How Does the Wave Function Determine Particle Location Probability?

[madmike159]

I was reading part of a book which was explaining about the probability of finding a particle on a 1d line.
\int^{+\infty}_{-\infty}P(x) dx = 1
This sounds right because if the line was infinitely long then the particle must be on it.
You can them intergrate between a and b to find the probability of it being in a length 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.)

 

[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.

 

[f95toli]

But when you intergrate P(x) dx you get \frac{Px^{2}}{2}

?

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.

 

[madmike159]

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

 

[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.

 

[madmike159]

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