For E<0, inversion of the free operator gives G0(x,y)=me−κ∣x−y∣/(κℏ2), with κ=−2mE/ℏ. The bound-state equation is ψ=−G0Uψ, henceFor the separable kernel V(y,z)=−αF(y)F∗(z), defineThenAn overall normalization still has to be chosen after finding a permitted energy. A nontrivial bound state must have C=0: if C=0, the potential acts as zero and a square-integrable negative-energy free solution on the whole line cannot exist. Decay or square-integrability assumptions on F are needed to make these integrals well defined; smoothness alone would not ensure convergence.
ψ(x)=−κℏ2m∫∫e−κ∣x−y∣V(y,z)ψ(z)dydz.
C=∫−∞∞F∗(z)ψ(z)dz.
ψ(x)=κℏ2αmC∫e−κ∣x−y∣F(y)dy.
Ancestors (12)
- d
- 2
- S18
- 2018
- Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
- Short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
- Physics course of the University of Oxford
- Undergraduate course of the University of Oxford
- Course of the University of Oxford
- University of Oxford
- List of universities
- Home
Discussion (0)
New discussionThere are no discussions about this article yet.
Articles by others on the same topic (0)
There are currently no matching articles.