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If we have a harmonic oscillator and look at it on small scale the energy is quantized and we can calculate the different eigenstates. In general the energy eigenvalues are given by $$E_n = \left(\frac{1}{2}+n\right) \hbar \omega$$

Even if we can bring this system at $T=0$ into it's ground state, there will be zero point motion or quantum fluctuations remaining. Now if we heat a system, depending on its excitations we need Maxwell–Boltzmann, Bose-Einstein or Fermi statistics to calculate the occupation of each state. The resulting spectra are due to thermal excitations from the ground state.

Now if we remain at about absolute zero the system still has quantum fluctuations. From Heisenberg's principle the energy is uncertain but that does not tell me which is the current energy or eigenstate. How can one calculate the 'quantum' spectrum of an harmonic oscillator, or bluntly, how can one calculate the probabilities for observing the system at the state $n=1,2,...$ at $T=0$ if we prepare the system to be in $n=0$?

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The energy in the ground state is certain, it does not fluctuate. It is the particle momentum and coordinate who fluctuate, not its energy. When your particle is in $n=0$ state, it cannot be found in any other state (the corresponding probabilities are zero). Mathematically it is known as orthogonality of the Hamiltonian eigenstates: $$\langle n|m\rangle=\delta_{mn}.$$

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That is what I thought so far, now reading (not understanding) quantum field theory one can have a fluctuating field, which corresponds to a fluctuating number of 'particles'. Now why can the quantum fluctuations appear but the harmonic oscillator has zero probability for $n=1$? –  Alexander Nov 26 '12 at 20:26
Zero-particle and one-particle states in QFT are not fluctuating - they like the ground state. However two- or more-particle states can be "unstable" - the particle interaction does not conserve the number of particles so their number can become uncertain with time. It is like the oscillator energy - if it is uncertain (only average energy $\langle E\rangle$ is known), then it can be distributed over different levels differently, in many ways. In other words, at a given $\langle E\rangle$ the occupation numbers are uncertain too. –  Vladimir Kalitvianski Nov 26 '12 at 20:39
Vladimir, first you need to calculate the variance of the operator in whatever basis you are working in to get a measure of fluctuation or noise. The RMS value is given as $\sqrt{\langle\hat{O}^2\rangle-\langle\hat{O}\rangle^2}$. –  Antillar Maximus Nov 27 '12 at 1:38
@AntillarMaximus: You are right. But if the system state is an eigenstate of $\hat O$, then its RMS is zero - there are no fluctuations of $\hat O$ in this state. –  Vladimir Kalitvianski Nov 27 '12 at 8:12

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