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Quantum Dynamics

Quantum dynamics describes how a quantum state evolves in time. It is relevant whenever you need to predict the time-dependence of a system: the oscillation of a qubit in a magnetic field, the scattering of two particles, the tunneling of an electron through a barrier, or the propagation of a wave packet. The dynamics is governed by the Hamiltonian, which encodes the interactions (see Interactions).

The Schrödinger equation

Definition. The state ∣ψ(t)⟩|\psi(t)\rangle evolves according to iℏ ∂t∣ψ(t)⟩=H^∣ψ(t)⟩i\hbar\,\partial_t |\psi(t)\rangle = \hat{H}|\psi(t)\rangle, where H^\hat{H} is the Hamiltonian operator. The formal solution is ∣ψ(t)⟩=U^(t,t0)∣ψ(t0)⟩|\psi(t)\rangle = \hat{U}(t,t_0)|\psi(t_0)\rangle with U^(t,t0)=Texp⁡(−iℏ∫t0tH^(t′) dt′)\hat{U}(t,t_0) = \mathcal{T}\exp\left(-\frac{i}{\hbar}\int_{t_0}^{t}\hat{H}(t')\,dt'\right). For a time-independent Hamiltonian: U^(t)=e−iH^t/ℏ\hat{U}(t) = e^{-i\hat{H}t/\hbar}.

Details

From the classical Hamiltonian to the quantum one

In classical mechanics, the state evolves via Hamilton’s equations: q˙=∂H/∂p\dot{q} = \partial H/\partial p, p˙=−∂H/∂q\dot{p} = -\partial H/\partial q. In quantum mechanics, the state is a vector in Hilbert space (see Quantum states) and evolves via the Schrödinger equation. The classical Hamiltonian function H(q,p)H(q,p) becomes an operator H^\hat{H} by the correspondence q→q^q \to \hat{q}, p→p^=−iℏ ∂/∂xp \to \hat{p} = -i\hbar\,\partial/\partial x.

Step 1: Time-independent case.

If H^\hat{H} does not depend on time, the evolution operator is U^(t)=e−iH^t/ℏ\hat{U}(t) = e^{-i\hat{H}t/\hbar}. If the initial state is an energy eigenstate H^∣n⟩=En∣n⟩\hat{H}|n\rangle = E_n|n\rangle, then ∣ψ(t)⟩=e−iEnt/ℏ∣n⟩|\psi(t)\rangle = e^{-iE_n t/\hbar}|n\rangle — the state picks up only a global phase, and all measurable probabilities are constant. Superpositions of energy eigenstates, however, oscillate: ∣ψ(t)⟩=c1e−iE1t/ℏ∣1⟩+c2e−iE2t/ℏ∣2⟩|\psi(t)\rangle = c_1 e^{-iE_1 t/\hbar}|1\rangle + c_2 e^{-iE_2 t/\hbar}|2\rangle has a time-dependent relative phase e−i(E2−E1)t/ℏe^{-i(E_2-E_1)t/\hbar} that produces oscillating probabilities.

Step 2: Time-dependent case.

When H^\hat{H} depends on time (e.g., a time-varying electromagnetic field), the evolution operator involves a time-ordering T\mathcal{T}. There is no simple exponential form, and one must use perturbation theory or numerical methods.

Example: a two-level system in a resonant field.

A qubit with energy splitting ℏω0\hbar\omega_0 is driven by an oscillating field of frequency ω≈ω0\omega \approx \omega_0. In the rotating-wave approximation, the effective Hamiltonian in the interaction picture is H^eff=ℏ2(ΔΩΩ−Δ)\hat{H}_{\text{eff}} = \frac{\hbar}{2}\begin{pmatrix} \Delta & \Omega \\ \Omega & -\Delta \end{pmatrix} where Δ=ω−ω0\Delta = \omega - \omega_0 is the detuning and Ω\Omega is the Rabi frequency. Starting in ∣0⟩|0\rangle, the probability of finding the system in ∣1⟩|1\rangle oscillates as P1(t)=Ω2Ω2+Δ2sin⁡2(Ω2+Δ2 t2)P_1(t) = \frac{\Omega^2}{\Omega^2 + \Delta^2}\sin^2\left(\frac{\sqrt{\Omega^2+\Delta^2}\,t}{2}\right). At resonance (Δ=0\Delta = 0), full population transfer occurs at t=π/Ωt = \pi/\Omega (a π\pi-pulse).

P₁(t) t Rabi oscillations (resonant drive) π-pulse: full transfer

Example problem.

Problem

A spin-½ particle in a uniform magnetic field B=B0z^\mathbf{B} = B_0\hat{z} has Hamiltonian H^=−γB0S^z=−ℏω02σz\hat{H} = -\gamma B_0 \hat{S}_z = -\frac{\hbar\omega_0}{2}\sigma_z where ω0=γB0\omega_0 = \gamma B_0. If the initial state is ∣↑x⟩=(∣↑z⟩+∣↓z⟩)/2|\uparrow_x\rangle = (|\uparrow_z\rangle + |\downarrow_z\rangle)/\sqrt{2}, find ∣ψ(t)⟩|\psi(t)\rangle and the probability of measuring Sz=+ℏ/2S_z = +\hbar/2.

Solution

∣ψ(t)⟩=e−iH^t/ℏ∣↑x⟩=eiω0tσz/212(11)=12(eiω0t/2e−iω0t/2)|\psi(t)\rangle = e^{-i\hat{H}t/\hbar}|\uparrow_x\rangle = e^{i\omega_0 t\sigma_z/2}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} e^{i\omega_0 t/2} \\ e^{-i\omega_0 t/2} \end{pmatrix}

P(↑z)=∣⟨↑z∣ψ(t)⟩∣2=12∣eiω0t/2∣2=12P(\uparrow_z) = |\langle\uparrow_z|\psi(t)\rangle|^2 = \frac{1}{2}|e^{i\omega_0 t/2}|^2 = \frac{1}{2}

The probability is constant at 1/2 — the spin precesses around z at frequency ω0\omega_0, but the z-component remains equally likely to be up or down. However, ⟨Sx⟩(t)=ℏ2cos⁡(ω0t)\langle S_x\rangle(t) = \frac{\hbar}{2}\cos(\omega_0 t) oscillates, showing the precession.

Time-dependent perturbation theory

Definition. For H^=H^0+V^(t)\hat{H} = \hat{H}_0 + \hat{V}(t), the transition amplitude from state ∣i⟩|i\rangle to state ∣f⟩|f\rangle (first order) is cf(1)(t)=−iℏ∫t0t⟨f∣V^(t′)∣i⟩ eiωfit′ dt′c_f^{(1)}(t) = -\frac{i}{\hbar}\int_{t_0}^{t}\langle f|\hat{V}(t')|i\rangle\,e^{i\omega_{fi}t'}\,dt' where ωfi=(Ef−Ei)/ℏ\omega_{fi} = (E_f - E_i)/\hbar. Fermi’s golden rule gives the transition rate: Γi→f=2πℏ∣⟨f∣V^∣i⟩∣2ρ(Ef)\Gamma_{i\to f} = \frac{2\pi}{\hbar}|\langle f|\hat{V}|i\rangle|^2\rho(E_f).”

Details

When the Hamiltonian changes slowly

Many physical situations involve a small time-dependent perturbation: an atom in a weak electromagnetic wave, a particle encountering a short potential barrier, a qubit subjected to a control pulse. Exact solutions are rare; perturbation theory expands in powers of V^\hat{V}.

Step 1: First-order amplitude.

Assume the system starts in an eigenstate ∣i⟩|i\rangle of H^0\hat{H}_0 at t=−∞t = -\infty. The first-order correction to the amplitude of finding it in state ∣f⟩|f\rangle at time tt is an integral of the matrix element of V^\hat{V} weighted by the phase factor eiωfit′e^{i\omega_{fi}t'}. The integral is large when ωfi≈0\omega_{fi} \approx 0 (energy conservation) — this is the resonance condition.

Step 2: Fermi’s golden rule.

For a perturbation that is on for a long time TT, the transition probability grows as T2 sinc2(ωfiT/2)T^2\,\text{sinc}^2(\omega_{fi}T/2). The rate (probability per unit time) in the limit T→∞T \to \infty becomes proportional to 2πδ(Ef−Ei)2\pi\delta(E_f - E_i), which for a continuum of final states gives the density of states ρ(Ef)\rho(E_f). This is Fermi’s golden rule.

Example: spontaneous emission.

An excited atom coupled to the electromagnetic vacuum (a time-dependent perturbation from the quantized field) decays with rate Γ=ω03∣d∣23πϵ0ℏc3\Gamma = \frac{\omega_0^3|\mathbf{d}|^2}{3\pi\epsilon_0\hbar c^3} where d\mathbf{d} is the dipole matrix element. This is a first-order process in the field coupling.

Example problem.

Problem

A particle in a 1D infinite well of width aa is in the ground state ψ1(x)=2/asin⁡(πx/a)\psi_1(x) = \sqrt{2/a}\sin(\pi x/a). A time-dependent perturbation V^(t)=V0x e−t2/τ2\hat{V}(t) = V_0 x\,e^{-t^2/\tau^2} is applied. Write the first-order amplitude for transition to the first excited state ψ2(x)=2/asin⁡(2πx/a)\psi_2(x) = \sqrt{2/a}\sin(2\pi x/a).

Solution

The matrix element is:

⟨2∣x∣1⟩=2a∫0asin⁡(2πx/a) x sin⁡(πx/a) dx\langle 2|x|1\rangle = \frac{2}{a}\int_0^a \sin(2\pi x/a)\,x\,\sin(\pi x/a)\,dx

Using sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]:

=1a∫0ax[cos⁡(πx/a)−cos⁡(3πx/a)] dx= \frac{1}{a}\int_0^a x[\cos(\pi x/a) - \cos(3\pi x/a)]\,dx

=1a[a2πsin⁡(πx/a)∣0a−a23πsin⁡(3πx/a)∣0a+aπ(1π−13π)]= \frac{1}{a}\left[\frac{a^2}{\pi}\sin(\pi x/a)\bigg|_0^a - \frac{a^2}{3\pi}\sin(3\pi x/a)\bigg|_0^a + \frac{a}{\pi}\bigg(\frac{1}{\pi} - \frac{1}{3\pi}\bigg)\right]

Actually, evaluating ∫0axcos⁡(nπx/a) dx=anπsin⁡(nπ)−a2n2π2[cos⁡(nπ)−1]=a2n2π2(1−cos⁡(nπ))\int_0^a x\cos(n\pi x/a)\,dx = \frac{a}{n\pi}\sin(n\pi) - \frac{a^2}{n^2\pi^2}[\cos(n\pi)-1] = \frac{a^2}{n^2\pi^2}(1-\cos(n\pi)):

For n=1n=1: a2π2⋅2=2a2/π2\frac{a^2}{\pi^2}\cdot 2 = 2a^2/\pi^2 For n=3n=3: a29π2⋅2=2a2/(9π2)\frac{a^2}{9\pi^2}\cdot 2 = 2a^2/(9\pi^2)

⟨2∣x∣1⟩=1a(2a2π2−2a29π2)=2aπ2⋅89=16a9π2\langle 2|x|1\rangle = \frac{1}{a}\left(\frac{2a^2}{\pi^2} - \frac{2a^2}{9\pi^2}\right) = \frac{2a}{\pi^2}\cdot\frac{8}{9} = \frac{16a}{9\pi^2}

The first-order amplitude is:

c2(1)(t)=−iV0ℏ⋅16a9π2∫−∞te−t′2/τ2 eiω21t′ dt′c_2^{(1)}(t) = -\frac{iV_0}{\hbar}\cdot\frac{16a}{9\pi^2}\int_{-\infty}^{t} e^{-t'^2/\tau^2}\,e^{i\omega_{21}t'}\,dt'

where ω21=(E2−E1)/ℏ=3π2ℏ/(2ma2)\omega_{21} = (E_2-E_1)/\hbar = 3\pi^2\hbar/(2ma^2). For τ≫1/ω21\tau \gg 1/\omega_{21} (slow perturbation), the transition is suppressed by energy non-conservation.

Scattering theory

Definition. Scattering describes the asymptotic behavior of a wave packet that approaches, interacts with, and recedes from a potential. The S-matrix S^\hat{S} maps incoming to outgoing states: ∣ψout⟩=S^∣ψin⟩|\psi_{\text{out}}\rangle = \hat{S}|\psi_{\text{in}}\rangle. The differential cross-section is dσ/dΩ=∣f(θ,ϕ)∣2d\sigma/d\Omega = |f(\theta,\phi)|^2 where ff is the scattering amplitude.

Details

From wave packets to plane waves

A scattering experiment sends a particle toward a target and measures where it ends up. Quantum mechanically, the incoming state is a wave packet (or, in the idealized limit, a plane wave). The outgoing state consists of an undistorted forward component plus a spherical outgoing wave f(θ,ϕ) eikr/rf(\theta,\phi)\,e^{ikr}/r.

Step 1: The scattering amplitude.

For a spherically symmetric potential V(r)V(r), the scattering amplitude in the partial-wave expansion is:

f(θ)=1k∑ℓ=0∞(2ℓ+1)eiδℓsin⁡δℓ Pℓ(cos⁡θ)f(\theta) = \frac{1}{k}\sum_{\ell=0}^{\infty}(2\ell+1)e^{i\delta_\ell}\sin\delta_\ell\,P_\ell(\cos\theta)

where δℓ\delta_\ell are the phase shifts and PℓP_\ell are Legendre polynomials. Each partial wave ℓ\ell contributes a term weighted by (2ℓ+1)(2\ell+1) and the phase shift δℓ\delta_\ell.

Step 2: Low-energy (s-wave) scattering.

At low energies, only ℓ=0\ell = 0 contributes. The cross-section becomes isotropic: σ=4πa2\sigma = 4\pi a^2 where a=−tan⁡δ0/ka = -\tan\delta_0/k is the scattering length. This single parameter characterizes low-energy scattering and is central to ultracold atom physics and neutron scattering.

Connection to the Standard Model.

In relativistic quantum field theory (see Quantum field theory), scattering is described by Feynman diagrams and the S-matrix elements are computed from loop integrals. The non-relativistic scattering theory above is the low-energy limit of the full QFT calculation.

Example problem.

Problem

A neutron of wavelength λ=1\lambda = 1 Å scatters from a small spherical potential well of depth V0=100V_0 = 100 meV and radius R=1R = 1 Å. In the s-wave approximation, estimate the scattering length if kR≪1kR \ll 1.

Solution

k=2π/λ=2π/10−10=6.28×1010k = 2\pi/\lambda = 2\pi/10^{-10} = 6.28\times10^{10} m⁻¹. kR=6.28×1010×10−10=6.28kR = 6.28\times10^{10}\times10^{-10} = 6.28. This is not ≪1\ll 1, so the s-wave approximation is marginal. For a true low-energy estimate, we would need λ≫R\lambda \gg R.

For a hard sphere of radius RR (infinite potential), a=Ra = R and σ=4πR2=4π(10−10)2≈1.3×10−19\sigma = 4\pi R^2 = 4\pi(10^{-10})^2 \approx 1.3\times10^{-19} m² = 130 barns.

For a finite well, the scattering length depends on the bound-state condition: if the well supports a virtual state near zero energy, aa can be much larger than RR (resonant scattering).

Path integrals

Definition. The path integral (Feynman) formulation expresses the propagator as a sum over all paths: K(xf,tf;xi,ti)=∫D[x(t)] exp⁡(iℏ∫titfL(x,x˙) dt)K(x_f,t_f;x_i,t_i) = \int\mathcal{D}[x(t)]\,\exp\left(\frac{i}{\hbar}\int_{t_i}^{t_f}L(x,\dot{x})\,dt\right). Each path contributes a phase proportional to the classical action. In the limit ℏ→0\hbar \to 0 (or large action), the path of stationary action (the classical trajectory) dominates — this is the correspondence principle.

Details

A third formulation of quantum mechanics

Alongside matrix mechanics (Heisenberg) and wave mechanics (Schrödinger), the path integral (Feynman, 1948) provides a conceptually distinct formulation. Instead of evolving a state vector, one sums over all possible histories weighted by eiS/ℏe^{iS/\hbar}.

Step 1: The propagator as a sum over paths.

The amplitude for a particle to go from (xi,ti)(x_i, t_i) to (xf,tf)(x_f, t_f) is:

K=lim⁡N→∞(m2πiℏϵ)N/2∫∏n=1N−1dxn exp⁡[iℏ∑n=0N−1(m(xn+1−xn)22ϵ−ϵV(xn))]K = \lim_{N\to\infty}\left(\frac{m}{2\pi i\hbar\epsilon}\right)^{N/2}\int\prod_{n=1}^{N-1}dx_n\,\exp\left[\frac{i}{\hbar}\sum_{n=0}^{N-1}\left(\frac{m(x_{n+1}-x_n)^2}{2\epsilon} - \epsilon V(x_n)\right)\right]

where ϵ=(tf−ti)/N\epsilon = (t_f-t_i)/N. Each term in the sum is the Lagrangian evaluated along a discrete path.

Step 2: The classical limit.

When the action S≫ℏS \gg \hbar, paths near the classical trajectory (where δS=0\delta S = 0) contribute constructively (their phases align), while paths far from the classical one cancel by destructive interference. The result is that the particle follows the classical trajectory — the correspondence principle connecting to Mechanics.

Step 3: Why path integrals matter beyond computation.

Path integrals are not merely a calculational tool for QFT. They provide:

Example: the free-particle propagator.

For V=0V = 0, the path integral can be evaluated exactly. The result is:

K(xf,tf;xi,ti)=m2πiℏ(tf−ti)exp⁡[im(xf−xi)22ℏ(tf−ti)]K(x_f,t_f;x_i,t_i) = \sqrt{\frac{m}{2\pi i\hbar(t_f-t_i)}}\exp\left[\frac{im(x_f-x_i)^2}{2\hbar(t_f-t_i)}\right]

This is a Gaussian in (xf−xi)(x_f - x_i) with a complex phase — the quantum analogue of the classical free-particle trajectory.

Example problem.

Problem

Using the path integral, show that for a free particle, the propagator at tf−ti=Tt_f - t_i = T satisfies the composition law K(xf,T1+T2;xi)=∫dx K(xf,T1;x) K(x,T2;xi)K(x_f,T_1+T_2;x_i) = \int dx\,K(x_f,T_1;x)\,K(x,T_2;x_i).

Solution

∫dx m2πiℏT1m2πiℏT2exp⁡[im2ℏ((xf−x)2T1+(x−xi)2T2)]\int dx\,\sqrt{\frac{m}{2\pi i\hbar T_1}}\sqrt{\frac{m}{2\pi i\hbar T_2}}\exp\left[\frac{im}{2\hbar}\left(\frac{(x_f-x)^2}{T_1}+\frac{(x-x_i)^2}{T_2}\right)\right]

Complete the square in the exponent. The coefficient of x2x^2 is −im/(2ℏ)(1/T1+1/T2)=−im/(2ℏ)⋅(T1+T2)/(T1T2)-im/(2\hbar)(1/T_1+1/T_2) = -im/(2\hbar)\cdot(T_1+T_2)/(T_1T_2). The linear term gives the shift. After Gaussian integration, the result is:

m2πiℏ(T1+T2)exp⁡[im(xf−xi)22ℏ(T1+T2)]=K(xf,T1+T2;xi)\sqrt{\frac{m}{2\pi i\hbar(T_1+T_2)}}\exp\left[\frac{im(x_f-x_i)^2}{2\hbar(T_1+T_2)}\right] = K(x_f,T_1+T_2;x_i)

The composition law is satisfied. ✓

Quiz

Test your understanding

Problem. A qubit with Hamiltonian H^=ℏω2σz\hat{H} = \frac{\hbar\omega}{2}\sigma_z is initially in state ∣0⟩=∣↑z⟩|0\rangle = |\uparrow_z\rangle. At time t=0t = 0, a sudden perturbation V^=ℏΩ σx\hat{V} = \hbar\Omega\,\sigma_x is applied for a duration τ=π/(2Ω)\tau = \pi/(2\Omega) (a π/2\pi/2-pulse). Find the state after the pulse (to first order in Ω/ω\Omega/\omega, assuming Ω≪ω\Omega \ll \omega).

Solution

During the pulse, the total Hamiltonian is H^+V^=ℏω2σz+ℏΩσx\hat{H} + \hat{V} = \frac{\hbar\omega}{2}\sigma_z + \hbar\Omega\sigma_x. To first order in Ω/ω\Omega/\omega, the dominant evolution is from H^\hat{H}: ∣ψ(t)⟩≈e−iωt/2∣0⟩|\psi(t)\rangle \approx e^{-i\omega t/2}|0\rangle. The perturbation gives a first-order correction:

∣ψ(τ)⟩≈e−iωτ/2[∣0⟩−iΩ∫0τdt eiωtσx∣0⟩]|\psi(\tau)\rangle \approx e^{-i\omega\tau/2}\left[|0\rangle - i\Omega\int_0^\tau dt\,e^{i\omega t}\sigma_x|0\rangle\right]

σx∣0⟩=∣1⟩\sigma_x|0\rangle = |1\rangle, so:

∣ψ(τ)⟩≈e−iωτ/2[∣0⟩−iΩ eiωτ/2eiωτ−1iω∣1⟩]|\psi(\tau)\rangle \approx e^{-i\omega\tau/2}\left[|0\rangle - i\Omega\,e^{i\omega\tau/2}\frac{e^{i\omega\tau}-1}{i\omega}|1\rangle\right]

For Ω≪ω\Omega \ll \omega and τ=π/(2Ω)\tau = \pi/(2\Omega), the integral ∫0τeiωtdt=(eiωτ−1)/(iω)≈−1/(iω)\int_0^\tau e^{i\omega t}dt = (e^{i\omega\tau}-1)/(i\omega) \approx -1/(i\omega) (since ωτ≫1\omega\tau \gg 1 and the oscillatory integral is small). So the correction is of order Ω/ω≪1\Omega/\omega \ll 1.

To leading order, the state remains approximately ∣0⟩|0\rangle (up to a phase). The π/2\pi/2-pulse would fully transfer population only if ω≈0\omega \approx 0 (resonant drive) — here the large detuning ω≫Ω\omega \gg \Omega suppresses the transition. This illustrates the importance of the resonance condition in time-dependent perturbation theory.

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