Negative energy of free particle: classical and quantum picture 
*

*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?

*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? 
 A: *

*Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the transformations connected to the identity (the proper orthochronous ones) to be physically relevant transformations in the sense that they go into another frame (since non-orthochronous ones reverse time). Therefore, special relativity indeed would allow you to fix the sign of the energy as positive: If you can't have negative energy solution in some frame, you are allowed to say there aren't any in any frame.

*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".
A: 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.
A: we can see negative energy solution as anti particle travelling in the -p direction in momentum space.creation operator have coefficents e^-ipx so it will create anti particle in the -p direction. here p is a four vector.four vector can be negative or positive.so we have solution in the positive p direction and solution in the - ve p direction.
