For E<0, impose decay and continuity on G=Ce−κ∣x−x′∣. Its derivative jump is −2κC, so −ℏ2(G+′−G−′)/(2m)=1 requiresThere is no nonzero free solution decaying at both infinities that can be added without spoiling the derivative condition.
G(x,x′)=κℏ2me−κ∣x−x′∣,κ=−2mE/ℏ.
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