# Can someone verify my work

1. Nov 17, 2009

### Void123

1. The problem statement, all variables and given/known data

I have an infinite square well and I am asked to show why E = 0 and E < 0 does not satisfy the Schrodinger's equation. I must do this by applying the boundary conditions.

For E = 0:

I argued that the second derivative of the wave function is zero.

So,

$$\Psi(x) = A + Bx$$

By imposing the boundary conditions $$\Psi (0) = \Psi (a) = 0$$ I get:

$$\Psi(x) = Bx$$

and

$$\Psi (a) = Ba = 0$$

Therefore I concluded that:

(1) $$B$$ cannot be zero, or else we get $$\Psi(x) = 0$$ which is physically unacceptable
(2) $$a\neq0$$ since $$a$$ is the upper bound.

Perhaps before presenting my second solution to E < 0 I should make sure all the above is correct.

Is it valid?

2. Relevant equations

...

3. The attempt at a solution

...

2. Nov 18, 2009

### Redbelly98

Staff Emeritus
Looks good so far.

3. Nov 18, 2009

### Void123

E < 0:

The second derivative of the wave function I set equal to $$k^{2}\Psi(x)$$

The solution is $$\Psi(x) = Ae^{-kx} + Be^{kx}$$, where the first term on the right cancels since it blows up at infinitely large values of $$x$$.

Thus, $$\Psi(0) = Be^{k0} = 0$$ (after imposing the boundary conditions), which does not give us a satisfying solution for $$B$$ or $$e^{0}$$. Neither does it work for $$\Psi(a) = 0$$.

So, the boundary conditions don't work.

4. Nov 18, 2009

### Redbelly98

Staff Emeritus
Note that the solution and coefficients A and B are only valid within 0<x<a. So it is okay if this Ψ(x) expression blows up at infinity, since the actual Ψ(x) will be a different expression for x<0 or x>a.

Use both boundary conditions (at x=0 and a), and the expression you wrote for Ψ(x).