How one can deduce the existence of antiparticles

[tommy01]

Hi together ...

I wonder how one can deduce the existence of antiparticles from the Klein-Gordon equation.
Starting from (\frac{\partial^2}{\partial t^2} - \nabla^2 + m^2) \Psi(t,\vec{x})=0

one gets solutions \Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x})) leading to E^2=p^2 + m^2. You need the plus/minus in the exponential to get a complete system of solution functions but why you can't just ignore the negative root of the enegry-momentum relation and why you interpret the exp(- ...) as negative momentum and negative energy? why you can't just say the energy is always positive but the functional dependence is exp(+ ...) respectively exp(- ...)?

thanks and merry christmas.

 

[mathman]

Since anti-particles have been found experimentally, what is the point of your assertion?

 

[tommy01]

I wonder how one can deduce the existence of antiparticles from the Klein-Gordon equation.

that was my point ...

I know that they have been found experimentally. But it is always quoted as a great triumph of theoretical physics that antiparticles were predicted on grounds of the Klein-Gordon equation or the Dirac-Equation respectively.

 

[Dickfore]

The general solution of the KGE is:

<br /> \Psi(t, \vec{x}) = \int{\frac{d^{3}k}{(2\pi)^{3} 2 \omega} \left[a(\vec{k}) \exp\left(\iota (\vec{k} \cdot \vec{x} - \omega t)\right) + a^{\dagger}(-\vec{k}) \exp\left(\iota (\vec{k} \cdot \vec{x} + \omega t)\right)\right] }, \; \omega = +c \sqrt{k^{2} + \left(\frac{m c}{\hbar}\right)^{2}}<br />

If we interpret \Psi(t, \vec{x}) = \Psi^{\dagger}(t, \vec{x}) as a field operator creating an excitation of the real scalar field at the space-time point (t, \vec{x}), then a(\vec{k}) creates a particle with momentum \hbar \vec{k} and energy \hbar \omega, while a^{\dagger}(-\vec{k}) creates an antiparticle with the same energy and momentum.

 

[tommy01]

Thanks. Thats a linear superposition of the fundamental solutions i cited above, but as you wrote \omega and thefore the energy is positive and there is no need for negative energy states called antiparticles. Or did i misunderstand something?

edit:
Back when they dealt with the Klein-Gordon equation. They didn't quantise the field, so there is no interpretation in terms of creation and annihilation operators.

 

[Dickfore]

notice the sign in front of t in both of the exponentials.

 

[tommy01]

yes, but as i mentioned earlier, why does this sign carry over to the energy? why don't say the energy is positive but the functional dependence, to get linearly independent solutions, is exp(- ...) respectively exp(+ ...)?
thanks for your help and please excuse my problems understanding that point.

 

[Dickfore]

As far as I know, according to a fundamental paper by Feynman, antiparticles can be regarded as particles with positive energy, but moving backwards in time.

 

[tommy01]

thank you. i also heard this statement, but it's the same story i think. only this time the sign is carried over to the paramter t instead of the energy.

i'll try to make my point clear:
although the solutions are
<br /> \Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))<br />
you don't say the energy can take on imaginary values only because there is an i in the exponential. it's simply part of the solution function. so why is a minus sign treated differently?

 

[Dickfore]

I'm afraid I don't understand what you are trying to say.

 

[tommy01]

my point is:

you say that there are negative energy solutions because, as you showed, the general solution bears a plus AND a minus sign in the exponent. but it seems to me it would be the same to say the energy can be imaginary because there is an i in the exponent. i simply don't get the point why you have to interpret the minus sign as part of the energy expression and not just as a part of the functional dependece of the solution like the i.

greetings.

 

[Dickfore]

IF you have a time dependence like this:

<br /> \exp{(-i \, \omega \, t)}<br />

then \hbar \omega is the energy associated with such propagating perturbation.

 

[tommy01]

O.k. so everything in the exponent multiplied by t except the -i is the frequency \omega. As E=\hbar \omega the negative energy comes from the negative frequency. Mhh i think i probably got it although there are many cases i remember where the time dependence was \exp(i \omega t) and no one thought of a negative energy.
nonetheless thanks for your patience and help.

 

[Dickfore]

In Physics, the convention is to use \exp{(-i \omega t)} as time dependence. In Electrical Engineering, it usually is \exp(2 \pi j f t) (they use j as the imaginary unit). Regardless of the convention, there are two terms in the expression - one with a plus sign and one with a minus sign. One of them must correspond to negative frequency. However, notice that the sign in the expansion coefficient a^{\dagger}(-\vec{k}) is also reversed. Thus, instead of interpreting this as negative energy, people interpreted it as moving backwards in time.