Assignment 7

Problem 1:

Consider a highly excited He atom. One electron is in a state with n₁ = 4, l₁ = 3 and the other in a state with n₂ = 5, l₂ = 4
(a) Find the possible values of l (total orbital angular momentum quantum number) for the system.
(b) Find the possible values of s (total spin angular momentum quantum number) for the system.
(c) Find the possible values of j (total angular momentum quantum number) for the system.
(d) How many distinct angular momentum states with j = 1 are there?

Solution:

Concepts:
Addition of angular momentum
Reasoning:
We are supposed to add the orbital and spin angular momentum.
Details of the calculation:
(a) The quantum numbers associated with the total orbital angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
l_max = l₁ + l₂ = 3 + 4 = 7.
l_min = |l₁ - l₂| = |3 - 4| = 1.
The possible values for L are 1, 2, 3, 4, 5, 6, and 7.
(b) The quantum numbers associated with the total spin angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
s_max = s₁ + s₂ = ½ + ½ = 1.
s_min = |s₁ - s₂| = |½ - ½| = 0.
The possible values for s are 0 and 1.
(c) The largest possible value for j will be found by adding together the maximum values for both l and s. The minimum value for j will be found by subtracting the largest possible value of s from the smallest possible value for l.
j_smax = l_max + s_max = 7 + 1 = 8.
j_min = |l_min - s_max| = |1 - 1| = 0.
j = 0, 1, 2, 3, 4, 5, 6, 7, 8.
(d) A state with quantum number j = 1 can have l = 1, s = 0, or l = 2, s = 1, or l = 1, s = 1
Each of these states can have m_j = -1, 0, +1.
So there are 9 distinct states with j = 1.

Problem 2:

(a) Write down Maxwell's equations for a conducting medium with conductivity σ, permittivity ε₀ and permeability μ₀.
(b) A plane wave of low frequency ω << σ/ε₀ is propagating in the z-direction inside the conducting medium.
Let E = E₀exp(i(kz - ωt)), B = B₀exp(i(kz - ωt)), where E₀, B₀, and k are complex.
Use Maxwell's equations to show that k = k₁ + ik₂, k₁ ≈ k₂ ≈ (μ₀ωσ/2)^½.
Calculate the ratio of the complex amplitude of the two fields, E₀/B₀ (magnitude and phase).
(c) Calculate the energy flux (time averaged Poynting vector) in the conducting medium.

Solution:

Concepts:
Maxwell's equations
Reasoning:
In regions with ρ_f= 0 and j_f = σE Maxwell's equations can be used to show that both E and B satisfy the damped wave equation.
Details of the calculation:
(a) ∇∙E = ρ/ε₀, ∇×E = -∂B/∂t, ∇∙B = 0, ∇×B = μ₀j + (1/c²)∂E/∂t.
Assume ε = ε₀, μ = μ₀, ρ_f= 0 and j_f= σE in the conductor. Then
∇×B = μ₀σE + μ₀ε₀∂E/∂t.
∇×(∇×B) = ∇(∇∙B) - ∇²B = μ₀σ(∇×E) + μ₀ε₀∂(∇×E)/∂t.
∇²B - μ₀σ∂B/∂t - μ₀ε₀∂²B/∂t² = 0.
Similarly: ∇²E - μ₀σ∂E/∂t - μ₀ε₀∂²E/∂t² = 0.
Both E and B satisfy the damped wave equation.

(b) Let E(r,t) = E₀exp(ikz)exp(-iωt), E ⊥ k.
We have a plane wave propagating into the z-direction.
Then k² = iμ₀σω + μ₀ε₀ω² = μ₀ε₀ω²(1 + iσ/(ε₀ω)).
k² = μ₀ε₀ω²(1 + σ²/(ε₀ω)²)^½e^iφ. tanφ = σ/(ε₀ω).
k = (μ₀ε₀)^½ω(1 + σ²/(ε₀ω)²)^¼e^iφ/2 = |k|e^iφ/2.
Assume σ/(ε₀ω) >> 1. tanφ --> ∞. φ -- > 90^o. φ/2 --> 45^o, and (1 + σ²/(ε₀ω)²)^¼ --> (σ/(ε₀ω))^½.
k = (μ₀σω)^½e^iπ/4= (μ₀σω)^½(cos(π/4) + i cos(π/4)) = (μ₀σω/2)^½(1 + i) = k₁ + ik₂,
k₁ ≈ k₂ ≈ (μ₀ωσ/2)^½.
∇×E = ik×E = -∂B/∂t.
B(r,t) = B(r)exp(-iωt). Therefore ik×E = iωB. ik(z/z)×E = iωB.
E₀/B₀ = ω/k = ω/|k| e^-iπ/4 = (ω/(μ₀σ))^½ e^-iπ/4.
With a convenient choice of the zero of t we have
Re(E) =|E₀| exp(-k₂z) cos(k₁z - ωt), Re(B) = |B₀| exp(-k₂z) cos(k₁z - ωt + π/4). "B lags behind E" when plotting E and B for fixed z as a function of time.

(c) S = (1/μ₀)[Re(E) × Re(B)] = k |E₀|²(σ/(ωμ₀))^½ exp(-2k₂z) cos(k₁z - ωt) cos(k₁z - ωt + π/4)
= k |E₀|²(σ/(2ωμ₀))^½ exp(-2k₂z) (cos²(k₁z - ωt) - cos(k₁z - ωt) sin(k₁z - ωt)).
<S> = k ½|E₀|²(σ/(2ωμ₀))^½ exp(-(2μ₀σω)^½z).

Problem 3:

Consider a two-level system |Φ_a>, |Φ_b> with <Φ_i|Φ_j> = δ_ij. Show that an "entangled", two-particle state of the form
α|Φ_a(1)>|Φ_b(2)> + β|Φ_b(1)>|Φ_a(2)>, α, β ≠ 0,
CANNOT be written as a product state |ψ_r(1)>|ψ_s(2)> for any one particle states |ψ_r> and |ψ_s>.

Solution:

Concepts:
Tensor product states, entangled states
Reasoning:
Let |Ψ> = α|Φ_a(1)>|Φ_b(2)> + β|Φ_b(1)>|Φ_a(2)> = α|Φ_aΦ_b> + β|Φ_bΦ_a>.
If |Ψ> is a tensor product state, then it can be written as
|Ψ> = |ψ_r(1)>|ψ_s(2)> = (a₁|Φ_a(1)> + a₂|Φ₂(1)>) ⊗ (b₁|Φ_a(2)> + b₂|Φ_b(2)>),
or
|Ψ> = a₁b₁|Φ_aΦ_a> + a₁b₂|Φ_aΦ_b> + a₂b₁|Φ_bΦ_a> + a₂b₂|Φ_bΦ_b>.
Details of the calculation:
For |Ψ> to be a tensor product state we need a₁b₁ = 0, a₁b₂ = α, a₂b₁ = β, a₂b₂ = 0.
a₁b₂ = α --> a₁, b₂ ≠ 0, a₂b₁ = β --> a₂, b₁ ≠ 0.
Therefore a₁b₁ ≠ 0, a₂b₂ ≠ 0.
This is inconsistent with what we need, no solution exists, |Ψ> is not a product state.

Problem 4:

A beam of monochromatic light of wavelength λ in vacuum is incident normally on a nonmagnetic dielectric film of refractive index n and thickness d. Calculate the fraction of the incident energy that is reflected.

Solution:

Concepts:
Maxwell's equations, boundary conditions for the electric and magnetic fields
Reasoning:
The intensities of the reflected and transmitted waves are proportional to the squares of the corresponding electric field amplitudes. We find the ratio of these amplitudes to the incident amplitude by applying boundary conditions.
Details of the calculation:
Let the film lie between the planes z = 0 and z = d.
At the interfaces the tangential components of E and H are continuous.
Let the wave be incident from z = -∞ and the polarization direction be the x-direction.
Let k = 2π/λ, ω = 2πc/λ. Define the positive direction for E as shown.

k_i = k_r = k_t = k, k₁ = k₂ = nk.
For z < 0 we have
E(r,t) = i [E_iexp(i(kz - ωt)) - E_r exp(i(-kz - ωt))],
B(r,t) = j [E_iexp(i(kz - ωt)) + E_rexp(i(-kz - ωt))]/c,
H(r,t) = B(r,t)/μ₀.
In the film for 0 < z < d we have
E(r,t) = i [E₁exp(i(nkz - ωt)) - E₂exp(i(-nkz - ωt))],
B(r,t) = j [E₁exp(i(nkz - ωt)) + E₁exp(i(-nkz - ωt))](n/c), H(r,t) = B(r,t)/μ₀.
For z > d we have
E(r,t) = i E_texp(i(kz - ωt)),
B(r,t) = j E_texp(i(kz - ωt))/c. H(r,t) = B(r,t)/μ₀.
The boundary conditions at z = 0 for the tangential component of E yield
E_i - E_r = E₁ - E₂.The boundary conditions at z = 0 for the tangential component of H yield
H_i + H_r = H₁ + H₂. H = B/μ₀ in all regions, B = En/c in the film, B = E/c outside the film.
E_i + E_r = n(E₁ + E₂).
Measuring all field strength in units of E_i we have express E₁ and E₂ in terns of E_r.
E_r = -E₁ + E₂ + 1, E_r = nE₁ + nE₂ - 1. E₂ = ((n + 1)E_r - n + 1)/(2n), E₁ = (n + 1 - (n - 1)E_r)/2n.
The boundary conditions at z = d for the tangential component of E yield
E₁exp(inkd) - E₂exp(-inkd) = E_texp(ikd).
The boundary conditions at z = d for the tangential component of H yield
H₁exp(inkd) + H₂exp(-inkd) = H_texp(ikd).
nE₁exp(inkd) + nE₂exp(-inkd) = E_texp(ikd).
-(n + 1)E₂exp(-inkd) = (n - 1)E₁exp(inkd).
Substituting E₁ and E₂ in terms of E_r from above we get
-(n + 1)((n + 1)E_r - n + 1)exp(-inkd) = (n - 1)(n + 1 - (n - 1)E_r)exp(inkd).
E_r[(n + 1)²exp(-inkd) - (n - 1)²exp(inkd)] = (n² - 1)exp(-inkd) - (n² - 1)exp(inkd).
E_r[4n cos(nkd) + i2(n² + 1) sin(kd)] = -i(n² - 1)2sin(kd).
R = |E_r|² = 4(n² - 1)²sin²(kd)/[16n² cos²(nkd) + 4(n² + 1)² sin²(kd)].
R = (n² - 1)²/[4n² cot²(nkd) + (n² + 1)²]
is the reflectance, i.e. the fraction of the incident energy that is reflected.