Complete set and Klein-Gordon equation In http://www.physics.ucdavis.edu/~cheng/teaching/230A-s07/rqm2_rev.pdf, it says that when there is some external potential, the Klein-Gordon equation is altered, and it says the following:

The solution $\Psi$ can always be expressed as a superposition of free particle solutions,
  provided that the latter form a complete set. They from a complete set only if
  the negative energy components are retained, so they cannot be simply discarded. 

Can anyone explain what this is talking about? First of all, what is a complete set? And why does forming a complete set require negative energy components being retained? Why is the equation (solutions, I guess) need to be a complete set?
 A: This statement is false. The positive energy solutions of the free Klein Gordon equation form a complete set. What this means is that any wavefunction of the form $\psi(x)$ where x ranges over space not time, can be written as a sum of free-particle solutions.
In this case, the way you do it is by Fourier transform: you write $\psi(x)$ as
$$ \psi(x) = \int e^{ik\cdot x} \psi(k) dk $$
and this expands an arbitrary function, and then you make it time dependent by replacing $e^{ik\cdot x}$ with $e^{ik\cdot x - i E_k t}$ where $E_k = \sqrt{k^2 + m^2}$. The result gives the positive energy time evolution of any initial wavefunction data.
This is no good not because it isn't complete, but because it isn't local-- the single particle positive energy solutions go places faster than light. This requires making a multiparticle theory to fix the formalism.
The completeness is not completeness in the sense of expressing every solution of the Klein Gordon equation in terms of initial data, because the initial data for KG is the function $\psi$ and it's first time derivative. This is not the data you specify for a quantum mechanical wavefunction, it's the data for a classical field.
The book is confused on this issue, and you should skip this part.
