Particle in a Box: Solving for Acceptable Wave Function with Boundary Conditions

[thomashpotato]

Homework Statement

V(x) = 0 if $\frac{-L}{2}$<x<$\frac{L}{2}$ and $\infty$ otherwise.

Is the wave function $\Psi$ = (2/L)$^{1/2}$ (sin ($\pi$x/L) an acceptable solution to this? Explain

Homework Equations

H$\Psi$= E$\Psi$ , normalization: 1 = $\int$ wavefunction^{2}dx

The Attempt at a Solution

My logic is that I have to come up with a wave function that satisfy the boundary condition. Therefore $\Psi$ (x= $\frac{-L}{2}$) = $\Psi$ (x = $\frac{L}{2}$) = 0

My initial answer was that it's not because I was thinking that the wavefunction itself has to be A[/itex]cosine(bx), where b=2n$\pi$/L. n = 1/4, 3/4 , 5/4 ...

I am not quite sure if that's correct. Also when I try to calculate the normalization factor (A), it turns out to be 1 = $\frac{A^{2}L}{2}$ + $\frac{A^{2}L}{n\pi}$ sin ($\frac{n\pi}{2}$)

 

[vela]

Homework Statement

V(x) = 0 if $\frac{-L}{2}$<x<$\frac{L}{2}$ and $\infty$ otherwise.

Is the wave function $\Psi$ = (2/L)$^{1/2}$ (sin ($\pi$x/L) an acceptable solution to this? Explain

Homework Equations

H$\Psi$= E$\Psi$ , normalization: 1 = $\int$ wavefunction^{2}dx

The Attempt at a Solution

My logic is that I have to come up with a wave function that satisfy the boundary condition. Therefore $\Psi$ (x= $\frac{-L}{2}$) = $\Psi$ (x = $\frac{L}{2}$) = 0

This is right. Does the proposed wave function satisfy these conditions?

My initial answer was that it's not because I was thinking that the wavefunction itself has to be A[/itex]cosine(bx), where b=2n$\pi$/L. n = 1/4, 3/4 , 5/4 ...

You'll actually get both sine and cosine solutions if you solve the infinite square well problem completely, so just because this wave function has a sine in it isn't reason enough to exclude it.

I am not quite sure if that's correct. Also when I try to calculate the normalization factor (A), it turns out to be 1 = $\frac{A^{2}L}{2}$ + $\frac{A^{2}L}{n\pi}$ sin ($\frac{n\pi}{2}$)

 

[Dick]

I think your reasoning is correct. The wave function doesn't vanish at the boundaries. So it's not good. I wouldn't worry about whether the normalization is correct if the boundary conditions aren't correct.