# Particle bound by quadratic potential

1. Nov 5, 2008

### kasse

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

A particle that can move in one dimension and that is in a stationary state, is bound by a potential V(x) = (1/2)kx^2. The wave function is $$\Psi$$(x,t) = $$\psi$$(x)exp(-iEt/$$\hbar$$)

We look at a state in which $$\psi$$(x) = Aexp(-x^2/2a^2a^2), where a is a constant and A is the normalisation constant. Determine a so that $$\psi$$(x) is an energy state. What is the energ of the particle?

3. The attempt at a solution

I don't really know what to do here, but setting up the wave equation tends to be a good start:

H$$\psi$$(x) = E$$\psi$$(x), where H = -($$\hbar$$^2/2m)(d^2/dx^2) + kx^2/2

How can I determine a so that $$\psi$$(x) is an energy state?

Last edited: Nov 6, 2008
2. Nov 5, 2008

### borgwal

Just calculate the left-hand and right-hand sides, and set them equal.

3. Nov 5, 2008

### kasse

That the wave function is an energy state simply means that it's described by the Schrödinger equation?

4. Nov 5, 2008

### borgwal

Yes, it's a solution to the *time-independent* SE, to be precise.

5. Nov 5, 2008

### kasse

OK, but I still don't see how to find a when E is unknown. My expression after differentiating twice is:

($$\hbar$$^2/2ma^2) (A - x^2/a^2) + kx^2/2 = E

6. Nov 5, 2008

### borgwal

First, you didn't calculate it right.

Second, the correct equation you'll get will have to be true for every x.

7. Nov 6, 2008

### kasse

Now I ended up with

($$\hbar$$^2/2ma^2)(1 - x^2/a^2) + kx^2/2 = E

What I should do here is to eliminate x so that the equation holds for all x-es? In that case, a must be ($$\hbar$$^2/km)^0.25

8. Nov 6, 2008

### borgwal

Your equation is now of the form (if you move E to the left-hand side)

C+Dx^2=0,

where C and D are constants (independent of x).
If this is to be true for all x, then both C and D must be zero.

9. Nov 6, 2008

### kasse

I got it right now, didn't I?

My next problem is to calculate the expectation value. That's not hard, I just wonder why the integral is written

$$\int$$dx $$\psi$$*(x)x$$\psi$$(x)

and not

$$\int$$ $$\psi$$*(x)x$$\psi$$(x) dx

Last edited: Nov 6, 2008
10. Nov 6, 2008

### f95toli

In this case it doesn't matter where you put the "dx", it is the same thing.

(note that there are cases where it DOES matter; e.g. if dx is a vector)