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Finite Square Well Potential For V finite outside the well Solutions to S E inside the well the same Have different outside The boundary conditions wavefunction and its derivative continuous give quantization for E V 0 longer wavelength lower Energy Finite number of energy levels Outside k1 2 mE 2 2 d 2 m dx 2 E V E V sin k0 x cos k0 and k 0 2 2m 2 m E V E V k0 E V e k2 x k2 2 m V E P460 square well 1 Boundary Condition Want wavefunction and its derivative to be continuous often a symmetry such that solution at a also gives one at a Often can do the ratio see book and that can simplify the algebra boundary boundary boundary boundary x x 1 boundary 1 boundary x x P460 square well 2 Finite Square Well Potential inside x A sin k1 x B cos k1 x x a 2 x Ce k 2 x De k 2 x x a 2 x Fe k 2 x Ge k 2 x as 0 as x D F 0 Equate wave function at boundaries A sin k21a B cos k21a Ge k2 a 2 A sin And derivative k1a 2 B cos k1a 2 Ce k2 a 2 Ak1 cos k1a 2 Bk1 sin k1a 2 Ck 2 e k 2 a 2 Ak1 cos k1a 2 Bk1 sin k1a 2 Gk2 e k 2 a 2 P460 square well 3 Finite Square Well Potential E R does algebra 2 classes Solve numerically I x A sin k1 x in well x A sin k2a e k 1 k1 cot k1a 2 k 2 2a 2 e k2 x x a 2 quantizati on II x B cos k1 x in well x B cos k2a e k 1 k1 tan k1a 2 k2 2a 2 e k2 x x a 2 quantizati on k1 and k2 both depend on E Quantization sets allowed energy levels En 2 2 n 2 2 ma 2 P460 square well V0 4 Finite Square Well Potential Number of bound states is finite Calculate assuming infinite well energies Get n Add 1 En Electron V 100 eV width 0 2 nm 2 2 n 2 2 ma 2 n 2 V0 n 2 2 ma 2V0 2 2 2 51 MeV 2 nm 2 100 eV 197 ev nm 2 3 14 2 10 7 N 4 number of levels Deuteron p n bound state Binding energy 2 2 MeV radius 2 1 F really need 3D S E n 2 2 940 MeV 2 1 F 2 2 2 MeV 197 MeV F 2 3 14 2 0 1 N 1 only 1 bound state P460 square well 5 Finite Square Well Potential Can do an approximation by guessing at the penetration distance into the forbidden region Use to estimate wavelength 1 b En 2 m V E n 2 2 2



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