A rigorous self-contained note covering the historical motivation, the mathematical framework, and the canonical exactly-solvable models of non-relativistic quantum mechanics. Every derivation is shown step by step.
Towards the end of the 19th century, classical physics seemed complete. Two phenomena shattered that confidence: the spectrum of radiation emitted by a hot object (blackbody radiation) and the photoelectric effect.
A blackbody is an idealised object that absorbs all incident radiation and re-emits it purely as a function of its temperature \(T\). Experimentally, the energy density \(u(\nu, T)\) (energy per unit volume per unit frequency) is a smooth curve peaked at a frequency proportional to \(T\) (Wien's displacement law).
Classical statistical mechanics treats each mode of the electromagnetic field as a harmonic oscillator. The equipartition theorem assigns average energy \(k_BT\) to each mode (kinetic + potential). The number of modes per unit volume in the frequency interval \([\nu, \nu+d\nu]\) is \(g(\nu) = 8\pi\nu^2/c^3\), giving:
This agrees with experiment at low frequencies but diverges as \(\nu\to\infty\). Integrating over all \(\nu\) gives infinite total energy — the ultraviolet catastrophe.
Max Planck proposed that each oscillator of frequency \(\nu\) can only hold energy in discrete multiples of a minimum quantum:
Using the Boltzmann factor, the probability of the oscillator being in state \(n\) is \(P_n = e^{-nh\nu/k_BT}/Z\), with partition function
\[ Z = \sum_{n=0}^\infty e^{-nh\nu/k_BT} = \frac{1}{1 - e^{-h\nu/k_BT}}. \]The average energy is then:
\[ \langle E \rangle = -\frac{\partial \ln Z}{\partial \beta} \quad \bigl(\beta = 1/k_BT\bigr) = \frac{h\nu}{e^{h\nu/k_BT} - 1}. \]Multiplying by the mode density:
\[ \boxed{u(\nu, T) = \frac{8\pi h\nu^3}{c^3}\cdot\frac{1}{e^{h\nu/k_BT}-1}.} \]Shining light on a metal surface ejects electrons. Classical wave theory predicts: (i) any frequency should eventually eject electrons if the intensity is high enough, (ii) the kinetic energy of emitted electrons should grow with intensity. Both predictions are wrong.
Einstein proposed that light consists of discrete quanta (photons), each carrying energy \(E = h\nu\). When a photon is absorbed by the metal, it gives all its energy to a single electron. Part of that energy is used to overcome the binding energy \(\phi\) (the work function), and the rest becomes kinetic energy:
To measure \(K_\text{max}\), apply a retarding voltage \(V_s\) (stopping potential) until the most energetic electrons are just stopped:
\[ eV_s = K_\text{max} = h\nu - \phi. \] Rearranging: \[ V_s = \frac{h}{e}\,\nu - \frac{\phi}{e}. \]This is a linear relation between \(V_s\) and \(\nu\). The slope \(h/e\) is universal (same for all metals) and gives a direct measurement of \(h\). The intercept \(-\phi/e\) depends on the metal.
Arthur Compton (1923) scattered X-rays off free electrons and found that the scattered X-rays had a longer wavelength than the incident ones. Classical wave theory predicts no change in wavelength. The photon picture explains it perfectly via an elastic collision.
Let the incident photon have wavelength \(\lambda\) and travel along the \(x\)-axis. After scattering, the photon leaves at angle \(\theta\) with wavelength \(\lambda'\), and the electron recoils at angle \(\phi\).
Photon four-momenta.
\[ E_\gamma = \frac{hc}{\lambda}, \quad p_\gamma = \frac{h}{\lambda}; \qquad E_{\gamma'} = \frac{hc}{\lambda'}, \quad p_{\gamma'} = \frac{h}{\lambda'}. \]Conservation of energy (electron initially at rest, so \(E_e = m_e c^2\)):
\[ \frac{hc}{\lambda} + m_e c^2 = \frac{hc}{\lambda'} + E_e'. \tag{1} \]Conservation of momentum (\(x\)- and \(y\)-components):
\[ \frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + p_e'\cos\phi, \tag{2} \] \[ 0 = \frac{h}{\lambda'}\sin\theta - p_e'\sin\phi. \tag{3} \]Eliminate the electron. From (2) and (3):
\[ p_e'\cos\phi = \frac{h}{\lambda} - \frac{h}{\lambda'}\cos\theta, \qquad p_e'\sin\phi = \frac{h}{\lambda'}\sin\theta. \] Squaring and adding: \[ (p_e'c)^2 = \left(\frac{hc}{\lambda}\right)^2 - 2\frac{h^2c^2}{\lambda\lambda'}\cos\theta + \left(\frac{hc}{\lambda'}\right)^2. \tag{4} \]From (1): \(E_e' = hc/\lambda - hc/\lambda' + m_ec^2\). Using \((E_e')^2 = (p_e'c)^2 + (m_ec^2)^2\):
\[ (p_e'c)^2 = (E_e')^2 - (m_ec^2)^2 = \left(\frac{hc}{\lambda} - \frac{hc}{\lambda'} + m_ec^2\right)^2 - (m_ec^2)^2. \tag{5} \]Expanding (5) and equating with (4) (the \((hc/\lambda)^2\) and \((hc/\lambda')^2\) terms cancel):
\[ -2\frac{h^2c^2}{\lambda\lambda'}\cos\theta = -2\frac{h^2c^2}{\lambda\lambda'} + 2m_ec^2\left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right). \] Dividing by \(2hc\): \[ -\frac{h}{\lambda\lambda'}\cos\theta = -\frac{h}{\lambda\lambda'} + m_ec\!\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right). \] Rearranging: \[ \boxed{\Delta\lambda = \lambda' - \lambda = \frac{h}{m_ec}(1-\cos\theta) = \lambda_C(1-\cos\theta).} \]The maximum shift is at \(\theta=180°\): \(\Delta\lambda_\text{max} = 2\lambda_C \approx 4.85\ \text{pm}\).
If light (classically a wave) behaves like a particle, perhaps particles (classically point masses) behave like waves. Louis de Broglie proposed this symmetry in his 1924 PhD thesis.
By analogy with the photon (\(E=h\nu\) and \(p=h/\lambda\)), de Broglie proposed for any particle:
Davisson and Germer (1927) fired electrons at a crystal and observed diffraction patterns — direct evidence that electrons have a wavelength. The same experiment has since been done with neutrons, atoms, and even large molecules (C\(_{60}\) buckyballs).
A free particle is associated with a wave packet: a superposition of plane waves centred around wave number \(k_0\). Two velocities matter:
The group velocity of the quantum wave packet equals the classical particle velocity. The phase velocity has no direct physical meaning.
A quantum particle is described by a complex-valued wave function \(\Psi(x,t)\). What is its physical meaning?
Since the particle must be somewhere, the total probability must equal one:
Define the probability current \(j(x,t)\) via the continuity equation:
Differentiating \(\rho = |\Psi|^2 = \Psi^*\Psi\) with respect to time and using the Schrödinger equation (derived in §6):
\[ \frac{\partial\rho}{\partial t} = \frac{\partial\Psi^*}{\partial t}\Psi + \Psi^*\frac{\partial\Psi}{\partial t}. \] From the TDSE: \(i\hbar\,\partial\Psi/\partial t = -(\hbar^2/2m)\,\partial^2\Psi/\partial x^2 + V\Psi\), so \(\partial\Psi/\partial t = \frac{i\hbar}{2m}\partial^2\Psi/\partial x^2 - \frac{i}{\hbar}V\Psi\) (and conjugate for \(\Psi^*\)). Substituting: \[ \frac{\partial\rho}{\partial t} = \frac{i\hbar}{2m}\!\left( \Psi^*\frac{\partial^2\Psi}{\partial x^2} - \Psi\frac{\partial^2\Psi^*}{\partial x^2} \right) = \frac{\partial}{\partial x}\!\left[ \frac{i\hbar}{2m}\!\left( \Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x} \right) \right]. \] (The \(V\) terms cancel because \(V\) is real.) This is the continuity equation \(\partial\rho/\partial t + \partial j/\partial x = 0\), with:The simplest quantum state is a free particle moving with definite momentum \(p\) and energy \(E = p^2/(2m)\). By de Broglie, it should behave as a plane wave:
Time derivative:
\[ \frac{\partial\Psi}{\partial t} = -i\omega\,\Psi \implies i\hbar\frac{\partial\Psi}{\partial t} = \hbar\omega\,\Psi = E\,\Psi. \] So the operation \(i\hbar\,\partial/\partial t\) acting on \(\Psi\) gives the energy \(E\) times \(\Psi\).First spatial derivative:
\[ \frac{\partial\Psi}{\partial x} = ik\,\Psi \implies -i\hbar\frac{\partial\Psi}{\partial x} = \hbar k\,\Psi = p\,\Psi. \] So \(-i\hbar\,\partial/\partial x\) acting on \(\Psi\) returns the momentum \(p\).Second spatial derivative:
\[ \frac{\partial^2\Psi}{\partial x^2} = -k^2\Psi \implies -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{\hbar^2 k^2}{2m}\Psi = \frac{p^2}{2m}\Psi = K\,\Psi, \] where \(K = p^2/(2m)\) is the kinetic energy.The classical energy conservation law is \(E = K + V = p^2/(2m) + V\). Replacing each side by its operator action on \(\Psi\):
When \(V = V(x)\) (no explicit time dependence), we look for solutions of the form \(\Psi(x,t) = \psi(x)\,\phi(t)\):
Substitute into TDSE and divide both sides by \(\psi\phi\):
\[ i\hbar\frac{\phi'(t)}{\phi(t)} = \frac{1}{\psi(x)}\!\left[-\frac{\hbar^2}{2m}\psi''(x) + V(x)\psi(x)\right]. \]The left side depends only on \(t\); the right side depends only on \(x\). Both must equal the same constant — call it \(E\):
\[ i\hbar\phi' = E\phi \quad\Longrightarrow\quad \phi(t) = e^{-iEt/\hbar}. \] \[ -\frac{\hbar^2}{2m}\psi'' + V\psi = E\psi. \]This is an eigenvalue equation: \(\psi\) is an eigenfunction of \(\hat H\) with eigenvalue \(E\). The time-dependence is always the simple phase factor \(e^{-iEt/\hbar}\).
In quantum mechanics, every physical observable is represented by a linear operator acting on the wave function. The measurement postulate states that the only possible outcomes of a measurement are the eigenvalues of the corresponding operator.
For a plane wave \(\psi = e^{ipx/\hbar}\) (a state of definite momentum \(p\)):
\[ -i\hbar\frac{\partial}{\partial x}e^{ipx/\hbar} = -i\hbar \cdot \frac{ip}{\hbar}\,e^{ipx/\hbar} = p\,e^{ipx/\hbar}. \]So \(\hat{p}\,\psi = p\,\psi\): the operator returns the momentum as its eigenvalue. ✓
The expectation value of an observable \(Q\) in state \(\Psi\) is the average result over many identical measurements:
The commutator \([\hat{x},\hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x}\). Let it act on an arbitrary function \(\psi(x)\):
\[ [\hat{x},\hat{p}]\,\psi = x\!\left(-i\hbar\frac{\partial\psi}{\partial x}\right) - \left(-i\hbar\frac{\partial}{\partial x}\right)(x\,\psi). \] Using the product rule on the second term: \[ = -i\hbar x\psi' + i\hbar(\psi + x\psi') = i\hbar\,\psi. \] Since this holds for all \(\psi\):For observable \(\hat{A}\), define \(\sigma_A^2 = \langle\hat{A}^2\rangle - \langle\hat{A}\rangle^2\). The generalised uncertainty principle states:
Define shifted operators \(\hat{f} = \hat{A} - \langle A\rangle\), \(\hat{g} = \hat{B} - \langle B\rangle\). Note \(\sigma_A^2 = \langle\hat{f}^2\rangle = \langle f|f\rangle\) and similarly for \(B\).
Cauchy–Schwarz inequality: For any two vectors \(|f\rangle, |g\rangle\) in a Hilbert space: \(\langle f|f\rangle\langle g|g\rangle \ge |\langle f|g\rangle|^2\). So:
\[ \sigma_A^2\sigma_B^2 \ge |\langle f|g\rangle|^2. \]Write \(\langle f|g\rangle = \frac{1}{2}\langle[\hat f,\hat g]\rangle + \frac{1}{2}\langle\{\hat f,\hat g\}\rangle\), where the commutator part is purely imaginary and the anti-commutator part is real. Taking the modulus:
\[ |\langle f|g\rangle|^2 \ge \left|\frac{1}{2}\langle[\hat f,\hat g]\rangle\right|^2 = \left|\frac{1}{2}\langle[\hat A,\hat B]\rangle\right|^2. \] Taking the square root gives the result.Applying to \(\hat{A}=\hat{x}\), \(\hat{B}=\hat{p}\) with \([\hat{x},\hat{p}]=i\hbar\):
Equality holds in the uncertainty principle when \(\hat g|\psi\rangle = ic\,\hat f|\psi\rangle\) for some real \(c\). For \(\hat x\) and \(\hat p\) this gives:
\[ \left(-i\hbar\frac{d}{dx} - \langle p\rangle\right)\psi = ic\bigl(x - \langle x\rangle\bigr)\psi. \] Solving: \(\psi(x) \propto \exp\!\left[-\frac{(x-\langle x\rangle)^2}{4\sigma_x^2} + \frac{i\langle p\rangle x}{\hbar}\right]\) — a Gaussian wave packet is the minimum-uncertainty state.The simplest bound-state problem: a particle confined to a box of length \(L\), free inside and forbidden from escaping.
The wave function must be continuous everywhere. Since \(V=+\infty\) outside the well, we must have \(\psi(x)=0\) for \(x\le 0\) and \(x\ge L\). Continuity then requires:
\[ \psi(0) = 0, \qquad \psi(L) = 0. \]Inside (\(0 < x < L\)), \(V=0\), so the TISE is:
\[ -\frac{\hbar^2}{2m}\psi'' = E\psi \implies \psi'' = -k^2\psi, \quad k = \frac{\sqrt{2mE}}{\hbar} \quad (E > 0). \]General solution:
\[ \psi(x) = A\sin(kx) + B\cos(kx). \]Apply BC at \(x=0\): \(\psi(0) = B = 0\), so \(B=0\).
Apply BC at \(x=L\): \(\psi(L) = A\sin(kL) = 0\).
Since \(A\ne 0\) (otherwise \(\psi=0\) everywhere — not normalisable), we need:
\[ \sin(kL) = 0 \implies kL = n\pi,\ n = 1, 2, 3, \ldots \implies k_n = \frac{n\pi}{L}. \] (\(n=0\) is excluded since it gives \(\psi\equiv 0\). Negative \(n\) give the same states.)The energy is quantised — only discrete values are allowed. \(E_1 > 0\) is the zero-point energy: the particle cannot be at rest inside the well, consistent with the uncertainty principle.
An arbitrary initial wave function in the well can be expanded as:
The harmonic oscillator — a particle in a parabolic potential \(V = \frac{1}{2}m\omega^2 x^2\) — is arguably the most important exactly-solvable problem in all of physics. It models vibrations of molecules, phonons in solids, photons in a cavity, and more.
Define two operators:
Compute \(\hat{a}_-\hat{a}_+\):
\[ \hat{a}_-\hat{a}_+ = \frac{1}{2m\hbar\omega}(i\hat{p} + m\omega\hat{x})(-i\hat{p} + m\omega\hat{x}) = \frac{1}{2m\hbar\omega}\!\left(\hat{p}^2 + m^2\omega^2\hat{x}^2 + im\omega[\hat{x},\hat{p}]\right). \] Using \([\hat{x},\hat{p}]=i\hbar\): \[ = \frac{1}{\hbar\omega}\!\left(\frac{\hat{p}^2}{2m} + \frac{m\omega^2\hat{x}^2}{2}\right) - \frac{1}{2} = \frac{\hat{H}}{\hbar\omega} - \frac{1}{2}. \] Similarly, \(\hat{a}_+\hat{a}_- = \hat{H}/(\hbar\omega) + 1/2\). Therefore: \[ \hat{H} = \hbar\omega\!\left(\hat{a}_-\hat{a}_+ - \frac{1}{2}\right) = \hbar\omega\!\left(\hat{a}_+\hat{a}_- + \frac{1}{2}\right) = \hbar\omega\!\left(\hat{N} + \frac{1}{2}\right), \] where \(\hat{N} = \hat{a}_+\hat{a}_-\) is the number operator. The commutator of the ladder operators: \[ [\hat{a}_-,\hat{a}_+] = 1. \]If \(\hat{H}\psi = E\psi\), consider \(\hat{a}_+\psi\):
\[ \hat{H}(\hat{a}_+\psi) = \hbar\omega\!\left(\hat{a}_+\hat{a}_- + \tfrac{1}{2}\right)\hat{a}_+\psi = \hbar\omega\,\hat{a}_+\!\left(\hat{a}_-\hat{a}_+ + \tfrac{1}{2}\right)\psi + \hbar\omega\,\hat{a}_+[\hat{a}_-,\hat{a}_+]\psi. \] Wait — let's use the simpler route: use \([\hat H, \hat a_+] = \hbar\omega\,\hat a_+\) (provable from ladder algebra): \[ \hat{H}(\hat{a}_+\psi) = ([\hat{H},\hat{a}_+] + \hat{a}_+\hat{H})\psi = \hbar\omega\,\hat{a}_+\psi + \hat{a}_+(E\psi) = (E + \hbar\omega)\,(\hat{a}_+\psi). \] So \(\hat{a}_+\psi\) is an eigenstate of \(\hat{H}\) with energy \(E + \hbar\omega\). ✓ Similarly, \(\hat{a}_-\psi\) has energy \(E - \hbar\omega\).Lowering cannot go on forever — the energy is bounded below by \(V_\text{min}=0\). There must be a ground state \(\psi_0\) annihilated by \(\hat{a}_-\):
A classical particle with energy \(E < V_0\) approaching a potential barrier of height \(V_0\) is always reflected. In quantum mechanics, the wave function penetrates the barrier and there is a non-zero probability of the particle emerging on the other side — tunnelling.
Region I (\(x < 0\)): \(V=0\), so \(\psi'' = -k^2\psi\) with \(k=\sqrt{2mE}/\hbar\). Incident wave (rightward) plus reflected wave (leftward):
\[ \psi_\text{I}(x) = A\,e^{ikx} + B\,e^{-ikx}. \]Region II (\(0 \le x \le a\)): \(V=V_0 > E\), so \(\psi'' = +\kappa^2\psi\) with \(\kappa = \sqrt{2m(V_0-E)}/\hbar > 0\). Real exponentials:
\[ \psi_\text{II}(x) = C\,e^{\kappa x} + D\,e^{-\kappa x}. \]Region III (\(x > a\)): \(V=0\) again. Only rightward-travelling wave (no source of a leftward wave on the right):
\[ \psi_\text{III}(x) = F\,e^{ikx}. \]At \(x=0\): \(\psi_\text{I}=\psi_\text{II}\) and \(\psi_\text{I}'=\psi_\text{II}'\):
\[ A + B = C + D, \tag{i} \] \[ ik(A - B) = \kappa(C - D). \tag{ii} \]At \(x=a\): \(\psi_\text{II}=\psi_\text{III}\) and \(\psi_\text{II}'=\psi_\text{III}'\):
\[ Ce^{\kappa a} + De^{-\kappa a} = Fe^{ika}, \tag{iii} \] \[ \kappa(Ce^{\kappa a} - De^{-\kappa a}) = ikFe^{ika}. \tag{iv} \]Set \(A=1\) (unit incident amplitude). The four equations determine \(B,C,D,F\). The algebra is lengthy; the result for the transmission amplitude is:
\[ F = \frac{e^{-ika}}{\cosh(\kappa a) + \frac{i}{2}\!\left(\frac{\kappa}{k}-\frac{k}{\kappa}\right)\!\sinh(\kappa a)}. \]When \(\kappa a \gg 1\), \(\sinh(\kappa a) \approx \frac{1}{2}e^{\kappa a}\), so:
\[ T \approx \frac{16k^2\kappa^2}{(k^2+\kappa^2)^2}\,e^{-2\kappa a} = 16\,\frac{E}{V_0}\!\left(1-\frac{E}{V_0}\right) e^{-2\kappa a}. \]The dominant factor is the exponential decay \(e^{-2\kappa a}\), where \(\kappa = \sqrt{2m(V_0-E)}/\hbar\). The prefactor is slowly varying and \(\sim 1\).
| Concept | Formula | What it says |
|---|---|---|
| Planck quantisation | \(E_n = nh\nu\) | Oscillator energy is discrete |
| Photon energy | \(E = h\nu = \hbar\omega\) | Light comes in quanta |
| Photon momentum | \(p = h/\lambda = \hbar k\) | Photon carries momentum |
| Photoelectric effect | \(K_\text{max} = h\nu - \phi\) | Energy conservation with photon |
| Compton shift | \(\Delta\lambda = \lambda_C(1-\cos\theta)\) | Photon–electron elastic collision |
| de Broglie wavelength | \(\lambda = h/p\) | Particles have wave character |
| Born rule | \(P(x,t)dx = |\Psi|^2\,dx\) | Probability from amplitude squared |
| TDSE | \(i\hbar\,\partial_t\Psi = \hat{H}\Psi\) | Quantum equation of motion |
| TISE | \(\hat{H}\psi = E\psi\) | Energy eigenvalue equation |
| Momentum operator | \(\hat{p} = -i\hbar\,\partial_x\) | Momentum as differential operator |
| Canonical commutator | \([\hat{x},\hat{p}] = i\hbar\) | Foundation of uncertainty |
| Heisenberg uncertainty | \(\sigma_x\sigma_p \ge \hbar/2\) | Fundamental limit on precision |
| Infinite well energies | \(E_n = n^2\pi^2\hbar^2/(2mL^2)\) | Quantised levels in a box |
| Harmonic oscillator energies | \(E_n = (n+\tfrac{1}{2})\hbar\omega\) | Equally spaced spectrum |
| Tunnelling (thick barrier) | \(T \approx e^{-2\kappa a}\) | Exponential barrier penetration |