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I am learning quantum field theory.

Now I am considering this case:

Suppose a spin-0 particle which obeys the Klein-Gordon field equation and its anti-particle obeying the same equation do not have the same solution to the equation. This means that the field should be a complex scalar field. Hence the field operator should be :

$$\hat{\psi}=\int_{p} \text{e}^{-\text{i}px}\hat{a}_{p}+\text{e}^{\text{i}px}\hat{b}^{+}_{p}$$

$\hat{a}$ is annihilation operator for a particle and $\hat{b}^{+}$ the creation operator for the corresponding anti-particle.

If the physical process we're interested in, for example, is a low-energy process, where particle numbers are conserved and antiparticles are not involved, can I simply treat the field as a real scalar field? i.e:

$$\hat{\psi}=\int_{p} \text{e}^{-\text{i}px}\hat{a}_{p}+\text{e}^{\text{i}px}\hat{a}^{+}_{p}$$

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  • $\begingroup$ Why would you neglect anti-particles, they have the same mass, therefore same energy scale? $\endgroup$
    – ohneVal
    Commented Aug 11, 2020 at 7:26
  • $\begingroup$ For example, I would like to consider two-particle collision, in which the total energy is small. Thus no extra particles would be created? $\endgroup$
    – xiang sun
    Commented Aug 11, 2020 at 7:37
  • $\begingroup$ Do you mean 1 or 2 real fields? $\endgroup$
    – Qmechanic
    Commented Aug 11, 2020 at 8:01
  • $\begingroup$ I mean can I just drop the part correspoding to antipaticle and rewrite the complex scalar field as a real scalar field. In my umderstanding, a complex scalar field is equivalent to 2 real scalar fields. Hence dropping the antiparticle part should work. Yet I am not sure. $\endgroup$
    – xiang sun
    Commented Aug 11, 2020 at 12:26

1 Answer 1

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A single real Klein-Gordon field has no electromagnetic interactions. Your particles therefore would not be charged if they were modeled with a real field, which would be a very incorrect description even at low energies. The only way I can think of that you might pull this off is by introducing self-interactions mediated by another field. You might be able to do that in such a way that it approximates the electromagnetic interaction in some circumstances. However, true E&M, which is not a high energy effect, comes from a gauge summery that is not present for a single real field.

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  • $\begingroup$ Great, I've never thought from E&M perspective. $\endgroup$
    – xiang sun
    Commented Aug 11, 2020 at 13:13

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