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 SchrÃ¶dinger 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)