# Delta function potential

1. Oct 28, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Why does it make sense that a negative delta function potential represents a highly localized attractive force and a positive delta function potential represents a highly localized repulsive force?

How do you explain that using

-dV/dx = f(x)

?

I guess I am confused about potentials.

The reason an infinite square is like a wall for the particle is that -dV(x)/dx is infinite at the boundary, right?

So, the particle has an infinite force pushing it away, right?

But isn't it minus infinity at the left boundary?

I am so confused.
2. Relevant equations

3. The attempt at a solution

Last edited: Oct 28, 2007
2. Oct 28, 2007

### Poop-Loops

3. Oct 29, 2007

### CompuChip

Basically, a delta function is the limit of a peak with area one, in the limit that the width goes to zero. So instead of a delta function, let me consider for example $V(x) = \pm e^{-x^2}$, with the + or - sign depending on if you want it do be attractive or repulsive of course.
Then
$$f(x) = - \frac{dV}{dx} = -(- 2 x) (\pm e^{-x^2}) = \pm 2 x e^{-x^2}.$$
Now $e^{-x^2}$ is always positive. So if you are somewhere at x > 0, then the force will be in the direction of the origin if the sign is - (it will be negative) and away from it if the sign is +. Similar reasoning (or the symmetry of the problem) will give you the same for x < 0. The same argument holds for the delta function (to get the idea, you could take exp(-a x^2) instead and take a to infinity, though technically this will not give you the delta function, so don't be alarmed if you get infinities; they're not there for the real delta function).

Last edited: Oct 29, 2007
4. Oct 29, 2007

### ehrenfest

I guess. My question is really about potentials in general. If I have a finite square well defined as

V(x) = { 0 when x < -a, -V_0 when abs(x) < a, 0 when x > a}

why does that represent an attractive potential? How do you explain the attraction using -dV(x)/dx = f(x) ?

The force is only nonzero at two places and it is infinite in both of those places, right?

5. Oct 29, 2007

### CompuChip

First of all, it always helps me to think of the potential as a sort of surface on which a free particle will move. So if you drop a particle somewhere inside the well, it can roll freely between x = -a and x = a, but it cannot get outside (so it's actually like you're dropping a ball inside a square well).

So suppose the particle is at x with |x| < a. At the left boundary of the well, there is an infinite downward slope in the potential so if the particle hits it, a force will work to the right (negative slope + the minus sign = positive x-direction). If you want to be precise, the derivative is not infinity, but it is the derivative of a (negative) delta peak. I suppose you can consider it as if there is an instantaneous infinite force to the right, pushing the particle away from the wall. But if you want to know the mathematical details, perhaps someone with knowledge of distributions can better explain it to you.

I don't know if what I'm about to say is valid -- maybe it's just gibberish, but the work done by this force to move a particle from -a to 0 will be
$$\int_{-a}^0 -V'(x) \, dx = -V(0) + V(-a) = \delta(0)$$
which is clearly not infinity, though it's hard to say what it is.

Last edited: Oct 29, 2007
6. Oct 29, 2007

### Jimmy Snyder

On a sheet of paper, draw a parabola opening upward. Think of this parabola as the surface upon which a ball rolls. It turns out that the shape of the parabola is the shape of the potential function. That is, as you climb the walls of the parabola, you increase gravitational potential energy. If you place the ball somewhere along the parabola, it will experience a gravitational force. Compare this force to the slope of the parabola at the point where the ball is and you will see qualitatively that when the slope is positive, the force is in the negative direction and vice versa.

By the way, since you are only concerned with the slope of the potential, the point of zero potential can be located at any point convenient to your calculations. Think of this example when you try to understand gauge invariance.

7. Oct 30, 2007

### ehrenfest

Thanks. I see.

8. Oct 30, 2007

### ehrenfest

Actually wait. If only changes in potential have physical significance why would the infinite square well have different solutions than the the delta potential barriers-on-both-sides well?

Is this what you were getting at, CompuChip, with the discussion about how the integral over the derivative of the delta function is not really infinite?

9. Oct 30, 2007

### CompuChip

For one, because with two delta barriers there would be three allowed areas (x < -a, -a < x < a and x > a) while in a square well there is just one (-a < x < a).