4
$\begingroup$

Brownian motion

In classical physics, Brownian motion can be modelled using the so-called Wiener process $W_t$, a continuous-time stochastic process. If we use $X_t$ to signify the position of a particle at time $t$, we can write: $$ \mathrm{d}X_t = \sigma \, \mathrm{d}W_t $$ where $\sigma$ is the deviation from mean position. The time evolution of the probability density $\mathrm{E}[X_t]$ of the particle's position satisfies the heat equation: $$ \frac{\partial}{\partial t} \, \mathrm{E}[X_t] \,+\, \frac{1}{2} \, \sigma^2 \, \Delta \, \mathrm{E}[X_t] = 0 $$ Furthermore, there is an integral on the space $\Gamma_\sigma$ of Wiener processes with deviation $\sigma$: $$ \int_{\Gamma_\sigma} F[X] \;\mathrm{d}w_\sigma(X) $$ and the time evolution of $\mathrm{E}[X_t]$ can be rewritten in terms of this integral using the Feynman-Kac formula.

Quantum Mechanics

Meanwhile in Quantum Mechanics, the wavefunction $\psi(x)$ of a free particle, ie. the probability amplitude of a particle being in position $x$, has an evolution which satisfies the time-dependent Schrödinger equation: $$ \mathrm{i} \hbar \, \frac{\partial}{\partial t} \, \psi = -\frac{\hbar^2}{2m} \, \Delta \, \psi \\ \Updownarrow \\[5pt] \frac{\partial}{\partial t} \, \psi - \frac{\mathrm{i} \, \hbar}{2m} \, \Delta \, \psi = 0 $$ This is a heat equation with an imaginary diffusion coefficient.

The time-evolution of $\psi$ can be alternatively described using the Feynman path integral over the set $\Omega_{x,t}$ of all classical trajectories that end in time $t$ and position $x$: $$ \psi(x, t) = \int_{\Omega_{x,t}} \mathrm{e}^{\mathrm{i} \, S[X] / \hbar} \; \psi(X_0, 0) \; \mathcal{D}X $$ Since there isn't a reasonable measure on $\Omega_{x,t}$, as proved by the Cameron theorem (source: Blank-Exner-Havlíček, 2008), the path integral is often defined this way:

  1. calculate the Wiener integral $f(\sigma) = \int_{\Gamma_\sigma} F[X] \, \mathrm{d}w_\sigma(X)$ for a suitable $F$
  2. construct an analytic continuation of $f$ to $\mathbb{C}$
  3. evaluate at $\sigma = \mathrm{i}/m$

The Question

The stated results seem to strongly indicate that Brownian trajectories and trajectories in Quantum Mechanics (in the sense of path integral) are closely related. Is it possible to describe these elusive “quantum trajectories” as stochastic processes? Would it make sense for a stochastic process $X_t$ to satisfy the equation $$ \mathrm{d}X_t = \frac{\mathrm{i}}{m} \, \mathrm{d}W_t $$ where $m$ is real? Then, could it be said that a typical trajectory of a quantum particle (again, in the sense of the path integral) is one described by this stochastic process?

$\endgroup$
2
+50
$\begingroup$

Both classical distributions describing stochastic processes, and quantum wavefunctions can be obtained by summing over trajectories given some initial conditions. You have already noted this in your question.

In the classical case the weight associated with each trajectory is a probability and hence a non-negative real number. As a result trajectories with large probabilities necessarily contribute a large amount to the final distribution, and may be considered "typical".

In quantum mechanics no analogous statement applies, due to the complex nature of the weights (or amplitudes) associated to each trajectory. A quantum trajectory my contribute a significant amplitude to a specific outcome, only for it to be cancelled by amplitudes of a different phase contributed by other trajectories. Thus, by considering a single trajectory in isolation, it cannot be determined whether it makes a significant contribution an outcome which has a high probability. The canonical two slits experiment is a straightforward demonstration of this.

As a separate point, in the context of open quantum systems, "unravelling" the master equation can provide stochastic quantum trajectories of the type you wrote down. An introduction to this perspective is provided in "An Open Systems Approach to Quantum Optics" by Carmichael, and reviewed in Plenio and Knight, Rev. Mod. Phys. 70, 101 (1991).

Edit: As the final paragraph was of interest to the OP, I also add that "Quantum State Diffusion", by Percival provides a pedagogical perspective on the older literature on quantum stochastic trajectories, while Daley, Adv. Phys. 63.2 (2014) provides a more modern overview focussed on connecting with AMO experiments.

$\endgroup$
2
  • $\begingroup$ Thank you, it looks like the two books contain exactly what I was looking for! $\endgroup$ – m93a Mar 9 at 14:05
  • 1
    $\begingroup$ Ok great. I added two other references since this particular part of the literature was of interest to you. $\endgroup$ – ComptonScattering Mar 9 at 17:37
0
$\begingroup$

Path integrals can be used for describing Brownian motion - the corresponding action/Lagrangian is called Onsager-Matchlup functional. One could read more about it in Risken's book on the Fokker-Planck equation or in the Kleinert's book in the path integral. Such description of the stochastic processes is particularly popular in finance applications.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer! However I don't think it really adresses my question. Since Feynman integral and Wiener integral are essentially just a Wick rotation of each other, it's not surprising that Feynman integral can be used to study Brownian motion (ie. Wiener processes). However the “quantum trajectories” that are integrated over in QM (and which I am interested in) are not Wiener processes. And they aren't discussed at all in your answer, if I understand it correctly. $\endgroup$ – m93a Mar 1 at 20:52
  • $\begingroup$ @m93a do you mean the difference between the classical probability and a wave function? $\endgroup$ – Roger Vadim Mar 2 at 2:05
  • 1
    $\begingroup$ No, I do not. I am specifically asking about stochastic processes. Trajectories in path integral have a lot in common with Wiener processes – my question is: how far does the analogy go? Is there a stochastic process associated with the quantum trajectories? $\endgroup$ – m93a Mar 2 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.