Review

Problem 1:

In the circuit shown the voltage varies sinusoidally, with V₀  = 9 V and ω = 440/s.
(a)  Calculate the peak value of the current I.
(b)  Calculate the phase of I relative to the applied voltage and state whether I(t) leads or trails V(t).

Solution:

• Concepts:
  AC circuits
• Reasoning:
  We are asked to analyze an AC circuit.
• Details of the calculation:
  (a)  V = IZ.  Let V = V₀exp(i(ωt + φ_V)).
  Z = R + iωL + 1/(iωC) = [40 + i*(440*0.5) - i/(440*10^-5)] Ω
  = [40 - i7.27] Ω = 40.66 exp(iφ) Ω with tanφ = -0.18,  φ = -0.178.
  I = V/Z = (9 V)exp(i(ωt + φ_V))/(40.66 exp(-i0.18)) Ω
  = 0.22 A exp(i(ωt + φ_V + 0.178)) = I₀exp(i(ωt + φ_I))
  The peak value of the current is 0.22 A.
  (b)  φ_I - φ_V = 0.178.  I(t) leads V(t).

Problem 2:

The operators a and a^† when acting on the energy eigenstates of the harmonic oscillator, denoted by |n>, have the property
a^†|n> = (n + 1)^½|n + 1>,  a|n> = n^½|n - 1>.
We have x = (a + a^†)/(2α),  p = -i(a - a^†)/(2β), where α =√(mω/(2ħ)),  β =1/√(2mωħ).
Find the mean value and root mean square deviation of p², when the oscillator is in the energy eigenstate |n>.

Solution:

• Concepts:
  Fundamental assumptions of QM, the eigenstates of the harmonic oscillator Hamiltonian, the raising and lowering operators
• Reasoning:
  For any operator A we have ∆A = (<(A - <A>)²>)^½ = (<A²> - <A>²)^½.
• Details of the calculation:
  p² = -(mħω/2)(a^†a^† - a^†a - aa^† + aa).
  <p²> = -(mħω/2)<n|(a^†a^† - a^†a - aa^†  + aa)|n> = (mħω/2)(2n + 1).
  The mean value is <p²> = (mħω/2)(2n + 1).
  
  p⁴ = (mħω/2)²(a^†a^† - a^†a - aa^† + aa) (a^†a^† - a^†a - aa^† + aa).
  <p⁴> = (mħω/2)²<n|(a^†a^† aa + a^†a a^†a + a^†a aa^† + aa^† a^†a + aa^† aa^†  + aaa^†a^†)|n>
  = (mħω/2)²(6n² + 6n + 3).
  <p⁴> - <p²>² = (mħω/2)²(2n² + 2n + 2).
  ∆p² = (mħω/2)( 2n² + 2n + 2)^½ is root mean square deviation of p².

Problem 3:

A capacitor composed of two parallel infinite conducting sheets separated by a distance d is connected to a battery.  The lower plate is maintained at some potential V₁ and the upper plate is maintained at some potential V.  A small hemispherical boss of radius a << d is introduced on the lower plate.  State the boundary conditions for this problem.  (Hint: Consider the limit as the distance between the plates becomes very large.)  Find the potential between the plates and the surface charge density on the plates.

Solution:

• Concepts:
  Boundary value problems, azimuthal symmetry.
• Reasoning:
  Put the origin of the coordinate system at the center of a sphere, the upper half of which is the hemispherical boss. Let the z-axis point up.
  Then Φ(r,θ) = ∑_n=0^∞[A_nrⁿ + B_n/rⁿ⁺¹]P_n(cosθ) is the most general solution in the region between the plates, since the charge density ρ = 0 there.  To find the specific solution we apply boundary conditions.
• Details of the calculation:
  The boundary conditions are Φ = V₁ on the lower plate and the boss and Φ = V on the upper plate.
  Φ = V₁ + A₁'rcosθ  + (B₁'/r²)cosθ  between the plates outside the boss.
  We assume all other coefficients are zero.  If we find a solution with this assumption, it is the only solution.
  Boundary conditions:
  (i)  Φ is continuous at r = a.  0 = A₁'a + B₁'/a².  B₁' = -A₁'a³.
  (ii)  E = -∇Φ = E₀ at r >> a,  i.e. far away from the hemispherical boss.
  -A₁'cosθ = E₀cosθ,  A₁'sinθ = -E₀sinθ,  A₁' = -E₀,  B₁' = E₀a³.  E₀ = (V - V₁)/d.
  Therefore
  Φ = V₁ - E₀rcosθ  + (E₀a³/r²)cosθ.
  E_r = E₀cosθ  + 2(E₀a³/r³)cosθ,  E_θ = -E₀sinθ + (E₀a³/r³)sinθ.
  E = E₀ + 2(E₀a³/r³)cosθ (r/r) + (E₀a³/r³)sinθ (θ/θ).
  E = external field + field due to polarized sphere.
  
  Surface charge on the plates: σ = ε₀E·n.
  E·n = -E₀ on the upper plate, σ = -ε₀E₀.
  E·n = -E_θ = E₀ - (E₀a³/r³) on the flat part of the lower plate, σ = ε₀E₀ - ε₀(E₀a³/r³).
  E·n = E_r = 3E₀cosθ on the boss, σ = ε₀3E₀cosθ.

Problem 4:

Through some coincidence, the Balmer lines from singly ionized helium in a distant star happen to overlap with the Balmer lines from hydrogen in the sun.  Assuming that the relative motion is along the line of sight, how fast is that star receding from us?

Solution:

• Concepts:
  Doppler shift, hydrogenic atoms
• Reasoning:
  f' = f [(c - v)/(c + v)]^½.
• Details of the calculation:
  Balmer lines from H:               hf_H = 13.6 eV(1/4 - 1/n_i²).
  Balmer lines from He+:           hf_He+ = 4*13.6 eV(1/4 - 1/n_i²).
  f_H = f_He+ [(c - v)/(c + v)]^½.
  1/4 = [(c - v)/(c + v)]^½.
  v = c(1 - 1/16)/(1 + 1/16) = 2.65*10⁸ m/s.

Problem 5:

Consider a quantum description of a non-relativistic 2D electron gas confined to the x−y plane of a Cartesian coordinate system, with a magnetic field B pointed in the z direction.  The effect of electronic coupling to the field may be included by modifying the momentum operator p = (p_x, p_y) in the Hamiltonian to include a term depending on the vector potential A associated with the magnetic field:

H = (1/(2m)) ((ħ/i)∇ - q_eA(r,t))²,

where q_e is the electron charge.   Show that the different (gauge) choices for the vector potential
A = (0, Bx, 0) (Landau gauge),
and
A = ½B(-y, x, 0) (symmetric gauge),
both give the required magnetic field.

Solution:

• Concepts:
  ∇∙B = 0 --> B = ∇×A.
• Reasoning:
  A is not unique.  A' = A + ∇ψ + C, with ψ an arbitrary scalar field and C an arbitrary constant vector is also a vector potential for the same field.
• Details of the calculation:
  ∇×A = (∂A_z/∂y - ∂A_y/∂z,  ∂A_x/∂z  - ∂A_z/∂x,  ∂A_y/∂x - ∂A_x/∂y).
  ∇×A₁ = (∂A_y/∂x)k = Bk = B.
  ∇×A₂ = ∂A_y/∂x - ∂A_x/∂y) k +  (½B + ½B) k  = Bk = B.
  Both vector potentials are associated with the same magnetic field B.

Problem 6:

Centered at the origin of coordinates is an insulated, conducting sphere of radius a.  A positive point charge q is located a distance d (d > a) from the center.  What is the smallest positive charge, which must be on the sphere in order that the surface charge density will nowhere be negative?

Solution:

• Concepts:
  The method of images
• Reasoning:
  The method of images is the preferred method for finding the potential of a charge distribution outside a conducting sphere.
• Details of the calculation:
  Assume that the point charge q is located on the z axis at z = d.  Place an image charge q' = -aq/d on the z-axis at z' = a²/d.  This will keep the sphere at zero potential.  Assume the sphere must hold a total charge Q = q' + q'' in order that the surface charge density will nowhere be negative.  Place q'' at the center of the sphere.
  σ = ε₀E_r is the surface charge density.
  E_r is the radial component of the electric field on the surface.
  E_r(ar/r) = [1/(4πε₀)][q(ar/r - dk)∙(r/r)/|ar/r - dk|³
  - (aq/d)(ar/r - (a²/d)k)∙(r/r)/|ar/r - (a²/d)k|³ + (q''/a²)]
  = [1/(4πε₀)][q(a - dcosθ)/(a² + d² - 2adcosθ)^3/2
  - (aq/d)(a - a²cosθ/d)/(a² + a²/d² - 2a³cosθ/d)^3/2 + (q''/a²)]
  
  = [1/(4πε₀)][q(a - dcosθ - d²/a + dcosθ)/(a² + d² - 2dacosθ)^3/2 + (q''/a²)]
  = [q/(4πε₀a²)][(1 - d²/a²)/(1 + d²/a² - 2(d/a)cosθ)^3/2 + (q''/q)].
  
  σ = ε₀E_r = [q/(4πa²)][(1 - d²/a²)/(1 + d²/a² - 2(d/a)cosθ)^3/2 + (q''/q)].
  
  (1 - d²/a²)/(1 + d²/a² - 2(d/a)cosθ)^3/2 is a negative number.  Its magnitude is largest when the denominator is smallest.
  σ_min = ε₀E_r = [q/(4πa²)][(1 - d²/a²)/(1 + d²/a² - 2(d/a))^3/2 + (q''/q)].
  We want σ_min = 0.
  q'' = q(d²/a² - 1)/(d/a - 1)³ = q(d/a + 1)/(d/a - 1)².
  The smallest positive charge, which must be on the sphere in order that the surface charge density will nowhere be negative is
  Q = q'' + q' = q'' - (aq/d) = q[(d/a + 1)/(d/a - 1)² - a/d].
  
  A simpler way of finding q'':
  σ = ε₀E_r is the surface charge density.  It will have its lowest value at z = a, because q' is closest to that surface point.
  E_r(z = a) = E(z = a) = [1/(4πε₀)][-q/(d - a)² - (aq/d)/(a - a²/d)² + q''/a²].
  σ(z = a) = [q/(4πa²)][-1/(d/a - 1)² - (d/a) /(d/a - 1)² + q''/a²]
  = [q/(4πa²)][-(d/a + 1)/(d/a - 1)² + q''].
  σ(z = a) is negative unless q'' ≥ (d/a + 1)/(d/a - 1)².
  For σ_min = 0 we need q'' = (d/a + 1)/(d/a - 1)².