Negative energy of free particle: classical and quantum picture

1. Classically, the energy of a free particle consists of only the kinetic energy given by $E=\frac{|\textbf{p}|^2}{2m}$ Since $|\textbf{p}|$is real and $m>0$, $E\geq 0$. However, since $$E^2=|\textbf{p}|^2+m^2$$ in special relativity. Mathematically therefore, $$E=\pm\sqrt{ |\textbf{p}|^2+m^2}$$ But can we have both positive and negative energies for a free particle classically? I mean shouldn’t we reject the negative root as unphysical? I think energy of a free particle is always positive because when we measure the energy of a free relativistic particle we always get a positive number. Right?

2. The next question is if we can disregard the negative energy solution classically then why can’t we do the same quantum mechanically? Why is in relativistic quantum mechanics, for example, in Klein-Gordan equation, both positive and negative energies are retained?

Addition : While "deriving" KG equation we directly use $E^2=|\textbf{p}|^2+m^2$. We can not use $E=+\sqrt{|\textbf{p}|^2+m^2}$ because in that case we get an infinite order differential operator which makes the theory non-local. Therefore, in some sense we allow for both positive and negative root into the equation right from the beginning. Is this the only reason that we cannot discard negative energy solutions of the theory?

2. That we may not discard the "negative energy" solution is not a quantum phenomenon. Nothing about the Klein-Gordon equation is a quantum equation - it is the classical equation of motion of a relativistic scalar field. If we write down the appropriate action for a scalar field $$S[\phi] = \int_{\mathbb{R}^4}(\partial_\mu\phi\partial^\mu\phi + m^2\phi^2)\mathrm{d}^4x$$ you may derive the actual energy density associated to this field by looking at the $00$-component of the stress-energy tensor. The actual form is rather unimportant except for the fact that the field always occurs quadratically, and, in particular, the energy density has only $\phi^*\phi$ and $\partial_0\phi^*\partial_0\phi$ as summands in it (and only positive factors in front). Hence even a negative $k_0$ in a plane wave Klein-Gordon solution yields a positive contribution to the energy density, and we have no physical ground to discard this solution as having "negative energy".
• @SRS: Wait, what problems? QFT has no problems with these "negative energy solutions" at all, and since the quantum field is an operator-valued distribution, not something that has "energy", you should not interpret anything as "negative energy" here. If you are talking about the issues taking $\phi$ to just be a relativistic wavefunction has, well, then the answer is - stop doing relativistic QM, and do QFT already! – ACuriousMind May 22 '15 at 16:14
• @ ACuriousMind- You are absolutely right. But my question is why not relativistic quantum mechanics excluding negative energies. I mean is there a mathematical criterion which prevents us to exclude negative energies? I don't know but can it be that the set of functions $\sim\{e^{ik_\mu x^\mu}\}$ (stationary solutions of KG equation) do not form a complete set if we exclude negative energies? – SRS May 23 '15 at 3:57
Following the discussion with AcuriousMind, my current understanding is that the set of functions $\{e^{ik\cdot x}\}$ (stationary solutions of KG equation) do not form a complete set if we exclude negative energies by hand.