# Schrödinger equation: P(r)>1 ?

1. Nov 19, 2005

### Kruger

Schrödinger equation: P(r)>1 ???

I have the solution of the Schrödinger equation for the ground state of the hydrogen electron. The solution ist:

u100(r)=sqrt(1/(pi*a^3))*exp(-r/a)

If I want to calculate some probabilty values I do this with:

P(r)=4*pi*r^2*|u100|^2

If I set r=10^-13 I get a value that is greater than 1, I get P(10^-13)=10^34.

This cannot be. Whats wrong with my probability formula?

2. Nov 19, 2005

### Hurkyl

Staff Emeritus
P is a probability... density. To get the probability that it lies in a region, you need to integrate over that region.

3. Nov 19, 2005

### Kruger

With what function are the graphs plotted? (I mean one like this:
http://panda.unm.edu/courses/finley/P262/Hydrogen/img106.gif)

But to calculate the charge distribution of the electron I can take:

Charge(r)=-e*abs(u100)^2

thanks for your help

4. Nov 19, 2005

### Hurkyl

Staff Emeritus
I'm not familiar with those graphs -- I might look at the Bernoulli or Airy functions, but if it's not them, I don't think I could guess. (Also, I think I've read that the Airy functions appear in dealing with the hydrogen atom)