Review

Problem 1:

A hydrogen atom is known to be in a state characterized the quantum numbers n = 4, l = 2.
(a)    Give the allowed values of j.
(b)    For each of the allowed values of j, calculate the square of the magnitude of the total angular momentum.

Solution:

• Concepts:
  Addition of angular momentum
• Reasoning:
  We are supposed to add the orbital and spin angular momentum of the electron in the hydrogen atom.
• Details of the calculation:
  (a)  l = 2, s = ½, j = 5/2, 3/2.
  (b)  J²|jm> = j(j + 1)ħ²|jm>.
  For j = 5/2, j(j + 1)ħ² = (35/4)ħ².
  For j = 3/2, j(j + 1)ħ² = (15/4)ħ².

Problem 2:

An electromagnetic wave with frequency 65.0 Hz travels in an insulating magnetic material that has dielectric constant κ_e = 3.64 and relative permeability κ_m = 5.18 at this frequency.  The electric field has amplitude 7.20*10⁻³ V/m.
(a)  What is the speed of propagation of the wave?
(b)  What is the wavelength of the wave?
(c)  What is the amplitude of the magnetic field?

Solution:

• Concepts:
  EM waves
• Reasoning:
  We are supposed to recall some basic properties of EM waves.
• Details of the calculation:
  (a)  v = c/(κ_eκ_m)^½ = 6.91*10⁷ m/s.
  (b)  λ = v/f = 1.06*10⁶ m.
  (c)  B_max = E_max/v = 1.04*10^-10 T.

Problem 3:

Consider an excited Nitrogen atom with a an electronic configuration (1s)², (2s)², (2p)², (3s).
Find the spectroscopic terms ^2S+1L_J that characterize the system.
Solution:

• Concepts:
  The Pauli exclusion principle, energy levels of multi-electron atoms, addition of angular momentum
• Reasoning:
  The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers.  We use it to find the allowed terms ^2S+1L.
• Details of the calculation:
  For identical fermions the total wave function must be antisymmetric under exchange of any two particles.
  Let (o) denote antisymmetric or odd, (e) denote symmetric or even.
  First add the angular momentum of the two 2p electrons.

  (3p)2		Possible terms
  L12	S12
  0 (e)	0 (o)	1S
  1 (o)	1 (e)	3P
  2 (e)	0 (o)	1D

  Now add to this the angular momentum of the 3s electron.

  		(3p)2(3s)				Possible terms
  L12	S12	l	s	L, (L = L12 + l)	S, (S = S12 + s)
  0	0	0	½	0	½	2S
  1	1	0	½	1	½, 3/2	2P, 4P
  2	0	0	½	2	½	2D

  Spectroscopic terms:

  2S+1LJ

  2S1/2

  2P3/2, 1/2

  4P5/2, 3/2, 1/2

  2D5/2, 3/2

Problem 4: 

A particle with mass m is confined to a 3D box.
V(x, y, z) = 0 if 0 ≤ x ≤ a AND 0 ≤ y ≤ b AND  0 ≤ y ≤ c,
V(x, y, z) = ∞ otherwise.
(a)  Write down expressions for the energy eigenfunctions and eigenvalues for this particle.
(b)  Assume that the particle is in the ground state of the system.  At time t = 0, all dimensions of the box suddenly double, i.e. a --> 2a, b --> 2b, c --> 2c.  A measurement of the energy of the particle is made just after the dimensions increase. (Assume that all this happens so quickly so the spatial wave function of the particle does not change).  What is the probability that the energy measurement yield a value EXACTLY the same as the energy of the ground state of original system?

Solution:

• Concepts:
  The particle in a 3-dimensional box, separation of variables
• Reasoning:
  We are asked to find the eigenfunctions and eigenvalues of the Hamiltonian of a particle in a 3D box.
• Details of the calculation:
  (a)  Φ_nx,ny,nz(x,y,z) = (8/(abc))^½sin(n_xπx/a) sin(n_yπy/b) sin(n_zπz/c).
  E_nx,ny,nz = π²ħ²[n_x²/(2ma²) + n_y²/(2mb²) + n_z²/(2mc²)].
  (b)  For the original system E_ground = E_1,1,1 = π²ħ²[1/(2ma²) + 1/(2mb²) + 1/(2mc²)].
  For the new system we have
  Φ^'_nx,ny,nz(x,y,z) = (1/(abc))^½sin(n_xπx/(2a)) sin(n_yπy/(2b)) sin(n_zπz/(2c)).
  E^'_nx,ny,nz = π²ħ²[n_x²/(8ma²) + n_y²/(8ma²) + n_z²/(8ma²)].
  An energy measurement yield a value EXACTLY the same as the energy of the ground state of original system if the system is in the found in the state Φ^'_2,2,2(x,y,z).
  The probability of finding the system in that state is |<Φ_1,1,1|Φ^'_2,2,2>|².
  |<Φ_1,1,1|Φ^'_2,2,2>|² = (8/(abc)²)|∫₀^asin²(πx/a)dx ∫₀^bsin²(πy/b)dy ∫₀^csin²(πz/a)dz|²
  = 1/8.
  The probability that the energy measurement yield a value EXACTLY the same as the energy of the ground state of original system is 1/8 = 12.5%.

Problem 5:

A non-conducting sphere of radius R = 0.1 m, mass M = 10 kg and uniform mass density carries a surface charge density σ = σ₀cosθ, with σ₀ = 10 microCoulomb.
(a)  Find the dipole moment p₀ of the sphere.
(b)  Assume that at t = 0 the sphere receives an angular impulse and starts rotating about the x-axis with angular velocity ω₀ = 1000/s.  Calculate the power radiated by the sphere when it rotates with angular frequency ω.
(c)  Estimate the time it takes for the rotation rate of the sphere to decrease by a factor of 2.

Solution:

• Concepts:
  Dipole moment, the Larmor formula
• Reasoning:
  The y and z components of the dipole moment are oscillating with angular frequency ω.  The rotating electric dipole will radiate energy at a rate P_rad = <(d²p/dt²)²>/(6πε₀c³).
• Details of the calculation:
  (a)  Dipole moment: p = ∫σ(r)r dA.  Symmetry dictates that p = p k.
  p = k ∫₀^πσ₀cosθ Rcosθ 2πR²sinθdθ = k σ₀(4π/3)R³ = k p₀.
  p₀ = 4.19*10^-8 Cm.
  
  (b)  p(t) = p₀cos(ωt) k + p₀sin(ωt) j.
  d²p/dt² = -ω²p,  <(d²p/dt²)²> = ω⁴p₀².
  P_rad = <(d²p/dt²)²>/(6πε₀c³) = ω⁴p₀²/(6πε₀c³).
  <P> = ω⁴*3.9*10^-31 W s⁴.
  For ω = ω₀ = 1000/s we have <P> = 3.9*10^-19 W.
  
  (c)  <P> = -dE/dt.  E = ½Iω² = (1/5)MR²ω².
  -dE/dt = -(2/5)MR²ω dω/dt  = ω⁴p₀²/(6πε₀c³).
  -5p₀²dt/(12 MR²πε₀c³) = dω/ω³.
   [-5p₀²/(12 MR²πε₀c³)] ∫₀^T dt = ∫_ω0 ^ω0/2 dω/ω³.
  [5p₀²/(12 MR²πε₀c³)] T = ½((ω₀/2)^-2 - ω₀^-2).
  T ≈ 10²³ s.

Problem 6:

Starting with the transformation of the electromagnetic fields under a Lorentz transformation show that
(a)  if E is normal to B in an inertial frame, it must be true in all other inertial frames, and
(b)  if |E| > c|B| in an inertial frame, it must be true in all other inertial frames.

Solution:

• Concepts:
  Vector products
• Reasoning:
  Lorentz transformation of the electromagnetic fields:
  E'_|| = E_||,  B'_|| = B_||,
  E'_⊥ = γ(E + v×B)_⊥,
  B'_⊥ = γ(B - (v/c²)×E)_⊥.
  We can show that (E∙B)² and E² - c²B² are invariant under a Lorentz transformation.
• Details of the calculation:
  (a)  We can choose a Cartesian coordinate system and evaluate the products using Cartesian coordinate.
  Consider to reference frames, K, and K'.  Let the direction of the relative velocity of K' with respect to K be the z-direction, v = vk.
  Then E_|| = E_z k and  E_⊥ = E_x i + E_y j,  B_|| = B_z k and  B_⊥ = Bxi + B_y j, 
  E∙B = E_xB_x + E_yB_y + E_zB_z.
  E'∙B' = E'_||∙B'_|| + E'_⊥∙B'_⊥.  E'_||∙B'_|| = E'_zB'_z = E_zB_z.
  E'_⊥ = E'_x i + E'_y j.  B'_⊥ = B'_x i + B'_y j.
  E'_x= γ(E_x - vB_y),  E'_y= γ(E_y + vB_x),  B'_x= γ(B_x + (v/c²)E_y),  B'_y= γ(B_y - (v/c²)E_x).
  E'_xB'_x = γ²(E_xB_x + (v/c²)E_xE_y - vB_yB_x - (v²/c²)B_yE_y).
  E'_yB'_y = γ²(E_yB_y - (v/c²)E_yE_x + vB_xB_y - (v²/c²)B_xE_x).
  E'_xB'_x + E'_yB'_y = γ²(1 - v²/c²)(E_xB_x + E_yB_y) = E_xB_x + E_yB_y.
  (E∙B)² is invariant under a Lorentz transformation.
  If E∙B is zero in one frame, it is zero in every frame.
  
  (b)  E² - c²B² = E_xE_x + E_yE_y + E_zE_z - c²(B_xB_x + B_yB_y + B_zB_z).
  E'_zE'_z - c²B'_zB'_z = E_zE_z - c²B_zB_z.
  E'_xE'_x = γ²(E_x² - 2vE_xB_y + v²B_y²).
  E'_yE'_y = γ²(E_y² + 2vE_yB_x + v²B_x²).
  B'_xB'_x = γ²(B_x² + 2(v/c²)E_yB_x + (v²/c⁴)E_y²).
  B'_yB'_y = γ²(B_y² - 2(v/c²)E_xB_y + (v²/c⁴)E_x²).
  E'_xE'_x + E_y'E'_y - c²(B'_xB'_x + B'_yB'_y)
  = γ²(E_x² + v²B_y² + E_y² + v²B_x² - c²B_x² - (v²/c²)E_y² - c²B_y² - (v²/c²)E_x²)
  = γ²(1 - v²/c²)(E_x² + E_y²) - γ²(c² - v²)(B_x² + B_y²) = E_xE_x + E_yE_y - c²(B_xB_x + B_yB_y).
  E² - c²B² is invariant under a Lorentz transformation.
  If |E| > c|B| in one inertial frame then |E| > c|B| in every inertial frame.