Problem 1:
An operator A has two normalized eigenstates ψ_{1} and ψ_{2},
with eigenvalues a_{1} and a_{2}, respectively. An operator
B, has two
normalized eigenstates, φ_{1} and φ_{2}, with eigenvalues b_{1}
and b_{2}, respectively. The eigenstates are related by
ψ_{1} = (3φ_{1} + 4φ_{2})/5, ψ_{2} = (4φ_{1}
- 3φ_{2})/5.
(a) Observable A is measured, and the value a_{1} 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 a_{1}?
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 a_{n} is given by P(a_{n}) = Σ_{i=0}^{gn}|<u_{n}^{i}|ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an orthonormal basis in the eigensubspace E_{n} associated with the eigenvalue a_{n}. If a measurement on a system in the state |ψ> gives the result a_{n}, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with a_{n}. | |
Details of the calculation: (a) After measuring a_{1}, the system is in the eigenstate ψ_{1}. (b) If B is now measured, the probability of obtaining b_{1} is 9/25 and the probability of obtaining b_{2} is 16/25. (c) There are two path. If b_{1} is measured the system is in the state φ_{1} = (3ψ_{1} + 4ψ_{2})/5. The probability of measuring a_{1} after measuring b_{1} is 9/25. If b_{2} is measured the system is in the state φ_{2} = (4ψ_{1} - 3ψ_{2})/5. The probability of measuring a_{1} after measuring b_{2} is 16/25. The probability of getting a_{1} 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 ± (¼T^{2} – 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 = -m^{2}, λ_{±} = ½M ± (¼M^{2} + m^{2}) = ½M ± ½M(1 + 4m^{2}/M^{2})^{½} ≈ ½M ± ½M(1 + 2m^{2}/M^{2}) = ½M ± (½M + m^{2}/M). The eigenvalues therefore are λ_{+} ≈ M, λ_{-} ≈ m^{2}/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|^{2}a/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 (ε_{1} <
ε_{2})
(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^{iγ},
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 = cos^{2}α + sin^{2}α = 1. Using ν = |ν|e^{iγ} and inserting the proposed eigenstate into the Schroedinger equation yields ε_{1 }cosα e^{iφ} + |ν|e^{iγ} sinα e^{-iφ} = E_{±} cosα e^{iφ}, |ν|e^{-iγ} cosα e^{iφ} + ε_{2} sinα e^{-iφ} = E_{±} sinα e^{-iφ}. These equations are only satisfied if γ = 2φ. We then have ε_{1 }cosα + |ν| sinα = E_{±} cosα, |ν| cosα + ε_{2} sinα = E_{±} sinα. Multiplying the first equation with cosα, and the second with sinα, and subtracting them yields ε_{1 }cos^{2}α - ε_{2} sin^{2}α = E_{±} (cos^{2}α - sin^{2}α). (E_{±} - ε_{1}) cos^{2}α = (E_{±} - ε_{2}) sin^{2}α, tan^{2}α = (E_{±} - ε_{1})/( E_{±} - ε_{2}). Alternatively, multiplying the first equation with sinα, and the second with cosα, and subtracting them yields |ν|/(ε_{1 }- ε_{2}) = sinα cosα/(cos^{2}α – sin^{2}α) = ½ sin(2α)/cos(2α) = tan(2α)/2. tan(2α) = 2|ν|/(ε_{1 }- ε_{2}). 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) = U_{0}, -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), E_{min} = Δp^{2}/(2m) + U_{0} = ħ^{2}/(8ma^{2}) + U_{0}. (b) Estimate: Δp = ħ/(2Δx), E = ħ^{2}/(8m|Δx|^{2}) + k|Δx|. dE/d(|Δx|) = -ħ^{2}/(4m|Δx|^{3}) + k = 0, |Δx|^{3} = ħ^{2}/(4mk). Inserting |Δx|^{3} = ħ^{2}/(4mk) into the equation for E yields E_{min} = 0.94 (kħ)^{2/3}/m^{1/3}. |
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 v_{max} 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 v_{max} can be estimated from ΔxΔp ~ h. | |
Details of the calculation: (a) The relativistic energy is E = (p^{2}c^{2} + m^{2}c^{4})^{½}. The rest energy of the electron is mc^{2} = 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 v_{max} must be on the order of Δv = h/(mΔx) = (6.63*10^{-34} Js)/(9.11*10^{-31} kg*10^{-10} m) = 7.28*10^{6} m/s. |