Quantum Physics Notes FE Sem 1 Applied Physics | MU

Electrons, protons and atoms behave as waves on the atomic scale, and this wave nature explains how electron microscopes, tunnel diodes and quantum computers work. This module starts with de Broglie's hypothesis and the wavelength of an accelerated electron, then defines the wave function, its probability density and the conditions it must satisfy. It explains why Schrödinger's equation was needed, derives its time independent and time dependent forms, and solves it for a particle in a rigid box to obtain quantised energies. It ends with quantum mechanical tunnelling and the qubit, the basic unit of quantum computing. It maps to course outcome CO5 and carries the most teaching hours (6) of the syllabus.

Summary

de Broglie hypothesis: Since light shows both wave and particle nature, de Broglie proposed in 1924 that material particles also have a wave associated with them. A particle of momentum p has a wavelength \(\lambda = h/p = h/mv\), called the de Broglie wavelength. In terms of kinetic energy E, \(\lambda = h/\sqrt{2mE}\). The idea was confirmed by the Davisson-Germer electron diffraction experiment. For everyday objects the wavelength is far too small to observe, so wave effects matter only for microscopic particles.

de Broglie wavelength of an electron: An electron accelerated from rest through a potential difference V gains kinetic energy eV, so \(\lambda = h/\sqrt{2meV}\). Substituting \(h = 6.63\times10^{-34}\) J s, \(m = 9.1\times10^{-31}\) kg and \(e = 1.6\times10^{-19}\) C gives \(\lambda = 12.27/\sqrt{V}\) Å (V in volts). For 100 V, \(\lambda \approx 1.23\) Å, comparable to atomic spacing in crystals, which is why electrons are diffracted by crystals and why electron microscopes have far better resolution than optical ones.

Properties of matter waves: (1) the wavelength is inversely proportional to mass and velocity, so lighter and slower particles have longer wavelengths; (2) they are associated with any moving particle, charged or neutral, so they are not electromagnetic waves; (3) a particle at rest has no matter wave (infinite wavelength); (4) the phase velocity of a matter wave can exceed the speed of light, but it carries no energy; the particle moves with the group velocity of the wave packet, which equals the particle velocity; (5) the amplitude of a matter wave has no direct physical meaning, but its square gives the probability of finding the particle; (6) matter waves show diffraction and interference.

Wave function and probability density: The wave function \(\psi(x,y,z,t)\) describes the state of a quantum particle; it is in general complex and has no direct physical meaning. Born's interpretation: \(|\psi|^2 = \psi^*\psi\) is the probability density, so \(|\psi|^2\,dV\) is the probability of finding the particle in the volume element dV at time t. Since the particle must be somewhere, \(\int |\psi|^2\,dV = 1\) over all space; this is the normalisation condition. Mathematical conditions for an acceptable (well behaved) wave function: it must be finite everywhere, single valued, continuous, have a continuous first derivative (where the potential is finite), and be square integrable so that it can be normalised.

Need and significance of Schrödinger's equation: Newton's laws cannot predict the behaviour of particles whose wave nature is important, and the de Broglie relation gives only the wavelength, not how the wave changes in space and time in a potential. Schrödinger's equation is the basic equation of quantum mechanics that plays the role of Newton's second law: given the potential energy V, its solutions give the allowed wave functions and energies. It naturally explains energy quantisation, atomic spectra, tunnelling and the behaviour of electrons in solids.

Schrödinger's time independent and time dependent equations: Combining the classical wave equation with \(\lambda = h/mv\) and \(E = \tfrac{1}{2}mv^2 + V\) gives the time independent equation \(d^2\psi/dx^2 + (2m/\hbar^2)(E - V)\psi = 0\), or in three dimensions \(\nabla^2\psi + (8\pi^2m/h^2)(E - V)\psi = 0\); it applies to stationary states of fixed energy. Using a wave function \(\Psi = Ae^{-i(Et - px)/\hbar}\) and the energy relation gives the time dependent equation \(i\hbar\,\partial\Psi/\partial t = -(\hbar^2/2m)\partial^2\Psi/\partial x^2 + V\Psi\), which describes how a state evolves in time. Here \(\hbar = h/2\pi\).

Particle in a rigid box: A particle of mass m moves freely (V = 0) between rigid walls at x = 0 and x = L, where V is infinite. Solving the time independent equation with \(\psi(0) = \psi(L) = 0\) gives \(\psi_n = \sqrt{2/L}\,\sin(n\pi x/L)\) and \(E_n = n^2h^2/8mL^2\), n = 1, 2, 3. So the energy is quantised, levels increase as n squared, the lowest energy (zero point energy) is not zero, and the probability density \(|\psi_n|^2\) has n humps with nodes at the walls. Typical problems ask for energies of an electron in a box of a few angstroms (tens of eV) and wavelengths of photons emitted in transitions.

Quantum mechanical tunnelling and qubits: Classically a particle of energy E cannot cross a barrier of height \(V_0 > E\). Quantum mechanically the wave function decays exponentially inside the barrier but is not zero at its far side, so if the barrier is thin there is a finite probability, roughly \(T \approx e^{-2\kappa a}\) with \(\kappa = \sqrt{2m(V_0 - E)}/\hbar\), that the particle appears on the other side. This tunnelling explains alpha decay and is used in the tunnel diode, scanning tunnelling microscope and flash memory. In quantum computing the basic unit is the qubit, a two-level quantum system (electron spin, photon polarisation, superconducting circuit) whose state is a superposition \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\) with \(|\alpha|^2 + |\beta|^2 = 1\). Measurement gives 0 with probability \(|\alpha|^2\) and 1 with probability \(|\beta|^2\). Superposition and entanglement let n qubits represent \(2^n\) states at once, enabling certain problems such as factoring and simulation of molecules to be solved much faster.

Mind map

Particle in a rigid box: solving Schrödinger's equation

Important definitions

Formulas

Key points to remember

• λ = h/mv: smaller mass or speed means larger wavelength; matter waves are significant only for microscopic particles.
• Electron accelerated through V volts: λ = 12.27/√V Å.
• Matter waves are not electromagnetic; they exist for neutral particles too.
• ψ itself is not measurable; |ψ|² gives probability density (Born interpretation).
• Well behaved ψ: finite, single valued, continuous, continuous first derivative, normalisable.
• Time independent equation is used for stationary states; time dependent equation gives time evolution.
• Particle in a box: E ∝ n² and E ∝ 1/L²; zero point energy is not zero.
• Energy spacing between levels increases with n: E2 − E1 = 3E1, E3 − E2 = 5E1.
• Tunnelling probability falls sharply with barrier width and with (V0 − E) and particle mass.
• A qubit holds a superposition of 0 and 1; n qubits span 2^n states.

Solved examples

Example 1. Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.

\(\lambda = h/\sqrt{2meV} = 6.63\times10^{-34}/\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100} = 6.63\times10^{-34}/5.40\times10^{-24} = 1.23\times10^{-10}\) m = 1.23 Å. Short cut: \(12.27/\sqrt{100} = 1.227\) Å.

Example 2. Find the de Broglie wavelength of an electron moving with a speed of \(10^6\) m/s, and of a 1 kg ball moving at 10 m/s. Comment.

Electron: \(\lambda = h/mv = 6.63\times10^{-34}/(9.1\times10^{-31}\times10^{6}) = 7.29\times10^{-10}\) m = 7.29 Å. Ball: \(\lambda = 6.63\times10^{-34}/(1\times10) = 6.63\times10^{-35}\) m. The ball's wavelength is immeasurably small, so wave nature is not observed for macroscopic objects.

Example 3. What potential difference must an electron be accelerated through to have a de Broglie wavelength of 0.5 Å?

\(\lambda = 12.27/\sqrt{V}\) Å, so \(\sqrt{V} = 12.27/0.5 = 24.54\) and \(V = 602\) V.

Example 4. An electron and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths (mp = 1.67 x 10^-27 kg).

\(\lambda = h/\sqrt{2mE}\), so for equal E, \(\lambda_e/\lambda_p = \sqrt{m_p/m_e} = \sqrt{1.67\times10^{-27}/9.1\times10^{-31}} = \sqrt{1835} = 42.8\). The electron's wavelength is about 43 times larger.

Example 5. An electron is confined in a rigid box of width 1 Å. Find the first three energy levels in eV.

\(E_1 = h^2/8mL^2 = (6.63\times10^{-34})^2/(8\times9.1\times10^{-31}\times(10^{-10})^2) = 4.396\times10^{-67}/7.28\times10^{-50} = 6.04\times10^{-18}\) J = 37.7 eV. \(E_2 = 4E_1 = 151\) eV, \(E_3 = 9E_1 = 340\) eV.

Example 6. An electron in a box of width 1 nm drops from n = 2 to n = 1. Find the wavelength of the emitted photon.

\(E_1 = h^2/8mL^2 = 4.396\times10^{-67}/(8\times9.1\times10^{-31}\times10^{-18}) = 6.04\times10^{-20}\) J (0.377 eV). \(\Delta E = E_2 - E_1 = 3E_1 = 1.81\times10^{-19}\) J. \(\lambda = hc/\Delta E = 6.63\times10^{-34}\times3\times10^{8}/1.81\times10^{-19} = 1.10\times10^{-6}\) m = 1.10 μm (infrared).

Example 7. A particle is in the ground state of a rigid box of width L. Find the probability of finding it between x = L/4 and x = 3L/4.

\(P = \int_{L/4}^{3L/4} (2/L)\sin^2(\pi x/L)\,dx = (1/L)\int_{L/4}^{3L/4}[1 - \cos(2\pi x/L)]\,dx\). First term: \((1/L)(L/2) = 1/2\). Second term: \(-(1/L)(L/2\pi)[\sin(2\pi x/L)]_{L/4}^{3L/4} = -(1/2\pi)(-1 - 1) = 1/\pi\). So \(P = 1/2 + 1/\pi = 0.818\), i.e. about 82 percent; the particle is most likely near the centre.

Example 8. Find the de Broglie wavelength of an electron with kinetic energy 10 eV.

\(\lambda = h/\sqrt{2mE} = 6.63\times10^{-34}/\sqrt{2\times9.1\times10^{-31}\times10\times1.6\times10^{-19}} = 6.63\times10^{-34}/1.706\times10^{-24} = 3.89\times10^{-10}\) m = 3.89 Å (check: \(12.27/\sqrt{10} = 3.88\) Å).

Important questions and answers

Q1. State de Broglie hypothesis and derive the expression for the de Broglie wavelength of an electron accelerated through a potential V. (5 marks)

de Broglie hypothesis: every moving material particle has a wave associated with it, of wavelength \(\lambda = h/p\). Reasoning: for a photon \(E = h\nu\) and \(E = mc^2\), so \(mc = h\nu/c = h/\lambda\), i.e. \(\lambda = h/p\); de Broglie extended this to particles: \(\lambda = h/mv\). Electron: accelerated from rest through V, it gains energy \(eV = \tfrac{1}{2}mv^2 = p^2/2m\), so \(p = \sqrt{2meV}\) and \(\lambda = h/\sqrt{2meV}\). Substituting \(h = 6.63\times10^{-34}\), \(m = 9.1\times10^{-31}\), \(e = 1.6\times10^{-19}\): \(\lambda = 1.227\times10^{-9}/\sqrt{V}\) m = \(12.27/\sqrt{V}\) Å.

Q2. State the properties of matter waves. (4 marks)

(1) \(\lambda = h/mv\): lighter, slower particles have longer wavelengths. (2) Associated with every moving particle, charged or uncharged, so they are not electromagnetic. (3) A particle at rest has no wave (λ infinite). (4) Phase velocity of the wave is \(c^2/v\), greater than c, but it carries no information or energy. (5) The particle travels with the group velocity of the wave packet, equal to the particle velocity v. (6) The amplitude is a probability amplitude; its square gives the probability of finding the particle. (7) They show diffraction, as in the Davisson-Germer experiment.

Q3. What is a wave function? Give its physical significance and the conditions for a well behaved wave function. (5 marks)

The wave function ψ(x, y, z, t) is a mathematical function, generally complex, that describes the quantum state of a particle; it is the amplitude of the matter wave. Significance: ψ itself is not measurable. By Born's interpretation \(|\psi|^2 = \psi\psi^*\) is the probability density, and \(|\psi|^2dV\) is the probability of finding the particle in volume dV. Since the particle must be somewhere, \(\int|\psi|^2dV = 1\) (normalisation). Conditions: (1) ψ must be finite everywhere; (2) single valued, so that probability is unique; (3) continuous; (4) its first derivative must be continuous wherever the potential is finite; (5) square integrable, so that it can be normalised; (6) it must vanish at infinity for bound states.

Q4. Derive Schrödinger's time independent wave equation. (6 marks)

Represent the matter wave by \(\psi = \psi_0 e^{-i\omega t}\), where \(\psi_0\) depends on position. The classical wave equation \(\nabla^2\psi = (1/u^2)\partial^2\psi/\partial t^2\) with \(\partial^2\psi/\partial t^2 = -\omega^2\psi\) gives \(\nabla^2\psi + (\omega^2/u^2)\psi = 0\). Since \(\omega = 2\pi\nu\) and \(u = \nu\lambda\), \(\omega/u = 2\pi/\lambda\), so \(\nabla^2\psi + (4\pi^2/\lambda^2)\psi = 0\). Using de Broglie \(\lambda = h/mv\): \(1/\lambda^2 = m^2v^2/h^2\). Total energy \(E = \tfrac{1}{2}mv^2 + V\), so \(m^2v^2 = 2m(E - V)\). Substituting: \(\nabla^2\psi + (8\pi^2m/h^2)(E - V)\psi = 0\), or with \(\hbar = h/2\pi\), \(\nabla^2\psi + (2m/\hbar^2)(E - V)\psi = 0\). In one dimension \(d^2\psi/dx^2 + (2m/\hbar^2)(E - V)\psi = 0\). This is the time independent Schrödinger equation, valid for stationary states.

Q5. Derive Schrödinger's time dependent wave equation. (5 marks)

A free particle wave in one dimension: \(\Psi = Ae^{-i(Et - px)/\hbar}\), using \(E = \hbar\omega\) and \(p = \hbar k\). Differentiating with respect to x twice: \(\partial^2\Psi/\partial x^2 = -(p^2/\hbar^2)\Psi\), so \(p^2\Psi = -\hbar^2\partial^2\Psi/\partial x^2\). Differentiating with respect to t: \(\partial\Psi/\partial t = -(iE/\hbar)\Psi\), so \(E\Psi = i\hbar\,\partial\Psi/\partial t\). Total energy \(E = p^2/2m + V\); multiplying by Ψ: \(E\Psi = (p^2/2m)\Psi + V\Psi\). Substituting: \(i\hbar\,\partial\Psi/\partial t = -(\hbar^2/2m)\partial^2\Psi/\partial x^2 + V\Psi\). This is the time dependent Schrödinger equation; in 3D replace the second derivative by \(\nabla^2\). In operator form \(\hat{H}\Psi = E\Psi\), with \(\hat{H} = -(\hbar^2/2m)\nabla^2 + V\) and \(E \to i\hbar\,\partial/\partial t\).

Q6. Obtain the energy eigenvalues and eigenfunctions of a particle in a one dimensional rigid box. (6 marks)

Potential: V = 0 for 0 < x < L, V infinite outside, so ψ = 0 at x = 0 and x = L. Inside, \(d^2\psi/dx^2 + k^2\psi = 0\), \(k^2 = 2mE/\hbar^2\). Solution: \(\psi = A\sin kx + B\cos kx\). At x = 0, ψ = 0 so B = 0. At x = L, \(A\sin kL = 0\); A cannot be zero, so \(kL = n\pi\), n = 1, 2, 3. Energy: \(E_n = \hbar^2k^2/2m = n^2\pi^2\hbar^2/2mL^2 = n^2h^2/8mL^2\). Normalisation: \(\int_0^L A^2\sin^2(n\pi x/L)dx = A^2L/2 = 1\) so \(A = \sqrt{2/L}\). Eigenfunctions: \(\psi_n = \sqrt{2/L}\sin(n\pi x/L)\). Conclusions: energy is discrete; \(E_n \propto n^2\); the lowest energy \(h^2/8mL^2\) is non-zero (zero point energy); \(|\psi_n|^2\) shows n maxima, e.g. for n = 1 the particle is most likely at the centre, unlike the uniform classical probability. Sketch ψ and |ψ|² for n = 1, 2, 3.

Q7. Explain the need and significance of Schrödinger's equation. (3 marks)

Need: classical mechanics fails for microscopic particles that show wave nature (atomic stability, discrete spectra, electron diffraction). The de Broglie relation gives the wavelength but not how the matter wave behaves in a potential. A wave equation for matter was needed, playing the role of Newton's law. Significance: Schrödinger's equation, with the given potential, gives the wave functions (probabilities) and allowed energies; it naturally yields quantised energy levels (box, hydrogen atom), explains tunnelling, chemical bonding and band theory of solids, and is the foundation of quantum mechanics and modern devices.

Q8. What is quantum mechanical tunnelling? Give its applications. (4 marks)

When a particle of energy E meets a potential barrier of height \(V_0 > E\) and finite width a, classical mechanics says it is always reflected. In quantum mechanics the solution of Schrödinger's equation inside the barrier is an exponentially decaying function \(e^{-\kappa x}\), \(\kappa = \sqrt{2m(V_0 - E)}/\hbar\), not zero; at the far edge it joins an oscillating wave, so there is a finite transmission probability \(T \approx e^{-2\kappa a}\). This penetration is called tunnelling. It is significant only for light particles and thin barriers (nanometres). Applications: alpha decay of nuclei; tunnel diode with very fast switching; scanning tunnelling microscope that images single atoms; flash memory and EEPROM (Fowler-Nordheim tunnelling); Josephson junctions used in superconducting qubits.

Q9. Explain the concept of a qubit and how it differs from a classical bit. (4 marks)

A classical bit is either 0 or 1. A qubit is a two-level quantum system (spin up or down of an electron, horizontal or vertical polarisation of a photon, two energy states of an ion or superconducting circuit) whose basis states are written |0⟩ and |1⟩. Because of superposition it can be in \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\), with complex α and β and \(|\alpha|^2 + |\beta|^2 = 1\). On measurement the superposition collapses to 0 with probability \(|\alpha|^2\) or to 1 with probability \(|\beta|^2\). The state can be pictured as a point on the Bloch sphere. n qubits can exist in a superposition of all \(2^n\) combinations and can be entangled, which lets quantum algorithms (Shor's factoring, Grover's search) work in parallel. Qubits are fragile and lose coherence through interaction with the environment (decoherence).

Q10. Calculate the lowest energy of an electron in a box of width 1 Å. (3 marks)

\(E_1 = h^2/8mL^2 = (6.63\times10^{-34})^2/(8\times9.1\times10^{-31}\times10^{-20}) = 6.04\times10^{-18}\) J = 37.7 eV.

Exam tips

• Memorise λ = 12.27/√V Å for electrons, but show the full formula h/√(2meV) in the answer before using it.
• Convert eV to joule (x 1.6 x 10^-19) before using λ = h/√(2mE).
• In the particle in a box answer, clearly state the boundary conditions and normalisation; sketch ψ and |ψ|² for the first three levels.
• Learn both derivations of Schrödinger's equation; the time independent one is asked more often.
• Write the conditions for a well behaved wave function as a numbered list.
• For tunnelling questions, at least name three applications: STM, tunnel diode, alpha decay.
• For the qubit, write the superposition state with the normalisation condition.

FAQs

Why don't we see the wave nature of a cricket ball?

Its mass is so large that λ = h/mv is around 10^-34 m, far smaller than any slit or obstacle, so diffraction and interference effects are impossible to observe.

Why can't a particle in a box have zero energy?

n = 0 would make ψ zero everywhere, meaning no particle. The lowest allowed state is n = 1 with E1 = h²/8mL². This also agrees with the uncertainty principle: confining the particle gives it a minimum momentum spread.

Is the wave function a physical wave like sound?

No. It is a probability amplitude that may be complex. Only |ψ|² has a physical meaning, as probability density.

Does a qubit store infinitely many values?

No. Although α and β can take continuous values, a measurement gives only 0 or 1. The advantage comes from interference and entanglement during computation, not from reading many values.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Physics) and were prepared with AI help and checked.