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Klein-Gordon equation in quantum field theory is known to suffer from the possibility of negative probability. So, the question is, despite this, Klein-Gordon describes spin-zero field. So, how can negative probability and scalar field co-exist?

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In QFT, we reinterpret the probability density as the probability charge density. In other words, negative probabilities correspond to antiparticles.

In fact, the Dirac equation which describes spin-1/2 also has this property, and it led to the prediction of the positron as the antiparticle of the electron.

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    $\begingroup$ but Wikipedia and physics.stackexchange.com/q/39224/2451 say that negative probability density do not make sense, and this is the reason for developing Dirac equation... So.. what is that and this? $\endgroup$ – War Oct 14 '12 at 22:52
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    $\begingroup$ Yes, negative probabilities do not make sense. What physicists discovered was that they were not calculating the probability density, but ~ probability density * charge. So if charge is opposite, they get a negative answer, but the actual probability is positive definite. $\endgroup$ – hwlin Oct 15 '12 at 0:25
  • $\begingroup$ I thought the Dirac density $j^{0} = \bar{\psi} \gamma^{0} \psi $ was positive definite... what do you mean by "The Dirac equation also has this property"? $\endgroup$ – P. C. Spaniel Feb 12 '17 at 22:22
  • $\begingroup$ @Spaniel: This issue is "circumvented" or fixed when the solutions of the Dirac equation are promoted to field-operators with anti-commutator rules. In that case if $j^0$ (as operator) is applied on a anti-particle state, is gets an negative eigenvalue, whereas applied on particle states, the eigenvalue is positive. $\endgroup$ – Frederic Thomas Sep 11 '17 at 17:48

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