Zero Point Energy of Coupled Dipoles

[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)

 

[dextercioby]

What's the Hamiltonian...?

Daniel.

 

[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

 

[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.

 

[secret2]

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.

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?

 

[dextercioby]

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

Daniel.

 

[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?

 

[dextercioby]

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

Daniel.

 

[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.

 

[Creator]

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)

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

Creator

 

[secret2]

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

Creator

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

 

[Creator]

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

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

 

[Creator]

Please note I've made an addition to the previous post to Secret2 for clarity.

 

[secret2]

Thank you!

 

[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 :)

 

[what_are_electrons]

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)

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