Lecture 6: Stationary Solutions in ECE 162A - Quantum Mechanics, Study notes of Materials science

Download Lecture 6: Stationary Solutions in ECE 162A - Quantum Mechanics and more Study notes Materials science in PDF only on Docsity! ECE 162A Mat 162A Lecture #6: Stationary Solutions Read Chapter 5,6 of Eisberg,Resnick John Bowers Bowers@ece.ucsb.edu Eigenvalue Equation • Using operators, Schroedinger’s equation can be expressed as an eigenvalue equation ψψ EEop =where V dx d m Eop +−= 2 22 2 h • The solution of the equation involves finding the particular solutions ψn, called eigenfunctions and E called eigenvalues n . Step Potential V0 • Sketch solutions for E>V0 and E<V0 0 Step potential EVxd )()()()( 22 ψh VEForSolution xxx dxm : 0 2 2 > =+− ψψ xikBxikAxxFor )exp()exp()(0 11 −+=< ψ VEkEkwhere xikDxikCxxFor )exp()exp()(0 2 2 22 1 2 22 −== −+=> ψ hh conditionsBoundary mm 22 0 xdxd xx )0()0( )0()0( +− +− →→ →=→ ψψ ψψ dxdx = The wave is entering from the left. Step Potential (E>V0) xikDxikCxxFor xikBxikAxxFor )exp()exp()(0 )exp()exp()(0 11 −+=> −+=< ψ ψ DleftfromWave 0: 22 = CikBAikdxd CBA )(:/ : 21 =− =+ ψ ψ C k kA )1( 2 1 1 2+= C k kB )1( 2 1 1 2−= • Sketch the solution • Explain the difference between E>V0 and E<V0 Problem • Particle in an infinite box otherwiseV axforV ∞= <<= 00 Solutions: ψ(x) = A sin kx + B cos kx 0 a Boundary Conditions: 321 0)()0( kL a ==ψψ Eigenvalues: 2 22 8 ,...,, hnE nn n = == π ma Square Well V022 0 2222 −== EV m E m k κhh 0 /2 /2)()i ()( 2/< kBkA axFor -a a 2/ coss n −< += axFor xxxψ 0: )exp()exp()( = −+= DconditionBoundary xDxCx κκψ )exp()exp()( 2/ −+= > xGxFx axFor κκψ 0: =FconditionBoundary