Solving Schrödinger equation via Tartaglia/Pascal triangle: a possible link between stochastic processing and quantum mechanics

Abstract

In a recent paper (Farina et al. in Signal Image Video Process 1–16, 2011), it was shown a clean connection between the solution of the classical heat equation and the Tartaglia/Pascal/Yang-Hui triangle. When the time variable in the heat equation is substituted with the imaginary time, the heat equation becomes the Schrödinger equation of the quantum mechanics. So, a conjecture was put forward about a connection between the solution of the Schrödinger equation and a suitable generalization of the Tartaglia triangle. This paper proves that this conjecture is true and shows a new—as far as the authors are aware—result concerning the generalization of the classical Tartaglia triangle by introducing the “complex-valued Tartaglia triangle.” A “complex-valued Tartaglia triangle” is just the square root of an ordinary Tartaglia triangle, with a suitable phase factor calculated via a discretized version of the ordinary continuous case of the Schrödinger equation. So, taking the square of this complex-valued Tartaglia triangle gives back exactly the probability distribution of a discrete random walk. We also discuss about potential connections between the theories of stochastic processes and quantum mechanics: a connection debated since the inception of the theories and still lively hot today.

Appendix

The Schrödinger and the heat equations have a common mathematical structure that makes them easy to manage using a Fourier transform. Indeed, let us consider the general parabolic differential equation

$$\begin{aligned} \frac{\partial \phi }{\partial t}=k\frac{\partial ^{2}\phi }{\partial t^{2}} \end{aligned}$$

(35)

with $k\in \mathfrak R$ for the heat equation and $k=\frac{i\hbar }{2m}$ for the Schrödinger equation. This equation is linear and can be treated with a Fourier transform.

So, let us define the mode $p$

$$\begin{aligned} \hat{{\phi }}_p =\mathfrak I (\phi )=\int \limits _{-\infty }^{+\infty } {\phi (x)\text{ e }^{-ipx}\text{ d }x} \end{aligned}$$

(36)

for the Fourier transform in the space coordinate. The given equation becomes

$$\begin{aligned} \frac{\partial \hat{{\phi }}_p (t)}{\partial t}=-kp^{2}\hat{{\phi }}_p(t) \end{aligned}$$

(37)

that can be immediately solved to give

$$\begin{aligned} \hat{{\phi }}_p (t)=\hat{{\phi }}_p (0)\text{ e }^{-kp^{2}t} \end{aligned}$$

(38)

Now, depending on $k$, we get two completely different behaviors: for a real $k$, this solution is decaying in time and so represents an irreversible behavior of the solution for the mode $p$. But the behavior of the Schrödinger equation is quite different due to the presence of the $i=\sqrt{-1}$ factor. In this case, also the solution with a reversed time is physically meaningful and can also be obtained but taking the conjugate of the given solution. Indeed, this is exactly the effect of the time reversal operator in quantum mechanics, and the Schrödinger equation is invariant for time reversal, contrarily to the case of the heat equation.

Now, we are in a position to get the kernel solution for both equations. If we have a point-like source centered at $x = 0$, then its Fourier transform is a constant, and we have to evaluate the integral

$$\begin{aligned} \phi (x,t)=\frac{1}{2\pi }\int \limits _{-\infty }^{+\infty } {e^{-ipx}\text{ e }^{-kp^{2}t}\text{ d }p} \end{aligned}$$

(39)

This is a Gaussian integral that can be readily evaluated to give the kernel for both equations. In the case of the Schrödinger equation, we have to cope with Fresnel integrals but the result is the same, and there is no convergence problem [18]. Indeed, in the current literature (e.g., [18]), these integrals are always dubbed “Gaussian” and are generally straightforward to evaluate.

Finally, let us consider the case when the source is not point like. So, we will get

$$\begin{aligned} \phi (x,t)=\frac{1}{2\pi }\int \limits _{-\infty }^{+\infty } {\phi _p(0)\text{ e }^{-ipx}\text{ e }^{-kp^{2}t}\text{ d }p} \end{aligned}$$

(40)

that can be seen as the Fourier transform of a convolution between the kernel and the initial value of the solution. In this way, we have recovered all the equations given in (5)–(6) and (10)–(12) of this paper.

We are now in a position to show the way a Gaussian wave packet spreads during its time evolution. For our aims, it is enough to take $\varphi _{p}$(0) as a Gaussian and we get

$$\begin{aligned} \phi (x,t)&= \left( {\frac{a^{2}}{2\pi ^{3}}} \right) ^{\frac{1}{4}}\int \limits _{-\infty }^{+\infty } {\text{ e }^{\left\{ {-\left( {ka} \right) ^{2}-\frac{i\hbar k^{2}t}{2m}+ikt} \right\} }\text{ d }k}\nonumber \\&= \left( {2\pi } \right) ^{-\frac{1}{4}}\left( {a^{2}+\frac{i\hbar t}{2ma}} \right) ^{-\frac{1}{2}}\text{ e }^{-\frac{x^{2}}{4a^{2}+\frac{2i\hbar t}{m}}} \end{aligned}$$

(41)

The position probability density of the Gaussian wave packet is then

$$\begin{aligned} \left| {\phi (x,t)} \right| ^{2}=\left\{ {2\pi \left[ {a^{2}+\frac{\hbar ^{2}t^{2}}{4m^{2}a^{2}}} \right] } \right\} ^{-\frac{1}{2}}\text{ e }^{-\frac{x^{2}}{2\left[ {a^{2}+\frac{\hbar ^{2}t^{2}}{4m^{2}a^{2}}} \right] }} \end{aligned}$$

(42)

that is Eq.(15) using the given definition for the Heisenberg time.