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 (\pix/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 (\pix/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.