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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$\begingroup$ More on Klein-Gordon equation and negative probabilities in quantum mechanics: physics.stackexchange.com/q/39224/2451 $\endgroup$– Qmechanic ♦Commented Oct 14, 2012 at 18:20
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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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2$\begingroup$ 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? $\endgroup$– WarCommented Oct 14, 2012 at 22:52
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1$\begingroup$ 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. $\endgroup$– hwlinCommented Oct 15, 2012 at 0:25
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1$\begingroup$ 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"? $\endgroup$ Commented Feb 12, 2017 at 22:22
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$\begingroup$ @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. $\endgroup$ Commented Sep 11, 2017 at 17:48