# What is a delta potential?

I know how a delta potential is described mathematically but how can a delta potential be a 'well'? Does it have particles outside the 'well' and 'bind' it or does it somehow have particle inside it?

Also, if we consider three dimensions then won't the potential be present at only one point in space meaning there being no particles inside it?

For example, suppose a situation where at origin we have a delta potential. Now, if a particle is at say the right side of the origin and moving towards it then how will it interact with the potential? Is this interpretation of the potential even correct?

I am sorry if I am unclear but I really don't understand the concept of delta potentials.

I think that a helpful way to think of delta potential (and maybe on the delta function in general) is through a limit process: we start with a finite square well of width $$a$$ and depth $$U=\lambda/a$$, and ask ourselves "what happens when we take $$a\to 0^+$$?" This can happen when we are interested, for example, in scales that are much larger than the width of the well, so we want to somehow make an approximation to zeroth order in $$a$$, but still keep the effects of the well. A nice thing here is that there are many different limit processes that can lead all to the same result, which is a very general expression of the potential as $$\lambda\delta(x)$$.

Note, that even though the width of the delta function is zero, it still has effect as it has non-zero measure $$\int\! dx \delta(x) = 1$$, which is quite obvious from the limit process that we introduced. Because we make sure to make it deep, we still keep its effects on whatever comes near it.

A particle can be "trapped" in the sense that any finite potential well can trap a particle - it has a probability to be found outside the well, as its the wave function decay exponentially outside the well for $$E<0$$. Now the particle has higher probability to be found near $$x=0$$ than far from it, in contrast to a free particle which spreads throughout the entire volume.

• Now suppose a particle is outside the well, will the well have any effect on it? – Korra Feb 21 at 10:47
• what do you mean by "outside"? there can be two possible interpretations for this statement - 1) the wave function of the particle is localized outside the well (let's say a narrow wave packet with its center far from the well) 2) a free particle with energy $E>0$. In both cases the well will have an effect over time, just like finite wells have effect on particles outside of them, because it will affect the wave function through the scattering process. In the first case, though, we can engineer it in such a way that the effects will be negligible (e.g. the packet travels away from the well). – yu-v Feb 21 at 10:56
• To add to what @yu-v said: If $\psi(x,t)=0$ around the point $x=L$, the potential will not affect the value of $\psi(x,t+dt)$. However it is a property of the schrödinger equation that $\psi$ quickly spreads to all real numbers, so in the next moment $\psi$ will be nonzero at the delta point and its motion will be affected (slightly, if $\psi$ is small). – doublefelix Feb 21 at 11:37
• @yu-v it's the first one and with E>0.its like if the potential is at say L then the particle has E>0 and is at a point x<L – Korra Feb 21 at 13:00

Usually $$\delta$$-potentials are a mathematical model of interactions with very short range.

The advantage of such models is that on one hand they are "exactly solvable", e.g. the spectrum and eigenvectors are explicitly known, on the other hand many interesting physical features are retained despite the simplification involved in approximating short-range with zero-range.

For example, ionization processes can be suitably well described, at least to some extent, by time-dependent point ($$\delta$$) interactions.

• How do you know that ionization is well-described by time-dependent $\delta$ interactions? That is a broad statement. It depends exactly what you want to know about the ionization; for certain things a $\delta$ potential will suffice but for more exact questions it will not. – doublefelix Feb 21 at 13:16
• @doublefelix Sure, as I said it is a simplification that retains useful physical features, but not all of them. – yuggib Feb 22 at 8:47