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From the measurement postulate, we are told that every observable quantity is represented by a Hermitian operator $\textbf{A}$, where the eigenvalues of said operator gives the possible measurement outcomes. In many body physics it often the goal to find the ground state of the system. For exmaple a magnetic lattice of localized spins with isotropic ferromagnetic exchange $J$ between nearest neighbour sites, \begin{equation} \hat{\mathrm{H}} = \sum_{i,i+1}^{N}J\hat{S}_{i}\cdot\hat{S}_{i+1}. \end{equation} The ground state is \begin{equation} \bigotimes_{i}\uparrow_{i}, \end{equation} which is an eigenvalue of the above Hamiltonian. My question is how can a system be said to be in it's low energy ground state eigenvalue, when such is a probabilistic outcome of a measurement of the observable, in this case the Hamiltonian. For example why can the system even at $T=0$ be in any of its higher energy states, as there is a finite probability of a measurement of $\hat{H}$ returning such an eigenvalue? Doesn't it go against the measurement postulate to assume at low energies the system tends to be in its ground state?

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There are a few misunderstandings here. It is best to look at this in terms of all the postulates of quantum mechanics to see exactly where you are misunderstanding this. The postulates are generally stated as some slight variant of the following set of statements:

  1. The state of a quantum system is represented by a normalized vector $|\psi\rangle$ belonging to some Hilbert space $\mathcal{H}$
  2. To each classical observable $A$, there corresponds a Hermitian operator $\hat{A}$ acting on $\mathcal{H}$
  3. A measurement of an observable A always yields an eigenvalue of $\hat{A}$
  4. If the system is in a state $|\psi\rangle$ the probability that a measurement of $A$ will yield a particular non-degenerate eigenvalue $a_{n}$ is $$P(a_{n}) = |\langle a_{n}|\psi\rangle|^2$$

So in light of this, the state of a system can indeed by any normalized vector lying in its Hilbert space. This vector does not necessarily have to be one of the eigenstates of its Hamiltonian. It could be some linear combination of the eigenstates of the Hamiltonian. It could also just be one of the eigenstates! That is a valid state of the system too.

You pose the question: "how can a system be said to be in it's low energy ground state eigenvalue, when such is a probabilistic outcome of a measurement of the observable, in this case the Hamiltonian." The ground state is just one of the vectors in the system's Hilbert Space. Of course it can be in that state! If the system is in it's ground state, all this means is that any measurement of the energy of the system (the Hamiltonian), will give precisely the ground state energy. A system could on the other hand be in a linear combination of the ground state and some of it's excited states. For example, suppose the system is in the state $|\psi\rangle = c_{1}\phi_{GS} + c_{2}\phi_{\text{excited}}$ where $\phi_{GS}$ is the ground state and $\phi_{\text{excited}}$ is some excited state. Then a measurement of the energy of the system will return the ground state energy with probability $|c_{1}|^2$ or the excited state energy with probability $|c_{2}|^2$. Post measurement, the system will have collapsed to the eigenstate associated with that eigenvalue.

As a further point here, at $T=0$ a system would most definitely be in its ground state.

A lot of these points are rather subtle and it is very important to get the postulates correct. I suggest for you to sit down and think through them systematically and attempt to internalize them.

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