# Reason for considering the positive root

In eqn. (3.11) of Srednicki's QFT book only the positive root is considered; i.e.,

$\omega = + \sqrt{(k^2 + m^2 )}$

Why the negative root is not considered? And what is the $\omega$?

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I'm pretty sure that you have to consider the negative root too, and reinterpret it as antimatter. That puzzled a lot of people at the time (Bohr, Heisenberg, ...). –  Learning is a mess Feb 24 '13 at 17:28

The negative root is also included, since your expression occurs within the plane wave solution to the Klein-Gordon equation, given by

$\varphi(x_i,t)\propto e^{ik_ix_i\pm i\omega t}.$

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I don't have the book right now but this is $\omega$ the particle energy, i.e the time component of the (on-shell) 4-momentum of some excited mode of the field you are considering. A negative root would correspond either to a particle moving backwards in time or to an antiparticle moving forwards in time.

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Instead of writing $\omega_1$ and $\omega_2=-\omega_1$, he defines one positive omega and writes both roots as $\pm\omega$.

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So, the plane wave solutions are: exp(ik**·x + iω_1t) and exp(ik**·x + iω_2t) –  Ome Feb 24 '13 at 17:55
@Ome: That's right! Exactly! –  Vladimir Kalitvianski Feb 24 '13 at 18:01
Thanks a lot! It has really helped. I have become completely confused at this point. –  Ome Feb 24 '13 at 18:08
@ Vladimir Kalitvianski: Are there any derivation of the general solution? Or it is just an ansatz? Can you suggest me any references? –  Ome Feb 24 '13 at 19:13
@Ome: Normally a linear differential equations has a superposition of independent solutions as its general solution. So you see a sum over $\vec{k}$ of different exponentials with different $\vec{k}\vec{x}-\omega(\vec{k})t$ and different coefficients like $a_{\vec{k}}$. It is know from the theory of differential equations. –  Vladimir Kalitvianski Feb 24 '13 at 19:24