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I read the following text form physics forum. How do you formally derive Schrodinger equation this way?

(The Feynman-Kac formula)

A Wiener process represents Brownian motion. Brownian motion has two terms: viscosity and N-dimensional Gaussian noise. Viscosity we set to zero. The variance of the N-Gaussian we calculate from Planck's constant. Often something called drift is added in. This is a constant momentum and may be set to whatever you like.

A Wiener process with no drift has a spectral representation as a sine series whose coefficients are independent N(0,1) random variables. So that's where you get the "uncertainty." Now combine this random spectrum with the drift, the mean momentum term we added in. Take the inverse Fourier transform using the drift time as the variable of integration and you get the Schroedinger wave function. Cool!

Reference https://www.physicsforums.com/threads/schroedinger-equation-and-random-walks.545822/

High GPA
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  • The process described above does not lead to the Schrodinger equation, nor does it claim to do so. It does claim to give you "the Schrodinger wave function," but that, as far as I know, is not a rigorously-defined concept. – probably_someone Jul 15 '17 at 06:53
  • Related/possible duplicates: https://physics.stackexchange.com/q/142169/50583, https://physics.stackexchange.com/q/46015/50583, https://physics.stackexchange.com/q/30537/50583, https://physics.stackexchange.com/q/323937/50583 – ACuriousMind Jul 15 '17 at 09:13

1 Answers1

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It is not possible to derive the Schrödinger equation from a classical Wiener process. I am not exactly sure what they are talking about in the text you quoted, but it seems like they are drawing a very vague analogy.

The core of the problem is: Brownian motion gives you "uncertainty" for the position and the momentum of the particle, $X$ and $P$ are random variables. However, you can never get Heisenberg's uncertainty relation that way.

If $X$ and $P$ are classical random variables, they have a cumulative joint distribution function $$ F(x, p) = \mu_t(X \leq x, P \leq p) $$ (where $\mu_t$ is the probability measure at time $t$). From it we can calculate, for every $(k_1, k_2)$, the distribution function of $A(\vec k) = k_1 X + k_2 P$: $$ F_{\vec k}(r) = \mu_t(A(\vec k) \leq r) . \tag{*}$$

In quantum mechanics, we can write down for every observable a random variable which has the same distribution function. However, Nelson's Theorem states: Given two operators $\hat X$ and $\hat P$, we can only find a classical probability space such that (*) is the distribution function corresponding to the operator $k_1 \hat X + k_2 \hat P$ (for all $\vec k$ and in all states $|\psi(t)\rangle$) if and only if $[\hat X, \hat P] = 0$.

(For more details I recommend section 2.1.4 of "The theory of open quantum systems" by Breuer and Petruccione.)

Noiralef
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  • Thanks I will definitely take a look at Breuer's book. Is Nelson's theorem included in that book? May I have reference to the Nelson's theorem? – High GPA Jul 24 '17 at 19:57