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If I have a Hamiltonian $H$ that only has kinetic energy and no potential energy, do the energy eigenvalues have to be non-negative? Could the ground state of Hamiltonian have negative eigenenergy? If so, is the negative value results from potential? What's the physical meaning of negative eigen-energy?

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  • $\begingroup$ What do you think negative energy means in classical physics, and why do you think quantum physics would be any different? $\endgroup$
    – ACuriousMind
    Sep 24, 2020 at 15:55
  • $\begingroup$ Which Hamiltonian are you considering? If it's a relativistic Hamiltonian, then negative energy states are an important issue. $\endgroup$ Sep 24, 2020 at 18:08
  • $\begingroup$ @JoshuaTS I'm considering the Hamiltonian in some scenarios like quantum oscillator. Why are the negative energy states important in relativistic Hamiltonian? Thanks! $\endgroup$
    – ZR-
    Sep 24, 2020 at 18:32

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In general, we have to always define some reference energy to which we compare other energies to.

For example in classical physics when talking about gravitational potential energy, one usually defines $E=0$ to be at the ground, and then when I rise up a height, $h$, my energy becomes $E=mgh$. If I wanted to then convert this energy to kinetic energy (for example by letting myself fall) the resultant kinetic energy when I reach the ground would be the (positive) difference between these two energies, and then $\frac{1}{2}mv^2=mgh\;\Rightarrow\;v=\sqrt{2gh}$ etc.

But there is no reason we have to set $E=0$ at the ground. I can define the zero of energy to be at height $h$, and then my potential energy when I'm on the ground would be $-mgh$. Still, if I wanted to work out what my kinetic energy would be when falling from $h$ to the ground, I would still get $mgh$, which is the difference between these energies. In this way, energy differences are more important than absolute energies.

Moving to quantum mechanics, exactly the same arguments apply. I can define my "zero of energy" to be whatever I like, often something that is convenient to me. For example if we were just considering a particle moving with no potential, I would define my $E=0$ to be when its kinetic energy is 0, for convenience.

If you're getting negative energy eigenvalues, it just means that the energy of that particular state is less than some reference energy we have defined somewhere in the problem. Again, this comparison of energies is important.

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