# QM: Harmonic Oscillator wave function

1. Mar 20, 2013

### Stan12

1. The problem statement, all variables and given/known data
For the n = 1 harmonic oscillator wave function, find the probability p that, in an experiment which measures position, the particle will be found within a distance d = (mk)-1/4√ħ/2 of the origin. (Hint: Assume that the value of the integral α = ∫01/2 x2e-x2/2 dx is known and express your result as a function of α)

2. Relevant equations
distance from 0 to d; d = (mk)-1/4√ħ/2

Normalize condition: Cn = 1/ (π2√2nn!)

Harmonic Oscillator wave function for n = 1 ψ1 = C1(2s)e-s2/2

Probability density
∫ψn(x)*ψ(x)

s = (km)1/41/2 x

3. The attempt at a solution
I first plugged in s and normalized condition into the harmonic oscillator wave function.
ψ(s) = (km/π)1/42x/√2ħ e-√(km)x2/2ħ

∫ψ(s)*ψ(s) ?
I'm not sure if this is the right approach to tackling this problem.

2. Mar 20, 2013

### fzero

$\psi^*(x)\psi(x)$ is the probability density. This means that it is the probability to find the particle in the interval from $x$ to $x+dx$. Can you think of a way to use this to determine the probability to find the particle in a certain interval?

3. Mar 21, 2013

### Stan12

the intervals would be from ∫-dd. Is there a way to simplify down the
distance given?
d = (mk)-1/4√h-bar/2

s = [(km)1/4/√h-bar] x

d = x/s√2 ; -d = -x/s√2 this doesn't seem right.

the integral, α = ∫-dd x2e-x2/2

Last edited: Mar 21, 2013
4. Mar 21, 2013

### fzero

I think you're confusing yourself by having too many variables defined. The distance d is given in the units of the x variable, so set the probability up as an integral over x. Then you can make a change of variable to one defined like the s variable. Your limits of integration will become very simple and you should be able to express the result in terms of the $\alpha$ integral.

5. Mar 21, 2013

### Stan12

-ss ψ*(x)ψ(x) dx ?

P = C24s2e-s2/2 + eis2/2

x = [√h-bar/mk1/4] * 1/√2
s = [(km)1/4/√h-bar] x
s = 1/√2
so...
this doesn't seem right, am I suppose to get a function out of probability density?

Last edited: Mar 21, 2013
6. Mar 21, 2013

### EEnerd

I am looking for the answer for this one too, I had it on my exam few days ago

7. Mar 21, 2013

### fzero

Why does your first line have s as the limit of integration? Where did P in the 2nd line come from?

As you deduced in an earlier post, the limits of integration are given by d. Then we can use the probability density to determine the desired probability:

$$P = \int_{-d}^d \psi^*(x) \psi(x) dx.$$

You should rewrite this expression using the normalized $\psi(x)$. It should be possible to find a change of variable such that

$$P = A \int_{-1/2}^{1/2} y^2 e^{-y^2} dy,$$

where $A$ is a constant that we can express in terms of $m,k,\hbar$.

8. Mar 22, 2013

### Stan12

-So we have ψ1 = C1 2se-s2/2
as our wave function of this specific harmonic oscillator n = 1

- Probability density is given as ψ*ψ and within the intervals of -d to d
giving us ∫d-d ψ*(x)ψ(x)dx

ψ*ψ = (C1(2s)es2/2)(C1(2s)e-s2/2) = C2(2s)2

I'm still not getting this problem..and how the probability P is rewritten in the form of
A∫1/2-1/2y2e-y2dy

Last edited: Mar 22, 2013
9. Mar 22, 2013

### fzero

This wavefunction is real. In particular

$$\left( e^{-s^2/2} \right)^* = e^{-s^2/2}$$

so the exponentials do not cancel out.

Go back and try this again with the correct value of $\psi^* \psi$. I meant what I said when I suggested to write out the expression in terms of $x$. It seems pointless for me to keep offering suggestions when you won't bother following any of them.

10. Mar 22, 2013

### Stan12

ψ(x) = C2xe-x2/2
ψ*ψ = |C2xe-x2/2|2

∫|C2xe-x2/2|2 dx
= C2∫4x2ex4/4 is this correct, all terms in x

so only when an exponential expression is cancelled out using pd is when it function is imaginary :ex: e-ix

Now by doing what I did above is the pdf corrected? I squared the function and set it under the integral of x.

Last edited: Mar 22, 2013
11. Mar 22, 2013

### vela

Staff Emeritus
No, it's not correct. You have a definition for s in terms of x. You need to use that. You don't simply replace s by x. You're making it way too complicated. This is just basic algebra.

12. Mar 22, 2013

### Stan12

I will try this from scratch

13. Mar 22, 2013

### fzero

Hopefully it will help to note the following. When you wrote

back in the OP, you were already confusing the s and x variables. The 1st excited state wavefunction is given in terms of x by (I am assuming that your normalization and the rest of your formula is correct, please check with your text to make sure.)

$$\psi(x) = \left(\frac{km}{\pi}\right)^{1/4} \frac{2x}{\sqrt{2}\hbar} \exp\left[ - \frac{\sqrt{km} x^2 }{2\hbar} \right].$$

This is the expression you want to start with. The change of variable to s is defined to make the argument of the exponential in the probability density come out as $-s^2$, but you can verify this yourself.

14. Mar 22, 2013

### Stan12

Now with this expression ψ(x) I can find the probability density (ψ*(x)ψ(x)) = √(km/π) 2x2/hbar2 exp[-(km)x4/4hbar2

Last edited: Mar 22, 2013
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