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 = h ν E = h\nu E = h ν where h = 6.626 × 10 − 34 h = 6.626 \times 10^{-34} h = 6.626 × 1 0 − 34 J⋅ \cdot ⋅ s is Planck's constant.
Einstein's photon (1905). Light consists of photons, each carrying energy E = h ν E = h\nu E = h ν and momentum
p = h / λ = h ν / c p = h/\lambda = h\nu/c p = h / λ = h ν / c . The photoelectric effect: E k = h ν − ϕ E_k = h\nu - \phi E k = h ν − ϕ where ϕ \phi ϕ is the work
function.
Compton scattering (1923). X-rays scattered off electrons show a wavelength shift:
Δ λ = h m e c ( 1 − cos θ ) \Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) Δ λ = m e c h ( 1 − cos θ )
This confirms that photons carry momentum p = h / λ p = h/\lambda p = h / λ .
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 λ = 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 ∣ ψ ⟩ in a complex Hilbert space H \mathcal{H} H .
Postulate 2 (Observables). Every measurable quantity (observable) is represented by a Hermitian
(self-adjoint) operator A ^ = A ^ † \hat{A} = \hat{A}^\dagger A ^ = A ^ † acting on H \mathcal{H} H .
Postulate 3 (Measurement). A measurement of observable A ^ \hat{A} A ^ yields one of the eigenvalues
a n a_n a n of A ^ \hat{A} A ^ . The probability of measuring a n a_n a n when the system is in state ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ is
P ( a n ) = ∣ ⟨ a n ∣ ψ ⟩ ∣ 2 P(a_n) = |\langle a_n | \psi \rangle|^2 P ( a n ) = ∣ ⟨ a n ∣ ψ ⟩ ∣ 2
where ∣ a n ⟩ |a_n\rangle ∣ a n ⟩ is the eigenstate corresponding to a n a_n a n . After measurement, the state collapses
to ∣ a n ⟩ |a_n\rangle ∣ a n ⟩ .
Postulate 4 (Time Evolution). The time evolution of the state is governed by the time-dependent
Schrodinger equation :
i ℏ ∂ ∂ t ∣ ψ ( t ) ⟩ = H ^ ∣ ψ ( t ) ⟩ i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle = \hat{H}|\psi(t)\rangle i ℏ ∂ t ∂ ∣ ψ ( t )⟩ = H ^ ∣ ψ ( t )⟩
where H ^ \hat{H} 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 ⟩ = ∑ n c n ∣ a n ⟩ |\psi\rangle = \sum_n c_n |a_n\rangle ∣ ψ ⟩ = ∑ n c n ∣ a n ⟩ .
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) ψ ( r , t ) ,
where ∣ ψ ( r , t ) ∣ 2 |\psi(\mathbf{r}, t)|^2 ∣ ψ ( r , t ) ∣ 2 is the probability density:
P ( r ∈ [ r , r + d r ] ) = ∣ ψ ( r , t ) ∣ 2 d 3 r P(\mathbf{r} \in [\mathbf{r}, \mathbf{r} + d\mathbf{r}]) = |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r} P ( r ∈ [ r , r + d r ]) = ∣ ψ ( r , t ) ∣ 2 d 3 r
Normalisation: ∫ − ∞ ∞ ∣ ψ ( r , t ) ∣ 2 d 3 r = 1 \int_{-\infty}^{\infty} |\psi(\mathbf{r}, t)|^2\, d^3\mathbf{r} = 1 ∫ − ∞ ∞ ∣ ψ ( r , t ) ∣ 2 d 3 r = 1 .
3.2 Time-Dependent Schrodinger Equation
i ℏ ∂ ψ ∂ t = H ^ ψ = ( − ℏ 2 2 m ∇ 2 + V ( r , t ) ) ψ i\hbar \frac{\partial \psi}{\partial t} = \hat{H}\psi = \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r}, t)\right)\psi i ℏ ∂ t ∂ ψ = H ^ ψ = ( − 2 m ℏ 2 ∇ 2 + V ( r , t ) ) ψ
3.3 Time-Independent Schrodinger Equation
For time-independent potentials V ( r ) V(\mathbf{r}) V ( r ) , separate variables: ψ ( r , t ) = ϕ ( r ) e − i E t / ℏ \psi(\mathbf{r}, t) = \phi(\mathbf{r}) e^{-iEt/\hbar} ψ ( r , t ) = ϕ ( r ) e − i E t /ℏ :
H ^ ϕ = E ϕ i . e . , − ℏ 2 2 m ∇ 2 ϕ + V ϕ = E ϕ \hat{H}\phi = E\phi \quad \mathrm{i.e.,} \quad -\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi H ^ ϕ = E ϕ i.e. , − 2 m ℏ 2 ∇ 2 ϕ + V ϕ = E ϕ
This is an eigenvalue problem: E E E is the energy eigenvalue, ϕ \phi ϕ is the energy eigenstate.
3.4 Probability Current
The probability current density is
J = ℏ 2 m i ( ψ ∗ ∇ ψ − ψ ∇ ψ ∗ ) \mathbf{J} = \frac{\hbar}{2mi}(\psi^* \nabla\psi - \psi \nabla\psi^*) J = 2 mi ℏ ( ψ ∗ ∇ ψ − ψ ∇ ψ ∗ )
It satisfies the continuity equation: ∂ ∣ ψ ∣ 2 ∂ t + ∇ ⋅ J = 0 \frac{\partial |\psi|^2}{\partial t} + \nabla \cdot \mathbf{J} = 0 ∂ t ∂ ∣ ψ ∣ 2 + ∇ ⋅ J = 0 ,
expressing conservation of probability.
4. Operators and Observables
4.1 Position and Momentum Operators
In the position representation:
x ^ = x , p ^ = − i ℏ ∂ ∂ x \hat{x} = x, \quad \hat{p} = -i\hbar\frac{\partial}{\partial x} x ^ = x , p ^ = − i ℏ ∂ x ∂
These satisfy the canonical commutation relation :
[ x ^ , p ^ ] = i ℏ [\hat{x}, \hat{p}] = i\hbar [ x ^ , p ^ ] = i ℏ
4.2 General Properties of Operators
Hermitian operators have real eigenvalues and orthogonal eigenstates -- essential for observables.
Theorem 4.1. If A ^ \hat{A} 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} [ A ^ , B ^ ] = A ^ B ^ − B ^ A ^ .
Theorem 4.2 (Generalised Uncertainty Principle). For observables A ^ \hat{A} A ^ and B ^ \hat{B} B ^ :
σ A σ B ≥ 1 2 ∣ ⟨ [ A ^ , B ^ ] ⟩ ∣ \sigma_A \sigma_B \geq \frac{1}{2}|\langle[\hat{A}, \hat{B}]\rangle| σ A σ B ≥ 2 1 ∣ ⟨[ A ^ , B ^ ]⟩ ∣
Corollary 4.3 (Heisenberg Uncertainty Principle). σ x σ p ≥ ℏ / 2 \sigma_x \sigma_p \geq \hbar/2 σ x σ p ≥ ℏ/2 .
Proof. This follows from the generalised uncertainty principle with [ x ^ , p ^ ] = i ℏ [\hat{x}, \hat{p}] = i\hbar [ x ^ , p ^ ] = i ℏ :
σ x σ p ≥ 1 2 ∣ ⟨ i ℏ ⟩ ∣ = ℏ 2 \sigma_x \sigma_p \geq \frac{1}{2}|\langle i\hbar \rangle| = \frac{\hbar}{2} σ x σ p ≥ 2 1 ∣ ⟨ i ℏ ⟩ ∣ = 2 ℏ
■ \blacksquare ■
4.4 Expectation Values
The expectation value of an observable A ^ \hat{A} A ^ in state ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ :
⟨ A ⟩ = ⟨ ψ ∣ A ^ ∣ ψ ⟩ = ∫ ψ ∗ A ^ ψ d x \langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \int \psi^* \hat{A} \psi\, dx ⟨ A ⟩ = ⟨ ψ ∣ A ^ ∣ ψ ⟩ = ∫ ψ ∗ A ^ ψ d x
Theorem 4.4 (Ehrenfest's Theorem). Quantum expectation values obey classical equations of motion:
d ⟨ x ^ ⟩ d t = ⟨ p ^ ⟩ m , d ⟨ p ^ ⟩ d t = − ⟨ ∂ V ∂ x ⟩ \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 d t d ⟨ x ^ ⟩ = m ⟨ p ^ ⟩ , d t d ⟨ p ^ ⟩ = − ⟨ ∂ x ∂ V ⟩
5. One-Dimensional Problems
5.1 The Infinite Square Well
A particle of mass m m m in a potential V ( x ) = 0 V(x) = 0 V ( x ) = 0 for 0 < x < L 0 \lt x \lt L 0 < x < L and V ( x ) = ∞ V(x) = \infty V ( x ) = ∞ otherwise.
Boundary conditions: ψ ( 0 ) = ψ ( L ) = 0 \psi(0) = \psi(L) = 0 ψ ( 0 ) = ψ ( L ) = 0 .
Solutions:
ϕ n ( x ) = 2 L sin ( n π x L ) , E n = n 2 π 2 ℏ 2 2 m L 2 , 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 ϕ n ( x ) = L 2 sin ( L nπ x ) , E n = 2 m L 2 n 2 π 2 ℏ 2 , n = 1 , 2 , 3 , …
Properties:
The ground state (n = 1 n = 1 n = 1 ) has the lowest energy E 1 > 0 E_1 > 0 E 1 > 0 (zero-point energy ).
Energy levels are not equally spaced; E n ∝ n 2 E_n \propto n^2 E n ∝ n 2 .
There are ( n − 1 ) (n - 1) ( n − 1 ) nodes in the n n n -th eigenstate.
5.2 The Quantum Harmonic Oscillator
V ( x ) = 1 2 m ω 2 x 2 V(x) = \frac{1}{2}m\omega^2 x^2 V ( x ) = 2 1 m ω 2 x 2 . The energy eigenvalues are:
E n = ℏ ω ( n + 1 2 ) , n = 0 , 1 , 2 , … E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots E n = ℏ ω ( n + 2 1 ) , n = 0 , 1 , 2 , …
The eigenfunctions involve Hermite polynomials H n H_n H n :
ϕ n ( x ) = ( m ω π ℏ ) 1 / 4 1 2 n n ! H n ( m ω ℏ x ) e − m ω x 2 / ( 2 ℏ ) \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)} ϕ n ( x ) = ( π ℏ mω ) 1/4 2 n n ! 1 H n ( ℏ mω x ) e − mω x 2 / ( 2ℏ )
Ladder operators:
a ^ = m ω 2 ℏ ( x ^ + i p ^ m ω ) , a ^ † = m ω 2 ℏ ( x ^ − i p ^ m ω ) \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 ^ = 2ℏ mω ( x ^ + mω i p ^ ) , a ^ † = 2ℏ mω ( x ^ − mω i p ^ )
a ^ ∣ n ⟩ = n ∣ n − 1 ⟩ , a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle, \quad \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle a ^ ∣ n ⟩ = n ∣ n − 1 ⟩ , a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩
H ^ = ℏ ω ( a ^ † a ^ + 1 2 ) \hat{H} = \hbar\omega\left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\right) H ^ = ℏ ω ( a ^ † a ^ + 2 1 )
5.3 The Free Particle
V ( x ) = 0 V(x) = 0 V ( x ) = 0 everywhere. The Schrodinger equation:
− ℏ 2 2 m d 2 ϕ d x 2 = E ϕ -\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi − 2 m ℏ 2 d x 2 d 2 ϕ = E ϕ
Solutions: ϕ k ( x ) = 1 2 π e i k x \phi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx} ϕ k ( x ) = 2 π 1 e ik x with E = ℏ 2 k 2 2 m E = \frac{\hbar^2 k^2}{2m} E = 2 m ℏ 2 k 2 .
The energy spectrum is continuous (all E ≥ 0 E \geq 0 E ≥ 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 ℏ ( y ∂ ∂ z − z ∂ ∂ y ) , L ^ y = − i ℏ ( z ∂ ∂ x − x ∂ ∂ z ) , L ^ z = − i ℏ ( x ∂ ∂ y − y ∂ ∂ x ) \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) L ^ x = − i ℏ ( y ∂ z ∂ − z ∂ y ∂ ) , L ^ y = − i ℏ ( z ∂ x ∂ − x ∂ z ∂ ) , L ^ z = − i ℏ ( x ∂ y ∂ − y ∂ x ∂ )
Commutation relations:
[ L ^ x , L ^ y ] = i ℏ L ^ z , [ L ^ y , L ^ z ] = i ℏ L ^ x , [ L ^ z , L ^ x ] = i ℏ L ^ 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 ^ x , L ^ y ] = i ℏ L ^ z , [ L ^ y , L ^ z ] = i ℏ L ^ x , [ L ^ z , L ^ x ] = i ℏ L ^ y
[ L ^ 2 , L ^ i ] = 0 f o r a l l i [\hat{L}^2, \hat{L}_i] = 0 \quad \mathrm{for all } i [ L ^ 2 , L ^ i ] = 0 forall i
Simultaneous eigenstates: ∣ l , m ⟩ |l, m\rangle ∣ l , m ⟩ with
L ^ 2 ∣ l , m ⟩ = ℏ 2 l ( l + 1 ) ∣ l , m ⟩ , L ^ z ∣ l , m ⟩ = ℏ m ∣ l , 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 L ^ 2 ∣ l , m ⟩ = ℏ 2 l ( l + 1 ) ∣ l , m ⟩ , L ^ z ∣ l , m ⟩ = ℏ m ∣ l , m ⟩
where l = 0 , 1 , 2 , … l = 0, 1, 2, \ldots l = 0 , 1 , 2 , … and m = − l , − l + 1 , … , l − 1 , l m = -l, -l+1, \ldots, l-1, l m = − l , − l + 1 , … , l − 1 , l .
6.2 The Hydrogen Atom
The Hamiltonian for hydrogen (electron of mass m e m_e m e and charge − e -e − e , proton of charge + e +e + e ):
H ^ = − ℏ 2 2 m e ∇ 2 − e 2 4 π ε 0 r \hat{H} = -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r} H ^ = − 2 m e ℏ 2 ∇ 2 − 4 π ε 0 r e 2
Energy eigenvalues:
E n = − m e e 4 2 ( 4 π ε 0 ) 2 ℏ 2 ⋅ 1 n 2 = − 13.6 e V n 2 , 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 E n = − 2 ( 4 π ε 0 ) 2 ℏ 2 m e e 4 ⋅ n 2 1 = − n 2 13.6 eV , n = 1 , 2 , 3 , …
Degeneracy: Each energy level E n E_n E n has degeneracy n 2 n^2 n 2 (ignoring spin). The quantum numbers are:
Principal: n = 1 , 2 , 3 , … n = 1, 2, 3, \ldots n = 1 , 2 , 3 , …
Orbital angular momentum: l = 0 , 1 , … , n − 1 l = 0, 1, \ldots, n - 1 l = 0 , 1 , … , n − 1
Magnetic: m l = − l , … , l m_l = -l, \ldots, l m l = − l , … , l
The ground state wave function (n = 1 , l = 0 , m l = 0 n = 1, l = 0, m_l = 0 n = 1 , l = 0 , m l = 0 ):
ψ 100 ( r , θ , ϕ ) = 1 π a 0 3 e − r / a 0 \psi_{100}(r, \theta, \phi) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} ψ 100 ( r , θ , ϕ ) = π a 0 3 1 e − r / a 0
where a 0 = 4 π ε 0 ℏ 2 m e e 2 ≈ 0.529 A ˚ a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \approx 0.529\,\mathrm{\AA} a 0 = m e e 2 4 π ε 0 ℏ 2 ≈ 0.529 A ˚ 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 / 2 1/2 1/2 particles
(e.g., electrons):
S ^ x = ℏ 2 σ x , S ^ y = ℏ 2 σ y , S ^ z = ℏ 2 σ z \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 S ^ x = 2 ℏ σ x , S ^ y = 2 ℏ σ y , S ^ z = 2 ℏ σ z
where σ x , σ y , σ z \sigma_x, \sigma_y, \sigma_z σ x , σ y , σ z are the Pauli matrices :
σ x = ( 0 1 1 0 ) , σ y = ( 0 − i i 0 ) , σ z = ( 1 0 0 − 1 ) \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} σ x = ( 0 1 1 0 ) , σ y = ( 0 i − i 0 ) , σ z = ( 1 0 0 − 1 )
7.2 Properties of Pauli Matrices
σ i 2 = I , σ i σ j = i ϵ i j k σ k ( i ≠ j ) \sigma_i^2 = I, \quad \sigma_i \sigma_j = i\epsilon_{ijk}\sigma_k \quad (i \neq j) σ i 2 = I , σ i σ j = i ϵ ij k σ k ( i = j )
[ σ i , σ j ] = 2 i ϵ i j k σ k , { σ i , σ j } = 2 δ i j I [\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k, \quad \{\sigma_i, \sigma_j\} = 2\delta_{ij}I [ σ i , σ j ] = 2 i ϵ ij k σ k , { σ i , σ j } = 2 δ ij I
Spin states: ∣ ↑ ⟩ = ( 1 0 ) |\uparrow\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} ∣ ↑ ⟩ = ( 1 0 ) (spin up, m s = + 1 / 2 m_s = +1/2 m s = + 1/2 ) and
∣ ↓ ⟩ = ( 0 1 ) |\downarrow\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} ∣ ↓ ⟩ = ( 0 1 ) (spin down, m s = − 1 / 2 m_s = -1/2 m 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 + λ H ^ ′ \hat{H} = \hat{H}_0 + \lambda \hat{H}' H ^ = H ^ 0 + λ H ^ ′ where H ^ ′ \hat{H}' H ^ ′ is "small" and H ^ 0 \hat{H}_0 H ^ 0
has known eigenstates ∣ n ( 0 ) ⟩ |n^{(0)}\rangle ∣ n ( 0 ) ⟩ and eigenvalues E n ( 0 ) E_n^{(0)} E n ( 0 ) .
First-order energy correction:
E n ( 1 ) = ⟨ n ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩ E_n^{(1)} = \langle n^{(0)} | \hat{H}' | n^{(0)} \rangle E n ( 1 ) = ⟨ n ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩
Second-order energy correction:
E n ( 2 ) = ∑ m ≠ n ∣ ⟨ m ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩ ∣ 2 E n ( 0 ) − E m ( 0 ) E_n^{(2)} = \sum_{m \neq n} \frac{|\langle m^{(0)} | \hat{H}' | n^{(0)} \rangle|^2}{E_n^{(0)} - E_m^{(0)}} E n ( 2 ) = ∑ m = n E n ( 0 ) − E m ( 0 ) ∣ ⟨ m ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩ ∣ 2
First-order state correction:
∣ n ( 1 ) ⟩ = ∑ m ≠ n ⟨ m ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩ E n ( 0 ) − E m ( 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 ∣ n ( 1 ) ⟩ = ∑ m = n E n ( 0 ) − E m ( 0 ) ⟨ m ( 0 ) ∣ H ^ ′ ∣ n ( 0 ) ⟩ ∣ m ( 0 ) ⟩
8.2 Degenerate Perturbation Theory
When E n ( 0 ) E_n^{(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
W i j = ⟨ n i ( 0 ) ∣ H ^ ′ ∣ n j ( 0 ) ⟩ W_{ij} = \langle n_i^{(0)} | \hat{H}' | n_j^{(0)} \rangle W ij = ⟨ n i ( 0 ) ∣ H ^ ′ ∣ n j ( 0 ) ⟩ within the degenerate subspace.
8.3 Worked Example
Problem. A one-dimensional infinite square well of width L L L has a small perturbation
H ′ = V 0 H' = V_0 H ′ = V 0 for 0 < x < L / 2 0 \lt x \lt L/2 0 < x < L /2 and H ′ = 0 H' = 0 H ′ = 0 for L / 2 < x < L L/2 \lt x \lt L L /2 < x < L . Find the first-order energy
corrections.
Solution. The unperturbed states are ϕ n ( 0 ) ( x ) = 2 / L sin ( n π x / L ) \phi_n^{(0)}(x) = \sqrt{2/L}\sin(n\pi x/L) ϕ n ( 0 ) ( x ) = 2/ L sin ( nπ x / L ) .
E n ( 1 ) = ⟨ n ( 0 ) ∣ H ′ ∣ n ( 0 ) ⟩ = ∫ 0 L / 2 V 0 2 L sin 2 ( n π x L ) d x E_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 E n ( 1 ) = ⟨ n ( 0 ) ∣ H ′ ∣ n ( 0 ) ⟩ = ∫ 0 L /2 V 0 L 2 sin 2 ( L nπ x ) d x
= 2 V 0 L ∫ 0 L / 2 1 − cos ( 2 n π x / L ) 2 d x = V 0 L [ L 2 − L 4 n π sin ( n π ) ] = V 0 2 = \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} = L 2 V 0 ∫ 0 L /2 2 1 − c o s ( 2 nπ x / L ) d x = L V 0 [ 2 L − 4 nπ L sin ( nπ ) ] = 2 V 0
The first-order correction is E n ( 1 ) = V 0 / 2 E_n^{(1)} = V_0/2 E n ( 1 ) = V 0 /2 for all n n n . ■ \blacksquare ■
Perturbation theory assumes the perturbation is "small" compared to the level spacing. If
∣ ⟨ m ∣ H ′ ∣ n ⟩ ∣ ∼ ∣ E n ( 0 ) − E m ( 0 ) ∣ |\langle m | H' | n \rangle| \sim |E_n^{(0)} - E_m^{(0)}| ∣ ⟨ m ∣ H ′ ∣ n ⟩ ∣ ∼ ∣ 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.