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The total energy of a mode in a quantum mechanical resonator is given by $E_n ~=~ (n+ 1/2)hf$ where $n$ is the number of modes. So when there are no modes or vibrations, i.e. $n=0$, the energy is called the zero-point energy.

What I don't understand is, if there are no modes, then what is this energy associated with? So what exactly is a mode? Also, is there a way of measuring this zero-point fluctuations? Why is the 1/2 photon introduced?

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The average energy is zero, but the variance need not be zero. –  Antillar Maximus Oct 31 '12 at 20:28
    
If you like this question you may also enjoy reading this post. –  Qmechanic Oct 31 '12 at 21:18
    
Great username. –  Emilio Pisanty Nov 1 '12 at 2:55
    
@EmilioPisanty thanks! :) –  Spaceman Spiff Nov 2 '12 at 6:52
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3 Answers

up vote 2 down vote accepted

The zero-point energy for the quantum mechanical harmonic oscillator can be related to the Heisenberg uncertainty principle (HUP). A bit oversimplified, the point is intuitively that if the mechanical energy

$$H~=~\frac{p^2}{2m}+\frac{1}{2}m\omega^2 q^2, \qquad \omega ~:=~ 2\pi f,$$

is zero, then the position $q$ and the momentum $p$ must also both be zero, in contradiction with the HUP. Instead both $q$ and $p$ "fluctuate/are fuzzy" in the ground state, thereby producing the zero-point energy.

The zero-point energy is e.g. important to theoretically explain the Casimir effect, which in turn has been experimentally observed.

In cosmology, zero-point energy is related to the cosmological constant.

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Qmechanic and @Foster Boondoggle- Thanks. It is starting to make sense to an extent. [calphysics.org/zpe.html] This helped as well. Thank you so much. –  Spaceman Spiff Oct 31 '12 at 21:02
    
@Spaceman Spiff: Be aware that the Calphysics.org site, you linked to, is controversial non-mainstream reading. Read e.g. Wikipedia instead. –  Qmechanic Nov 1 '12 at 2:27
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It's associated with the confinement of the particle. As the coefficient of the quadratic term in the harmonic potential goes to zero, the particle becomes less & less localized and the the zero point energy drops accordingly. It can be as small as you want.

When you speak of photons, now you're referring to QFT and the energy in a mode of the EM field. There's no 1/2 a photon. There's an arbitrary normalization of the zero point of the vacuum field energy but no one knows how this relates to the cosmological constant -- in QFT it's arbitrary, and there are only semiclassical arguments of questionable rigor for how it should couple to the curvature tensor in GM.

Incidentally, you can also derive the Casimir effect without any reference to zero modes of the field, as the analog of the van der Waals force between polarizable atoms/molecules. The wikipedia page has the details.

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Choice of the zero-point energy is similar to choice of zero of the potential - it is rather arbitrary. One can safely put it equal to zero. However one should not understand it as a "zero-motion". The motion in the ground state exists (HUP) and it determines the size of systems created by bound particles (like atomic size, etc).

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