Normalizing wave function, factor of 2 out

[Frinkz]

Homework Statement

Consider the wave function

\Psi(x, t) = Ae^{-\lambda|x|}e^{-i\omega t}

where A, \lambda and \omega are positive real constants.

Normalize \Psi

Homework Equations

\int |\Psi(x, t)|^{2} dx = 1

|\Psi(x, t)|^{2} = \Psi^{*}\Psi

The Attempt at a Solution

I have a model solution - with a step missing, I think my error is in complex conjugate math...

1, Finding |\Psi(x, t)|^{2}

\Psi^{*}\Psi = (Ae^{-\lambda|x|}e^{i\omega t}) (Ae^{-\lambda|x|}e^{-i\omega t})

= A^{2}e^{-2\lambda|x|}e^{i\omega t}e^{-i\omega t}
= A^{2}e^{-2\lambda|x|}e^{0} = A^{2}e^{-2\lambda|x|}

I think this is where my problem is, I am told that

|\Psi(x, t)|^{2} = 2|A|^{2}e^{-2\lambda|x|}So I am missing a factor of 2?

Is there a complex conjugate rule somewhere I am missing?

 

[thebigstar25]

your steps looks fine to me, I am not sure from where they got the factor of 2?.. and there is no other complex conjugate rule you miss ..

is their answer says that psi^2 is after obtaining the normalized wavefunction? or is it just the step that follows the one you did ?..

 

[Frinkz]

Thanks for checking.


Turns out, the model answers were a bit badly written.


They had put in the factor of two, because the integral was changed from -infinity to +infinity, to 0 to +infinity, so you can make use of a definite integral

\int_0^{\infty} \! e^{-\lambda x} = \frac{1}{2}\sqrt{\frac{\pi}{\lambda}}

That step was just omitted from the solution I was trying to understand, but I get it now :)

 

[thebigstar25]

the integral you mentioned should gives 1/2lamda ...

check this link ..

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ExpIntegrals.htm

 

[Frinkz]

Sorry, that was a typo.

Should have been e^(-lambda x^2)