Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $\omega$?

share|improve this question
    
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
add comment

3 Answers

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

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
    
So, the plane wave solutions are: exp(i**k**·x + iω_1t) and exp(i**k**·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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.