When looking back at my courses of quantum mechanics, I noticed that assumptions about the ground state of a quantum mechanical system where rather vague and unprecise. It is always assumed that a ground state exists and that it had a finite energy. So my questions are the following:
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Should every quantum mechanical system have a ground state? And how can we be sure of this?
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Should this ground-state have a finite energy?
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Or is an energy of $-\infty$ also allowed?
A neat proof for the ground state is given in "Introduction to Quantum Mechanics" by D. J. Griffiths for the simplest system (problem 2.2). There it is shown that if you have an energy-eigenstate $\psi(x)$ (working in the position space for simplicity) with energy $E$, so:
$$\hat{H}\psi(x)=E\psi(x),$$ where we consider a simple non-relativistic point-particle, so $$\hat{H}=\hat{T}+\hat{V}.$$By applying this on this on the equation for the eigenstate $\psi(x)$ this yields:$$\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x)=E\psi(x),$$ or (after some simple rewriting) as:$$\frac{\partial^2\psi(x)}{\partial x^2}=\frac{2m}{\hbar^2}\left[V(x)-E\right]\psi(x).$$In the case that we would have an energy which is lower than the minimum of $V(x)$ (assuming that the potential has a minimum), the wave-function would be non-normizable since $\psi(x)$ and $\partial_{xx}\psi(x)$ whould have the same sign. This is because of the fact that $\psi(x)$ can only have a minimum if it's positive and a maximum if it's negative, which yields the non-normalizable character.
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So this still leaves me with the question on how they know it for sure for potentials of the Coulomb-kind?
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and how they do this in quantum field theory (and, by extension, classical physics)?
I know this question is in the same line of reasoning as (Why does a system try to minimize potential energy?) kindoff. But what I'm looking for is not "Why does everything tend to a minimal energy". But more "Why do we assume such a minimal energy-state exists?".
Best Answer
The Hamiltonian of good physical systems is always assumed to be bounded from below, that is, there is a state with lowest energy, the ground-state. Because you can always shift all the energies (in non-relativistic QM), you could shift all energies by $-\infty$ in principle, though the spacing between the ground-state and the excited states would stay the same.
Not having a ground-state would imply some kind of instability in the system. For instance, non-interacting bosons in the grand-canonical ensemble at zero-temperature do not form a good system at positive chemical potential: you can put an infinite number of bosons in the state of zero momentum, which implies an infinite density and an infinite (negative) energy. This is not a sound state of matter.