Physics help?

A free particle has the initial wave function:

phi(x,0) = Ae^(-a|x|)

A and a are constants,

Normalise phi(x,0)

I got A = root a, but I doubt this is right somehow...

Yh, thats right.

Now stop wasting our time.

Hmm, I'm not getting a real answer for A, since the wave function isn't complex:

Since phi(x,0) = Ae^{-a|x|}, P(x) = A^2 e^{-2a|x|}

Normalisation:
int_{0}^{inf} A^2 e^(-2a|x|) = \frac{A^2}{-2a} = 1 (integration is not evaluated from -inf to inf because of the absolute value of x)

Solving for A:
\frac{A^2}{-2a} = 1
A = \sqrt{\frac{1}{-2a}} -> nonreal value

I'm confuss. It's been a while though, I could be completely off :P

>>3
Your integral limits are wrong. |x| is x on 0 to inf and -x on -inf to 0. You have to split the integral up to evaluate, but you can't just ignore one half.

>>4
You're right! However, that way I don't get a good answer at all:
A^2 \left [ \int_{0}^{\infty}e^{-2ax}\,dx + \int_{0}^{\infty}e^{2ax}\,dx \right ] = A^2 \left [ -\frac{1}{2a} + \frac{1}{2a} \right ] = 1 ....?

>>5
aw crap, the limits of the second integral should be from -infty to 0, off course...

>>3
\int_{0}^{\infty} A^2 e^{-2ax}\,dx = \frac{A^2e^{-2ax}}{-2a}|_{0}^{\infty} = (\frac{A^2}{-2a})(e^{-\infty} - e^0) = (\frac{A^2}{-2a})(0 - 1) = \frac{A^2}{2a}

You probably flipped your limits; e^x is positive definite.  Of course, the integral from -infty to 0 is the same thing by symmetry.

>>5
>>7
How do you do formulae properly like that???

>>8
TeX syntax between [m ath] and [/m ath] (without the spaces, of course)