Strength of Dirac Delta Potential

[dyn]

When talking about the strength of a delta potential , the delta potential is multiplied by a parameter ie α but how does a delta potential have a strength ? It is zero everywhere and infinite at x = 0. The parameter makes no difference to zero or infinity.

 

[stevendaryl]

When talking about the strength of a delta potential , the delta potential is multiplied by a parameter ie α but how does a delta potential have a strength ? It is zero everywhere and infinite at x = 0. The parameter makes no difference to zero or infinity.

You can think of $f(x) = - \alpha \delta(x)$ as a limiting case of a step function:

$x < \frac{-\epsilon}{2} \longrightarrow f_\epsilon(x) = 0$
$\frac{-\epsilon}{2} < x < \frac{+\epsilon}{2} \longrightarrow f_\epsilon(x) = -\frac{\alpha}{\epsilon}$
$+\epsilon < x \longrightarrow f_\epsilon(x) = 0$

Then for any other smooth function $\psi(x)$, you will have: (for any $A, B$ such that $A < B$)

1. $lim_{\epsilon \rightarrow 0} \int_{A}^{B} f_\epsilon(x) \psi(x) dx = -\alpha \psi(0)$ (if $A < 0 < B$)
2. $lim_{\epsilon \rightarrow 0} \int_{A}^{B} f_\epsilon(x) \psi(x) dx = 0$(otherwise)

So you can think of solving the wave equation for a delta-function potential as solving it for a square well, and then taking the limit as the width $\epsilon$ goes to zero.

 

[dyn]

So does the potential have a value at x=0 ?

 

[stevendaryl]

So does the potential have a value at x=0 ?

The delta function is undefined at $x=0$, except indirectly through the integral law:

$\int \delta(x) \psi(x) dx = \psi(0)$

No real function has that property, but as I said, you can think of it as some kind of limiting case of a step function.