# Interpretation of induced force between two Dirac delta potential wells

This problem is based on MIT OCW course 8.04 problem set 6 question 5(e). https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/assignments/MIT8_04S13_ps6.pdf

Consider two Dirac delta potential wells separated by a distance $$L$$. There is only one bound state for a particle in this system. One can solve for the energy $$E$$ as a function of $$L$$ numerically.

If we calculate $$-\frac{dE}{dL}$$, we would obtain a negative quantity, so the system has a tendency to minimize its energy by decreasing $$L$$. However, if we interpret this derivative as a "force", on what object is this force acting? What physical mechanism gives rise to this force? Furthermore, what about the case of 3 delta potential wells?

• Forces and gradients of energy are just two words for the same thing. How could you have one but not the other? – knzhou Jun 29 at 20:01
• @knzhou But what is the object the force is acting on? Does it make sense to say that there is an attractive force between two Dirac delta potentials? If yes, then what about the case in which there are three delta potentials, how should this gradient of energy be interpreted? – Leo L. Jun 29 at 20:04
• IMHO, OP has a point. If you were to do the same calculation for the hydrogen atom, $E\propto -a^{-2}$, you would get $dE/da>0$. Does this mean there is a positive force? Does the system tend to maximise the Bohr radius? Well, I'd say no: as far as this problem is concerned, $a$ is a constant (and so is $L$), so I don't think it makes sense to interpret $dE/da$ as a force. – AccidentalFourierTransform Jun 29 at 20:13
• That's true. In this case, you should think of the delta functions as sourced by some other objects, that attract the particle you're dealing with. The interpretation is that the particle causes an attraction between the objects in turn. – knzhou Jun 29 at 20:33
• @AccidentalFourierTransform For the hydrogen atom example, if we rely on this toy model, then this positive "force" balances the negative Coulomb force (at Bohr radius I intuit, but I haven't calculated it). So we could interpret $\frac{dE}{da}$ as a force. But I don't understand the physical meaning of this force. – Leo L. Jun 29 at 20:49