Harmonic Oscillator wave function

[diegoarmando]

Uncertainty - Harmonic Oscillator

The Wave function for the ground state of a quantum harmonic oscillator is
$$\psi=(\alpha/\pi)^{1/4}e^{-\alpha x^2/2}$$ where $$\alpha = \sqrt{ mk/ \hbar^2}$$.

Compute $$\Delta x \Delta p$$known:
Heisenberg Uncertainty Principle:
$$\Delta p \Delta x >= \hbar/2$$

In order to compute $$\Delta x \Delta p$$, what do I need to do? any integral?

 

[mgb_phys]

Well done - but I think that is already known to science.
Or did you have a further question about it?

 

[diegoarmando]

My question is how to find $$\Delta x \Delta p$$

 

[Dick]

You'll want to find the expectation values of x^2 and p^2 (since the expectation values of x and p are zero), and take their square roots. These are your deltas. So yes, you have to integrate to find <psi|x^2|psi> and <psi|p^2|psi>.

 

[genneth]

Just to be pedantic, $$\Delta x^2 = \langle x^2 \rangle - \langle x \rangle^2$$. Just that in this case, $$\langle x \rangle = 0$$.

 

[diegoarmando]

Thanks guys,
but how the <x> and <p> are zero, could you please help me for integral part, what is the limits of integral in this case?

 

[Dick]

The limits on the integral are +/- infinity. And both integrals have the general form of an integral of x*exp(-K*x^2). So the integrand is an odd function. It's integral is zero.

 

[diegoarmando]

ok, I find $$<x^2> =1/2\alpha$$ what should I do for <p^2>
p=-i*hbar ?

 

[Dick]

Uh, p=-i*hbar*d/dx. You need to apply that operator twice to psi since you are finding <psi|p^2|psi>. Your <x^2> looks good.

 

[diegoarmando]

I found $$\Delta x \Delta p=\hbar/\sqrt {2}$$
which means $$\Delta x \Delta p$$ is independent from the value of alpha, what do you think? somehow I think I should have gotten $$\Delta x \Delta p=\hbar/2$$

 

[Dick]

You did really well, except yes, you should have gotten hbar/2. I did. Can you find the missing sqrt(2)? I can check intermediate results if you want to post them.

 

[diegoarmando]

Thanks for the reply, I think I found the missing sqrt(2)

 

[Dick]

Thanks for the reply, I think I found the missing sqrt(2)

Well done.