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I am looking for a qualitative explanation, if possible, of the following passage from Peskin & Schroeder, bottom of page 26, concerning the quantised real Klein-Gordon field:

A positive frequency solution of the field equation has as its coefficient the operator that destroys a particle in that single particle wavefunction. A negative frequency solution of the field equation, being the Hermitian conjugate of a positive frequency solution, has as its coefficient the operator that creates a particle in that positive energy single particle wavefunction. In this way, the fact that relativistic wave equations have both positive and negative frequency solutions is reconciled with the requirement that a sensible quantum theory contains only positive excitation energies.

I am having a hard time deciphering the last sentence here. A negative frequency solution carries with it the creation operator: does this create a field excitation/particle that "cancels out" the negative frequency particle? I am quite lost as to what exactly is going on, physically, here. I'd like a picture in my head if possible, rather than equations!

Qmechanic
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shadowbiscuit
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1 Answers1

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There is an isomorphism $ i \to -i $ Finally, it is related to take the most simple wave like $ e^{ikx-\omega t}$. You have a wave with positive momentum visually travel to the right. But there is a tolally mathematical isomorphism. It have more consequences, of course, the $-i$ in the momentum operator, the $+i$ in the differential operator in time in Schrödinger equation, but it has no importance, but when we visualize and make simulations, of course, we are the habit, for example, related with this, that a particle with $x$ positive component in $x$ velocity is moving to the right. In a complete plane wave it have no sense in special relativity because space is infinite, but if you have a packet of waves, the description is more "natural". That is not a profound physical question really, but it's right you have the curiosity. It is sure there is no problem in any physical equation, because the automorphism is purelly mathematical

Ting-Kai Hsu
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Raul
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  • So you're essentially saying that it doesn't make a difference to the physics which operator is associated with which positive/negative solution, it's purely convention? – shadowbiscuit May 31 '20 at 13:43