# Energy of the harmonic oscillator

1. Jun 30, 2006

### Repetit

Hey!

Can someone explain to why the energy of the harmonic oscillator must be at least:

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

I mean, $$\Delta x$$ and $$\Delta p$$ represents the uncertainty in the position and momentum, and therefore it does not really have anything to do with the actual true value of the position and momentum does it? If you don't understand what I mean please let me know.

Thanks!

2. Jun 30, 2006

### maverick280857

The Hamiltonian is given by

$$H = \frac{1}{2m}\left[p^2 + (m\omega x)^2\right]$$

If I understand your problem correctly::

There is no "true" value that is measurable. The observables are the expectation values of $x$ and $p$ and the uncertainty principle gives you an inequality relating the standard deviations (in statistical language, what you call uncertainties).

Last edited: Jun 30, 2006
3. Jun 30, 2006

### Repetit

Well, what I want to do is derive the minimum energy of the harmonic oscillator. I start with:

$$<E>=\frac{<p^2>}{2m} + \frac{1}{2}m \omega^2 <x^2>$$

...and use the fact that:

$$(\Delta x)^2 = <x^2> - <x>^2$$

and

$$(\Delta p)^2 = <p^2> - <p>^2$$

To rewrite the formula for the energy expectation value into the one in my first post I assume that $$<p>$$ and $$<x>$$ are zero so that $$(\Delta x)^2=<x^2>$$ and $$(\Delta p)^2=<p^2>$$. Now, my question is how can I assume that the expectation value of position and momentum is zero? Is it because the particle in the harmonic oscillator potential spends equal amounts of time in the two outer positions? And for the momentum, is it because the velocity of the particle reverses sign in the two outermost positions just as in a classical oscillator?

Thanks!

4. Jun 30, 2006

### maverick280857

To convince yourself you could calculate $<x>$ and $<p>$ for the n-th harmonic ocillator state though I assure you that the integration won't be straightforward (you could use the hermitian adjoint operators...the ladder operators to ease a bit).

For n = 1, its just

$$\psi(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}exp\left(-\frac{m\omega^2}{2\hbar}\right)$$

Try calculating the expectation values with this (note that you don't need to tack the time dependence for expectation values).

When you get the result, you can think of it as being analogous to the classical harmonic oscillator (as you say) but I wouldn't like to stretch the analogy too far because I don't have the time dependence of $x$ here--in QM it makes no sense--to predict how long it remains in a particular part of the region $-A \leq x \leq A$. All my computations are based on how $\psi(x)$ behaves and not how $x(t)$ behaves (the second function has no meaning in QM).

Last edited: Jun 30, 2006
5. Jul 2, 2006

### Eye_in_the_Sky

Look again!

Look again, Repetit. The answer you seek is staring you in the face from behind no more than a wispy veil.

From the first three relations of your second post, it follows that

<E> = E + <p>2/2m + mω2<x>2/2 ,

where I have defined E as the expression you gave in your opening post.

Clearly, <E> ≥ E .

6. Jul 3, 2006

### pmb_phy

That is not quite right. When the energy of a system is measured the only possible measured results are eigenvalues. Each eigenvalue can be determined to within an arbitrarily small value. Therefore a "true" value is measureable and has a well defined meaning. Observables are not the expectation values of an operator. See

http://www.geocities.com/physics_world/qm/harmonic_oscillator.htm

for the details.

Pete

7. Jul 4, 2006

### maverick280857

pmb_phy: The site you gave doesn't work (maybe geocities is down right now or something).

Interesting you should say that...when I was reading Griffiths QM, I learnt that it is the expectation values that are measured and this idea was confirmed by Eisberg/Resnick. Could you please enlighten me about what you want to show through that website?

Last edited: Jul 4, 2006