P460 - square well1 Square Well Potential Start with the simplest potential Boundary condition is that  is continuous:give: V 0 -a/2 a/2

P460 - square well2 Infinite Square Well Potential Solve S.E. where V=0 Boundary condition quanitizes k/E, 2 classes Even  =Bcos(k n x) k n =n  /a n=1,3,5...  (x)=  (-x) Odd  =Asin(k n x) k n =n  /a n=2,4,6...  (x)=-  (-x)

P460 - square well3 Infinite Square Well Potential Need to normalize the wavefunction. Look up in integral tables What is the minimum energy of an electron confined to a nucleus? Let a = m = 10 F

P460 - square well4 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:

P460 - square well5 Finite Square Well Potential Equate wave function at boundaries And derivative

P460 - square well6 Finite Square Well Potential Book does algebra. 2 classes. Solve numerically k1 and k2 both depend on E. Quntization sets allowed energy levels

P460 - square well7 Finite Square Well Potential Number of bound states is finite. Calculate assuming “infinte” well energies. Get n. Add 1 Electron V=100 eV width=0.2 nm Deuteron p-n bound state. Binding energy 2.2 MeV radius = 2.1 F (really need 3D S.E……...

P460 - square well8 Finite Square Well Potential Can do an approximation by guessing at the penetration distance into the “forbidden” region. Use to estimate wavelength Electron V=100 eV width=0.2 nm 