Ensemble Interpretation of Quantum Mechanics and the Two-Slit Experiment

Abstract

An evolution equation model is provided for the two-slit experiment of quantum mechanics. The state variable of the equation is the probability density function of particle positions. The equation has a local diffusion term corresponding to stochastic variation of particles, and a nonlocal dispersion term corresponding to oscillation of particles in the transverse direction perpendicular to their forward motion. The model supports the ensemble interpretation of quantum mechanics and gives descriptive agreement with the Schrödinger equation model of the experiment.

Appendix

Let $X = C[-\infty ,\infty ]$, the space of bounded uniformly continuous functions on $(-\infty ,\infty )$ with norm $\Vert f \Vert = \sup _{-\infty< x < \infty }|f(x)|$. For $\sigma > 0$ let

$$\begin{aligned}&(T_{\sigma }(t)f)(x) = \nonumber \\&\qquad \frac{1}{2 \sqrt{\sigma t}} \int _{-\infty }^{\infty } \exp \bigg (- \frac{(x - y)^2}{4 \sigma t}\bigg ) f(y) \, dy, \quad f \in X, \ t > 0, \ -\infty< x < \infty . \end{aligned}$$

(13)

Then $T_{\sigma }(t), t \ge 0$ is a strongly continuous holomorphic semigroup of positive linear operators in X with infinitesimal generator $(A_{\sigma } f)(x) = \sigma d^2 f(x) / d x^2$ satisfying $| T_{\sigma }(t) | \le 1, t \ge 0$ [41, Chap. IX]. For $t > 0$, $A_{\sigma } T_{\sigma }(t)$ is bounded in X, and there exists $M_{\sigma } > 0$ such that $| A_{\sigma } T_{\sigma }(t) | \le M_{\sigma } / t, t > 0$ [16, Part 2]. Further, $(T_{\sigma }(t)f)(x)$ is the strong solution in X to the diffusion equation

$$\begin{aligned} \frac{\partial }{\partial x}u(x,t) = \sigma \frac{\partial ^2}{\partial x^2} u(x,t), \quad u(x,0) = f(x), \ t > 0, \ -\infty< x < \infty , \ f \in X. \end{aligned}$$

(14)

For $\beta > 0$ define the bounded linear operator B in X by

$$ (Bf)(x) = \beta \left( f(x+1) - 2 f(x) + f(x-1) \right) , \quad f \in X, \ -\infty< x < \infty . $$

Define

$$\begin{aligned} (\exp (t B) f )(x) = \bigg ( \sum _{n=0}^{\infty } \frac{(t B)^n}{n!} f \bigg )(x) = e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} f(x +m -n) \end{aligned}$$

(15)

for $f \in X$, $t \ge 0$, $-\infty< x < \infty$. Then $\exp (t B)$, $t \ge 0$, is a group of positive linear operators in X with infinitesimal generator B satisfying $|\exp (t B) | \le 1$, $t \ge 0$ (see [41, p. 244]).

Theorem 1

Let $\beta > 0$, $s = 1$, $f \in X \cap L_+^1(-\infty ,\infty )$, and let $\int _{-\infty }^{\infty } f(x) dx = 1$. Then for $t \ge 0$,

$$\begin{aligned} \int _{-\infty }^{\infty } (\exp (t B)f)(x) dx = \int _{-\infty }^{\infty } f(x) dx = 1. \end{aligned}$$

(16)

If also, $x^2 f(x) \in L^1(-\infty ,\infty )$, then the mean of $\exp (t B)f =$ the mean of f and the variance of $\exp (t B)f =$ the variance of $f + 2 t \beta$.

Proof

From (15), we obtain (16), since for $t \ge 0$

$$ \begin{aligned} \int _{-\infty }^{\infty } (\exp (t B) f)(x) dx&= \int _{-\infty }^{\infty } e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} f(x + m - n) dx \\&= e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \int _{-\infty }^{\infty } f(x ) dx = \int _{-\infty }^{\infty } f(x) dx. \end{aligned} $$

Let $\mu _f$ be the mean of f and $\mu _{\exp (tB) f}$ the mean of $\exp (tB) f$. Then

$$\begin{aligned} \mu _{\exp (tB) f}&= \int _{-\infty }^{\infty } x (\exp (t B) f)(x) dx \nonumber \\&= \int _{-\infty }^{\infty } e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} x f(x + m - n) dx \nonumber \\&= e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \int _{-\infty }^{\infty } (x +m -n)f(x ) dx \nonumber \\&= e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \left( \int _{-\infty }^{\infty } x f(x ) dx +\int _{-\infty }^{\infty } (m-n) f(x ) dx\right) \nonumber \\&= \mu _f + e^{- 2 t \beta } \left( \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} m - \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} n \right) \nonumber \\&= \mu _f + e^{- 2 t \beta } \left( \sum _{n=0}^{\infty } \sum _{m=1}^{\infty } \frac{(t \beta )^{n+m}}{n! (m-1)!} - \sum _{n=1}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{(n-1)! m!} \right) = \mu _f. \end{aligned}$$

(17)

Let $\nu _f$ be the variance of f and $\nu _{\exp (tB) f}$ the variance of $\exp (tB) f$. A calculation similar to (17) yields

$$\begin{aligned}&\nu _{\exp (tB) f} = \int _{-\infty }^{\infty } x^2 (\exp (t B) f)(x) dx \nonumber \\&= \int _{-\infty }^{\infty } e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} x^2 f(x + m - n) dx \nonumber \\&= e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \int _{-\infty }^{\infty } (x +m -n)^2 f(x) dx \nonumber \\&= e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \left( \int _{-\infty }^{\infty } x^2 f(x ) dx + 2 (m-n) \int _{-\infty }^{\infty } x f(x ) dx + (m-n)^2 \right) \nonumber \\&= \nu _f + e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} \left( 2 (m - n) \mu _f + (m-n)^2 \right) \nonumber \\&= \nu _f + e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} (m^2 - 2 m n + n^2) = \nu _f + 2 t \beta , \end{aligned}$$

(18)

since

$$ e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} m^2 = \beta t (1 + \beta t), \quad e^{- 2 t \beta } \sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{(t \beta )^{n+m}}{n! m!} 2 m n = 2 (\beta t )^2. $$

$\square$

Theorem 2

Let $\alpha > 0$, $\beta > 0$, and $s = 1$. For $\omega _0 \in X \cap L_+^1(-\infty ,\infty )$ such that $\int _{-\infty }^{\infty } \omega _0(x) dx = 1$, the unique solution $\omega (x,z) = \omega (x,t)$ of (9) is

$$ (T_{\alpha }(t) \exp (t B) \omega _0)(x), \quad t > 0, \ -\infty< x < \infty , $$

(where we have identified $z = t$ in (9)). Further, there exists a constant C such that

$$\begin{aligned} | \omega (x,t) - (T_{\alpha + \beta }(t) \omega _0)(x) | \le \frac{C}{t} \Vert \omega _0 \Vert , \quad t > M_{\alpha }, \ -\infty< x < \infty . \end{aligned}$$

(19)

Remark 1

If $\omega _0$ also has compact support, then (19) implies that uniformly on bounded sets of $x \in (-\infty ,\infty )$

$$ \lim _{t \rightarrow \infty } \sqrt{t} \omega (x,t) = \frac{1}{2 \sqrt{\alpha + \beta }}, \quad \text{ since } \lim _{t \rightarrow \infty } \sqrt{t} (T_{\alpha + \beta }(t) \omega _0)(x) = \frac{1}{2 \sqrt{\alpha + \beta }}. $$

Proof

Since $A_{\sigma }$ and B commute, $T_{\sigma }(t)$ and $\exp (t B)$ commute for $t \ge 0$, and $T_{\sigma }(t) \exp (t B), t \ge 0$ is the semigroup of operators for the solutions of (9). Since $T_{\alpha }(t), t \ge 0$ is a holomorphic semigroup in X, $T_{\alpha }(t) \exp (t B) \omega _0$ is the unique strong solution of (9) in X for $t > 0$, $\omega _0 \in X$. If $f \in X$ is analytic, then

$$ f(x\pm 1) = f(x) + \sum _{n=1}^{\infty } \frac{(\pm 1)^n}{n!} f^{(n)}(x) $$

and

$$ f(x+1) - 2f(x) + f(x-1) = f^{(2)}(x) + 2 \sum _{n=2}^{\infty } \frac{1}{(2n)!} f^{(2n)}(x). $$

For $g \in X$, $t > 0$, $T_{\alpha }(t) g$ is analytic, and so

$$ B T_{\alpha }(t) g= \beta A_1 T_{\alpha }(t) g + 2 \beta \sum _{n=2}^{\infty } \frac{1}{(2n)!} A_1^{n} T_{\alpha }(t) g $$

Then,

$$\begin{aligned}&\frac{d}{d t} T_{\alpha }(t) \exp (t B) \omega _0 = (A_{\alpha } + B) T_{\alpha }(t) \exp (t B) \omega _0 \\&\qquad \qquad \qquad \quad = (\alpha + \beta ) A_1 T_{\alpha }(t) \exp (t B) \omega _0 + 2 \beta \sum _{n=2}^{\infty } \frac{1}{(2n)!} A_1^{n}T_{\alpha }(t) \exp (t B) \omega _0. \end{aligned}$$

From [20, Chap. 9] the solution of this nonhomogeneous equation satisfies

$$\begin{aligned}&T_{\alpha }(t) \exp (t B) \omega _0 \\&\quad =T_{\alpha + \beta }(t) \omega _0 + 2 \beta \int _0^t T_{\alpha + \beta }(s) \sum _{n=2}^{\infty } \frac{1}{(2n)!} A_1^{n} T_{\alpha }(t-s) \exp ((t-s)B) \omega _0 ds \\&\qquad =T_{\alpha + \beta }(t) \omega _0 + 2 \beta \int _0^t T_{\beta }(s) \sum _{n=2}^{\infty } \frac{1}{(2n)! \alpha ^n} A_{\alpha }^{n} T_{\alpha }(t) \exp ((t-s)B) \omega _0 ds. \end{aligned}$$

Thus, (19) follows, since for $t \ge M_{\alpha }$,

$$\begin{aligned} \Vert T_{\alpha }(t) \exp (t B) \omega _0&- T_{\alpha + \beta }(t) \omega _0 \Vert \le 2 \beta t \sum _{n=2}^{\infty } \frac{(M_{\alpha } / (\alpha t))^n}{(2n!)} \Vert \omega _0\Vert \\&\qquad \qquad = 2 \beta t \left( \cosh (\sqrt{M_{\alpha } / (\alpha t)}) - 1 - M_{\alpha } / (2 \alpha t) \right) \Vert \omega _0\Vert \end{aligned}$$

and by L’Hospital’s Rule

$$ \lim _{t \rightarrow \infty } \frac{\cosh (\sqrt{M_{\alpha } / (\alpha t)}) - 1 - M_{\alpha } / (2 \alpha t)}{(M_{\alpha } / (\alpha t))^2} = \frac{1}{4!}. $$

$\square$