Quantum harmonic oscillator

1. Sep 12, 2009

kof9595995

2. Sep 12, 2009

xepma

The kinetic energy is at a minimum when the momentum is precisely zero. On the other hand, the potential energy is at a minimum when the coordinate x is precisely zero. But they cannot be zero at the same time - Heisenberg's uncertainty principle forbids it. So instead we replace the quantities by their "fluctuation" around the values $$x=0$$ and $$p=0$$. The fluctuations are just $$\Delta p$$ and $$\Delta x$$. Heisenberg's uncertainty principle simply states that these fluctuations (or uncertainties) are related by:

$$\Delta p\Delta x \geq \hbar/2$$

Which again tells us that the fluctuations cannot both be zero at the same time. If we set the momentum precisely zero (so also no fluctuations) then the kinetic energy is zero as well. But then the fluctuations of the position blows up, and the potential energy does so too. So the situations where the momentum is precisely zero certainly does not correspond to the minimum energy.

If I'm not clear enough, just let me know :)

3. Sep 12, 2009

kof9595995

Well，thanks. But I'm still confused, because as far as I can understand, even around p=0 $$\Delta p\$$ is not p, uncertainty is the standard deviation of the measurements ,right? so when uncertainty is $$\Delta p\$$, the p can still much smaller than $$\Delta p\$$ , so why can't the actual energy be smaller than that?

4. Sep 12, 2009

kof9595995

I think something is wrong with my latex code, please just ignore the funny brackets after delta p

5. Sep 12, 2009

Snarky Fellow

Well, around p=0 one needs to distinguish $$\Delta p$$ and p, but in fact we are interested only in $$\langle p^2 \rangle$$ that is equal to $$\langle\Delta p^2 \rangle$$ in case the mean momentum is 0.

6. Sep 12, 2009

kof9595995

So the so called minimum energy is still under the statistical meaning, right?

7. Sep 12, 2009

kof9595995

Snarky Fellow, is this your first post? Welcome to the forum!

8. Sep 12, 2009

elduderino

$$E = \frac{{{(p+\Delta p)^2}}}{{2m}} + \frac{1}{2}m{\omega ^2}{(x+\Delta x)^2}$$

For minimum energy, you would classically put p=0 and x=0. Which is what you do above, only that there are uncertainties associated with p and x and the Heisenberg principle does not allow them to be simultaneously zero. For example, if $$\Delta p=0$$ then $$\Delta x$$ would blow up- which does not correspond to the minimum energy.

9. Sep 13, 2009

kof9595995

Actually in harmonic oscillator <p> should be always 0 because of the symmetry, so $$\langle p^2 \rangle$$=$$\Delta p$$, am I right?

10. Sep 13, 2009

kof9595995

This formula does not seem quite right to me; I think it should be just
$$E = \frac{{\left\langle {{p^2}} \right\rangle }}{{2m}} + \frac{1}{2}m{\omega ^2}\left\langle {{x^2}} \right\rangle$$
And what Snarky fellow said is why we can substitute $$\Delta p^2$$=$$\langle p^2 \rangle$$

11. Sep 13, 2009

Snarky Fellow

If we deal with an eigen-state of hamiltonian, which is static in some sense, than due to the symmetry the mean momentum will be equal to zero - you're right. Of course, we can take any superposition of eigen-states, where the symmetry could be broken by our choice, but it's out of interest in this problem.

To say honestly, the derivation of minimal energy using the uncertainty principle isn't very strict - it's just an assessment. It proves only that the minimal energy cannot be less than that we've found. The fact they're equal is a mere coincidence. As far as I know, it works only in the case of a harmonic potential. Usually states with a well-determined energy are not the states with the least uncertainty. Thus a good derivation involves ladder-operator technique or solving the Schroedinger's equation. So I suppose, the main idea of this derivation is to show that classical solution in the bottom of the potential condradicts the uncertainty principle.

12. Sep 13, 2009

kof9595995

I think I get the idea, thanks.