How one can deduce the existence of antiparticles

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  • #1
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Hi together ...

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

one gets solutions [tex]\Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))[/tex] leading to [tex]E^2=p^2 + m^2 [/tex]. 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.
 

Answers and Replies

  • #2
mathman
Science Advisor
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Re: Antiparticles

Since anti-particles have been found experimentally, what is the point of your assertion?
 
  • #3
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Re: Antiparticles

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.
 
  • #4
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Re: Antiparticles

The general solution of the KGE is:

[tex]
\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}}
[/tex]

If we interpret [itex]\Psi(t, \vec{x}) = \Psi^{\dagger}(t, \vec{x})[/itex] as a field operator creating an excitation of the real scalar field at the space-time point [itex](t, \vec{x})[/itex], then [itex]a(\vec{k})[/itex] creates a particle with momentum [itex]\hbar \vec{k}[/itex] and energy [itex]\hbar \omega[/itex], while [itex]a^{\dagger}(-\vec{k})[/itex] creates an antiparticle with the same energy and momentum.
 
  • #5
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Re: Antiparticles

Thanks. Thats a linear superposition of the fundamental solutions i cited above, but as you wrote [tex]\omega[/tex] 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.
 
  • #6
2,956
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Re: Antiparticles

notice the sign in front of t in both of the exponentials.
 
  • #7
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Re: Antiparticles

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.
 
  • #8
2,956
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Re: Antiparticles

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.
 
  • #9
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Re: Antiparticles

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
[tex]
\Psi(t,\vec{x})=\exp(\pm i (- E t + \vec{p} \cdot \vec{x}))
[/tex]
you don't say the energy can take on imaginary values only because there is an [tex]i[/tex] in the exponential. it's simply part of the solution function. so why is a minus sign treated differently?
 
  • #10
2,956
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Re: Antiparticles

I'm afraid I don't understand what you are trying to say.
 
  • #11
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Re: Antiparticles

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 [tex]i[/tex] 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 [tex]i[/tex].

greetings.
 
  • #12
2,956
5
Re: Antiparticles

IF you have a time dependence like this:

[tex]
\exp{(-i \, \omega \, t)}
[/tex]

then [itex]\hbar \omega[/itex] is the energy associated with such propagating perturbation.
 
  • #13
40
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Re: Antiparticles

O.k. so everything in the exponent multiplied by t except the [tex]-i[/tex] is the frequency [tex]\omega[/tex]. As [tex]E=\hbar \omega[/tex] 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 [tex]\exp(i \omega t) [/tex] and no one thought of a negative energy.
nonetheless thanks for your patience and help.
 
  • #14
2,956
5
Re: Antiparticles

In Physics, the convention is to use [itex]\exp{(-i \omega t)}[/itex] as time dependence. In Electrical Engineering, it usually is [itex]\exp(2 \pi j f t)[/itex] (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 [itex]a^{\dagger}(-\vec{k})[/itex] is also reversed. Thus, instead of interpreting this as negative energy, people interpreted it as moving backwards in time.
 

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