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Quantum Mechanics: Ground States for 2 Charged Particles in the 1D Infinite Square Well

In this blog post I want to have a look at the Coulomb interaction, the governing equation of electrostatics, in the context of quantum mechanics. In my undergraduate QM courses, when discussing multi-particle systems, the first thing textbooks tended to do was to turn off the Coulomb interaction to render the problems analytically tractable. The hydrogen atom is an important exception, but it treats the proton classically and considers the motion of an electron in a classical field. This aroused my interest in the quantum mechanics of interacting stationary electric charges, all of which are treated quantum mechanically, which can be investigated numerically. I use my implementation of the quantum Monte Carlo (QMC) algorithm for this purpose – I may describe this algorithm and my code in more detail in the future, if people are interested. In this introductory post, I look at the ground state wave functions in a qualitative way, without worrying about the energies.

Description of the Problem

We will look at the quantum mechanics of two electrically charged particles of unit mass, the product of whose charges is $Q$, trapped in an infinite square well of extent $L$ (this is roughly analogous to two stationary charges confined to a thin conducting wire of length $L$). The particles are described by a wave function of two spatial dimensions, $\psi(x_1,x_2,t)$, $x_1$ and $x_2$ being the particle position variables, whose range is $\left[-L/2,L/2\right]$. We are making approximations by neglecting magnetic fields and by treating the scalar potential as static. We justify these by considering that we are in a 1D space (no magnetism) and that for ground states, the momentum expectation value is low (so relativistic effects can be ignored). Therefore, non-relativistic quantum mechanics will suffice.

For a quick refresher on two-particle states in QM, see this excellent page.

The objective is to find the system’s ground state, $\psi_{n=0}(x_1,x_2)$, and its corresponding energy expectation value, $E_{n=0}$, where $n$ represents the states’ quantum number(s). This can be achieved by solving the time-independent Schrödinger equation (TISE), which for this system reads (adopting natural units such that $\hbar=c=\epsilon_0=1$):

$-\frac{1}{2} \left[\frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} \right] \psi_{n} + \frac{Q}{4 \pi } \frac{1}{ \lvert x_1 - x_2 \rvert } \psi_{n}=E_{n} \psi_{n}$.

The presence of the infinite square well potential forces the following boundary condition on $\psi_n$:

$\psi_n(\pm L/2, x_2)=\psi_n(x_1, \pm L/2)=0$.

The TISE cannot be solved analytically for this case, due to the Coulomb interaction term – this is why we’ll turn to the QMC algorithm.

Note that if the product of the charges is negative, then the two particles carry charges of opposite sign, which means that the Coulomb force between them is attractive, and that the particles are distinguishable. If the product of the charges is positive, then the particles are of like charge, and the force is repulsive. Since the masses are identical, particles of like charge are indistinguishable and their wave function $\psi_{n}$ must be symmetric (for bosons) or antisymmetric (for fermions).

Also note that although the two particles live in a 1D space, their wave function is in fact two-dimensional. This is par for the course in multi-particle quantum mechanics; the dimensionality of the configuration space in which $\psi$ is described depends on the number of particles under consideration. This is a topic worthy of a separate blog post, but for now, we can just keep in mind that $\psi$ exists in a two-dimensional configuration space on which we can define a two-dimensional potential:

$V(x_1,x_2) = \frac{Q}{4 \pi } \frac{1}{ \lvert x_1 - x_2 \rvert }$.

The next step is to enter the above description into my QMC code and plot the results.

Results of the QMC Simulations

The QMC algorithm starts with a trial wave function which then slowly converges to the ground state as the algorithm runs. For my trial wave functions, I simply pick a constant wave function:

$\psi_{trial} = 1/L^2,$

where $L^2$ is the area of the two-dimensional configuration space in which $\psi$ lives.

Let’s start off with the limiting case $Q=0$ (i.e. the particles don’t interact), which is equivalent to the case of a single particle in a two-dimensional infinite square well. If the particles are distinguishable and/or are bosons (so that their wave function can be symmetric), we end up with the result plotted in Fig. 1 (note that in this simulation, $L=16$):

Figure 1 – Ground state of two non-interacting particles in the ISW.

Note that this simple case can be solved analytically, and we have

$\psi_{0,Q=0}=A \cos(\pi x / L) \cos(\pi y/L),$

Where $A$ is a normalization constant. Note also that, if the two particles were indistinguishable fermions, this ground state would be forbidden (since it does not vanish on the line $x_1=x_2$, i.e. there is a non-zero chance of finding both particles at the same position). To see what happens in that case, I again refer you to this page.

Next, let’s move on to the more interesting case $Q<0$. This time there will be an attractive electrical interaction between the particles; as discussed before, the particles are necessarily distinguishable (since the charges are different). The result is plotted in Fig. 2:

Figure 2 – Ground state of two particles of opposite charge (i.e. attractive Coulomb force) in the ISW. Note that the wave function peaks on the line $x_1=x_2$ – the particles are more likely to be found near each other.

As one might expect intuitively, the particles now tend to cluster together. The resulting ground state shares some similarities with the ground state of hydrogen; $\psi$ exhibits a ‘kink’ on the line $x_1=x_2$. On this line, the potential diverges (the Coulomb force does that when two particles are infinitely close together); in QM, a diverging potential ensures a diverging first spatial derivative of the wave function (in a way, these two divergences ‘cancel out’ to produce a finite energy – see the ground state of the hydrogen atom!)

Finally, let’s see what happens when $Q>0$, i.e. when there is a repulsive force between the two particles. If the particles are of like charge, they are indistinguishable (since we ignore spin, the particles have no qualities other than mass and charge); thus, the wave function must be (anti)symmetric.

Figure 3 – Ground state of two particles of like charge (i.e. repulsive Coulomb force) in the ISW. Note that the wave function vanishes on the line $x_1=x_2$; the particles are more likely to be found far away from each other.

We see that the wave function is symmetric as currently plotted (so the two particles are bosons), and we have $\psi(x_1,x_2) = \psi(x_2,x_1)$. To find the corresponding wave function for fermions, we can simply flip the sign of $\psi$ on either side of the line $x_1=x_2$, so that $\psi(x_1,x_2) = -\psi(x_2,x_1)$. Note that we could not perform the same trick on the case where $Q<0$! If we tried to flip the sign of $\psi_{Q<0}$ on either side of the line $x_1=x_2$ we would end up with a discontinuous wave function. Fortunately, we are saved from this problem by the fact that attracting particles necessarily have different charges and are thus distinguishable.

That’s all I wanted to cover for now – if you have any questions or feedback, don’t hesitate to leave a comment!