University Quantum Mechanics II · Composite Systems, Entanglement and Density Operators · 4.6

The von Neumann equation, and thermal states

The density operator gets its equation of motion here, and its sharpest consequence is a prohibition: no Hamiltonian, however it varies in time, changes how mixed a closed system's state is. The thermal state is the mixture that needs no motion at all, and a spin in a field shows how small its polarisation usually is.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An ensemble is a list of kets |ψₖ⟩ with weights pₖ, and ρ = Σ pₖ|ψₖ⟩⟨ψₖ|. Nothing inside a closed system reaches in and relabels its members, so the weights stay fixed while every ket obeys the Schrödinger equation; differentiating term by term gives iħ dρ/dt = [Ĥ, ρ], solved by ρ(t) = Ûρ(0)Û†. That solution is a similarity transformation, and it carries the whole argument.

The eigenvalues of ρ never change, so Tr ρ², the entropy and every other function of the spectrum are constants of the motion: a pure state stays pure and a mixture cannot be purified, whatever Ĥ is and however it varies in time. For a spin the Bloch vector turns about the field at fixed length, Unit 3's Larmor precession carried into the inside of the ball. States that commute with Ĥ do not move at all, and the canonical state e^(−βĤ)/Z, quoted from statistical physics, is one of them; for a spin-½ moment μ in a field B it has polarisation tanh(βμB), 0.22 per cent for an electron at 1 T and 300 K.

The model stops exactly where a real spin starts to relax. Getting from a pure, precessing state to the thermal one means changing the spectrum of ρ, which no unitary can do: it needs coupling to an environment, and this lesson neither derives the canonical state nor models the approach to it.

Simple definition: The von Neumann equation iħ dρ/dt = [Ĥ, ρ] moves a density operator as the Schrödinger equation moves a ket, turning ρ without changing its eigenvalues, so the states that never move are exactly those that commute with Ĥ, the thermal state e^(−βĤ)/Z among them.
Example: An electron spin at 1.00 T and 300 K has μB/(k_B T) = 9.285 × 10⁻²⁴ J ÷ 4.142 × 10⁻²¹ J = 2.24 × 10⁻³, so its thermal Bloch vector has length tanh(2.24 × 10⁻³) = 0.224% of the ball's radius and Tr ρ² = 0.5000025, a hair from fully mixed.

The von Neumann equationiħ dρ/dt = [Ĥ, ρ]

Derived by letting every member ket obey the Schrödinger equation with its weight pₖ fixed. The sign is opposite to Unit 3's dÂ_H/dt = (i/ħ)[Ĥ, Â_H].

ρ dimensionless, Ĥ in J, t in s, ħ = 1.055 × 10⁻³⁴ J s, so both sides are energies

Solution by the evolution operatorρ(t) = Û(t) ρ(0) Û†(t)

Derived by differentiating. A similarity transformation, so the eigenvalues of ρ, and with them Tr ρ², Tr ρⁿ and the entropy −Tr(ρ ln ρ), never change. Pure stays pure.

Û = e^(−iĤt/ħ) for a constant Ĥ, otherwise the solution of iħ dÛ/dt = ĤÛ, dimensionless and unitary

Matrix elements in the energy basisρₘₙ(t) = ρₘₙ(0) e^{(−i(Eₘ − Eₙ)t/ħ)}

Derived from the commutator in the energy basis. Each coherence turns at its Bohr frequency (Eₘ − Eₙ)/ħ in rad s⁻¹, so ρ is stationary exactly when it commutes with Ĥ.

For a constant Ĥ. Eₘ, Eₙ in J, t in s. Populations ρₙₙ never change, coherences keep their modulus

Bloch vector in a static fielddr/dt = Ω × r, for Ĥ = (ħ/2) Ω⋅σ

Derived from the Pauli algebra: rotation about Ω at fixed |r|, Larmor precession for mixed states too. Electron: Ω along B, |Ω|/B = 1.761 × 10¹¹ rad s⁻¹ T⁻¹, so |Ω|/(2πB) = 28.02 GHz T⁻¹.

r dimensionless with |r| ≤ 1, Ω in rad s⁻¹. For a moment γS in a field B, Ω = −γB

The canonical state, quotedρₜₕ = e^(−βĤ)/Z, with Z = Tr e^(−βĤ) and β = 1/(k_B T)

Taken from statistical physics as the state of a system held at temperature T, not derived here. A function of Ĥ, it commutes with Ĥ and so never moves.

β in J⁻¹, k_B = 1.381 × 10⁻²³ J K⁻¹, T in K, Z dimensionless

Thermal polarisation of a spin-½ momentP = tanh(μB/(k_B T))

Derived from the quoted canonical state. Tends to μB/(k_B T), Curie's law, when μB ≪ k_B T, and to 1 as T → 0. With 2μB = hf: 0.224% for an electron at 1 T and 300 K.

μ in J T⁻¹, measured: electron 9.285 × 10⁻²⁴, proton 1.411 × 10⁻²⁶. B in T, T in K, P = |r| dimensionless

Every member obeys the Schrödinger equation

Take ρ = Σ pₖ|ψₖ⟩⟨ψₖ| and ask what a closed system does to it. Each ket obeys iħ d|ψₖ⟩/dt = Ĥ|ψₖ⟩, and its adjoint obeys −iħ d⟨ψₖ|/dt = ⟨ψₖ|Ĥ because Ĥ is Hermitian. The weights record how the ensemble was assembled, and nothing inside a closed system relabels its members, so dpₖ/dt = 0. The product rule gives iħ d(|ψₖ⟩⟨ψₖ|)/dt = Ĥ|ψₖ⟩⟨ψₖ| − |ψₖ⟩⟨ψₖ|Ĥ, and summing with the weights, iħ dρ/dt = Ĥρ − ρĤ = [Ĥ, ρ]. Dimensions: ħ over a time is J s ÷ s = J, and [Ĥ, ρ] is an energy times a pure number. With a single member, ρ = |ψ⟩⟨ψ|, it is the Schrödinger equation again. The solution follows from iħ dÛ/dt = ĤÛ and its adjoint −iħ dÛ†/dt = Û†Ĥ: differentiating Ûρ(0)Û† gives ĤÛρ(0)Û† − Ûρ(0)Û†Ĥ, the same commutator, and at t = 0 it returns ρ(0). So ρ(t) = Ûρ(0)Û†, with Û = e^(−iĤt/ħ) when Ĥ is constant.

Why the sign is opposite to Heisenberg's

Unit 3's Heisenberg equation reads dÂ_H/dt = (i/ħ)[Ĥ, Â_H]; the von Neumann equation, divided through by iħ, reads dρ/dt = −(i/ħ)[Ĥ, ρ]. Both are right, because they keep opposite books. A Heisenberg-picture observable is Â_H(t) = Û†ÂÛ, with the adjoint on the left, while ρ is built from kets and carries Û on the left. The prediction is the same either way: ⟨A⟩(t) = Tr(Ûρ(0)Û†Â) = Tr(ρ(0)Û†ÂÛ) by the cyclic property of the trace, so a picture moves ρ or Â, never both. ρ is a Hermitian operator, but it describes the state rather than a quantity being measured, and in the Heisenberg picture it stays frozen at ρ(0). The sign is easy to test on a spin. For the electron in a field along +z, Unit 3 found ⟨S⟩ turning from +x towards +y. Written with ρ = (I + r⋅σ)/2, the correct sign turns r the same way; the Heisenberg sign sends it from +x towards −y, a precession in the wrong sense.

The spectrum of ρ is a constant of the motion

Ûρ(0)Û† has the eigenvalues of ρ(0): if ρ(0)|n⟩ = λₙ|n⟩, then Ûρ(0)Û† acting on Û|n⟩ gives λₙÛ|n⟩. Evolution turns the eigenvectors and leaves the eigenvalues alone, so everything built from the spectrum is conserved: Tr ρ² = Σλₙ², every Tr ρⁿ, and the entropy −Σλₙ ln λₙ. The purity can be checked directly too: d(Tr ρ²)/dt = 2 Tr(ρ dρ/dt) = −(2i/ħ) Tr(ρĤρ − ρρĤ) = 0 by cyclicity. Nothing here assumed Ĥ constant, so no pulse, laser or switched field acting on the system alone turns a pure state, Tr ρ² = 1, into a mixture, or distils a mixture into a pure state. An ensemble with eigenvalues 0.80 and 0.20 keeps Tr ρ² = 0.64 + 0.04 = 0.68 for ever, even though its populations in a fixed basis can swing anywhere between 0.20 and 0.80. Purity changes only through a step the equation leaves out: a measurement, whether its results are discarded or used to select a sub-ensemble, or coupling to a second system that the description then ignores, the case topic 04.07 treats with the partial trace.

A mixed spin precesses exactly as a pure one does

Write ρ = (I + r⋅σ)/2, as topic 04.05 did, and Ĥ = (ħ/2)Ω⋅σ, which is −γB⋅S for a moment γS in a field B, so Ω = −γB. The Pauli identity [a⋅σ, b⋅σ] = 2i(a × b)⋅σ gives [Ĥ, ρ] = (iħ/2)(Ω × r)⋅σ, while iħ dρ/dt = (iħ/2)(dr/dt)⋅σ. Matching the coefficients of σ: dr/dt = Ω × r. The vector turns about Ω at angular frequency |Ω| with its length fixed, so a vector of length 0.3 turns exactly as the pure states Unit 3 treated, and r = 0, the unpolarised ensemble, does not move at all. For the electron γ = −1.761 × 10¹¹ rad s⁻¹ T⁻¹, so Ω points along B and in 1 T the vector turns at 28.02 GHz; for the proton γ is positive and it turns the other way, at 42.58 MHz. The component of r along the field, and with it ⟨Ĥ⟩ = (ħ/2)Ω⋅r, stay constant. The measured electron rate is 0.116% above the 27.99 GHz T⁻¹ that g = 2, the Dirac value, would give: the anomaly α/2π, first order in α and quoted from quantum electrodynamics.

What commutes with Ĥ stands still: the canonical state

For a constant Ĥ, in an orthonormal basis of its energy eigenstates, ⟨m|[Ĥ, ρ]|n⟩ = (Eₘ − Eₙ)ρₘₙ, so each element evolves on its own: ρₘₙ(t) = ρₘₙ(0) e^{(−i(Eₘ − Eₙ)t/ħ)}. Populations never change, and each coherence keeps its modulus while it turns at its Bohr frequency. So ρ is stationary exactly when it commutes with Ĥ, that is, when it has no coherence between different energies. The canonical state ρₜₕ = e^(−βĤ)/Z, with β = 1/(k_B T), is quoted from statistical physics, which obtains it for a system exchanging energy with a large reservoir at temperature T. It is a function of Ĥ, diagonal in the energy basis with populations e^(−βEₙ)/Z, so the von Neumann equation leaves it alone. Its limits are the right ones: as T → ∞ every population tends to 1/d and ρ → I/d, the most mixed state; as T → 0 only the ground level survives and, if that level is not degenerate, ρ becomes pure. Stationarity does not single it out, though. Every function of Ĥ is stationary, including one with its populations upside down.

A spin-½ in a field: polarisation tanh(βμB)

A moment μ along B has energy −μB, and against it +μB. The canonical populations are e^(±βμB)/Z with Z = 2 cosh(βμB), so the polarisation, the excess fraction in the lower level, is P = tanh(βμB), and the thermal Bloch vector lies on the field axis with |r| = P: along B for the proton, against B for the electron, whose moment opposes its spin. For an electron at 1.00 T and 300 K, βμB = 9.285 × 10⁻²⁴ J ÷ 4.142 × 10⁻²¹ J = 2.24 × 10⁻³, so P = 0.224% and Tr ρ² = (1 + P²)/2 = 0.5000025: thermal spins at room temperature are almost completely mixed. The proton's P in the same conditions is 3.4 × 10⁻⁶. When βμB ≪ 1, P ≈ βμB, the 1/T of Curie's law measured in dilute paramagnets, and as T → 0, P → 1. The splitting 2μB is the photon energy hf of the measured resonance, so P = tanh(hf/(2k_B T)) needs only the measured f, 28.02 GHz at 1 T, and the temperature. How a spin reaches this state, relaxation with its time T₁, lies outside the von Neumann equation and is not modelled.

Change one variable at a time

Make the relationship visible.

Interactive model

Field B1.0 T

Temperature T1 K

Time t after the 90° pulse10.0 ps

At 1 K and 1 T, step t: the solid arrow runs round its orbit at 28 GHz per tesla and never leaves it, so it never becomes the dashed one. Drag T to 300 K and both shrink to 0.2% of the radius. At 10 T and 1 K, P saturates while the Curie line runs off above 1.

THERMAL POLARISATION P0.5866

PURITY Tr ρ²0.6721

LARMOR FREQUENCY28.02 GHz

TURNED THROUGH ω₀t101 °

Live interpretationTHERMAL POLARISATION P: 0.5866. PURITY Tr ρ²: 0.6721. LARMOR FREQUENCY: 28.02 GHz. TURNED THROUGH ω₀t: 101 °

Catch the common trap

Explain before calculating.

Misconception check

Leave an isolated spin in a field long enough, the belief goes, and it settles into the thermal state. It does not, and the von Neumann equation says why. Unitary evolution keeps the eigenvalues of ρ, so a spin tipped into the equatorial plane with |r| = 0.60 circles the field with |r| = 0.60 and Tr ρ² = 0.68 for ever, while the thermal state in 1 T at 300 K has |r| = 0.0022 and Tr ρ² = 0.5000025, a different spectrum that no Ĥ can reach. Nor does precession average the transverse vector away: averaging r over one turn leaves only its component along the field, but that average describes no ensemble at any instant, and the state itself keeps its full coherence. Real spins do relax, with times T₁ and T₂ set by their coupling to the lattice and to one another. That coupling is not in the spin's Ĥ, so relaxation has to be added to the equation, not found inside it.

Electron spins in a 0.50 T field along +z are given the Bloch vector r = (0.60, 0, 0) at t = 0 and then left isolated, so that only the field acts on them. Their Larmor frequency is 14.0 GHz. Which describes the ensemble at t = 1.00 μs, some 14 000 turns later?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 94.0 GHz ESR spectrometer detects free electron spins, g = 2.0023. Find the field its magnet must supply, then the thermal polarisation, the populations of the two levels and Tr ρ² at 4.2 K and at 300 K. Which way does the thermal Bloch vector point?

Resonance: hf = 2μB with the electron's rate 28.025 GHz T⁻¹, so B = 94.0 GHz ÷ 28.025 GHz T⁻¹ = 3.354 T.
The Boltzmann exponent is βμB = hf/(2k_B T). hf = 6.626 × 10⁻³⁴ J s × 94.0 × 10⁹ s⁻¹ = 6.228 × 10⁻²³ J, so hf/2 = 3.114 × 10⁻²³ J. At 4.2 K, k_B T = 5.799 × 10⁻²³ J and βμB = 0.537, a pure number as an exponent must be.
P = tanh(0.537) = 0.491. The lower level holds (1 + P)/2 = 0.745 of the spins and the upper (1 − P)/2 = 0.255, and Tr ρ² = (1 + P²)/2 = 0.620.
At 300 K, k_B T = 4.142 × 10⁻²¹ J and βμB = 7.52 × 10⁻³, so P ≈ βμB: P = 0.752%, populations 0.5038 and 0.4962, Tr ρ² = 0.50003. Cooling by a factor of 71 raised P only 65-fold, because tanh saturates.
Direction: the electron's moment opposes its spin, so the lower level has spin against B, and the thermal Bloch vector points along −B with length P.

AnswerB = 3.35 T. At 4.2 K: P = 0.491, populations 0.745 and 0.255, Tr ρ² = 0.620. At 300 K: P = 0.752%, Tr ρ² = 0.50003. In both cases r points against B.

MediumElectron spins in B = 0.350 T along +z have, at t = 0, equal populations 0.50 of |↑⟩ and |↓⟩ and the real coherence ⟨↑|ρ|↓⟩ = 0.30. Find r, the eigenvalues of ρ and Tr ρ². Use the von Neumann equation to find the coherence at time t and the first time r points along +y, and say whether the isolated ensemble ever reaches the thermal state at 300 K.

With ρ = (I + r⋅σ)/2, equal populations give r_z = 0, and ⟨↑|ρ|↓⟩ = (rₓ − i r_y)/2 = 0.30 gives r = (0.60, 0, 0). The eigenvalues are (1 ± |r|)/2 = 0.80 and 0.20, so Tr ρ² = 0.64 + 0.04 = 0.68.
For the electron Ĥ = (ħω₀/2)σ_z with ω₀ = |γ|B, so |↑⟩ is the upper level: f = 28.025 GHz T⁻¹ × 0.350 T = 9.809 GHz and ω₀ = 2πf = 6.163 × 10¹⁰ rad s⁻¹.
In the energy basis the populations stay 0.50, and ⟨↑|ρ|↓⟩(t) = 0.30 e^(−iω₀t) = 0.30 cos ω₀t − 0.30i sin ω₀t. Matching to (rₓ − i r_y)/2: rₓ = 0.60 cos ω₀t and r_y = 0.60 sin ω₀t, so r turns from +x towards +y at fixed length.
r first lies along +y when ω₀t = π/2: t = 1/(4f) = 1/(4 × 9.809 × 10⁹ s⁻¹) = 25.5 ps.
Check in the Heisenberg picture: dSₓ/dt = (i/ħ)[Ĥ, Sₓ] = −ω₀S_y and dS_y/dt = ω₀Sₓ give ⟨Sₓ⟩ = (ħ/2)(0.60) cos ω₀t, the value Tr(ρ(t)Sₓ) gives. Tr ρ² = Σ|ρₘₙ|² = 0.25 + 0.25 + 2 × 0.09 = 0.68 at every t.
The thermal state at 300 K has r = −7.8 × 10⁻⁴ ẑ and Tr ρ² = 0.5000003, a different spectrum. Since |r| stays 0.60, the isolated ensemble never gets there; only relaxation can.

Answerr = (0.60, 0, 0), eigenvalues 0.80 and 0.20, Tr ρ² = 0.68 at every t. ⟨↑|ρ|↓⟩ = 0.30 e^(−iω₀t) with f = 9.81 GHz, so r first lies along +y at 25.5 ps. The isolated ensemble never reaches the thermal |r| = 7.8 × 10⁻⁴.

HardElectron spins are in thermal equilibrium at 1.50 K in a 2.00 T field along +z. A resonant π pulse, fast compared with any relaxation, turns every Bloch vector through 180° about the x-axis. Find P and Tr ρ² before and after, decide whether the new state is stationary and whether it is thermal, and find the energy the pulse delivered per spin.

Before: βμB = (9.285 × 10⁻²⁴ J T⁻¹ × 2.00 T) ÷ (1.3806 × 10⁻²³ J K⁻¹ × 1.50 K) = 0.8967, so P = tanh(0.8967) = 0.7147, or 0.715. The thermal r = −0.715 ẑ, with 0.857 of the spins in the lower level |↓⟩ and 0.143 in |↑⟩, and Tr ρ² = (1 + 0.5108)/2 = 0.755.
The pulse is unitary: a 180° turn about x takes (rₓ, r_y, r_z) to (rₓ, −r_y, −r_z), so r becomes +0.715 ẑ at the same length. The eigenvalues stay 0.857 and 0.143 and Tr ρ² stays 0.755. What changes is which level holds the majority: 0.857 now sits in the upper level.
Stationary? The new ρ is diagonal in the energy basis, so it commutes with Ĥ, and once the pulse is over nothing moves: the populations are fixed and there is no coherence to precess.
Thermal? The canonical state at any positive T puts fewer spins in the upper level. Here upper over lower is (1 + P)/(1 − P) = 1.7147/0.2853 = 6.01 = e^(+2βμB), which e^(−βĤ)/Z gives only with β negative, a formal spin temperature of −1.50 K. Stationary is not the same as thermal.
Energy: ⟨Ĥ⟩ = −μBP before and +μBP after, with μB = 1.857 × 10⁻²³ J, so the pulse delivered 2μBP = 2 × 1.857 × 10⁻²³ J × 0.7147 = 2.65 × 10⁻²³ J = 0.166 meV per spin. Only relaxation, handing that energy to the lattice, returns r to −0.715 ẑ. Purcell and Pound made such an inverted nuclear-spin state in lithium fluoride in 1951, by reversing the field faster than the spins could follow.

AnswerP = 0.715 and Tr ρ² = 0.755 both before and after. The inverted state is stationary but not thermal at any positive temperature (formally −1.50 K). The pulse delivered 2.65 × 10⁻²³ J, 0.166 meV, per spin.