I know basically the difference between Klein-Gordon and Dirac field is spin. But I am not sure where we need to implement this info.
The solutions of both equations are the wave packets which includes the sum of creation and annihilation operators.
Ψ(x)=∫[ape−ipx+a†peipx]d3p.
where we use the spin information? Why should I use Klein Gordon for spin-0 and Dirac equation for 1/2?
Answer
The momentum decomposition you wrote is valid only for a scalar (spinless), real field, satisfying the Klein-Gordon equation.
When considering a field with spin, like a spin-1/2 field satisfying the Dirac equation, you must include the polarization vectors, obtaining something of the form ψα(x)=∑p,sNs,p[cs(p)uα(s,p)e−ipx+d†s(p)vα(s,p)eipx]
In other words, the spin information is encoded in the additional degrees of freedom of the field, in this case the spin-index which I denoted with α. This additional degrees of freedom do not evolve independently, as you can see from the (free) Dirac equation, which showing explicitly the spinor indices reads (iγμαβ∂μ−mδαβ)ψβ(x)=0,∀α=1,2,3,4,
For another example you can look at spin-1 fields (e.g. photons). In this case the quantum field is denoted by Aμ(x) with μ a vector index, which is the spin-index for spin-1 fields. The decomposition has now the form: Aμ(x)=∑p,sNs,p[a(s,p)εμ(s,p)e−ipx+a†(s,p)ε∗μ(s,p)eipx].
Finally, note that each spin component of the Dirac field satisfied the Klein-Gordon equation: (◻+m2)ψα(x)=0,∀α
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