Free electron or empty lattice schrodinger equation solution

[AL-Hassan Naser]

in the solution for free electron we start with

X(x) = A exp (ikx) + B exp (-ikx)

then using boundary conditions we eliminate B if the wave is traveling in the positive direction and vice versa
my questions are:
1. what is the boundary condition used?
2. is it X(-inf) = 0? because this would cancel out B and leave A
3. But then if we also assume that X(inf)=0 then A=0 so we have no wave function and cancel out the electron which is a fallacy

please help i have my midterm on sunday

 

[_superluminal]

As this is a free electron, remember that if we were to take the integral $\int_{-\infty}^{\infty} \mathrm{X}^{*}\mathrm{X} = 1$

 

[AL-Hassan Naser]

X=Aexp(ikx)+Bexp(-ikx)
X*=Aexp(-ikx)+Bexp(ikx)
X.X*=A^2+B^2+AB[exp(-2ikx)+exp(2ikx)]
<XIX>=1
X(A^2+B^2) +[AB/2ik]*[exp(2ikx)-exp(-2ikx)] = 1

how can I solve that?!

please help me i am losing time

 

[AL-Hassan Naser]

really guys I need help here

 

[DrClaude]

It's not so much a question of boundary conditions as one of initial condition. Taking k to be positive, then one case corresponds to a particle of wave number k traveling towards +x, the other to a particle of wave number k traveling towards -x. You then keep the solution that corresponds to the physical situation you want to describe, for instance a particle coming in from $x = -\infty$.

As for the normalization, I don't know what superluminal had in mind, but a plane wave is not square integrable, and hence can't describe a physical state. For that, you need to build a wave packet by summing over a few plane waves.