Problem 1:

An electromagnetic wave with circular frequency ω propagates in a medium of dielectric constant ε, magnetic permeability μ, and conductivity σ.
(a)  Show that there is a plane wave solution in which the amplitude of the E and B fields decreases exponentially along the direction of propagation, and find the characteristic decay length.
(b)  Simplify by assuming that σ is great enough so that σ/(εω) >> 1.

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/ε,  ∇×E = -∂B/∂t,  ∇∙B = 0,  ∇×B = μj + με∂E/∂t.
  Assume ρ_f = 0 and j_f = σE in the conductor.  Then
  ∇×B = μσE + με∂E/∂t.
  ∇×(∇×B) = ∇(∇∙B) - ∇²B = μσ(∇×E) + με∂(∇×E)/∂t.
  ∇²B - μσ∂B/∂t - με∂²B/∂t² = 0.
  Similarly:  ∇²E - μσ∂E/∂t - με∂²E/∂t² = 0.
  Both E and B satisfy the damped wave equation.
  
  Try solutions of the form E(r,t) = E₀ exp(i(k∙r - ωt)).
  Then k² = iμσω + μεω² = μεω²(1 + iσ/(εω)).
  k² = μεω²(1 + σ²/(εω)²)^½ e^iφ,  tanφ = σ/(εω).
  k = (με)^½ω(1 + σ_c²/(εω)²)^¼ e^iφ/2 = |k|e^iφ/2.
  k is a complex number, k = β + iα/2.
  To find the real and imaginary parts we have to find cos(φ/2) and sin(φ/2) in terms of tanφ.
  cos(φ/2) = ((1 + cosφ)/2)^½,  sin(φ/2) = ((1 - cosφ)/2)^½,  cosφ = (1 + tan²φ)^-½ in the first quadrant.
  Therefore cos(φ/2) = 2^-½(1 + (1 + (σ/(εω))²)^-½)^½, 
  sin(φ/2) = 2^-½(1 - (1 + (σ/(εω))²)^-½)^½.
  k = (με/2)^½ω[(1 + (σ/(εω))²)^½ + 1]^½ + i(με/2)^½ω[(1 + (σ/(εω))²)^-½ - 1]^½.
  β = (με/2)^½ω[(1 + (σ/(εω))²)^½ + 1]^½,  α/2 = (με/2)^½ω[(1 + (σ/(εω))²)^½ - 1]^½.
  The skin depth δ is the distance it takes to reduce the amplitude by a factor of 1/e.
  δ = 2/α. 
  (b)  Let σ/(εω) >> 1.
  Then α/2 ~ (με/2)^½ω(σ/(εω))^½ = (μσω/2)^½,
  β ~ α/2 ~ (μσω/2)^½ = 1/δ.

Problem 2:

The HeNe lasers in our undergraduate laboratories produce unpolarized light. 
(a)  Describe how you can produce a beam of right-hand or left-hand polarized light from such a HeNe laser.  How much of the original light intensity do you loose?
(b)  Explain how you can change the polarization axis of a linearly polarized laser beam without loosing intensity.

Solution:

• Concepts:
  Polarization
• Reasoning:
  Choose a coordinate system.  Let E_x = E_0xexp(i(kz - ωt)), E_y = E_0yexp(i(kz - ωt + φ)) be the electric field of the laser beam.
  Linearly polarized light:  φ = nπ, n = 0, 1, 2, ...
  Circularly polarized light:  E_0x = E_0y = E₀  φ = nπ/2, n = 1, 3, 5, ... .
  RHC:  φ = -π/2,  E = i E₀cos(kz - ωt) + j E₀sin(kz - ωt).
  LHC:  φ = π/2,  E = i E₀cos(kz - ωt) - j E₀sin(kz - ωt).
  Note:  This definition is not unique.
  Wave plates:  A thin plate of thickness d of a birefringent crystal cut parallel to the optic axis is known as a wave plate.  Assume linearly polarized light is entering the wave plate normally.  The components of E parallel and perpendicular to optic axis emerge with a phase difference δ between them given by δ = (2πd/λ)Δn.
• Details of the calculation:
  (a)  First produce linearly polarized light.  Use, for example, Polaroid film.  Let the direction of the transmission axis be (i + j)/√2.
  Half of the incident intensity is lost, the electric field of the laser beam after the Polaroid is
  E = i E₀exp(i(kz - ωt)) + j E₀exp(i(kz - ωt)). 
  (Note:  Phase shifts common to the x- and y-component are ignored. E₀² = (E_unpolarized)²/4.)
  Now use a quarter-wave plate with its fast axis oriented along the x axis.  Define Δn = n_slow- n_fast.  The electric field at the exit of the plate is
  E = i E₀exp(i((2π/λ)n_fastd - ωt)) + j E₀exp(i(((2π/λ)n_slowd - ωt)
  =  i E₀exp(i(k_fastd - ωt)) + j E₀exp(i(k_fastd - ωt + δ), with δ = (2πd/λ)Δn = π/2.
  Note d = λ/(4Δn).
  The emerging wave is LHC polarized.
  Orienting the fast axis along the y-axis produces RHC polarized light.
  An ideal quarter-wave plate does not reduce the intensity of the light, so overall, half of the intensity is lost.

  (b) A half-wave plate δ = π can be used to rotate the plane of linearly polarized light.  The angle of rotation is 2θ, where θ is the angle between the angle of polarization and the wave plate's fast axis.
  Let d = λ/(2Δn).
  Before the plate:  E = i E'cosθ + j E'sinθ.
  After the plate:  E = i E₀exp(i((2π/λ)n_fastd - ωt)) + j E₀exp(i(((2π/λ)n_slowd - ωt)
  =  i E₀exp(i(k_fastd - ωt)) + j E₀exp(i(k_fastd - ωt + δ), with δ = (2πd/λ)Δn = π.
  E = i E'cosθ - j E'sinθ.
  No intensity is lost.
  (An optically active material can also be used.)

Problem 3:

Consider a composite system made of two non-identical spin ½ particles.  For t < 0 the Hamiltonian does not depend on time and can be taken to be zero.  For t > 0 the Hamiltonian is given by  H = (4∆/ħ²)S₁∙S₂, where ∆ is a constant.  Suppose that the system is in the state |+-> for t ≤ 0.  Find, as a function of time, the probability for being in each of the states |++>, |+->, |-+>, and |-->,
(a)  by solving the problem exactly, using |Ψ(t)> = U(t, t₀)Ψ(t₀)> and
(b)  by solving the problem assuming the validity of first-order time-dependent perturbation theory with H as a perturbation which is switched on a t = 0.
(b)  Under what conditions does the perturbation calculation disagree with the exact solution and why?

Solution:

• Concepts:
  The state space of two spin ½ particles, time-dependent perturbation theory
• Reasoning:
  S₁∙S₂ = ½(S² - S₁² - S₂²).  The eigenstates of S₁∙S₂ are the singlet state and the triplet states, {|S, M_s>}, S = 0, 1.  These are common eigenfunctions of S² and S_z.
• Details of the calculation:
  (a)  The common eigenfunctions of S² and S_z are
  |1,1> = |++>,
  |1,0> = ½^½(|+-> + |-+>),
  |1,-1> =|-->,
  |0, 0> = ½^½(|+-> - |-+>).
  Therefore |+-> = ½^½(|0,0> + |1,0>), |-+> = ½^½(|0,0> - |1,0>).
  For the singlet state we have S₁∙S₂|0,0> = -(3/4)ħ²,  H|0,0> = -3∆|0,0>.
  For the triplet states we have S₁∙S₂|1,M_s> = (1/4)ħ²,  H|1,M_s> = ∆|1,M_s>.
  |ψ(0)> = |+-> = ½^½(|0,0> + |1,0>),
  |ψ(t)> = U(t, 0)|+-> = 2^-½(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>).
  P(t) of being in the state |++> = |1,1> = 0.
  P(t) of being in the state |--> = |1.-1> = 0.
  P(t) of being in the state |+-> = |<+-| ψ(t)>|².
  <+-|ψ(t)> = ½(<0,0| + <1,0|)(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>)
  = ½(exp(i3∆*t/ ħ) + exp(-i∆*t/ħ))
  = ½exp(i∆*t/ ħ) (exp(i2∆*t/ħ) + exp(-i2∆*t/ħ)) = exp(i∆*t/ħ) cos(2∆*t/ħ).
  P(t) of being in the state |+-> = cos²(2∆*t/ħ).
  For 2∆*t/ħ << 1, P(t) ~ 1.
  P(t) of being in the state |-+> = sin²(2∆*t/ħ), since the total probability of being in one of the 4 basis states must be equal to 1.
  For 2∆*t/ħ << 1, P(t) ~ (2∆*t/ħ)².
  
  (b)  Time-dependent perturbation theory
  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|².
  The initial and the final state must be different for this formula to be valid.  For i = f, P(t) ~ 1.
  W_fi = <f|H|i>.  Here W_fi = <f|H|+->. 
  W_fi = 0 for |f> = |++> and |f> = |-->, so P_if(t) = 0 for these states.
  For |f> = |-+> we have W_fi = 2^-½<-+|H|(|00> + |10>)
  = 2^-½[-3∆<-+|0,0> + ∆<-+|1,0>] = ½(3∆ + ∆) = 2∆.
  ω_fi = (E_f - E_i)/ω_fi = 0 for the unperturbed states.
  P_if(t) = (1/ħ²)|∫₀^t(2∆)dt'|² = 4∆²t²/ħ².
  
  (c)  The time-dependent perturbation theory calculation result agrees with the exact result as long as x = 2∆*t/ħ << 1, so that cos(x) = 1 and sin(x) = x are very good approximations.  For large x (or t) the probability of finding the particle in the state |-+> oscillates between 0 and 1.

Problem 4:

Submit this problem on Canvas as Assignment 8.  If you used an AI as a Socratic tutor, submit a copy of your session leading to your solution and reflect on your session.  If you did not need any help or worked with another student, explain your reasoning, do not just write down formulas. 

A pair of magnetic ions with individual spins s₁ and s₂ interact through the scaled Hamiltonian H = Cs₁∙s₂.  Let s(s + 1) be the eigenvalue of s², where s = s₁ + s₂ is the total angular momentum.  Note ħ = 1.
(a)  Show that the total angular momentum is a conserved quantity.
(b)  Find the ground state energy E₀ and the energy of the highest state E_max.
(c)  Now suppose a magnetic field is applied and the new Hamiltonian is H', where
H' = H - b(s_1z + s_2z).  What is the residual symmetry of the new Hamiltonian H', and what are the associated "good quantum numbers"?

Solution:

• Concepts:
  Addition of angular momentum, common eigenbasis for J₁², J₂², J², J_z.
• Reasoning:
  Consider two angular momentum operators, J₁² and J₂², operating in two different vector spaces E₁ and E₂.
  Let J = J₁ + J₂.  The operators  J₁², J₂², J², and J_z all commute and a common eigenbasis {|j₁,j₂;j,m>} exists.
  Here we have s = s₁ + s₂ and {|s₁,s₂;s,m_s>} is a common eigenbasis of s₁², s₂², s², and s_z.
  s₁∙s₂ = ½(s² - s₁² - s₂²),  s_1z + s_2z = s_z,  the vectors {|s₁,s₂;s,m_s>} are also eigenvectors of s₁∙s₂ and  s_1z + s_2z.
• Details of the calculation:
  (a)  s₁², s₂², s², and s_z are a set of commuting operators.  H = Cs₁∙s₂ commutes with s² and s_z.
  [Similarly {s₁², s₂², s², s_x} and {s₁², s₂², s², s_y} are sets of commuting operators and H commutes with s_x and s_y.]
  Therefore [H, s] = 0, H commutes with all components of s, the total angular momentum is a conserved quantity.
  (b)  With ħ = 1, H|s₁,s₂;s,m_s> = ½C[s(s+1) - s₁(s₁+1) - s₂(s₂+1)]|s₁,s₂;s,m_s>.
  s_max = s₁ + s₂, s_min = |s₁ - s₂|.
  E₀ = ½C[s_min(s_min+1) - s₁(s₁+1) - s₂(s₂+1)],
  E_max = ½C[s_max(s_max+1) - s₁(s₁+1) - s₂(s₂+1)].
  (c)  H' commutes with s², and s_z = s_1z + s_2z, but not with s_x and s_y;  s and m_s remain good quantum numbers.