# Infinite square potential well

1. Jan 28, 2016

### whatisreality

1. The problem statement, all variables and given/known data
I think this is a square well potential problem. The question asks me to sketch the ground-state probability density, for the following situation:

A quasielectron moves in a 'quantum dot' device. The potential V(x) = 0 for 0 ≤ x < L, and is infinite otherwise.

2. Relevant equations

3. The attempt at a solution
I'm going to need to solve the Schrodinger equation within the device. I have no idea what a quantum dot is, but presumably it doesn't actually matter! I think you can just treat it like an infinite square potential. The Schrodinger equation is

$-\frac{\hbar^2}{2m} \frac{\partial \psi}{\partial x^2} = E \psi$
Within the device. And I also need boundary conditions. One of these boundary conditions is $\psi(L)=0$, since the potential is infinite at L. Annoyingly, the potential is 0 at x=0, so I can't use the same reasoning there. So my first question is, how do I get my second boundary condition? I don't see how the potential being zero at x=0 in any way restricts what $\psi$ could be...

2. Jan 28, 2016

Staff Emeritus
A. I wouldn't solve anything. It says "sketch".

B. If I were going to solve something, I'd make the problem more symmetric.

3. Jan 28, 2016

### whatisreality

Don't I need to calculate the probability density in order to sketch it?

4. Jan 28, 2016

### PeroK

Specifying $V(x) = 0 \ (0 \le x < 1)$ is equivalent to $V(x) = 0 \ (0 < x < 1)$.

5. Jan 28, 2016

### whatisreality

Why? In the first inequality V(x) is zero at x = 0, but in the second inequality the potential is infinite at x = 0. So I can't say that at x = 0, the wavefunction is zero, and I don't know where else to get a second boundary condition from.

6. Jan 28, 2016

### PeroK

Whether the boundary points are in or out doesn't affect the functions or the integrals. If the wavefunction is not $0$ at $x=0$ then it is discontinuous when viewed as a function for all $x$.

7. Jan 28, 2016

### whatisreality

So if I said the wavefunction is zero at x = 0, then I'm excluding the electron from being at x = 0, even though at x = 0 the potential is zero... that's ok? I thought it would have something to do with continuity! I was trying to find a way to use the derivative of $\psi$.

Vanadium's post makes it sound like I don't actually need to solve for $\psi$ though to sketch the probability distribution. I know it's zero at the edges, but don't I need to integrate the magnitude between infinity and negative infinity to work out the probability density? How could I sketch it without an equation?

8. Jan 28, 2016

### PeroK

Yes. I suspect whover set the question didn't give it a second thought!

9. Jan 28, 2016

### whatisreality

I know I'd only actually have to integrate between L and 0, and it should equal 1. And it's zero at both edges. Where does the whole 'ground state' thing come into it, anyway?

If this is actually a really easy question, I must be missing some fairly basic information.

10. Jan 28, 2016

### PeroK

Are you supposed to know the ground state of the infinite square well? If not, then to sketch it, you'd have to solve the Schrodinger equation, I imagine.

11. Jan 28, 2016

### whatisreality

OK. We haven't done anything specifically about ground states of infinite square wells. I'd better solve the Schrodinger equation then!

12. Jan 28, 2016

### PeroK

It can't do any harm!