# Quantum Mechanical Two-Body Problem with Gaussian potential¶

Quantum Mechanical Two-Body Problem with Gaussian potential

Figure 1: Two heavy (M) particles and one light (m) particle in a three-body system

This application note is inspired by the work described in [Thies et al. 2022]. In this paper the authors develop a numerical method to calculate binding energies of a quantum mechanical three-body system efficiently. This three-body system is composed of two heavy and one light particle. In figure 1 the system is displayed with its two-body (heavy/light) subsystems marked by dashed ellipses. The routine to calculate binding energies for the three-body system first solves the two-body subsystem. This application note aims to reproduce the findings in for the two-body systems using an analog computer.

## Implementation¶

The Schrödinger equation for the two-body system is given by

\begin{align} \left[ -\frac{1}{2} \Delta_\xi - v_0 f(\xi) \right] \psi(\xi) &= E\psi(\xi),\tag{1} \label{eq:schrödinger} \end{align}

where $$\xi$$ is the distance between the two particles, $$E$$ their energy, $$\psi(\xi)$$ the wave function of the system, $$-v_0f(\xi)$$ an attractive potential between the particles, and $$\Delta_\xi$$ the Laplace operator. The depth of the potential is given by $$v_0$$ and the shape is defined by a Gaussian function

\begin{align} f(\xi) = \exp(-\xi^2). \tag{2} \end{align}

With a Gaussian potential equation (1) becomes symmetric under the transformation $$\xi \rightarrow -\xi$$ and the solutions are either even ($$\psi(\xi) = \psi(-\xi)$$) or odd ($$\psi(-\xi) = -\psi(\xi)$$). Hence, equation (1) can be solved for $$\xi>0$$ with initial conditions of either $$\psi(0)\neq 0$$ and $$\psi'(0) = 0$$ (even) or $$\psi(0) = 0$$ and $$\psi'(0)\neq 0$$ (odd).

Figure 2: Analog computer program solving the Schr ̈odinger equation with Gaussian potential.

In figure 2 the analog program to solve equation (1) is shown. In the upper half the Gaussian function is generated by solving the differential equation

\begin{align} \frac{\text{d}}{\text{d}\xi}f(\xi) = -2\xi f(\xi),\hspace{1cm}\text{and}\hspace{1cm} f(0)=1. \tag{3} \label{eq:diffGaussian}\end{align}

One has to be careful about the variable of interest in equations such as (3) because $$\frac{\text{d}}{\text{d}\xi} \neq \frac{\text{d}}{\text{d}t}$$ (all integrators integrate over time).

In the following $$\xi$$ is defined as $$\xi = \sqrt{\frac{\alpha}{2}} t$$, implying $$\frac{\text{d}}{\text{d}\xi} = \sqrt{\frac{2}{\alpha}} \frac{\text{d}}{\text{d}t}$$. With this eq. (3) can be rewritten as

\begin{split}\begin{align} \sqrt{\frac{2}{\alpha}} \frac{\text{d}}{\text{d}t}f(\xi) &= -2 \sqrt{\frac{\alpha}{2}} t f(\xi) \notag \\ \Leftrightarrow \frac{\text{d}}{\text{d}t}f(\xi) &= -\alpha t f(\xi) \hspace{1cm}\text{and}\hspace{1cm} f(0)=1. \tag{4} \label{eq:diffGaussian2}\end{align}\end{split}

The implementation of eq. (4) can directly be seen in the upper half of the analog program in figure 2. The lower half implements the two-body Schrödinger equation in eq. (1). To see this correspondence the equation can be rewritten:

\begin{split}\begin{align} \frac{\text{d}^2}{\text{d}\xi^2} \psi(\xi) &= -2 \left[v_0f(\xi) + E \right]\psi(\xi) \tag{5}\\ \Leftrightarrow \hspace{0.5cm} \left( \sqrt{\frac{2}{\alpha}} \right)^2 \frac{\text{d}^2}{\text{d}t^2} \psi(\xi) &= -2 \left[v_0f(\xi) + E \right]\psi(\xi) \tag{6}\\ \Leftrightarrow \hspace{2.2cm} \frac{\text{d}^2}{\text{d}t^2} \psi(\xi) &= -\alpha \left[v_0f(\xi) + E \right]\psi(\xi). \tag{7} \label{eq:schrödImpl}\end{align}\end{split}

The implementation of eq. (7) in the lower half of figure (2) is straightforward. The potentiometer for $$E$$ gets a negative reference input since for a positive potential depth $$v_0>0$$ the wave function $$\psi$$ is only bound if the energy is negative. The initial conditions for $$\psi$$ in figure (2) are set to generate even solutions.

## Calculation¶

In binding energies for the three-body system are calculated for values of the potential depth $$v_0$$ for which the two-body subsystem has specific energy values. So for a given energy one is interested in the value of $$v_0$$, or in other words the strength of the attractive force between the two particles, for which the two-body system is bound.

A system is in a bound state, if its wave function $$\psi$$ remains localized. This implies that for large values of $$\xi$$, $$\psi$$ tends to zero ($$\lim_{\xi\rightarrow \pm \infty} \psi(\xi) = 0$$). In the following two-body energies of $$E=-10^{-1},-10^{-2},-10^{-3}$$ are investigated. The potential depth $$v_0$$ required for the system to be in a bound state can be derived by varying $$v_0$$ until $$\psi$$ is localized.

This process in depicted in figure 3. The program is set up for $$E=-0.1$$ and $$\alpha=0.1$$ on an Analog Paradigm Model-1. All integrators have a time scale factor of $$k_0 = 10^4$$ with the exception of two integrators with an $$\alpha=0.1$$ scaling in front, which is absorbed into the time scale factor by setting $$k_0 = 10^3$$. With this setup the effect on $$\psi$$ by varying $$v_0$$ can be tested and a bound state of the system can be derived.

Figure 3: Two runs of the analog program for E = −0.1 and α = 0.1

In figure 3 it can be seen that even very slight changes of $$v_0$$ affect $$\psi$$. Both of the states are not bound states, because $$\lim_{\xi\rightarrow \pm \infty} \psi(\xi) \neq 0$$. However, the two states in figure 3 suggest that for some value of $$v_0$$ between $$0.342$$ and $$0.343$$ there is a bound state. With this process regions of $$v_0$$ for different values of $$E$$, in which the system is bound, can be derived.

Table 1: Values of v0 at different energies E. Results from the Model-1 analog computer are compared with results from [Thies et al. 2022].

## Results¶

In table 1 the results from the analog computer are compared with the results in . The values of $$v_0$$ derived by the analog computer setup are all close the theoretical values. For $$E=-0.1$$ and $$E=-0.01$$ the deviations are less than $$0.5\%$$ and for $$E=-10^{-3}$$ it is about $$5\%$$. The uncertainties given for values of $$v_0$$ from the Model-1 are derived from the variation of $$v_0$$ around the bounded state of $$\psi$$. Uncertainties of the analog program due to the limited precision of analog components are not analysed.

## References¶

[Thies et al. 2022] Jonas This, Moritz Travis Hof, Matthias Zimmermann, Maxim Efremov, “Tensor Product Scheme for Computing of Bound States of the Quantum Mechanical Three-Body Problem”, https://arxiv.org/pdf/2111.02534.pdf, 2022