Skip to main content

Quantum Mechanics

1. Historical Motivation

1.1 Failures of Classical Physics

By the late 19th century, classical physics could not explain several phenomena:

Blackbody radiation. The Rayleigh-Jeans law predicted infinite energy at short wavelengths (the "ultraviolet catastrophe"). Experiment showed a peak that shifts with temperature.

Photoelectric effect. Classical theory predicted that the kinetic energy of emitted electrons depends on the intensity of light. Experiment showed it depends on the frequency.

Atomic spectra. Atoms emit light at discrete frequencies, not the continuous spectrum predicted by classical electrodynamics.

Stability of atoms. Classical electrodynamics predicts orbiting electrons radiate energy and spiral into the nucleus.

1.2 Key Experiments

Planck's quantisation (1900). Blackbody radiation is explained by assuming energy is emitted in discrete quanta: E=hνE = h\nu where h=6.626×1034h = 6.626 \times 10^{-34} J\cdots is Planck's constant.

Einstein's photon (1905). Light consists of photons, each carrying energy E=hνE = h\nu and momentum p=h/λ=hν/cp = h/\lambda = h\nu/c. The photoelectric effect: Ek=hνϕE_k = h\nu - \phi where ϕ\phi is the work function.

Compton scattering (1923). X-rays scattered off electrons show a wavelength shift:

Δλ=hmec(1cosθ)\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)

This confirms that photons carry momentum p=h/λp = h/\lambda.

Davisson-Germer experiment (1927). Electrons scattered off a nickel crystal produce a diffraction pattern, confirming de Broglie's hypothesis that matter has wave properties: λ=h/p\lambda = h/p.

2. Postulates of Quantum Mechanics

2.1 The Postulates

Postulate 1 (State Space). The state of a quantum system is completely described by a normalised vector ψ|\psi\rangle in a complex Hilbert space H\mathcal{H}.

Postulate 2 (Observables). Every measurable quantity (observable) is represented by a Hermitian (self-adjoint) operator A^=A^\hat{A} = \hat{A}^\dagger acting on H\mathcal{H}.

Postulate 3 (Measurement). A measurement of observable A^\hat{A} yields one of the eigenvalues ana_n of A^\hat{A}. The probability of measuring ana_n when the system is in state ψ|\psi\rangle is

P(an)=anψ2P(a_n) = |\langle a_n | \psi \rangle|^2

where an|a_n\rangle is the eigenstate corresponding to ana_n. After measurement, the state collapses to an|a_n\rangle.

Postulate 4 (Time Evolution). The time evolution of the state is governed by the time-dependent Schrodinger equation:

itψ(t)=H^ψ(t)i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle = \hat{H}|\psi(t)\rangle

where H^\hat{H} is the Hamiltonian (energy operator).

Postulate 5 (Composite Systems). The state space of a composite system is the tensor product of the state spaces of the components.

2.2 Implications

  • Superposition: A system can be in a linear combination of eigenstates: ψ=ncnan|\psi\rangle = \sum_n c_n |a_n\rangle.
  • Uncertainty Principle: Non-commuting observables cannot be simultaneously measured with arbitrary precision.
  • Probabilistic Nature: Quantum mechanics predicts probabilities, not deterministic outcomes.

3. Wave Functions and the Schrodinger Equation

3.1 Wave Functions

In the position representation, the state is described by a wave function ψ(r,t)\psi(\mathbf{r}, t), where ψ(r,t)2|\psi(\mathbf{r}, t)|^2 is the probability density:

P(r[r,r+dr])=ψ(r,t)2d3rP(\mathbf{r} \in [\mathbf{r}, \mathbf{r} + d\mathbf{r}]) = |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r}

Normalisation: ψ(r,t)2d3r=1\int_{-\infty}^{\infty} |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r} = 1.

3.2 Time-Dependent Schrodinger Equation

iψt=H^ψ=(22m2+V(r,t))ψi\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi = \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}, t)\right)\psi

3.3 Time-Independent Schrodinger Equation

For time-independent potentials V(r)V(\mathbf{r}), separate variables: ψ(r,t)=ϕ(r)eiEt/\psi(\mathbf{r}, t) = \phi(\mathbf{r}) e^{-iEt/\hbar}:

H^ϕ=Eϕi.e.,22m2ϕ+Vϕ=Eϕ\hat{H}\phi = E\phi \quad \mathrm{i.e.,} \quad -\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi

This is an eigenvalue problem: EE is the energy eigenvalue, ϕ\phi is the energy eigenstate.

3.4 Probability Current

The probability current density is

J=2mi(ψψψψ)\mathbf{J} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi \nabla\psi^*)

It satisfies the continuity equation: ψ2t+J=0\frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0, expressing conservation of probability.

4. Operators and Observables

4.1 Position and Momentum Operators

In the position representation:

x^=x,p^=ix\hat{x} = x, \quad \hat{p} = -i\hbar\frac{\partial}{\partial x}

These satisfy the canonical commutation relation:

[x^,p^]=i[\hat{x}, \hat{p}] = i\hbar

4.2 General Properties of Operators

Hermitian operators have real eigenvalues and orthogonal eigenstates -- essential for observables.

Theorem 4.1. If A^\hat{A} is Hermitian, then:

  • All eigenvalues are real.
  • Eigenstates corresponding to distinct eigenvalues are orthogonal.
  • The eigenstates form a complete basis (for the space of physical states).

4.3 Commutators

The commutator of two operators is [A^,B^]=A^B^B^A^[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}.

Theorem 4.2 (Generalised Uncertainty Principle). For observables A^\hat{A} and B^\hat{B}:

σAσB12[A^,B^]\sigma_A \sigma_B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle|

Corollary 4.3 (Heisenberg Uncertainty Principle). σxσp/2\sigma_x \sigma_p \geq \hbar/2.

Proof. This follows from the generalised uncertainty principle with [x^,p^]=i[\hat{x}, \hat{p}] = i\hbar:

σxσp12i=2\sigma_x \sigma_p \geq \frac{1}{2}|\langle i\hbar \rangle| = \frac{\hbar}{2}

\blacksquare

4.4 Expectation Values

The expectation value of an observable A^\hat{A} in state ψ|\psi\rangle:

A=ψA^ψ=ψA^ψdx\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \int \psi^* \hat{A} \psi\, dx

Theorem 4.4 (Ehrenfest's Theorem). Quantum expectation values obey classical equations of motion:

dx^dt=p^m,dp^dt=Vx\frac{d\langle \hat{x} \rangle}{dt} = \frac{\langle \hat{p} \rangle}{m}, \quad \frac{d\langle \hat{p} \rangle}{dt} = -\left\langle \frac{\partial V}{\partial x}\right\rangle

5. One-Dimensional Problems

5.1 The Infinite Square Well

A particle of mass mm in a potential V(x)=0V(x) = 0 for 0<x<L0 \lt x \lt L and V(x)=V(x) = \infty otherwise.

Boundary conditions: ψ(0)=ψ(L)=0\psi(0) = \psi(L) = 0.

Solutions:

ϕn(x)=2Lsin(nπxL),En=n2π222mL2,n=1,2,3,\phi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right), \quad E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}, \quad n = 1, 2, 3, \ldots

Properties:

  • The ground state (n=1n = 1) has the lowest energy E1>0E_1 > 0 (zero-point energy).
  • Energy levels are not equally spaced; Enn2E_n \propto n^2.
  • There are (n1)(n - 1) nodes in the nn-th eigenstate.

5.2 The Quantum Harmonic Oscillator

V(x)=12mω2x2V(x) = \frac{1}{2}m\omega^2 x^2. The energy eigenvalues are:

En=ω(n+12),n=0,1,2,E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots

The eigenfunctions involve Hermite polynomials HnH_n:

ϕn(x)=(mωπ)1/412nn!Hn ⁣(mωx)emωx2/(2)\phi_n(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\,x\right) e^{-m\omega x^2/(2\hbar)}

Ladder operators:

a^=mω2(x^+ip^mω),a^=mω2(x^ip^mω)\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right), \quad \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)

a^n=nn1,a^n=n+1n+1\hat{a}|n\rangle = \sqrt{n}|n-1\rangle, \quad \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle

H^=ω(a^a^+12)\hat{H} = \hbar\omega\left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\right)

5.3 The Free Particle

V(x)=0V(x) = 0 everywhere. The Schrodinger equation:

22md2ϕdx2=Eϕ-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi

Solutions: ϕk(x)=12πeikx\phi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx} with E=2k22mE = \frac{\hbar^2 k^2}{2m}.

The energy spectrum is continuous (all E0E \geq 0). The eigenfunctions are not normalisable (plane waves); physical states are wave packets constructed by superposition.

6. Angular Momentum and the Hydrogen Atom

6.1 Angular Momentum Operators

L^x=i(yzzy),L^y=i(zxxz),L^z=i(xyyx)\hat{L}_x = -i\hbar\left(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}\right), \quad \hat{L}_y = -i\hbar\left(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}\right), \quad \hat{L}_z = -i\hbar\left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)

Commutation relations:

[L^x,L^y]=iL^z,[L^y,L^z]=iL^x,[L^z,L^x]=iL^y[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z, \quad [\hat{L}_y, \hat{L}_z] = i\hbar\hat{L}_x, \quad [\hat{L}_z, \hat{L}_x] = i\hbar\hat{L}_y

[L^2,L^i]=0foralli[\hat{L}^2, \hat{L}_i] = 0 \quad \mathrm{for all } i

Simultaneous eigenstates: l,m|l, m\rangle with

L^2l,m=2l(l+1)l,m,L^zl,m=ml,m\hat{L}^2|l,m\rangle = \hbar^2 l(l+1)|l,m\rangle, \quad \hat{L}_z|l,m\rangle = \hbar m|l,m\rangle

where l=0,1,2,l = 0, 1, 2, \ldots and m=l,l+1,,l1,lm = -l, -l+1, \ldots, l-1, l.

6.2 The Hydrogen Atom

The Hamiltonian for hydrogen (electron of mass mem_e and charge e-e, proton of charge +e+e):

H^=22me2e24πε0r\hat{H} = -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r}

Energy eigenvalues:

En=mee42(4πε0)221n2=13.6eVn2,n=1,2,3,E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} = -\frac{13.6\,\mathrm{eV}}{n^2}, \quad n = 1, 2, 3, \ldots

Degeneracy: Each energy level EnE_n has degeneracy n2n^2 (ignoring spin). The quantum numbers are:

  • Principal: n=1,2,3,n = 1, 2, 3, \ldots
  • Orbital angular momentum: l=0,1,,n1l = 0, 1, \ldots, n - 1
  • Magnetic: ml=l,,lm_l = -l, \ldots, l

The ground state wave function (n=1,l=0,ml=0n = 1, l = 0, m_l = 0):

ψ100(r,θ,ϕ)=1πa03er/a0\psi_{100}(r, \theta, \phi) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}

where a0=4πε02mee20.529A˚a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \approx 0.529\,\mathrm{\AA} is the Bohr radius.

7. Spin

7.1 The Spin Operators

Spin is an intrinsic form of angular momentum with no classical analogue. For spin-1/21/2 particles (e.g., electrons):

S^x=2σx,S^y=2σy,S^z=2σz\hat{S}_x = \frac{\hbar}{2}\sigma_x, \quad \hat{S}_y = \frac{\hbar}{2}\sigma_y, \quad \hat{S}_z = \frac{\hbar}{2}\sigma_z

where σx,σy,σz\sigma_x, \sigma_y, \sigma_z are the Pauli matrices:

σx=(0110),σy=(0ii0),σz=(1001)\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

7.2 Properties of Pauli Matrices

σi2=I,σiσj=iϵijkσk(ij)\sigma_i^2 = I, \quad \sigma_i \sigma_j = i\epsilon_{ijk}\sigma_k \quad (i \neq j)

[σi,σj]=2iϵijkσk,{σi,σj}=2δijI[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k, \quad \{\sigma_i, \sigma_j\} = 2\delta_{ij}I

Spin states: =(10)|\uparrow\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} (spin up, ms=+1/2m_s = +1/2) and =(01)|\downarrow\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} (spin down, ms=1/2m_s = -1/2).

7.3 Stern-Gerlach Experiment

A beam of silver atoms passes through an inhomogeneous magnetic field and splits into two beams, confirming the quantisation of angular momentum (spin-1/2 for the outer electron).

8. Perturbation Theory

8.1 Time-Independent Perturbation Theory

For a Hamiltonian H^=H^0+λH^\hat{H} = \hat{H}_0 + \lambda \hat{H}' where H^\hat{H}' is "small" and H^0\hat{H}_0 has known eigenstates n(0)|n^{(0)}\rangle and eigenvalues En(0)E_n^{(0)}.

First-order energy correction:

En(1)=n(0)H^n(0)E_n^{(1)} = \langle n^{(0)} | \hat{H}' | n^{(0)} \rangle

Second-order energy correction:

En(2)=mnm(0)H^n(0)2En(0)Em(0)E_n^{(2)} = \sum_{m \neq n} \frac{|\langle m^{(0)} | \hat{H}' | n^{(0)} \rangle|^2}{E_n^{(0)} - E_m^{(0)}}

First-order state correction:

n(1)=mnm(0)H^n(0)En(0)Em(0)m(0)|n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle m^{(0)} | \hat{H}' | n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} |m^{(0)}\rangle

8.2 Degenerate Perturbation Theory

When En(0)E_n^{(0)} is degenerate, the corrections are found by diagonalising the perturbation matrix in the degenerate subspace.

Theorem 8.1. The correct zeroth-order states are the eigenvectors of the matrix Wij=ni(0)H^nj(0)W_{ij} = \langle n_i^{(0)} | \hat{H}' | n_j^{(0)} \rangle within the degenerate subspace.

8.3 Worked Example

Problem. A one-dimensional infinite square well of width LL has a small perturbation H=V0H' = V_0 for 0<x<L/20 \lt x \lt L/2 and H=0H' = 0 for L/2<x<LL/2 \lt x \lt L. Find the first-order energy corrections.

Solution. The unperturbed states are ϕn(0)(x)=2/Lsin(nπx/L)\phi_n^{(0)}(x) = \sqrt{2/L}\sin(n\pi x/L).

En(1)=n(0)Hn(0)=0L/2V02Lsin2(nπxL)dxE_n^{(1)} = \langle n^{(0)} | H' | n^{(0)} \rangle = \int_0^{L/2} V_0 \frac{2}{L}\sin^2\left(\frac{n\pi x}{L}\right) dx

=2V0L0L/21cos(2nπx/L)2dx=V0L[L2L4nπsin(nπ)]=V02= \frac{2V_0}{L}\int_0^{L/2} \frac{1 - \cos(2n\pi x/L)}{2}\, dx = \frac{V_0}{L}\left[\frac{L}{2} - \frac{L}{4n\pi}\sin(n\pi)\right] = \frac{V_0}{2}

The first-order correction is En(1)=V0/2E_n^{(1)} = V_0/2 for all nn. \blacksquare

Common Pitfall

Perturbation theory assumes the perturbation is "small" compared to the level spacing. If mHnEn(0)Em(0)|\langle m | H' | n \rangle| \sim |E_n^{(0)} - E_m^{(0)}|, the perturbation series may diverge. The method also fails for systems where the unperturbed Hamiltonian has closely spaced or degenerate levels that are not handled correctly.