Finding potential for a given wave function

[Desh627]

Here's the sitch:

I am given an equation, A*e^{-a(mx2/h-bar+it)}

I need to find the value for A that will satisfy normalization, as well as find the Potential of the Schrödinger Equation using this value.

What do I do?

P.S. I have NOT learned gaussian integration, which is where I run into my first major problem, and this stops me from completing the problem.

Thanks guys,
Desh627

 

[Desh627]

I should also mention what I have solved so far...

psi (x,t) = A*e^{-a(mx2/h-bar+it)}

psi (x,0) = A*e^{-a(mx2/h-bar+i(0))}
psi (x,0) = A*e^{-a(mx2/h-bar)}

Integral from negative infinity to infinity dx (absolute value)A*e^{-a(mx2/h-bar)}(/absolute value)² = 1

A² Integral from negative infinity to infinity dx e^{-a(mx2/h-bar)}=1

A = sq. rt (1/Integral from negative infinity to infinity dx e^{-a(mx2/h-bar)})

and this is where I run out of room.

I should also mention that I have taken high school physics and ap calc, but no linear algebra or anything beyond ap calc.

 

[yaychemistry]

Hi,
Let's see if we can't figure this one out.
To figure out how to normalize your wavefunction, you need to know what the integral of a Gaussian function is, where a Gaussian function is any function of the form:
f\left(x\right) = \exp\left(-\frac{x^2}{2\sigma^2}\right)
I doubt you would know a good technique from AP calculus to solve this integral, so I would suggest reading this wikipedia article:
http://en.wikipedia.org/wiki/Gaussian_integral

For the second part, since your wavefunction already has time explicity in it, we can make a guess that it's a solution to the time dependent schrodinger equation:

\hat{H}\psi = i\hbar\frac{\partial}{\partial t}\psi

where H is the operator such that:
\hat{H}\psi\left(x\right) = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi\left(x\right) + V\left(x\right)\psi\left(x\right)
And V(x) is the potential. Try applying the time-dependent schrodinger equation and see if you can convert the problem into a differential equation in x only (and not t). From there it might be possible to figure out what V(x) is.