Away from x=x′, the free equation is G′′+k2G=0. The outgoing boundary condition selects G=Aeik∣x−x′∣, continuous at its source. Integrating the differential equation across the source gives the derivative jumpThe jump for the proposed form is 2ikA, so A=im/(kℏ2). ThusFor x far to either side, the x′-dependent phase is absorbed into the outgoing amplitude. The outgoing prescription is equivalently the boundary value of the resolvent with E→E+i0; changing that prescription would give a different Green function.
−2mℏ2[G′(x′+,x′)−G′(x′−,x′)]=1.
G(x,x′)=kℏ2imeik∣x−x′∣.
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