Assignment 6, solutions

Problem 1:

An operator A has two normalized eigenstates ψ1 and ψ2, with eigenvalues a1 and a2, respectively.  An operator B, has two normalized eigenstates, φ1 and φ2, with eigenvalues b1 and b2, respectively.   The eigenstates are related by
ψ1 = (3φ1 + 4φ2)/5,    ψ2 = (4φ1 - 3φ2)/5.
(a)  Observable A is measured, and the value a1 is obtained.  What is the state of the system immediately after this measurement?
(b)  If B is measured immediately afterwards, what are the possible results, and what are their probabilities?
(c)  If the result of the measurement of B is not recorded and right after the measurement of B, A is measured again, what is the probability of getting a1?

Solution:

 Concepts: The postulates of Quantum Mechanics Reasoning: When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue an is given by P(an) = Σi=0gn||2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an. If a measurement on a system in the state |ψ> gives the result an, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with an. Details of the calculation: (a)  After measuring a1, the system is in the eigenstate ψ1.(b)  If B is now measured, the probability of obtaining b1 is 9/25 and the probability of obtaining b2 is 16/25.(c)  There are two path.  If b1 is measured the system is in the state φ1 = (3ψ1 + 4ψ2)/5.The probability of measuring a1 after measuring b1 is 9/25.If b2 is measured the system is in the state φ2 = (4ψ1 - 3ψ2)/5.The probability of measuring a1 after measuring b2 is 16/25.The probability of getting a1 again after a measurement of B is (9/25)2 + (16/25)2 = 0.5392.

Problem 2:

(a)  Show that the eigenvalues of a general 2×2 matrix A can be expressed as

where D is the determinant of A and T is the trace of A (sum of diagonal elements).
Show that the matrix

with M >> m has two eigenvalues, with one much larger than the other.

(b)  Show that the most general form of a 2×2 unitary matrix U with unit determinant can be parameterized as

subject to the constraint aa* + bb* = 1, where * denotes complex conjugation.

Solution:

 Concepts: Mathematical foundations of quantum mechanics Reasoning: This is a linear algebra problem. Details of the calculation: (a)  To find the eigenvalues of an n×n matrix A requires solving the characteristic (also termed secular) equation det(A - λI) = 0 for λ, with I the n×n unit matrix,  For the general 2×2 matrix we have (a – λ)(d – λ) – bc = 0.λ± = ½[a + d ± ((a + d)2 – 4(ad - bc))½ = ½T ± (¼T2 – D)½, where T = Tr(A) = a + b and D = det(A) = ad – bc.Applying this to the matrix and assuming M >> m gives T = M,  D = -m2,λ± = ½M ± (¼M2 + m2) = ½M ± ½M(1 + 4m2/M2)½ ≈ ½M ± ½M(1 + 2m2/M2)= ½M ± (½M + m2/M).The eigenvalues therefore are λ+ ≈ M,  λ- ≈ m2/M,  |λ+| >> |λ-|.(b)  Parameterize the matrix in terms of complex parameters a, b, c, and d, and require the unitarity condition UU† = I. UU† = I --> aa* + bb* = 1, cc* + dd* =  1, ac* + bd* = 0.  b = -c*a/d*. det(U) = 1 --> ad – bc = 1,  ad + |c|2a/d* = 1,  a(|d|2 + |c|2) = d*,  a = d*, b = -c*.Therefore |a|2 + |b|2 = 1.

Problem 3:

A quantum system with two states has the Hamiltonian matrix (ε1 < ε2)

(a)  What are the two energies E± of the system?
(b)  It is of advantage to parameterize the eigenstates as

with real α, φ.  Show that this state is normalized.
Show that φ = γ/2 for the complex off-diagonal matrix element ν = |ν|e, and find the values of α for the two eigenstates.

Solution:

 Concepts: The eigenvalues and eigenvectors of a Hermitian operator Reasoning: H is a Hermitian operator.  The eigenvalues of a Hermitian operator are real.  Every Hermitian operator has at least one basis of orthonormal eigenvectors. Details of the calculation: (a) (b)  The state is normalized because ψ†ψ = cos2α + sin2α = 1. Using ν = |ν|eiγ and inserting the proposed eigenstate into the Schroedinger equation yieldsε1 cosα eiφ + |ν|eiγ sinα e-iφ = E± cosα eiφ, |ν|e-iγ cosα eiφ + ε2 sinα e-iφ = E± sinα e-iφ.These equations are only satisfied if γ = 2φ.  We then have ε1 cosα + |ν| sinα = E± cosα, |ν| cosα + ε2 sinα = E± sinα.Multiplying the first equation with cosα, and the second with sinα, and subtracting them yieldsε1 cos2α - ε2 sin2α = E± (cos2α - sin2α).    (E± - ε1) cos2α = (E± - ε2) sin2α,tan2α = (E± - ε1)/( E± - ε2). Alternatively, multiplying the first equation with sinα, and the second with cosα, and subtracting them yields |ν|/(ε1 - ε2) = sinα cosα/(cos2α – sin2α)  = ½ sin(2α)/cos(2α) = tan(2α)/2.tan(2α) = 2|ν|/(ε1 - ε2).In the interval 0 ≤ 2α < 2π, we have two solutions.

Problem 4:

Use the uncertainty principle, ΔxΔp ≥ ħ/2, to estimate the ground state energy of a particle in a one-dimensional well of the form
(a)  U(x) = U0,  -a/2 < x < a/2,  U(x) = infinite for all other values of x,
(b)  U(x) = k|x|.

Solution:

 Concepts: The uncertainty principle Reasoning: The ground state energy is the lowest possible energy.  For case (a) Δx is fixed.  E depends only on p.  We use  ΔxΔp = ħ/2 to express Δp in terms of the fixed Δx to estimate the ground state energy. For case (b) Δx changes with E.  E depends on both x and p.  We use  ΔxΔp = ħ/2 to express Δp in terms of the Δx, and the minimize or estimate for E with respect to Δx. Details of the calculation: (a)  Estimate:  Δx = a, Δp = ħ/(2a),  Emin = Δp2/(2m) + U0 = ħ2/(8ma2) + U0. (b)  Estimate:  Δp = ħ/(2Δx),  E = ħ2/(8m|Δx|2) +  k|Δx|. dE/d(|Δx|) = -ħ2/(4m|Δx|3) + k = 0,  |Δx|3 = ħ2/(4mk). Inserting |Δx|3 = ħ2/(4mk) into the equation for E yields Emin = 0.94 (kħ)2/3/m1/3.

Problem 5:

(a)  Find the relativistic energy (in MeV) of an electron with a deBroglie wavelength of 0.0012 nm.
(b)  Estimate the minimum value for the speed vmax of an electron in a hydrogen atom, confined within a distance of about 0.1 nm?

Solution:

 Concepts: The deBroglie wavelength, the uncertainty principle Reasoning: The deBroglie wavelength lets us estimate the momentum of the electron, p = h/λ. The minimum value for the speed vmax can be estimated from ΔxΔp ~ h. Details of the calculation: (a)  The relativistic energy is E = (p2c2 + m2c4)½.  The rest energy of the electron is mc2 = 8.19*10-14 J. p = h/λ = (6.63*10-34 Js)/(1.2*10-12 m) = 5.525*10-22 kg m/s. pc = 1.66*10-13 J = 1.04 MeV.  E = 1.15 MeV = 1.84*10-13 J. (b)  Δp = mΔv.   The minimum value for the speed vmax must be on the order of Δv = h/(mΔx) = (6.63*10-34 Js)/(9.11*10-31 kg*10-10 m) = 7.28*106 m/s.