# Intuition for the number of bound states to the double Dirac potential well

I was wondering about the bound states to the double Dirac potential well. As an example, we can consider $$V(x)=-\alpha\delta(x-a)-\alpha\delta(x+a)$$ as the potential function. I already know how to derive the solution mathematically.

In general, when $$a$$ are close, there is only one bound state, which is even. When $$a$$ are further apart, there are two bound states, one even and one odd. I have a couple questions about why we should expect this. It seems that when $$a$$ is further apart, we can consider the two wells "separately."

1. But why should we physically expect the odd state to disappear when $$a$$ becomes small? (And I guess this depends on $$\alpha$$ too; how would that work?).

2. Also, why should the odd state also have more energy than the even state, intuitively? I guess I am overall missing the how to think about these situations qualitatively.

• Commented Oct 9, 2021 at 15:56
• Study the small a limit of the odd solution, cf. WP and ref therein. Commented Oct 9, 2021 at 16:18

1. It follows from standard arguments that a single Dirac potential well has exactly 1 bound state, cf. e.g. this Phys.SE post.

2. On one hand, when the 2 wells sit on top of each other, it's really just 1 well, so there should be 1 bound states.

On the other hand, when we start separating the 2 wells, and the distance gets big, we expect the 2 wells to be independent of each other, i.e. there should be 2 bound states.

3. The energy of the bound state increase with the number of nodes, cf. e.g. this Phys.SE post. Hence the symmetric/even/node-less state is the ground state.

4. Further arguments:

Let us ignore the boundary conditions (BCs) at $$x=\pm\infty$$ and consider real wavefunction solutions $$x\mapsto \psi(x)$$ to the TISE as a function of energy $$E<0$$.

It turns out that the discontinuity of the relative slope $$\frac{ \psi^{\prime}(x+0^+)-\psi^{\prime}(x-0^+)}{\psi(x)}$$ at a well is a finite constant independent of $$E$$.

Also recall that in the bulk the wavefunction is a linear combination of exponentials, which bends away from from the $$x$$-axis.

Therefore, if the 2 wells are too close to each other, the wavefunction $$\psi$$ does not have sufficient space to develop a node, i.e. then there is only 1 bound state.

• What a great answer. Commented Oct 12, 2021 at 17:53
• @Sarthak: Thanks. Commented Oct 14, 2021 at 15:42

A single Dirac well supports ine bound state. Let's say its energy is $$-E_0$$. When we have two wells, the tunneling lifts the degeneracy, transforming the two $$-E_0$$ states into: $$-E_0\pm\Delta(a)$$.

The amplitude of the correction $$\Delta(a)$$ depends on how close the wells are. If they are very far apart, it may be very small and negligeable (in comparison to other problem parameters). If they are so close that $$\Delta>E_0$$, the asymmetric state might be "pushed" above the zero, i.e., it may stop to exist.

Source of intuition
The intuitive picture described above is based on an analogy with two degenerate levels coupled via a small perturbation (i.e., two levels in the each potential well coupled via tunneling). This is described by a Hamiltonia $$H=\begin{bmatrix} -E_0&\Delta\\\Delta&-E_0\end{bmatrix},$$ which is easily diagonalized to give solutions $$E_\pm=-E_0\pm \Delta.$$ The energy levels split with one state lowering in respect to the original energy level, and the other going up. In case of a double well, if $$-E_0 +\Delta >0$$, this state is pushed into the continuum, i.e., it ceases to be bound.

Exact solution
The problem with two Dirac wells can be easily solved, resulting in the following equation for energies: $$E=-\frac{\hbar^2 q^2}{2m}, (q-q_0)^2-q_0^2e^{-2q a}=0,$$ where the energy of the bound state in a single potential well is given by $$E_0=-\frac{\hbar^2 q_0^2}{2m},q_0=\frac{m\alpha}{\hbar^2}.$$

The equation for $$q$$ can be recast as two equations: $$q=q_0+q_0e^{-qa},\\ q=q_0-q_0e^{-qa}.$$ Solving these graphically, one can immediately see that the first equation always has a solution with $$q>q_0$$. This is the symmetric/bonding state that always remains bound. The second equation has a non-trivial solution only if the initial slope of the curve $$q_0-q_0e^{-qa}$$ is greater than $$1$$, i.e. $$q_0a >1$$ - this is the indication for the disappearance of the bound state when the separation between the two wells is too small.

Finally, for $$q_0a\gg 1$$ we can approximate the two solutions to order $$e^{-q_0a}$$ by $$q=q_0(1\pm e^{-q_0 a}),$$ which gives us energies $$E_\pm=E_0(1\pm e^{-q_0 a})^2=E_0(1+e^{-2q_0 a})\pm 2E_0 e^{-q_0 a}\approx E_0\pm 2E_0 e^{-q_0 a},$$ consistent with our initial intuition.

• Could you explain what you mean by being "pushed" above the zero? And what exactly is this $\Delta(a)$ correction? Commented Oct 9, 2021 at 14:21
• @Vasting since you have already done the math,, I suggest that you post the solution for energies (or an equation for them) - it will be easier to discuss then. Commented Oct 9, 2021 at 15:54