research28 min studyLesson 564 in Physics

Worldline quantum Monte Carlo

Map thermal traces into imaginary-time configurations and identify positivity, balance, and discretization assumptions.

Before you begin

Variational Monte Carlo estimators

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The question for today

What you’ll understand

Construct an imaginary-time representation of a thermal trace.
Separate Trotter error from sampling and finite-temperature effects.
Explain why worldline weights need not always be positive.

A quantum partition function is a trace of an operator exponential, not an ordinary sum of Boltzmann factors over simultaneous values of noncommuting observables. Worldline methods resolve that trace through sequences of basis states in imaginary time. This maps a quantum statistical problem into a larger configuration sum, which can be sampled when its weights define a suitable probability distribution. The added coordinate organizes thermal fluctuations and must not be interpreted as a movie of particles moving in laboratory time.

Insert basis states into a thermal trace

For temperature T, define inverse energy $\beta=1/(k_BT)$ . Splitting the imaginary interval into M slices gives $\delta\tau=\beta/M$ . Insert complete basis states between short-time propagators, with the trace imposing periodic closure of the sequence. Off-diagonal matrix elements connect configurations between slices; diagonal terms weight residence in configurations. A worldline picture is a representation of these matrix elements, and its apparent paths depend on the chosen basis.

If H separates into noncommuting groups K and V, a symmetric Trotter product approximates each short propagator with exponentials that are easier to evaluate. At fixed beta, its leading global discretization error is typically quadratic in slice size under the needed boundedness and regularity conditions. Increasing M is therefore a convergence test, not just a way to make a drawing smoother. Continuous-imaginary-time and series-expansion formulations can remove this particular time-slice error while retaining sampling and finite-temperature limitations.

Z=\operatorname{Tr}e^{-\beta H}=\sum_{s_1,\ldots,s_M}\prod_{j=1}^M\langle s_{j+1}|e^{-\delta\tau H}|s_j\rangle,\qquad s_{M+1}=s_1

Z is dimensionless; H has energy units and beta, delta tau have inverse-energy units. This identity uses exact short propagators; replacing them by a Trotter approximation introduces a separate error.↔ Scroll wide equations horizontally.

A configuration weight is not automatically a probability

When H has real nonpositive off-diagonal entries in a suitable basis, the short-time exponential has nonnegative entries for a finite matrix. Such a representation supports positive worldline weights. Frustration, fermionic permutations, or a less favorable basis can introduce negative or complex contributions. The quantum-to-classical rewriting alone does not solve that problem. Even with nonnegative weights, local updates may sample slowly near criticality or fail to change a winding or conserved sector when the chosen moves are incomplete.

Metropolis moves compare complete configuration weights and their proposal probabilities. Loop or cluster updates change correlated pieces of a worldline configuration and can improve mixing; their construction depends on the model’s local vertex weights. Detailed balance is a sufficient stationarity condition, not a guarantee of fast exploration. Verify equilibration and autocorrelations for the observables of interest. An update that preserves a topological sector can yield apparently stable averages with a systematic sampling error if the intended ensemble includes other sectors.

e^{-\delta\tau(K+V)}=e^{-\delta\tau K/2}e^{-\delta\tau V}e^{-\delta\tau K/2}+O(\delta\tau^3)

The one-step error is cubic, giving a leading quadratic global error at fixed beta for appropriate finite or controlled operators. Products are operator products, not commuting scalar Boltzmann factors.↔ Scroll wide equations horizontally.

Thermal observables and real-frequency questions differ

An estimator must be derived from the represented partition function. Energy can be obtained from a beta derivative of the log partition function; it is not generally the average of bare diagonal basis energies because off-diagonal processes contribute. Imaginary-time correlations provide information about excitations through an integral kernel, often a decaying exponential or a thermal variant. They do not directly provide oscillatory real-time trajectories. Recovering a sharp real-frequency spectrum from noisy imaginary-time data is an ill-conditioned inverse problem.

To approach a ground state, increase beta until observables stop changing at the desired precision, taking low-lying gaps and possible degeneracies into account. Then study spatial size and time discretization separately. At a closing gap, the required beta can grow with size; keeping beta fixed while increasing the lattice may change the physical regime. The two-state Markov lab illustrates a positive stationary distribution and mixing only. Its discrete steps are sampling updates, not imaginary-time slices or quantum worldline events.

E=-\frac{\partial\ln Z}{\partial\beta},\qquad C(\tau)=\int_0^\infty d\omega\,A(\omega)e^{-\tau\omega}\quad\text{in a simple zero-temperature case}

Here omega denotes excitation energy, so tau has inverse-energy units; A has the units required by the chosen correlation. Thermal kernels and operator order can differ. Inverting the decaying kernel amplifies noise and requires stated regularization assumptions.↔ Scroll wide equations horizontally.

Work it through

An exactly soluble quantum spin

For $H=-\Gamma\sigma_x$ with $\Gamma>0$ , compute Z and thermal energy at $\beta\Gamma=\ln2$ . Explain why averaging only the diagonal z-basis energies would fail.

The energy eigenvalues are $-\Gamma$ and $+\Gamma$ , so $Z=e^{\beta\Gamma}+e^{-\beta\Gamma}$ . At the specified inverse temperature these weights are two and one half, yielding $Z=2.5$ .
The thermal numerator is $(-\Gamma)(2)+(+\Gamma)(1/2)=-1.5\Gamma$ . Dividing by 2.5 gives $E=-0.6\Gamma$ , equivalently $-\Gamma\tanh(\ln2)$ .
Both z-basis diagonal entries of H vanish. Averaging those alone would give zero, missing all transverse-field energy. In a worldline representation, transitions between slices encode that off-diagonal contribution.

The result

The exact partition function is 2.5 and energy is −0.6Γ; a correct worldline estimator must retain off-diagonal physics.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

Positive weights and mixing in two states

Reset and set $p=0.2$ , $q=0.8$ . Predict stationary probability 0.2 in state one; one Step already reaches it from state zero because p plus q is one. Repeat from Reset with both controls $p=0.02$ , $q=0.08$ : the stationary law is unchanged but convergence is slower. The probabilities are exact, not sampled worldlines or imaginary-time dynamics.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn unchanged target distribution can coexist with very different sampling relaxation rates.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A cubic one-step error accumulates to quadratic order over a fixed imaginary interval. Positivity and sampling convergence remain separate issues.

Your turn

Try it yourself

A two-configuration thermal model has energies zero and $\Delta>0$ , with $\beta\Delta=\ln3$ . Find the upper-state probability and mean energy. For symmetric proposals, check the Metropolis acceptance factors in both directions.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The upper-to-lower Boltzmann weight ratio is one third; normalize the two weights before computing expectations.

Compare with the solution

The weights are one and one third, giving upper probability one quarter and mean energy $\Delta/4$ . A proposed upward move is accepted with probability one third, while a downward move is accepted with probability one. With equal proposal probabilities, stationary flows match: $(3/4)(1/3)=1/4$ and $(1/4)(1)=1/4$ . This diagonal toy model tests thermal weighting and balance, rather than demonstrating noncommuting quantum dynamics.

Keep the idea

Take it with you

Worldlines represent a thermal trace in imaginary time rather than real particle trajectories.
Positive weights and fast ergodic sampling are additional requirements after the formal mapping.
Time-slice, statistical, thermal, and spatial errors must be varied independently.

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