Zero Point Energy and Frequency

Most recent answer: 01/21/2014

Q:
I am confused about zero point energy. My question is that "how the zero point energy could be equal to 1/2hv (half of photon's energy), while according to quantum mechanics the energy always integral multiple of hv??" Please anyone give the solution...... Ayesha Riaz
- Ayesha Riaz (age 24)
Faisalabad
A:

That question gives a chance to clear up a point that can be confusing. The energy frequency-energy relation E=hv is universal, where v is the frequency that the quantum phase changes, rotating in the complex plane. In other words, that frequency v and E are really just the same thing measured in different units.

The frequency  that is associated with a particular oscillator, however, is not the same thing, although unfortunately it often goes by the same name. This frequency is the classical oscillation frequency of the oscillator. Here let's call it "f" instead of v. Then the possible quantum frequencies of the oscillator are f/2, 3f/2, 5f/2, 7f/2,.... . Notice that the differences between possible frequencies are f, 2f, 3f,.... So the possible energy differences between states are hf, 2hf, 3hf,.... Those then are the possible changes that can occur when the oscillator trades energy with something else. However the actual energies of the oscillator states are hf/2. 3hf/2,... The lowest possible one of those is called the zero-point energy. 

The zero-point energy isn't just some abstraction. When you want to figure out the energy of, for example, a hydrogen atom bonded to some silicon, you need to include the energy of the oscillator formed by the hydrogen mass and the springy bond. If you were to kep the same bond but double the mass (using deuterium) the oscillation frequency would go down and so would the zero-point energy. That means the deuterium would be bonded more tightly to the silicon. Exactly that principle was used by Joe Lyding and Karl Hess here at Illinois to invent a way to make more stable silicon devices, replacing hydrogen with deuterium.

Mike W.


(published on 01/21/2014)