3 Klein-Gordon EquationIn non-relativistic mechanics, the energy for a free particle isE =p22m. (35)To get quantum mechanics, we make the following substitutions:E → i~∂∂t, p → −i~∇, (36)and the Schr¨odinger equation for a free particle is−~22m∇2Ψ = i~∂Ψ∂t. (37)In relativistic mechanics, the energy of a free particle isE =pp2c2+ m2c4. (38)Making the same substitution we obtain√−~2c2∇2+ m2c2Ψ = i~∂Ψ∂t. (39)It’s difficult to interprete the operator on the left hand side, so instead we tryE2= p2c2+ m2c4(40)⇒i~∂∂t2Ψ = −~2c2∇2+ m2c4Ψ, (41)or1c2∂∂t2Ψ −∇2Ψ ≡ 2Ψ = −m2c2~2Ψ, (42)where2 =1c2∂∂t2− ∇2= ∂µ∂µ. (43)Plane-wave solutions are readily found by inspection,Ψ =1√Vexpi~p ·xexp−i~Et, (44)where E2= p2c2+ m2c4and thus E = ±pp2c2+ m2c4. Note that there is anegative energy solution as well as a positive energy solution for each value of p.Na¨ıvely one should just discard the negative energy solution. For a free particlein a positive energy state, there is no mechanism for it to make a transition to6the negative energy state. However, if there is some external potential, the Klein-Gordon equation is then altered by the usual replacements,E → E − eφ, p → p −ecA, (45)(i~∂t− eφ)2Ψ = c2(−i~∇ −ecA)2Ψ + m2c4Ψ. (46)The solution Ψ can always be expressed as a superposition of free particle solutions,provided that the latter form a complete set. They from a complete set only ifthe negative energy components are retained, so they cannot be simply discarded.Recall the probability density and current in Schr¨odinger equation. If we multiplythe Schr¨odinger equation by Ψ∗on the left and multiply the conjugate of theSchr¨odinger equation by Ψ, and then take the difference, we obtain−~22m(Ψ∗∇2Ψ − Ψ∇2Ψ∗) = i~(Ψ∗˙Ψ + Ψ˙Ψ∗)⇒ −~22m∇(Ψ∗∇Ψ − Ψ∇Ψ∗) = i~∂∂t(Ψ∗Ψ) (47)Using ρs= Ψ∗Ψ, js=~2mi(Ψ∗∇Ψ − Ψ∇Ψ∗), we then obtain the equation ofcontinuity,∂ρs∂t+ ∇ ·js= 0 (48)Now we can carry out the same procedure for the free-particle Klein-Gordon equa-tion:Ψ∗2Ψ = −m2c2~Ψ∗ΨΨ2Ψ∗= −m2c2~ΨΨ∗(49)Taking the difference, we obtainΨ∗2Ψ − Ψ2Ψ∗= ∂µ(Ψ∗∂µΨ −|psi∂µΨ∗) = 0. (50)This suggests that we can define a probability 4-current,jµ= α(Ψ∂µΨ − Ψ∂µΨ∗), where α is a constant (51)and it’s conserved, ∂µjµ= 0, jµ= (j0, j). To make j agree with js, α is chosen tobe α = −~2mi. So,ρ =j0c=i~2mc2Ψ∗∂Ψ∂t− Ψ∂Ψ∗∂t. (52)ρ does reduce to ρs= Ψ∗Ψ in the non-relativistic limit. However, ρ is not positive-definite and hence can not describe a probability density for a single particle.Pauli and Weisskopf in 1934 showed that Klein-Gordon equation describes a spin-0(scalar) field. ρ and j are interpreted as charge and current density of the particlesin the