# Wave Function Probability

1. Nov 6, 2008

### madmike159

I was reading part of a book which was explaining about the probability of finding a partical on a 1d line.
$$\int^{+\infty}_{-\infty}$$P(x) dx = 1
This sounds right because if the line was infinitely long then the partical must be on it.
You can them intergrate between a and b to find the probability of it being in a lenght and if a and b were the same the probability would be 0.
But when you intergrate P(x) dx you get $$\frac{Px^{2}}{2}$$
by putting the numbers in you get P$$\infty$$ - -P$$\infty$$
or P$$\infty$$ + P$$\infty$$ = P$$\infty$$
A probability can't be more than 1. I must be missing something or dealing with the infinities in the wrong way.
(Sorry it looks like P^infinity its P x infinity but I couldn't change it.)

2. Nov 6, 2008

### cks

$$\int_\infty^\infty P(x) dx=1$$ doesn't mean $$\int_\infty^\infty Px dx=1$$ P(x) means the probability in function of x not P a constant times x.

3. Nov 6, 2008

### f95toli

?

If you integrate P(x) dx you get the integral of P(x); I am not quite sure why you think you would get $$\frac{Px^{2}}{2}$$?
P(x) is a FUNCTION, not a constant; there is no way to integrate it unless you know what that function is.

4. Nov 6, 2008

### madmike159

Oh so P(x) is the wave function? I'm going to read the chapter again.

5. Nov 6, 2008

### Manilzin

No, P(x) is the probability function, which is the wavefunction squared (actually, absolute value squared)... So P(x)dx gives the probability for finding the particle on a bit of length dx at position x.

6. Nov 6, 2008

### madmike159

Yea, I re-read it and think I understand now. W(x) is the wave function and P(x) =|W(x)^2

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