# Calculate the RMSD for the odd solutions

1. Aug 2, 2007

### ehrenfest

Calculate the RMSD for the odd solutions of the infinite symmetric well problem.

The odd solutions have the form:

$$u_n(x) = 1/\sqrt{a} \cdot sin(n \pi x/a)$$

So, I should multiply u_n by x, then take the integral over the bottom of the well to get <x>. Then I should multiply u_n by x^2, take the integral over the bottom of the well to get <x^2>. From this I can get delta x.

Is all that correct?

2. Aug 2, 2007

### Gokul43201

Staff Emeritus
How is the expectation value of an observable defined?

3. Aug 2, 2007

### ehrenfest

I should take the integral over x times the modulus squared of u_n to get <x>. That better.

4. Aug 3, 2007

### Gokul43201

Staff Emeritus
Better, but it doesn't look like your wavefunctions have been normalized.

5. Aug 3, 2007

### ehrenfest

They have been. Remember this it represents a probability amplitude not a probability density and your region of integration should be -a to a.

6. Aug 3, 2007

### Gokul43201

Staff Emeritus
In that case, the wavefunction looks wrong to me.

7. Aug 4, 2007

### ehrenfest

Its my book that does it oddly. Instead of breaking it up into cases of n = 0,2,4,.. and n =1,3,5,... they used different arguments for sine and cosine so that they can use n = 1,2,3,... for both cases.

8. Aug 4, 2007

### Gokul43201

Staff Emeritus
I don't think that's particularly unusual.

If your well width is 2a I'd imagine your wavefunctions to be $\sqrt{1/a}~cos(n \pi x/2a)$, and if the well goes from 0 to a then you'd have $\phi_n(x) = \sqrt{2/a}~sin(n \pi x/a)$. The wavefunction you've written is neither of these.

Last edited: Aug 4, 2007
9. Aug 4, 2007

### ehrenfest

It goes from -a to a. The even solutions are of the form B*cos(k*x) , where k = sqrt(2mE/hbar^2). The boundary conditions give us that cosk(k*a) = 0, so set k_n*a = (n - 1/2)pi to get k_n = (n-1/2)pi/a. Plug that back into cosine and normalize.