I am totally a beginner in QFT, here's the problem that I got: for the real scalar fields, are there any elementary particles descriped by them. If so, how to understand the negative probability problem?
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$\begingroup$ What is the negative probability problem? (And yes, the Higgs particle is the excitation of a scalar field) $\endgroup$– ACuriousMind ♦Commented Jan 19, 2015 at 15:05
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$\begingroup$ QFT does not have a negative probability problem. On the other hand relativistic quantum mechanics (RQM), meaning taking wave function solutions rather than operator solutions to the KG equation, does have that problem. This is one of the reasons QFT was invented. $\endgroup$– BMSCommented Jan 19, 2015 at 17:45
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Quantum field theory solves the problem by giving a different interpretation to the "probability". In the case of complex fields, quantum field theory also introduces antiparticles.
In the first-quantized Klein-Gordon equation, the time component $j^0$ of the probability current vector $j^\mu$ may indeed be both positive and negative and negative probabilities are bad, as you point out.
However, $j^\mu$ is a bilinear expression constructed from the field $\phi$ and its derivatives, roughly $\phi\cdot \partial^\mu\phi$, and when $\phi$ becomes a quantum field (instead of a wave function), which is an operator (or operator-distribution), $j^\mu$ becomes an operator, too. There is nothing wrong about $j^0$ being positive or negative (indefinite) because it defines the charge density for a complex field $\phi$.
In quantum field theory, the wave functions that could have had both positive and negative probabilities are used as prefactors in formulae for quantum fields and the positive-energy (and positive-probability) and negative-energy (and negative-probability) solutions for the wave function are treated asymetrically. The latter must be multiplied by the creation operator and the former by the annihilation operators.
In effect, it means that in quantum field theory, we may "create" an arbitrary number of particles in wave functions that are allowed by the first-quantized (one-particle) quantum mechanical theory but we are only allowed to use the positive-energy (positive-probability) wave function to excite the vacuum. The negative-energy ones are multiplied by annihilation operators which annihilate the vacuum so we get no state.
Those things may only be properly understood along with the full mathematical apparatus of QFT and it is covered in every course or textbook on quantum field theory.
answered Jan 19, 2015 at 15:16
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$\begingroup$ so that's how it is.It sounds like I need more work on QFT. $\endgroup$– ShaneCommented Jan 20, 2015 at 14:05