^{i k x}) = A^2$.
  • NEET Tip: The probability density for a plane wave is constant; normalization requires a finite region. Common error: Assuming $\psi$ is real.

Solved Example: A JEE Advanced problem involves $\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right)$ for $0 \leq x \leq L$. Find $\langle x \rangle$.

• Solution:
  $\langle x \rangle = \int_0^L x \left(\frac{2}{L}\right) \sin^2\left(\frac{2\pi x}{L}\right) dx = \frac{2}{L} \int_0^L x \left(\frac{1 - \cos\left(\frac{4\pi x}{L}\right)}{2}\right) dx = \frac{L}{2}$.
  • JEE Tip: Symmetry of $\sin^2$ gives $\langle x \rangle$ at the midpoint; evaluate integrals carefully. Common error: Incorrect trigonometric identity.

Solved Example: A JEE Main problem involves $\psi(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right)$. Find the probability between $x = \frac{a}{4}$ and $x = \frac{3a}{4}$.

• Solution:
  $P = \int_{a/4}^{3a/4} \frac{2}{a} \sin^2\left(\frac{\pi x}{a}\right) dx = \frac{1}{a} \left[ x - \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) \right]_{a/4}^{3a/4} = \frac{1}{2} + \frac{1}{\pi} \approx 0.818$.
  • JEE Tip: Evaluate the definite integral over the given range; expect a probability between 0 and 1. Common error: Misapplying limits.

Application: Wave functions are used in quantum mechanics, semiconductor physics, and spacecraft nanotechnology (e.g., quantum sensors, aligning with your interest, April 19, 2025).

39.2 Particle in a Box

The particle in a box model introduces energy quantization, a core topic for JEE/NEET.

Particle in a Box Model

• A particle of mass $m$ is confined in a one-dimensional box of length $L$ with infinite potential walls (potential $V = 0$ inside, $V = \infty$ outside).
• Boundary conditions: $\psi(0) = \psi(L) = 0$.

Wave Function and Energy Levels

• The normalized wave function is:
  $$ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right), \quad n = 1, 2, 3, \ldots $$
• Energy levels are quantized:
  $$ E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}, \quad \hbar = \frac{h}{2\pi} $$

Probability Density

• Probability density:
  $$ |\psi_n(x)|^2 = \frac{2}{L} \sin^2\left(\frac{n \pi x}{L}\right) $$
• Nodes increase with $n$ (e.g., $n=1$ has no nodes inside, $n=2$ has one node at $x = \frac{L}{2}$).

Derivation: Energy Levels in a Box
The time-independent Schrödinger equation inside the box ($V = 0$) is:
$$ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi $$ Rearrange:
$$ \frac{d^2 \psi}{dx^2} + \frac{2m E}{\hbar^2} \psi = 0 $$ Let $k^2 = \frac{2m E}{\hbar^2}$, so $\frac{d^2 \psi}{dx^2} + k^2 \psi = 0$. The general solution is:
$$ \psi(x) = A \sin(kx) + B \cos(kx) $$ Apply boundary conditions: $\psi(0) = 0 \implies B = 0$, $\psi(L) = 0 \implies A \sin(kL) = 0 \implies kL = n \pi$, $k = \frac{n \pi}{L}$. Thus:
$$ E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2 (n \pi / L)^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2 m L^2} $$ Normalize $\psi(x) = A \sin\left(\frac{n \pi x}{L}\right)$: $\int_0^L A^2 \sin^2\left(\frac{n \pi x}{L}\right) dx = 1$, $A^2 \cdot \frac{L}{2} = 1$, $A = \sqrt{\frac{2}{L}}$.

Derivation: Energy in Rocket Quantum Systems
A spacecraft quantum system confines an electron ($m = 9.11 \times 10^{-31} , \text{kg}$) in a box $L = 1 , \text{nm}$. For $n=1$:
$$ E_1 = \frac{\pi^2 \hbar^2}{2 m L^2} = \frac{(3.1416)^2 (1.055 \times 10^{-34})^2}{2 (9.11 \times 10^{-31}) (10^{-9})^2} \approx 6.03 \times 10^{-20} , \text{J} \approx 0.376 , \text{eV} $$ This quantization aids quantum computing applications (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves an electron in a box of $L = 2 , \text{nm}$, $n=1$. Find $E_1$ in eV.

• Solution:
  $E_1 = \frac{\pi^2 \hbar^2}{2 m L^2}$, $\hbar = 1.055 \times 10^{-34} , \text{J·s}$, $m = 9.11 \times 10^{-31} , \text{kg}$, $L = 2 \times 10^{-9} , \text{m}$, $E_1 \approx 1.508 \times 10^{-20} , \text{J} \approx 0.0942 , \text{eV}$.
  • JEE Tip: Convert energy to eV (1 eV = $1.6 \times 10^{-19} , \text{J}$); energy scales as $\frac{1}{L^2}$. Common error: Incorrect units for $L$.

Solved Example: A NEET problem involves a particle in a box with $E_1 = 0.1 , \text{eV}$, $n=2$. Find $E_2$.

• Solution:
  $E_n = n^2 E_1$, for $n=1$, $E_1 = 0.1 , \text{eV}$, for $n=2$, $E_2 = (2)^2 \cdot 0.1 = 0.4 , \text{eV}$.
  • NEET Tip: Energy scales with $n^2$; use the ratio directly. Common error: Forgetting the $n^2$ dependence.

Solved Example: A JEE Advanced problem involves $\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{3\pi x}{L}\right)$. Find the number of nodes inside $(0, L)$.

• Solution:
  For $n=3$, $\sin\left(\frac{3\pi x}{L}\right) = 0$ at $x = \frac{L}{3}, \frac{2L}{3}$, so 2 nodes inside $(0, L)$.
  • JEE Tip: Nodes occur where $\psi = 0$; exclude boundaries. Common error: Counting boundary points.

Solved Example: A JEE Main problem involves an electron in a box $L = 0.1 , \text{nm}$, $n=1$. Find $E_1$ in eV.

• Solution:
  $E_1 \approx \frac{(3.1416)^2 (1.055 \times 10^{-34})^2}{2 (9.11 \times 10^{-31}) (0.1 \times 10^{-9})^2} \approx 6.03 \times 10^{-18} , \text{J} \approx 37.7 , \text{eV}$.
  • JEE Tip: Smaller $L$ increases energy; compute in Joules, then convert to eV. Common error: Unit mismatch.

Application: The particle in a box model applies to quantum dots, nanotechnology in spacecraft systems (e.g., quantum computing, aligning with your interest, April 19, 2025).

39.3 Quantum Tunneling

Quantum tunneling allows particles to penetrate barriers, a key concept for JEE/NEET.

Tunneling Phenomenon

• A particle can pass through a potential barrier even if its energy $E$ is less than the barrier height $V_0$.
• Classical physics forbids this, but quantum mechanics allows it due to the wave nature of particles.

Tunneling Probability

• For a rectangular barrier of width $a$ and height $V_0$, the tunneling probability $T$ is approximated as:
  $$ T \approx e^{-2 \kappa a}, \quad \kappa = \sqrt{\frac{2 m (V_0 - E)}{\hbar^2}} $$
• $T$ decreases with increasing barrier width $a$ and energy difference $V_0 - E$.

Applications

• Scanning Tunneling Microscopy (STM): Measures tunneling current to image surfaces at the atomic level.
• Nuclear Fusion: Tunneling enables fusion reactions in stars by allowing protons to overcome Coulomb repulsion.
• Semiconductors: Tunneling in diodes (e.g., tunnel diodes) enables fast switching.

Derivation: Tunneling Probability (Simplified)
Consider a particle of energy $E < V_0$ approaching a barrier of width $a$. Inside the barrier, the Schrödinger equation is:
$$ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V_0 \psi = E \psi \implies \frac{d^2 \psi}{dx^2} - \frac{2 m (V_0 - E)}{\hbar^2} \psi = 0 $$ Define $\kappa = \sqrt{\frac{2 m (V_0 - E)}{\hbar^2}}$, so $\frac{d^2 \psi}{dx^2} - \kappa^2 \psi = 0$. The solution is $\psi(x) = A e^{-\kappa x} + B e^{\kappa x}$. For a barrier, the decaying solution dominates, and the transmission probability is approximated as:
$$ T \approx e^{-2 \kappa a} $$

Derivation: Tunneling in Rocket Nanotechnology
A spacecraft uses tunneling in a quantum device: $m = 9.11 \times 10^{-31} , \text{kg}$, $V_0 - E = 1 , \text{eV} = 1.6 \times 10^{-19} , \text{J}$, $a = 0.5 , \text{nm}$, $\kappa = \sqrt{\frac{2 (9.11 \times 10^{-31}) (1.6 \times 10^{-19})}{(1.055 \times 10^{-34})^2}} \approx 5.11 \times 10^9 , \text{m}^{-1}$, $T \approx e^{-2 (5.11 \times 10^9) (0.5 \times 10^{-9})} \approx e^{-2.555} \approx 0.078$, enabling nanoscale switching (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves an electron tunneling through a barrier with $V_0 - E = 2 , \text{eV}$, $a = 0.2 , \text{nm}$. Find $T$.

• Solution:
  $\kappa = \sqrt{\frac{2 (9.11 \times 10^{-31}) (2 \times 1.6 \times 10^{-19})}{(1.055 \times 10^{-34})^2}} \approx 7.22 \times 10^9 , \text{m}^{-1}$, $T \approx e^{-2 (7.22 \times 10^9) (0.2 \times 10^{-9})} \approx e^{-2.888} \approx 0.056$.
  • JEE Tip: Convert energy to Joules, width to meters; $T$ is small for large $\kappa a$. Common error: Unit mismatch.

Solved Example: A NEET problem involves a proton tunneling with $V_0 - E = 1 , \text{eV}$, $a = 0.1 , \text{nm}$. Find $T$.

• Solution:
  $m = 1.67 \times 10^{-27} , \text{kg}$, $\kappa = \sqrt{\frac{2 (1.67 \times 10^{-27}) (1.6 \times 10^{-19})}{(1.055 \times 10^{-34})^2}} \approx 1.95 \times 10^9 , \text{m}^{-1}$, $T \approx e^{-2 (1.95 \times 10^9) (0.1 \times 10^{-9})} \approx e^{-0.39} \approx 0.677$.
  • NEET Tip: Larger mass reduces $\kappa$, increasing $T$; ensure SI units. Common error: Using electron mass.

Solved Example: A JEE Advanced problem involves tunneling with $T = 0.1$, $a = 0.3 , \text{nm}$, $m = 9.11 \times 10^{-31} , \text{kg}$. Find $V_0 - E$ in eV.

• Solution:
  $T = e^{-2 \kappa a}$, $\ln(0.1) = -2 \kappa (0.3 \times 10^{-9})$, $\kappa \approx 3.837 \times 10^9 , \text{m}^{-1}$, $(V_0 - E) = \frac{\kappa^2 \hbar^2}{2 m} \approx 0.915 , \text{eV}$.
  • JEE Tip: Solve for $\kappa$, then energy; convert to eV. Common error: Incorrect logarithm.

Solved Example: A JEE Main problem involves an electron tunneling with $V_0 - E = 0.5 , \text{eV}$, $a = 0.4 , \text{nm}$. Find $T$.

• Solution:
  $\kappa = \sqrt{\frac{2 (9.11 \times 10^{-31}) (0.5 \times 1.6 \times 10^{-19})}{(1.055 \times 10^{-34})^2}} \approx 3.61 \times 10^9 , \text{m}^{-1}$, $T \approx e^{-2 (3.61 \times 10^9) (0.4 \times 10^{-9})} \approx e^{-2.888} \approx 0.056$.
  • JEE Tip: Exponential decay dominates tunneling probability; compute $\kappa a$ carefully. Common error: Unit conversion errors.

Application: Quantum tunneling is used in STM, tunnel diodes, and spacecraft nanotechnology (e.g., quantum switching, aligning with your interest, April 19, 2025).

39.4 Uncertainty Principle and Applications

The uncertainty principle quantifies quantum limits, a crucial topic for JEE/NEET.

Heisenberg Uncertainty Principle

• Position and momentum cannot be measured simultaneously with infinite precision:
  $$ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} $$
• Similarly, energy and time:
  $$ \Delta E \cdot \Delta t \geq \frac{\hbar}{2} $$

Implications

• A particle’s wave nature (de Broglie waves, Chapter 38) leads to inherent uncertainties.
• Smaller $\Delta x$ (better position knowledge) increases $\Delta p$ (less momentum certainty).

Applications

• Electron Microscopy: Limits resolution due to uncertainty in electron position/momentum.
• Quantum Systems: Explains zero-point energy in the particle in a box (cannot have $\Delta p = 0$).
• Particle Physics: Uncertainty in energy allows virtual particles to exist briefly.

Derivation: Uncertainty Principle for Position and Momentum
A particle’s wavefunction is a wave packet with position spread $\Delta x$ and wave number spread $\Delta k$. From Fourier analysis, $\Delta k \approx \frac{1}{\Delta x}$. Since $p = \hbar k$, $\Delta p = \hbar \Delta k \approx \frac{\hbar}{\Delta x}$, so:
$$ \Delta x \cdot \Delta p \approx \hbar $$ More precisely, the minimum uncertainty product is:
$$ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} $$

Derivation: Uncertainty in Rocket Quantum Sensors
A spacecraft quantum sensor measures an electron with $\Delta x = 0.1 , \text{nm}$. Minimum $\Delta p$:
$$ \Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1.055 \times 10^{-34}}{2 \cdot 0.1 \times 10^{-9}} \approx 5.275 \times 10^{-25} , \text{kg·m/s} $$ This limits precision in quantum navigation systems (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves an electron with $\Delta x = 1 , \text{nm}$. Find the minimum $\Delta p$ in kg·m/s.

• Solution:
  $\Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{1.055 \times 10^{-34}}{2 \cdot 1 \times 10^{-9}} \approx 5.275 \times 10^{-26} , \text{kg·m/s}$.
  • JEE Tip: Use $\hbar = \frac{h}{2\pi}$; convert $\Delta x$ to meters. Common error: Using $h$ instead of $\hbar$.

Solved Example: A NEET problem involves a particle with $\Delta p = 1 \times 10^{-20} , \text{kg·m/s}$. Find the minimum $\Delta x$ in meters.

• Solution:
  $\Delta x \geq \frac{\hbar}{2 \Delta p} = \frac{1.055 \times 10^{-34}}{2 \cdot 1 \times 10^{-20}} \approx 5.275 \times 10^{-15} , \text{m}$.
  • NEET Tip: Solve for $\Delta x$; result is in meters. Common error: Forgetting the factor of 2.

Solved Example: A JEE Advanced problem involves an electron ($m = 9.11 \times 10^{-31} , \text{kg}$) with $\Delta x = 0.5 , \text{nm}$. Find $\Delta v$ in m/s.

• Solution:
  $\Delta p \geq \frac{1.055 \times 10^{-34}}{2 \cdot 0.5 \times 10^{-9}} \approx 1.055 \times 10^{-25} , \text{kg·m/s}$, $\Delta v = \frac{\Delta p}{m} \approx \frac{1.055 \times 10^{-25}}{9.11 \times 10^{-31}} \approx 1.158 \times 10^5 , \text{m/s}$.
  • JEE Tip: Use $\Delta p = m \Delta v$ to find velocity uncertainty. Common error: Incorrect mass units.

Solved Example: A JEE Main problem involves a particle with $\Delta E = 1 \times 10^{-19} , \text{J}$. Find the minimum $\Delta t$ in seconds.

• Solution:
  $\Delta t \geq \frac{\hbar}{2 \Delta E} = \frac{1.055 \times 10^{-34}}{2 \cdot 1 \times 10^{-19}} \approx 5.275 \times 10^{-16} , \text{s}$.
  • JEE Tip: Apply the energy-time uncertainty; ensure SI units. Common error: Using $h$ instead of $\hbar$.

Application: The uncertainty principle applies to quantum computing, electron microscopy, and spacecraft quantum systems (e.g., navigation sensors, aligning with your interest, April 19, 2025).

Summary and Quick Revision

• Wave Function: $\psi(x)$, probability density $|\psi(x)|^2$, $\int |\psi|^2 dx = 1$, $\langle x \rangle = \int x |\psi|^2 dx$.
• Particle in a Box: $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)$, $E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}$, nodes at $\frac{k L}{n}$.
• Quantum Tunneling: $T \approx e^{-2 \kappa a}$, $\kappa = \sqrt{\frac{2 m (V_0 - E)}{\hbar^2}}$, used in STM, fusion.
• Uncertainty Principle: $\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$, $\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$, limits precision in measurements.
• JEE/NEET Tips: Normalize wave functions, compute $E_n$ with $n^2$ scaling, use tunneling exponential decay, apply uncertainty limits, verify significant figures (April 14, 2025).
• SI Units: $E$ (J or eV), $p$ (kg·m/s), $\lambda$ (m), $\psi$ (m$^{-1/2}$), $\Delta x$ (m).

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Note: Content regularly updated to align with current JEE/NEET syllabi.