Harmonic Oscillator wave function

[diegoarmando]

Uncertainty - Harmonic Oscillator

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

Compute \Delta x \Delta pknown:
Heisenberg Uncertainty Principle:
\Delta p \Delta x &gt;= \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 &lt;x^2&gt; =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.