Engineering Physics (BT-201) Unit 1 Notes

How unit 1 is examined

This unit covers matter waves, operators, the Schrodinger equation and its box application, the wave function, wave packets, and the uncertainty principle; the Schrodinger derivations, the uncertainty principle and the particle in a box carry most marks.

Introduction to Quantum mechanics

Not asked since 2022

Definition. Quantum mechanics is the branch of physics that describes the behaviour of microscopic particles such as electrons, atoms and photons, whose energy is quantized and whose motion is governed by probability.

Key points.

Classical mechanics fails at atomic scale, for example in black-body radiation, the photoelectric effect and atomic spectra.
Energy is exchanged in quanta, $E = h\nu$, where $h = 6.63\times10^{-34}$ J s.
Particles show wave behaviour, with wavelength $\lambda = h/p$.
Only probabilities of outcomes can be predicted, using the wave function $\psi$.

Wave nature of Particles

Definition. ==Wave-particle duality means that radiation and matter show both wave and particle properties, linked by the de Broglie relation $\lambda = h/p = h/mv$.==

Key points.

Light shows wave nature in interference, diffraction and polarization.
Light shows particle nature in the photoelectric effect and Compton scattering, where it behaves as photons of energy $h\nu$ and momentum $h/\lambda$.
De Broglie extended this to matter: a particle of momentum $p$ has wavelength $\lambda = h/p$, and an electron accelerated through $V$ volts has $\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{1.227}{\sqrt V}$ nm.
Electron diffraction (Davisson-Germer) confirmed matter waves; wavelength is negligible for large bodies.

Example. At $V = 100$ V, $\lambda = 1.227/10 = 0.123$ nm.

Asked: [5 marks] (Dec 2023) Explain Dual nature of light.

Operators

Definition. An operator is a mathematical rule that acts on a wave function to give another function; every measurable quantity (observable) is represented by an operator.

Wave function. $\psi$ is the complex quantity describing a particle's state, and $|\psi|^2$ is the probability density. A well-behaved $\psi$ is single-valued, continuous, finite and square-integrable, with a continuous first derivative.

Derivation. Take the plane wave $\psi = Ae^{i(kx-\omega t)}$, with $p = \hbar k$ and $E = \hbar\omega$.

$$\frac{\partial\psi}{\partial x} = ik\psi = \frac{ip}{\hbar}\psi \;\Rightarrow\; -i\hbar\frac{\partial}{\partial x}\psi = p\psi$$

$$\frac{\partial\psi}{\partial t} = -i\omega\psi = -\frac{iE}{\hbar}\psi \;\Rightarrow\; i\hbar\frac{\partial}{\partial t}\psi = E\psi$$

Key points.

The momentum operator is $\hat p = -i\hbar\nabla$, or $-i\hbar\,\partial/\partial x$ in one dimension.
The energy operator is $\hat E = i\hbar\,\partial/\partial t$.
The kinetic energy operator is $\hat p^2/2m = -\dfrac{\hbar^2}{2m}\nabla^2$.
The Hamiltonian is $\hat H = -\dfrac{\hbar^2}{2m}\nabla^2 + V$, and $\hat H\psi = E\psi$ is the Schrodinger equation.

Answer frame. Open with the definition of an operator; for the 7-mark form first define $\psi$ and list its conditions; then derive $\hat p$ and $\hat E$ from the plane wave; close with the Hamiltonian.

Asked: [5 marks] (Dec 2023) Discuss the energy and momentum operator. Asked: [7 marks] (Jun 2023) Define wave function and state its properties. Derive energy and momentum operator.

Time-dependent and time-independent Schrodinger equation

Definition. ==The Schrodinger equation is the fundamental wave equation of quantum mechanics: $i\hbar\,\partial\psi/\partial t = -\dfrac{\hbar^2}{2m}\nabla^2\psi + V\psi$, which gives the wave function of a particle in potential $V$.==

Derivation of the time-dependent equation.

Write the plane matter wave for a free particle: $\psi = Ae^{i(px - Et)/\hbar}$.
Differentiate twice in $x$: $\dfrac{\partial^2\psi}{\partial x^2} = -\dfrac{p^2}{\hbar^2}\psi$, so $p^2\psi = -\hbar^2\dfrac{\partial^2\psi}{\partial x^2}$.
Differentiate once in $t$: $\dfrac{\partial\psi}{\partial t} = -\dfrac{iE}{\hbar}\psi$, so $E\psi = i\hbar\dfrac{\partial\psi}{\partial t}$.
Use the energy relation $E = \dfrac{p^2}{2m} + V$ and multiply by $\psi$.
Substitute steps 2 and 3:

$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi$$

Derivation of the time-independent equation.

For $V$ independent of time, separate variables: $\Psi(\vec r,t) = \psi(\vec r)\,\phi(t)$.
Dividing by $\psi\phi$: $\dfrac{i\hbar}{\phi}\dfrac{d\phi}{dt} = \dfrac{1}{\psi}\left[-\dfrac{\hbar^2}{2m}\nabla^2\psi + V\psi\right]$. The left side depends only on $t$ and the right only on $\vec r$, so both equal a constant $E$.
The time part gives $\phi = e^{-iEt/\hbar}$; the space part gives

$$\nabla^2\psi + \frac{2m}{\hbar^2}(E-V)\psi = 0$$

Key points.

The time-independent equation is $\hat H\psi = E\psi$, an eigenvalue equation whose eigenvalues $E$ are the allowed energies.
Solutions are stationary states with $|\Psi|^2 = |\psi|^2$, independent of time.
The equation is linear, so superposition of solutions is also a solution.

Answer frame. For the time-dependent form: open with the plane wave, draw no figure, do steps 1-5, and close with the boxed equation. For "derive both": add the separation of variables, then close with $\nabla^2\psi + \frac{2m}{\hbar^2}(E-V)\psi = 0$.

Pitfall: Do not forget the factor $-i/\hbar$ when differentiating in time; the sign error gives $-\hbar$ in the equation.

Asked: [7 marks] (Jun 2022, Dec 2023) Obtain the time dependent Schrodinger wave equation. Asked: [7 marks] (Nov 2022) Obtain the time independent Schrodinger wave equation. Asked: [7 marks] (Jun 2023, Jun 2025) Derive time dependent and time independent Schrodinger wave equation.

Application: Particle in a one dimensional box

Definition. ==A particle in a one-dimensional box is confined to $0<x<L$ by a potential $V=0$ inside and $V=\infty$ at and beyond the walls, so it cannot leave.==

Steps.

Inside the box the equation is $\dfrac{d^2\psi}{dx^2} + k^2\psi = 0$, with $k^2 = \dfrac{2mE}{\hbar^2}$.
The general solution is $\psi = C_1\cos kx + C_2\sin kx$.
Since $\psi(0)=0$, $C_1 = 0$. Since $\psi(L)=0$, $C_2\sin kL = 0$, so $kL = n\pi$ with $n = 1,2,3,\dots$
Hence $E_n = \dfrac{n^2\pi^2\hbar^2}{2mL^2} = \dfrac{n^2h^2}{8mL^2}$.
Normalize: $\int_0^L C_2^2\sin^2\dfrac{n\pi x}{L}dx = C_2^2\dfrac L2 = 1$, so $C_2 = \sqrt{2/L}$.

$$\psi_n(x) = \sqrt{\frac2L}\sin\frac{n\pi x}{L}$$

Key points.

Energy is quantized in the ratio $1:4:9$, and $n=0$ is not allowed because the particle would vanish.
The lowest (zero-point) energy $E_1 = h^2/8mL^2$ is non-zero.
The wave function has $n-1$ nodes inside the box.

Example. An electron in $L = 1$ nm gives $E_1 = \dfrac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(10^{-9})^2} = 6.03\times10^{-20}$ J $= \mathbf{0.376}$ eV.

Answer frame. Open by defining the potential well; draw the well with $V=\infty$ walls and the first three $\psi_n$; develop steps 1-5; close with $E_n$ and $\psi_n$.

Asked: [7 marks] (Jun 2022, Dec 2023, Dec 2024) Deduce the energy eigenvalues and wave function of a particle moving in a one dimensional box; prove $\psi_n = A\sin\frac{n\pi x}{L}$.

Born interpretation

Definition. ==Born's interpretation states that $|\psi(x,t)|^2 = \psi^*\psi$ is the probability density, so $|\psi|^2dV$ is the probability of finding the particle in volume $dV$ at that time.==

Key points.

The wave function $\psi$ is complex and has no direct physical meaning; it is the probability amplitude.
$|\psi|^2$ is real and positive, and is the measurable quantity.
The particle must be found somewhere, so the normalization condition is $\int|\psi|^2dV = 1$.
A well-behaved $\psi$ is single-valued, continuous, finite and square-integrable, with a continuous derivative.
Because the Schrodinger equation is linear, $A\psi$ is also a solution, and $A$ is fixed by normalization.

Answer frame. Open by defining $\psi$ as a complex probability amplitude; state Born's rule; list the conditions; close with the normalization condition.

Asked: [7 marks] (Jun 2022, Jun 2025) Discuss the physical significance of wave function.

Free-particle wavefunction and wave packets

Definition. ==A free particle ($V=0$) has the plane-wave function $\psi = Ae^{i(kx-\omega t)}$, and a wave packet is a superposition of such waves with slightly different $k$ that is localized in space.==

Key points.

A single plane wave has $|\psi|^2 = |A|^2$ everywhere, so it is spread over all space and cannot be normalized.
Adding waves with a small spread of $k$ and $\omega$ gives a localized packet whose envelope moves at the group velocity $v_g = d\omega/dk$.
The packet speed equals the classical particle velocity, $v_g = v$, while the phase velocity is $v_p = \omega/k$.
A narrower packet needs a wider spread of $k$, which leads to the uncertainty principle.

Asked: [7 marks] (Dec 2024) Discuss about the Free Particle Wave Function and Wave Packets.

vg and vp relation

Definition. ==Phase velocity $v_p = \omega/k$ is the speed of a single wave crest; group velocity $v_g = d\omega/dk$ is the speed of the wave packet.==

Derivation.

From $v_p = \omega/k$, write $\omega = kv_p$.
Differentiate with respect to $k$:

$$v_g = \frac{d\omega}{dk} = v_p + k\frac{dv_p}{dk}$$

Using $k = 2\pi/\lambda$, $dk = -\dfrac{2\pi}{\lambda^2}d\lambda$, so $k\dfrac{dv_p}{dk} = -\lambda\dfrac{dv_p}{d\lambda}$:

$$v_g = v_p - \lambda\frac{dv_p}{d\lambda}$$

Key points.

If $dv_p/d\lambda = 0$ (no dispersion), $v_g = v_p$.
In normal dispersion $dv_p/d\lambda > 0$, so $v_g < v_p$.
In anomalous dispersion $dv_p/d\lambda < 0$, so $v_g > v_p$.
For a de Broglie wave $v_p = c^2/v$ and $v_g = v$, so $v_pv_g = c^2$.

Answer frame. Open with the two definitions; do steps 1-3; discuss the three dispersion cases; close with $v_g = v_p - \lambda\,dv_p/d\lambda$.

Asked: [7 marks] (Nov 2022, Dec 2023) Deduce the relation between phase and group velocities.

Uncertainty principle

Definition. Heisenberg's uncertainty principle states that it is impossible to measure simultaneously both the position and the momentum of a microscopic particle exactly: $\Delta x\,\Delta p_x \ge \hbar/2$.

Formulae.

$$\Delta x\,\Delta p_x \ge \frac\hbar2,\qquad \Delta E\,\Delta t \ge \frac\hbar2,\qquad \Delta L\,\Delta\theta\ge\frac\hbar2$$

Elementary proof (wave packet).

A particle is represented by a wave packet of extent $\Delta x$ made of waves whose wave numbers spread over $\Delta k$.
For a packet to cancel outside width $\Delta x$, the spread must satisfy $\Delta x\,\Delta k \approx 1$ (at minimum, $\ge 1/2$).
Since $p = \hbar k$, $\Delta p = \hbar\Delta k$, so $\Delta x\,\Delta p \ge \hbar/2$.

Gamma-ray microscope. To see an electron, light of wavelength $\lambda$ must scatter from it into a lens of half-angle $\theta$. The microscope resolves $\Delta x \approx \lambda/(2\sin\theta)$. The scattered photon transfers momentum, so the electron's momentum is uncertain by $\Delta p_x \approx (2h/\lambda)\sin\theta$. Then $\Delta x\,\Delta p_x \approx h$.

Key points.

Shorter wavelength gives a sharper position but a larger momentum disturbance, so both cannot be exact.
The limit is a property of nature, not of the instrument.
It explains why an electron cannot exist inside the nucleus: confinement to about $10^{-14}$ m needs an energy of many MeV.
It gives a non-zero zero-point energy, for example in the box.
For macroscopic bodies $\hbar$ is negligible, so the principle has no visible effect.

Answer frame. Open with the statement and $\Delta x\,\Delta p \ge \hbar/2$; draw the microscope with electron, photon and lens; give the proof or thought experiment; close with significance and the macroscopic remark.

Pitfall: Some books write the bound as $\ge h$ or $\ge\hbar$; state the version you prove and keep it consistent.

Asked: [7 marks] (Nov 2022, Jun 2023, Dec 2023, Dec 2024) Explain Heisenberg's uncertainty principle in detail; state and prove (elementary proof) the uncertainty principle.

Last-minute revision

De Broglie wavelength: $\lambda = h/p$; for an electron, $\lambda = 1.227/\sqrt V$ nm.
Momentum operator $\hat p = -i\hbar\nabla$; energy operator $\hat E = i\hbar\,\partial/\partial t$.
Time-dependent equation: $i\hbar\,\partial\psi/\partial t = -\frac{\hbar^2}{2m}\nabla^2\psi + V\psi$.
Time-independent equation: $\nabla^2\psi + \frac{2m}{\hbar^2}(E-V)\psi = 0$.
Box: $E_n = n^2h^2/8mL^2$, $\psi_n = \sqrt{2/L}\sin(n\pi x/L)$; electron in 1 nm has $E_1 = 0.376$ eV.
Born: $|\psi|^2$ is the probability density and $\int|\psi|^2dV = 1$.
Well-behaved $\psi$: single-valued, continuous, finite, square-integrable.
$v_g = v_p - \lambda\,dv_p/d\lambda$; for matter waves $v_pv_g = c^2$.
Uncertainty: $\Delta x\Delta p \ge \hbar/2$ and $\Delta E\Delta t \ge \hbar/2$.

Memory hooks

"p is minus i h-bar del, E is plus i h-bar d-by-dt": the space derivative carries the minus sign.
Box: walls kill $\psi$, so $\sin$ survives and $kL = n\pi$.
$v_g$ is the group's speed: it is $v_p$ minus the dispersion term.
Born: "square the amplitude to get the chance".
Microscope: shorter light, sharper position, bigger kick.

Coverage checklist

Introduction to Quantum mechanics: definition and key points (unasked).
Wave nature of Particles: Dec 2023 dual nature of light.
operators: Dec 2023 energy and momentum operator; Jun 2023 wave function and operators.
Time-dependent and time-independent Schrodinger equation for wavefunction: Jun 2022, Nov 2022, Jun 2023, Dec 2023, Jun 2025 derivations.
Application: Particle in a One dimensional Box: Jun 2022, Dec 2023, Dec 2024 eigenvalues and wave function.
Born interpretation: Jun 2022, Jun 2025 significance of wave function.
Free-particle wavefunction and wave-packets: Dec 2024.
vg and vp relation: Nov 2022, Dec 2023.
Uncertainty principle: Nov 2022, Jun 2023, Dec 2023, Dec 2024.

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Engineering Physics (BT-201) - Unit 1 Short Notes

How unit 1 is examined

Introduction to Quantum mechanics

Wave nature of Particles

Operators

Time-dependent and time-independent Schrodinger equation

Application: Particle in a one dimensional box

Born interpretation

Free-particle wavefunction and wave packets

vg and vp relation

Uncertainty principle

Last-minute revision

Memory hooks

Coverage checklist