Solutions to Midterm Exam 2 - Quantum Mechanics | PHYS 324, Exams of Quantum Mechanics

Download Solutions to Midterm Exam 2 - Quantum Mechanics | PHYS 324 and more Exams Quantum Mechanics in PDF only on Docsity! 1 Physics 324 Soultion to Second Midterm Exam Autumn 2003 Part I. (12 pts) A particle of mass m and energy E moves in a potential V (x) as shown to the right: for x < 0, (Region I) V (x) = 0, and for x > 0, (Region II) V (x) = −V0). E and V0 are both positive constants as shown. 1. (4 pts) Find a general solution, ΦI(x), to the time independent Schrodinger equation for a particle entering from the left in Region I. Do not worry about the overall normalization of your solution. Be sure to define any constants you use to express your solution and show your work. Solution: In region I, the potential is 0, so our solution is the free particle solution: ΦI(x) = eikx + Be−ikx where k2 = 2mE h̄2 The normalization of the incoming wave (eikx) is chosen to be 1 for simplicity. 2. (3 pts) Find a solution, ΦII(x), to the time independent Schrodinger equation for a particle entering from the left in Region II. Be sure to define any constants you use to express your solution and show your work. Solution: In region II, we have: ∂Φ(x) ∂x2 = −2m(E + V0) h̄2 Φ(x) = −q2Φ(x) Again, we have plane wave solutions, but because there is no incoming wave from the right, we have: ΦII(x) = Ceiqx 3. (5 pts) Derive an expression for the coefficient of reflection, that is, the ratio of the probability for finding a wave travelling to the left in Region I (a scattered wave) to the probability of finding a wave travelling to the right in Region I (the incident wave). Solution: The coefficient of reflection, R, is given by |B|2 because we chose the incoming wave to have an amplitude of 1. To find B, we apply the boundary conditions on Φ at x = 0: ΦI(0) = ΦII(0) => 1 +B = C ∂ΦI(x) ∂x ∣∣∣ x=0 = ∂ΦII(x) ∂x ∣∣∣ x=0 => ik(1 −B) = iqC => ik(1 −B) = iq(1 +B) => k − q = (k + q)B => B = k − q k + q R = (k − q)2 (k + q)2 where k2 = 2mE h̄2 , q2 = 2m(E + V0) h̄2 2 Part II. (13 pts) For matrix M = ( 0 1 − i 1 + i 0 ) 1. (3 pts) Write the matrix form for the Hermitian conjugate, M†, of M. M† = M = ( 0 1 − i 1 + i 0 ) (ie M is a Hermitian matrix). 2. (3 pts) What are the eigenvalues of M ? Solution: We find the eigenvalues, λ, from the characteristic equation: det(M− λI) = 0 => det ( −λ 1 − i 1 + i −λ ) = 0 => λ2 − (1 + i)(1 − i) = 0 => λ2 − 2 = 0 => λ = ± √ 2 3. (4 pts) Find one of the eigenvectors of M. Soultion: Let’s find the eigenvector for the for the positive eigenvalue, λ = √ 2. The eigenvector, a, must satisfy the equation: Ma = λa => ( 0 1 − i 1 + i 0 ) ( a1 a2 ) = √ 2 ( a1 a2 ) => (1 − i)a2 = √ 2 a1 Choosing a2 = √ 2, we have a = ( 1 − i√ 2 ) 4. (3 pts) If we label the normalized eigenvectors of M as |a〉 and |b〉, write an expression for a wavefunction that has a probability of 2/3 of being found in state |a〉 and a probability of 1/3 of being found in state |b〉. Solution: The normalized eigenvectors of a Hermitian matrix form an orthonormal basis. We can write a general vector (wavefunction) in that basis and the probability of finding a particular eigenvalue is the squared magnitude of the amplitude of the corresponding eigenvector: |ψ〉 = √ 2 3 |a〉 + √ 1 3 |b〉 Part III. (12 pts) Assume that we have defined an operator, Ŝq , such that for all functions, f(x), Ŝq f(x) = e−iqx f(x)