# Zero point energy

1. Feb 10, 2005

### secret2

Imagine that a system with two coupled dipole is diagonalized, so that the symmetric and anti-symmetric states are chosen to be the bases. Why does the Zero point energy equal
$$\frac{1}{2}\hbar (\omega_s + \omega_a)$$

2. Feb 10, 2005

### dextercioby

What's the Hamiltonian...?

Daniel.

3. Feb 10, 2005

### secret2

$$H = H_0 - \frac {e^2 x_1 x_2}{2 \pi \epsilon_0 R^3}$$

where Ho is the sum of of two unperturbed SHO, x1 and x2 are respectively the distance between positive and negative ends of the dipoles, and R is the inter-dipole distance

4. Feb 10, 2005

### dextercioby

Compute the matrix element of the perturbation wrt fundamental states and show it is zero.Therefore,the vacuum (zero point energy) doesn't have 1-st order corrections.So the total energy is the one stated by you in the first post.

Daniel.

5. Feb 10, 2005

### secret2

I can see why the vacuum (zero point energy) doesn't have 1-st order corrections, but shouldn't half h-bar (omega)_s alone has lower energy than the one in the first post?

6. Feb 10, 2005

### dextercioby

Yes,of course,but why did u reject the contribution due to antisymmetric wave functions...?

Daniel.

7. Feb 10, 2005

### secret2

But isn't zero point energy defined as the lowest possible energy?
I am a little confused. Or maybe there is some good reference regarding this specific example?

8. Feb 10, 2005

### dextercioby

I wouldn't know,it's the first time i saw it.Where did u get it...?

Daniel.

9. Feb 10, 2005

### secret2

It's from one of the problem sheet for the condensed matter course. I just wonder if the zero point energy is defined to be as mentioned.

10. Feb 11, 2005

### Creator

Zero point energy of a single dipole oscillator is defined (derived by Planck) to be
$$\frac{1}{2}\hbar (\omega)$$

Creator

11. Feb 11, 2005

### secret2

Of course, of course.
And my question is why NOT exclude the anti-symmetric mode, which has higher energy than the symmetric mode alone?

12. Feb 11, 2005

### Creator

I think you had it right the first time.....
For a coupled harmonic oscillator the zero point energy is defined to be the sum of the lowest energy of each oscillator.

However, note:

$$\frac{1}{2}\hbar (\omega_s) \neq \frac{1}{2}\hbar(\omega_o)$$

$$\frac{1}{2}\hbar (\omega_a) \neq \frac{1}{2}\hbar(\omega_o)$$

...where $$\omega_o$$ is the frequency of the uncoupled oscillator. IOW, the lowest energy of each coupled oscillator is different from the lowest energy of each when uncoupled.

More specifically, $$\frac{1}{2}\hbar(\omega_s + \omega_a) \ll \frac{1}{2}\hbar(\omega_o) + \frac{1}{2}\hbar(\omega_o)$$

IOW, for the coupled oscillators the total zero point energy is lower than the sum of two uncoupled oscillators.

Creator

Last edited: Feb 11, 2005
13. Feb 11, 2005

### Creator

14. Feb 12, 2005

### secret2

Thank you!

15. Feb 12, 2005

### secret2

Now I think I understand. When diagonalizing the system, we turned the two INTERACTING SHO's into two non-physical NON-INTERACTING symmetric and anti-symmetric modes, and they just act like any other SHO's. The zero point energy of the system is the sum of the zero point energy of the two non-physical SHO's, and quantum mechanics dictates that none of the individual (ie $\frac{1}{2} \hbar \omega_s$ or $\frac{1}{2} \hbar \omega_a$） could be zero, hence the answer :)

Last edited: Feb 12, 2005
16. Feb 12, 2005

### what_are_electrons

A question: What was the proposed cause of the dipole in this question? Thanks.