Review of Quantum Physics

1.6. ANSWERS TO CHAPTER 1 PROBLEMS 211.6 Answers to Chapter 1 Problems1. Oscillating wavefunction where E > V, decaying wavefunction in forbidden regions E < V.2. (i) Linear operators obey Â(af 1(x)+bf 2 (x)) = aÂf 1(x)+bÂf 2(x)(ii) Orthogonal functions obey ∫ ba f 1(x) ∗ f 2 (x)dx = 0 for a particular range (a,b)(iii) Any arbitrary function ψ can be written as a unique sum of a complete set of functionsψ n : ψ = ∑ n c nψ n3. (i) ˆx = x(ii) ˆp x = −i¯h ddx(iii) ˆp y = −i¯h ddy(iv) ˆP = −i¯h∇(v) ˆP 2 = −¯h 2 ∇ 2(vi) ˆT = ˆp2 x2m = − ¯h22md 2dx 2(vii) ˆT = ˆP 22m = − ¯h22m ∇2(viii) ˆV = V(x)(ix) Ĥ = ˆT + ˆV = − ¯h22m dx+V(x) 2d 2(x) Ĥ = ˆT + ˆV = − ¯h22m ∇2 +V(x)4. (i) 〈ψ 1 |cψ 2 〉 = ∫ dxψ ∗ 1cψ 2 = c〈ψ 1 |ψ 2 〉(ii) 〈cψ 1 |ψ 2 〉 = ∫ dx(cψ 1 ) ∗ ψ 2 = c ∗ 〈ψ 1 |ψ 2 〉(iii) 〈ψ 3 |ψ 1 +ψ 2 〉 = ∫ dxψ ∗ 3 (ψ 1 +ψ 2 ) = 〈ψ 3 |ψ 1 〉+〈ψ 3 |ψ 2 〉(iv)ψ = ∑ nc n ψ n5. (a) If ψ = 1 √1−c 2 (−cφ 1 +φ 2 ) then 〈ψ|φ 1 〉 = 0(b) If ψ = 1 √ 2−2c(φ 1 −φ 2 ) then 〈ψ|φ 1 +φ 2 〉 = 0⇒ 〈ψ m |ψ〉 = ∑ c n 〈ψ m |ψ n 〉} {{ }nδ mn⇒ c m = 〈ψ m |ψ〉6. Linear:  2 ,Â3,Â4,Â6Hermitian:  3 ,Â4,Â6Eigenfunctions (Âψ = λψ): 3 : ψ(x) = e −iλx 4 : ψ(x) = δ(x−x 0 ) 4 : ψ(x) = Ae √ λx +Be −√ λx