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  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?

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3 Answers 3

  1. 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.

  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".

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@ ACuriousMind: I meant why not discard the negative energy solutions in the quantum KG equation so as to remedy the problems associated with it. – SRS May 22 at 16:10
@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 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 at 3:57
@SRS: You are also absolutely right ;) Excluding the "negative energy" solutions means you do not have a complete set of solutions anymore, i.e. you would discard actual solutions of the equation of motion just because you don't like them. – ACuriousMind May 24 at 13:02

Maybe a chemists perspective could be useful to understand why we take both energies in quantum mechanics? It's not an elegant answer but is easy to rationalise!

In computational chemistry we use the variational principle to compute the orbital mixing coefficients of molecular orbitals from an atomic orbital basis set. In doing so along the way we end up with a set of secular equations which to have a non-trivial solution must have a non-zero secular determinant. We solve this determinant in terms of coulomb, resonance and overlap integrals to give the energy as a squared term.

Taking the root gives us two solutions for the energy of a molecular orbital, the bonding and anti bonding orbitals.

Both are superposition solutions to the Hartree-Fock equations and both are physically accessible molecular orbitals!

Hence within the scope of quantum chemistry taking both positive and negative roots for the energies as solutions is the correct procedure.

You should note that with respect to a free particle both the energies are more negative, this is for reference to gauge how much more stable an atom is in a particular molecular environment with respect to the free atom say.

Since energy is a quantity I wouldn't say I have heard of negative energy (except at perhaps a yoga class) rather I have heard of relative differences in energy. The energy of a molecule can never be negative in terms of Absolute zero, but can be with respect to something less stable.

Hope that helps :)

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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 we have solution in the positive p direction and solution in the - ve p direction.

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