Start with a simple scalar field Lagrangian $\mathcal{L}(\phi)$ at zero temperature $T = 0$, which has a hidden symmetry and spontaneously break it. By the standard procedure a field $\phi$ is redefined
$$\phi \rightarrow \langle \phi \rangle + \phi',$$
where $\phi'$ is a quantum fluctuation around some constant value $\langle\phi\rangle$. The constant value $\langle \phi \rangle$ is called a condensate (or vacuum expectation value) of the field $\phi$. (For example, in the case of pions and sigma mesons ($\mathcal{L}$ is a linear sigma model Lagrangian) fluctuations $\phi'$ are physical pions and sigma mesons, with pion condensate equal to zero, and sigma meson condensate equal to $\langle \sigma \rangle = f_\pi$.)
The spontaneus symmetry breaking looks the same for $T \neq 0$ scalar field theory. Again, we redefine the field $\phi \rightarrow \langle \phi \rangle + \phi'$ and obtain physical particles $\phi'$ as a fluctuations around the condensate, which is now temperature dependent variable; and it can serve as an order parameter of the theory. (For example, in the case of sigma mesons and pions, the condensate $\langle \sigma \rangle $ will vanish at the chiral temperature point, displaying the existance of the chiral phase transition.)
So my question is, are the quantum fluctuations $\phi'$ (i.e. the physical particles) the same in $T = 0$ and $T\neq0$ field theory? Or are they somehow 'mixed', so they are both thermal and quantum fluctuations? In addition, the diagram here http://upload.wikimedia.org/wikipedia/commons/0/06/QuantumPhaseTransition.png basically says that quantum and classical (critical) behaviour is the same thing, which adds up to my confusion.
Of course, if I completely missed the point, I hope that someone can explain in a better way what is the concept of the symmetry breaking and emergence of a condensate (and physical particles).
Best Answer
A simple (and quite accurate) answer is that quantum fluctuations are the fluctuations that exist at zero temperature. What it means is that even at zero temperature, there might be fluctuations in the measurements of observables, which does not happen for classical systems at zero temperature, due to the non-commutativity of the dynamical and potential parts of the Hamiltonian. (Although frustrated classical systems and quantum systems with Hamiltonians described by only commutating operators slightly complicate this picture.)
Of course, at finite temperature, there will be both types of fluctuations, and how to take them apart (that is, know which part of the fluctuations are quantum or thermal) is still an active subject of research.
In a thermal QFT, one is dealing with fields $\phi(\tau,x)$ with $\tau$ the imaginary time that serves to encode the quantumness of the system (when constructing the path integral). In particular, one easily sees that if $\phi(\tau,x)$ is time-independent (but still depends on $x$), the field theory looks like a classical statistical field theory, and that's why people sometime says that this time-independent field (or the zero Matsubara frequency field) is the classical field. This, however, is not telling us which fluctuation is thermal or quantum. I don't think that is easy to tell (see above).
Concerning the phase diagram: at zero temperature, the transition is obviously driven by quantum fluctuations. A way to see that is that the frequencies are continuous (instead of discrete at finite temperature), so that there are always a large (in fact infinite) number of quantum modes that participate to the low energy physics, even when the low energy scale (say the gap) vanishes as one gets closer and closer to the transition.
For a finite temperature phase transition, there is always a value of the gap, smaller than the temperature, and therefore the quantum modes (with non-zero Matsubara frequencies) are gapped out by the temperature, and they cannot participate to the critical physics of the transition. They can be integrated out, and one is left with a classical (thermal) field theory with renormalized parameters.