Momentum-Space Wave Function

1. Oct 23, 2007

Rahmuss

1. The problem statement, all variables and given/known data
Find the momentum-space wave function, $$\Phi (p,t)$$, for a particle in the ground state of the harmonic oscillator. What is the probability (to 2 significant digits) that a measurement of $$p$$ on a particle in this state would yield a value outside the classical range (for the same energy)? Hint: Look in a math table under "Normal Distribution" or "Error Function" for the numerical part -- or use Mathematica.
(Problem 3.11 - Intro to QM, 2nd Edition, by Griffiths)

2. Relevant equations
$$\Phi (p,t) = \frac{1}{\sqrt{2\pi \hbar}} \int^{\infty}_{-\infty} exp[{\frac{-ipx}{\hbar}}] \Psi (x,t) dx$$

3. The attempt at a solution
The above equation is the only one that I know should be right. That being said, here is what I've tried so far:

$$\psi_{o} (x,t) = (\frac{m\omega}{\pi \hbar})^{\frac{1}{4}} exp[{\frac{-m\omega x^{2}}{2\hbar}}] exp[{\frac{-iE_{o}t}{\hbar}}]$$

With $$E_{o} = \frac {\hbar \omega}{2}$$

$$\psi_{o} (x,t) = (\frac{m\omega}{\pi \hbar})^{\frac{1}{4}} exp[{\frac{-m\omega x^{2}-i\hbar \omega t}{2\hbar}}]$$

$$\Phi (p,t) = \frac{1}{\sqrt{2\pi \hbar}} \int^{\infty}_{-\infty}exp[{\frac{-2ipx -m\omega x^{2} -i\hbar \omega t}{2\hbar}]$$.

And if that's right, then I'm not sure how to integrate that. Then after I get what that equals (lets call it ANS), then to find the probability outside the classical value, I need to square $$\Phi (p,t)$$ right? I'm guessing that I would have something like:

$$2\int^{\infty}_{ClassicalValue}ANS dp$$

I'm taking 2 times the integral because I'm guessing since it's an even function that I could take the negative classical value to -infinity as well; but it would equal this. But I'm not sure what to use for the classical value because of how the problem is worded. Am I on the right track?

2. Oct 23, 2007

Gokul43201

Staff Emeritus
Yes, on the right track.

To evaluate the integral, first take any constants out (note: you've omitted the integration variable), then use any tricks you may know for evaluating Gaussian integrals.

To find the classical range of values of momentum, assume the total energy of the classical oscillator is some E. What are the maximum and mimimum values of the KE? That leads to the classically allowed values of p in terms of E. Now make E=ground state energy of quantum HO.

3. Oct 23, 2007

Rahmuss

Gokul43201 - I'm not sure I know what you mean by the "integration variable". You mean, what we're integrating with respect to? I have that in there as $$dp$$.

So :

$$E = \frac{1}{2}kx^{2}$$
$$KE = \frac{1}{2}kx^{2}sin^{2}(\omega_{o}t + \phi)$$
$$PE = \frac{1}{2}kx^{2}cos^{2}(\omega_{o}t + \phi)$$

Would suggest at $$x = 0$$ would have a minimum and where $$cos^{2}(\omega_{o}t + \phi) = 0$$ would be another minimum.

For a max we have $$x = \infty$$ with $$cos^{2}(\omega_{o}t + \phi) = 1$$.

$$p = m\frac{dx}{dt}$$. So I guess I'm not quite sure what you mean. I'm probably just being an idiot here and it should be very simple.

4. Oct 23, 2007

Gokul43201

Staff Emeritus
I was talking about the previous integral - the one for $\Phi(p,t)$.

We are looking for the bounds of the KE. Since $0 \leq sin^2 \theta \leq 1$, the KE is bounded by 0 and E. Now we plug in the E for the ground state of the quantum harmonic oscillator, which is ...?

Since KE = p^2/2m, the bounds on KE tell us the bounds on p for the classical oscillator.

5. Oct 23, 2007

Rahmuss

$$E_{n} = (n + \frac{1}{2})\hbar \omega$$. And so the ground state for the QHO is:

$$E_{o} = \frac{\hbar \omega}{2}$$.

So for our integral we're looking for :

$$2*\int^{\infty}_{\frac{\hbar \omega}{2}} (ANS)^{2} dp$$. Is that right?

6. Oct 23, 2007

Gokul43201

Staff Emeritus
Almost!

$$E_{o} = \frac{\hbar \omega}{2} = \frac{p_{_{Cl}}^2}{2m}$$

$$\implies p_{_{Cl}} = \sqrt{m\hbar \omega}$$

7. Oct 23, 2007

Rahmuss

Oh, so:

$$2*\int^{\infty}_{\frac{p^{2}}{2m}}(ANS)^{2} dp$$.

Thank you very much for the help on this.

P.S. What does the subscript "Cl" represent on $$p_{Cl}^{2}$$?

8. Oct 23, 2007

Gokul43201

Staff Emeritus
No, you are integrating over momenta, not energies. So, the lower limit of the integral needs to be p_{Cl}.

That's the classical upper bound on the momentum.

9. Oct 23, 2007

Rahmuss

Oops. I already turned my homework in. Oh well. Live and learn. Thanks again for the help.