### The intimate relation of quantum mechanics and classical stochastics?

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6 months ago by
If one has ever written the Fokker-Planck equation
$\partial_t P(x, t) = D \partial^2_x P(x,t)$
next to the Schrödinger equation in the form
$\partial_t \Psi(x, t) = i D \partial^2_x \Psi(x,t),$
where $D = \hbar/2m$, one kind of keeps wondering what is really going on... The complex "constant of diffusion" in Schrödinger's equation makes our lives a lot harder. It is, however, also crucially responsible for the fact that one can get "interference of probabilities". You can find a large number of attempts out there to reconstruct Schrödinger's equation purely from classical stochastics. None of them has been wholly convincing so far, apparently. Still it is rather interesting to even see one of these attempts. The most recent and perhaps also the most simple one I have come accross so far is from Hagen Kleinert's path-integral bible cited below, chapter 18.24 in the 5th and latest edition. It is also possible to download single chapters from his homepage.

Starting point is the following Langevin equation in two dimensions,
$\dot{\boldsymbol{x}}(t) = \boldsymbol{x}(t) \times \boldsymbol{\omega} + \boldsymbol{\eta}(t),$
where $\boldsymbol{\omega} = (0, 0, \omega)^T$ and $\boldsymbol{\eta}(t) = (\eta(t), \eta(t), 0)^T$. This describes a rotation in the $xy$ plane combined with independent noise in both directions. Compared to the book, I have reversed the cross product, because I do not get the correct signs otherwise when I use the above definition of $\boldsymbol{\omega}$ (it is not explicitly defined in the book). Since this only affects the direction of the rotation in the equation of motion, it should not matter. The noise is characterised by
$\langle\eta(t)\eta(t')\rangle = \frac{\hbar}{2}\delta(t-t'),$
where I have added a factor of 1/2 because else I do not come out with the final result from the book (that factor also agrees with the "diffusion constant" above for unit mass).

Now any time-independent function $\boldsymbol{u}(\boldsymbol{x}) = (u_1(\boldsymbol{x}), u_2(\boldsymbol{x}))$ will acquire a time-dependence when evaluated on the stochastic process:
$\boldsymbol{u}_{\eta}(\boldsymbol{x}, t) := \boldsymbol{u}(\boldsymbol{x}(t)).$
By the rules of stochastic calculus, its time differential is then
\begin{align*}
\boldsymbol{u}_{\eta}(\boldsymbol{x}, \Delta t) - \boldsymbol{u}_{\eta}(\boldsymbol{x}, 0) &= \Delta t\; (\boldsymbol{x} \times \boldsymbol{\omega})\cdot \nabla\; \boldsymbol{u}_{\eta}(\boldsymbol{x}, 0) + \int_0^{\Delta t} \mathrm{d}t'\; \eta(t')\; ((1,1)\nabla)\; \boldsymbol{u}_{\eta}(\boldsymbol{x}, 0) \\
&+ \frac{1}{2} \int_0^{\Delta t} \mathrm{d}t'\;\int_0^{\Delta t} \mathrm{d}t''\; \eta(t')\eta(t'')\; ((1,1)\nabla)^2\; \boldsymbol{u}_{\eta}(\boldsymbol{x}, 0) + O(\Delta t^{3/2}).
\end{align*}
The terms $((1,1)\nabla)$ come from the dot product of the noise vector with the gradient. By taking the noise average $\boldsymbol{u}(\boldsymbol{x}, t) := \langle \boldsymbol{u}_{\eta}(\boldsymbol{x}, t) \rangle$, one finds that the function $\boldsymbol{u}(\boldsymbol{x}, t)$ obeys the differential equation
$\partial_t \boldsymbol{u}(\boldsymbol{x}, t) = \hat{H} \boldsymbol{u}(\boldsymbol{x}, t),$
with the time-independent operator $\hat{H} = (\boldsymbol{x} \times \boldsymbol{\omega})\cdot \nabla + \frac{\hbar}{4} ((1,1)\nabla)^2$. In this step we have used that the noise vanishes on the average, as well as its correlation function. Written out in components, the differential equation reads
$\partial_t (u_1, u_2)^T = (\omega y \partial_x - \omega x \partial_y) (u_1, u_2)^T + \frac{\hbar}{2} (\partial_x + \partial_y)^2 (u_1, u_2)^T.$
Now cometh the essential point of the whole derivation: if we assume that the original function $\boldsymbol{u}(\boldsymbol{x})$ is conformal, we have $\nabla^2 \boldsymbol{u}(\boldsymbol{x}) = 0$ as well as
$\nabla^2 \boldsymbol{u}(\boldsymbol{x}, t) = 0,$
since the Laplacian commutes with $\hat{H}$. This condition is equivalent to the Cauchy-Riemann equations (please have a look at the corresponding Wikipedia article)
$\partial_x u_1 = \partial_y u_2, \qquad \partial_y u_1 = -\partial_x u_2.$
When using these on the differential equation we have found, we find
\begin{align*}
\partial_t u_1 &= \omega y \partial_x u_1 + \omega x \partial_x u_2 -\frac{\hbar}{2}\partial_x^2 u_2, \\
\partial_t u_2 &= \omega y \partial_x u_2 - \omega x \partial_x u_1 +\frac{\hbar}{2} \partial_x^2 u_1.
\end{align*}
One can convince oneself by direct calculation that the following version of the Schrödinger equation of the harmonic oscillator
$i\hbar\partial_t\psi(x,t) = \left( -\frac{\hbar^2}{2}\partial_x^2 + \frac{\omega^2x^2}{2} - \frac{\hbar\omega}{2} \right)\psi(x,t)$
reduces to the above pair of equations for $y=0$ if one takes the ansatz
$\psi(x,t) = e^{-\omega x^2/2\hbar}(u_1(x,t)+ iu_2(x,t)),$
that is, plugging this into the Schrödinger equation will yield
\begin{align*}
\partial_t u_1 &= \omega x \partial_x u_2 -\frac{\hbar}{2}\partial_x^2 u_2, \\
\partial_t u_2 &= - \omega x \partial_x u_1 +\frac{\hbar}{2} \partial_x^2 u_1.
\end{align*}
Is this not rather amazing? Setting $y=0$ is not an obvious step, and the physical meaning of it all is entirely obscure. Yet it means that "the quantum mechanics of a harmonic oscillator" can be modelled on a subset of a classical stochastical process if one demands that the "wavefunction" be a conformal function of this process. Kleinert also generalizes this for arbitrary potentials. As he emphasizes, it remains totally unclear, however, what physics would be required to explain why we actually experience the square of the wavefunction as a probability for detecting a particle, for instance.

- Kleinert, Hagen. Path integrals in quantum mechanics, statistics, polymer physics, and financial markets. World scientific, 2009.