Calculating Expectation Value of Particle in Square Potential Well

[cscott]

I'm confused re a particle of energy E < V inside a square potential of width 'a' centered at x = 0 with depth V.

They give the wavefunction for outside the well as \Psi(x) = Ae^{k|x|} for |x| > a/2

and k^2 = -\frac{2ME}{\hbar^2} => k = i\frac{\sqrt{2ME}}{\hbar} ?

I need the probability that the particle is outside the potential well. So I integrate \int{\Psi(x)\Psi^*(x)dx} from a/2 to infinity if I take x to be positive and then multiply by 2 for symmetry?

But isn't \Psi(x)^*\Psi(x) = A? So my integral is infinity but shouldn't it be 0?

 

[nrqed]

I'm confused re a particle of energy E < V inside a square potential of width 'a' centered at x = 0 with depth V.

They give the wavefunction for outside the well as \Psi(x) = Ae^{k|x|} for |x| > a/2

and k^2 = -\frac{2ME}{\hbar^2}

I need the probability that the particle is outside the potential well. So I integrate \int{\Psi(x)\Psi^*(x)dx} from 0 to infinity if I take x to be positive and then multiply by 2 for symmetry?

But isn't \Psi(x)^*\Psi(x) = A? So my integral is infinity but shouldn't it be 0?

you must integrate from a/2 to infinity, not from 0.

Because the exponential is real then Psi Psi^* does not give a constant (that only happens for imaginary exponentials)

 

[cscott]

Sorry I meant so say from a/2 to infinity.

But doesn't k have to have an i in it because k squared is negative?

 

[nrqed]

Sorry I meant so say from a/2 to infinity.

But doesn't k have to have an i in it because k squared is negative?

I am a bit confused by your choice of zero for the potential. You are doing a bound state, right? If you set V =0 outside of the well, then it means E < 0 (but E > -V where I am assuming V is a positive number). Then k is real.