University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.1
The Heisenberg picture, conservation and Ehrenfest's theorem
Quantum Mechanics I moved the ket and held every operator still. This lesson flips the bookkeeping: the state is frozen and each observable carries its own equation of motion. Conservation laws become one-line commutator tests, and you can say exactly how far a quantum mean may stray from the classical path.
Build the model
Connect the measurement to the mechanism.
Nothing measurable is a ket. Every prediction is a mean ⟨ψ(t)|A|ψ(t)⟩ or a probability |⟨a|ψ(t)⟩|², and in both the evolution operator can be moved off the state and onto the observable: AH(t) = U†(t)AU(t), with the ket frozen at |ψ(0)⟩. Differentiating that definition with nothing but iħ dU/dt = HU gives dAH/dt = (i/ħ)[HH, AH] + (∂A/∂t)H, Hamilton's equations with a commutator standing where the Poisson bracket stood, and valid even when H depends on time.
Two payoffs follow. An observable with no explicit time dependence that commutes with H is conserved as an operator, so the whole distribution of its outcomes is frozen, not only its mean. And for H = p²/2m + V(x) the operator equations are exactly Newton's, dxH/dt = pH/m and dpH/dt = −V′(xH).
The model stops at the averaging step. d⟨p⟩/dt = −⟨V′(x)⟩ is the force averaged over the packet, which equals the force at the packet's centre for every packet only when V′ is linear in x. So the mean follows the classical path exactly for free, uniform-field and harmonic motion, and elsewhere only approximately, for a packet narrow on the scale over which the force bends and only until it spreads.
- Simple definition
- In the Heisenberg picture the state stays at |ψ(0)⟩ and each observable carries the time dependence, AH(t) = U†(t)AU(t), evolving by dAH/dt = (i/ħ)[HH, AH] + (∂A/∂t)H, with every prediction the same as in the Schrödinger picture.
- Example
- For a free electron dxH/dt = pH/m gives xH(t) = x + pt/m, so an unchirped packet 0.50 nm wide, with σₚ = ħ/2σₓ = 1.05 × 10⁻²⁵ kg m s⁻¹, is √(0.50² + 2.32²) nm = 2.37 nm wide after 20 fs, with no Fourier transform needed.
⟨ψ(t)|AS|ψ(t)⟩ = ⟨ψ(0)|AH(t)|ψ(0)⟩ and AH(t) keeps the eigenvalues of A, so every prediction is unchanged. Only the bookkeeping moves.
U is dimensionless and unitary, iħ dU/dt = HS U with U(0) = I; the ket stays at |ψ(0)⟩; AH has the SI unit of A
Derived from iħ dU/dt = HS U alone, so it holds for a time-dependent H. When H is time-independent, U commutes with it and HH = HS.
HH = U†HS U in J; ħ = 1.055 × 10⁻³⁴ J s; every term carries the unit of A per second
The operator stands still, so every outcome probability ⟨ψ(0)|Πₐ|ψ(0)⟩ is frozen, not just ⟨A⟩. A mean that happens to be constant in one state proves nothing.
For a time-dependent H the commutator must vanish at every t; Πₐ, the projector onto eigenvalue a, then commutes with H too
From [p, V(x)] = −iħ dV/dx. The right side is the force averaged over the packet; it equals the force at ⟨x⟩ for every state only when dV/dx is linear in x.
p in kg m s⁻¹, V in J, x in m; the brackets average over the frozen state, so it is exact for every state
The operator obeys the classical solution, so ⟨x⟩(t) = ⟨x⟩₀ cos ωt + (⟨p⟩₀/mω) sin ωt for every packet, however wide or oddly shaped.
ω = √(k/m) in rad s⁻¹; x and p are the operators at t = 0; ω → 0 recovers the free result x + pt/m
The 3⟨x⟩σₓ² and μ₃ terms are what a classical particle at ⟨x⟩ never feels. They vanish as λ → 0, and for a coherent state μ₃ = 0 while the width term is of order ħ.
V = ½kx² + λx⁴; k in N m⁻¹, λ in J m⁻⁴; σₓ² in m²; μ₃ = ⟨(x − ⟨x⟩)³⟩ in m³
Move U from the state onto the operator
No experiment reports a ket. It reports a mean ⟨ψ(t)|A|ψ(t)⟩ or a probability |⟨a|ψ(t)⟩|², and with |ψ(t)⟩ = U(t)|ψ(0)⟩ from Quantum Mechanics I the mean regroups as ⟨ψ(0)|U†AU|ψ(0)⟩. Define AH(t) = U†(t)AU(t): the ket stays at |ψ(0)⟩, the operator moves, and the two pictures agree at t = 0 and give the same number at every t. The spectrum survives because U†AU is a unitary similarity transform: AH(t) has the eigenvalues of A, with eigenkets U†(t)|a⟩, and the probability |⟨a|U(t)|ψ(0)⟩|² is the squared modulus of the same amplitude in either picture. Since UU† = I, products map to products, (AB)H = AH BH, so every commutator keeps its value, [xH(t), pH(t)] = iħ at all t, and a function of an operator follows its argument, V(x)H = V(xH). Nothing physical has changed. What has changed is that each observable now obeys an equation of motion, which can often be solved without ever finding a state.
The equation of motion needs only iħ dU/dt = HU
Differentiate AH = U†AU by the product rule. The evolution operator obeys iħ dU/dt = HS U, whose adjoint is −iħ dU†/dt = U†HS. Substituting, and inserting UU† = I on either side of A, gives dAH/dt = (i/ħ)(U†HS U U†AU − U†AU U†HS U) + U†(∂A/∂t)U, which is (i/ħ)[HH, AH] + (∂A/∂t)H. Nothing assumed U = exp(−iHt/ħ), so the result holds for a time-dependent H, whose U Unit 12 constructs; there HH = U†HS(t)U need not equal HS(t). For a time-independent H, U is a function of H and commutes with it, so HH = HS. Dimensions: [H, A] carries joules times the unit of A, and dividing by ħ in J s leaves the unit of A per second. Put A = H and the commutator vanishes: dHH/dt = (∂H/∂t)H, so energy is conserved exactly when H has no explicit time dependence, and a drive −F(t)x changes it at d⟨H⟩/dt = −(dF/dt)⟨x⟩. The combination (1/iħ)[A, H] stands exactly where Hamilton's mechanics has the Poisson bracket of A with H.
A conserved observable freezes its whole distribution
If ∂A/∂t = 0 and [H, A] = 0, then dAH/dt = 0 and AH(t) = A for every t and every state. That is stronger than a constant mean. The projector Πₐ onto the eigenvalue a is a function of A, so it commutes with H as well, and each outcome probability ⟨ψ(0)|Πₐ|ψ(0)⟩ is fixed: the whole distribution of A is frozen, every moment included. Each conserved quantity is a symmetry of H. The momentum of a free particle survives because V is unchanged by translation; L² and all three components of L survive in a central V; Lz alone survives when V is symmetric only about the z axis; parity survives when V(−x) = V(x); and H survives whenever it has no explicit time dependence. The converse fails. In an energy eigenstate every ⟨A⟩ is constant, ⟨x⟩ included, because U acts on that ket as a phase; that is a property of the state, not of x. And because commuting operators share an eigenbasis, conserved quantities label the energy eigenstates, which is why hydrogen's carry l and m beside n.
A term that breaks a symmetry sets how fast a quantity drifts
Add to H a term that fails to commute with A, and that commutator is the rate at which A changes. Hydrogen's spin-orbit term ξ(r)L⋅S is the standard case; it is quoted here from Unit 8, which shows it is of order α² times the Bohr energies. From [Lz, Lₓ] = iħLy and [Lz, Ly] = −iħLₓ, [Lz, L⋅S] = iħ(LySₓ − LₓSy), and the same steps for spin give [Sz, L⋅S] = −iħ(LySₓ − LₓSy). Neither Lz nor Sz is conserved, but Jz = Lz + Sz is: whatever the spin gains, the orbit loses. The size of the term sets the clock. The measured 2p₃/₂ to 2p₁/₂ interval is 10.97 GHz, or 4.54 × 10⁻⁵ eV, which is 1.334 × 10⁻⁵ of the 3.40 eV binding of n = 2, beside the α²/4 = 1.331 × 10⁻⁵ that Unit 8's fine-structure formula gives. So a 2p electron prepared with mₗ = 1 and mₛ = −½ keeps its mₗ only for times short compared with ħ/ΔE = 14.5 ps, while mⱼ stays sharp.
Ehrenfest: Newton's law for operators, not for averages
For H = p²/2m + V(x) two commutators do all the work. [x, p²] = 2iħp gives dxH/dt = pH/m, and [p, V(x)] = −iħ dV/dx, from p = −iħ d/dx acting on a product Vf, gives dpH/dt = −V′(xH). As operator equations these are Hamilton's equations exactly. Averaging over the frozen state gives Ehrenfest's theorem, d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨V′(x)⟩, true for every state. The catch is the last term: it is the force averaged over the packet, not the force at ⟨x⟩. Expand V′ about ⟨x⟩ and average term by term: ⟨V′⟩ = V′(⟨x⟩) + ½V‴(⟨x⟩)σₓ² + (1/6)V⁗(⟨x⟩)μ₃ + …, with μ₃ = ⟨(x − ⟨x⟩)³⟩. Every correction needs a derivative of V beyond the second, so for V at most quadratic, which covers free motion, a uniform field and a harmonic well, the mean obeys Newton's law for every packet. The uniform-field case is measured: Peters, Chung and Chu (1999) found g measured with interfering caesium atoms to agree with a falling corner cube's within about 1 part in 10⁸.
In a quartic well the packet's shape enters the force
Take V = ½kx² + λx⁴, so V′ = kx + 4λx³. Writing x = ⟨x⟩ + Δx and cubing gives ⟨x³⟩ = ⟨x⟩³ + 3⟨x⟩σₓ² + μ₃ exactly, so d⟨p⟩/dt = −k⟨x⟩ − 4λ(⟨x⟩³ + 3⟨x⟩σₓ² + μ₃). A classical particle at ⟨x⟩ feels only the first two terms. For λ > 0 the width term always adds restoring force: a Gaussian with σₓ = 0.30⟨x⟩ is pulled back by an extra 27% of the anharmonic force. The Ehrenfest equations also stop closing: d⟨p⟩/dt needs σₓ², whose rate needs ⟨xp + px⟩, whose rate needs ⟨x⁴⟩, without end, so ⟨x⟩ and ⟨p⟩ alone cannot predict ⟨x⟩(t). A wide packet samples a range of periods and dephases, and its mean oscillation collapses while ⟨H⟩ stays fixed, which no single classical orbit does. Tracking holds while σₓ²|V‴| ≪ |V′| at ⟨x⟩, and only until spreading breaks that. For a coherent state σₓ² = ħ/2mω, so the correction is of order ħ. Unit 11's stationary-phase argument gives a classical limit free of the quadratic restriction.
Change one variable at a time
Make the relationship visible.
Set β = 0 and widen σ: dot and ring never part, because a straight-line force averages to its value at the centre. Raise β and the ring moves beyond the dot, away from the axis: the packet is pulled back harder than a classical particle. At ⟨x⟩ = 0 both vanish, yet the relative gap stays 3βσ².
FORCE AT CENTRE F(⟨x⟩)-1.107 kℓ
MEAN FORCE ⟨F⟩ = d⟨p⟩/dt-1.467 kℓ
GAP ⟨F⟩ − F(⟨x⟩)-0.360 kℓ
RELATIVE GAP32.5 %
Live interpretationFORCE AT CENTRE F(⟨x⟩): −1.107 kℓ. MEAN FORCE ⟨F⟩ = d⟨p⟩/dt: −1.467 kℓ. GAP ⟨F⟩ − F(⟨x⟩): −0.360 kℓ. RELATIVE GAP: 32.5 %
Catch the common trap
Explain before calculating.
A Gaussian packet in the pure quartic well V = λx⁴ has ⟨x⟩ = 2.0 nm, σₓ = 0.80 nm and ⟨p⟩ = 0 at t = 0. What is the ratio of |d⟨p⟩/dt| at that instant to the magnitude of the force a classical particle at x = 2.0 nm would feel?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA ⁸⁷Rb atom of mass 1.443 × 10⁻²⁵ kg is released from rest in an unchirped Gaussian packet of width σₓ = 5.0 μm along the vertical x axis, pointing up, with σₓσₚ = ħ/2. With H = p²/2m + mgx and g = 9.81 m s⁻², use the Heisenberg equations to find how far ⟨x⟩ falls and what σₓ is after 0.20 s, and say which depends on g.
- Commutators: (i/ħ)[p²/2m, x] = p/m and (i/ħ)[mgx, p] = (i/ħ)(iħmg) = −mg, so dxH/dt = pH/m and dpH/dt = −mg. The second right side is a number times the identity, so pH(t) = p − mgt.
- Integrate again: xH(t) = x + pt/m − ½gt², the classical free-fall solution as an operator identity. Averaging, ⟨x⟩(t) = ⟨x⟩₀ + ⟨p⟩₀t/m − ½gt² for any packet, so ⟨x⟩ falls ½ × 9.81 m s⁻² × (0.20 s)² = 0.196 m, exactly as a dropped stone does.
- Spread: −½gt² shifts every outcome equally, so it drops out of ΔxH = Δx + Δp t/m. Squaring and averaging, σₓ(t)² = σₓ² + (t/m)⟨ΔxΔp + ΔpΔx⟩ + σₚ²t²/m², and the middle term is zero for an unchirped packet.
- σₚ = ħ/2σₓ = 1.0546 × 10⁻³⁴ J s ÷ (1.0 × 10⁻⁵ m) = 1.05 × 10⁻²⁹ kg m s⁻¹, so σₚ t/m = 1.05 × 10⁻²⁹ × 0.20 ÷ (1.443 × 10⁻²⁵) m = 14.6 μm.
- σₓ(0.20 s) = √(5.0² + 14.6²) μm = 15.4 μm. Checks: g appears nowhere in σₓ, ħ appears nowhere in ⟨x⟩, and t → 0 returns 5.0 μm.
Answer⟨x⟩ falls 0.196 m, the classical distance, and σₓ grows from 5.0 μm to 15.4 μm. Gravity moves only the mean; the spreading is the free-particle spreading and does not depend on g.
MediumA ⁴⁰Ca⁺ ion of mass 6.636 × 10⁻²⁶ kg sits in a harmonic trap with ω = 2π × 1.00 MHz. It is prepared at rest, ⟨p⟩ = 0, centred at ⟨x⟩ = 50.0 nm, in an unchirped Gaussian of width σₓ = 5.62 nm, half the ground-state width, with σₓσₚ = ħ/2. Solve the Heisenberg equations as operators, find ⟨x⟩, ⟨p⟩ and σₓ a quarter-period later, and show that ⟨H⟩ has not changed.
- dxH/dt = pH/m and dpH/dt = −mω²xH are linear, so they solve as operators: xH(t) = x cos ωt + (p/mω) sin ωt and pH(t) = p cos ωt − mωx sin ωt. Check: [xH(t), pH(t)] = iħ(cos²ωt + sin²ωt) = iħ, as unitarity requires.
- A quarter-period is T/4 = 0.250 μs, where ωt = π/2, so xH = p/mω and pH = −mωx. Then ⟨x⟩ = ⟨p⟩₀/mω = 0 and ⟨p⟩ = −mω⟨x⟩₀ = −(4.170 × 10⁻¹⁹ kg s⁻¹)(5.00 × 10⁻⁸ m) = −2.08 × 10⁻²⁶ kg m s⁻¹, a speed of ω⟨x⟩₀ = 0.314 m s⁻¹: the classical values, whatever the width.
- Width: σₓ(T/4) = σₚ/mω = ħ/(2σₓ mω). With ħ/mω = 2.529 × 10⁻¹⁶ m², σₓ(T/4) = 2.529 × 10⁻¹⁶ m² ÷ (2 × 5.62 × 10⁻⁹ m) = 22.5 nm. In general σₓ(t)² = σₓ² cos²ωt + (σₚ/mω)² sin²ωt, so the packet breathes at 2ω while its centre swings at ω.
- Energy at t = 0: ⟨x²⟩ = ⟨x⟩² + σₓ² = 2.532 × 10⁻¹⁵ m², so ½mω²⟨x²⟩ = 3.316 × 10⁻²⁷ J; σₚ = ħ/2σₓ = 9.38 × 10⁻²⁷ kg m s⁻¹, so ⟨p²⟩/2m = 6.63 × 10⁻²⁸ J. The total is 3.979 × 10⁻²⁷ J.
- At T/4, ⟨p²⟩ = m²ω²⟨x²⟩₀ and ⟨x²⟩ = σₚ²/m²ω², so the kinetic and potential shares have swapped and the total is still 3.979 × 10⁻²⁷ J = 6.01ħω. H has no explicit time dependence and commutes with itself, so this is required, not lucky.
AnswerAt t = 0.250 μs: ⟨x⟩ = 0, ⟨p⟩ = −2.08 × 10⁻²⁶ kg m s⁻¹ (0.314 m s⁻¹) and σₓ = 22.5 nm. ⟨H⟩ = 3.98 × 10⁻²⁷ J = 6.01ħω at both times.
HardFor H = p²/2m + ½mω²x² + λx⁴ with g = λħ/(m²ω³) = 0.0100, a coherent state (a ground-state Gaussian, σₓ = ℓ/√2 with ℓ = √(ħ/mω)) is centred at ⟨x⟩ = X with ⟨p⟩ = 0. Find d⟨p⟩/dt in units of ħω/ℓ, its excess over the classical force at X = 3.0ℓ, how that excess scales with X and ħ, and the X beyond which it falls below 1%.
- Scale out the units. With x = ℓu, ½mω²x² = ½ħω u² and λx⁴ = λℓ⁴u⁴ = għω u⁴, since λℓ⁴ = λħ²/(m²ω²) = għω. The force is −dV/dx = −(ħω/ℓ)(u + 4gu³), and ħω/ℓ is in J m⁻¹, a force, as it must be.
- Ehrenfest: d⟨p⟩/dt = −(ħω/ℓ)(⟨u⟩ + 4g⟨u³⟩), with ⟨u³⟩ = U³ + 3Uσᵤ² + μ₃ and U = X/ℓ. A Gaussian has μ₃ = 0, and here σᵤ² = ½, so d⟨p⟩/dt = −(ħω/ℓ)(U + 4gU³ + 6gU).
- At U = 3.0 the classical force is 3.000 + 4(0.0100)(27.0) = 4.080 in units of ħω/ℓ, and the width term adds 6(0.0100)(3.0) = 0.180, giving 4.260. The mean is pulled back 4.4% harder than a classical particle at 3.0ℓ, and the width term is 17% of the anharmonic force alone.
- Scaling: the excess is 6gU/(U + 4gU³) = 6g/(1 + 4gU²). Limits: g → 0 gives zero, the harmonic well where Ehrenfest closes; large U gives 3/(2U²) = 3ħ/(2mωX²), independent of λ and proportional to ħ, so it vanishes in the classical limit at fixed amplitude.
- Below 1%: 0.0600/(1 + 0.0400U²) < 0.0100 needs U² > 125, so X > 11.2ℓ, or 178 nm for the ℓ = 15.9 nm of the Medium example's trap. The packet starts only 0.71ℓ wide; what must be small is σₓ²/X², not σₓ itself, and only at this instant, since σₓ grows as the packet dephases.
Answerd⟨p⟩/dt = −(ħω/ℓ)(U + 4gU³ + 6gU) = −4.26ħω/ℓ at X = 3.0ℓ, 4.4% beyond the classical 4.08ħω/ℓ. The excess, 6g/(1 + 4gU²), falls below 1% for X > 11.2ℓ and scales as ħ at large amplitude.