Review

 Time-dependent approximations Addition of angular momentum, EM waves Producing electromagnetic radiation Relativistic E&M Identical particles

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 κe = 3.64 and relative permeability κm = 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/(κeκm)½ = 6.91*107 m/s.
(b)  λ = v/f = 1.06*106 m.
(c)  Bmax = Emax/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+1LJ 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)  Φnx,ny,nz(x,y,z) = (8/(abc))½sin(nxπx/a) sin(nyπy/b) sin(nzπz/c).
Enx,ny,nz = π2ħ2[nx2/(2ma2) + ny2/(2mb2) + nz2/(2mc2)].
(b)  For the original system Eground = E1,1,1 = π2ħ2[1/(2ma2) + 1/(2mb2) + 1/(2mc2)].
For the new system we have
Φ'nx,ny,nz(x,y,z) = (1/(abc))½sin(nxπx/(2a)) sin(nyπy/(2b)) sin(nzπz/(2c)).
E'nx,ny,nz = π2ħ2[nx2/(8ma2) + ny2/(8ma2) + nz2/(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 |<Φ1,1,1'2,2,2>|2.
|<Φ1,1,1'2,2,2>|2 = (8/(abc)2)|∫0asin2(πx/a)dx ∫0bsin2(πy/b)dy ∫0csin2(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 Prad = <(d2p/dt2)2>/(6πε0c3).
• Details of the calculation:
(a)  Dipole moment: p = ∫σ(r)r dA.  Symmetry dictates that p = p k.
p
= k0πσ0cosθ 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.
Prad = <(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)] ∫0T 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/c2E).
We can show that (EB)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|| = Ez k and  E = Ex i + Ey jB|| = Bz k and  B = Bxi + By j
EB = ExBx + EyBy + EzBz.
E'∙B' = E'||B'|| + E'B'E'||B'|| = E'zB'z = EzBz.
E' = E'x i + E'y j.  B' = B'x i + B'y j.
E'x= γ(Ex - vBy),  E'y= γ(Ey + vBx),  B'x= γ(Bx + (v/c2)Ey),  B'y= γ(By - (v/c2)Ex).
E'xB'x = γ2(ExBx + (v/c2)ExEy - vByBx - (v2/c2)ByEy).
E'yB'y = γ2(EyBy - (v/c2)EyEx + vBxBy - (v2/c2)BxEx).
E'xB'x + E'yB'y = γ2(1 - v2/c2)(ExBx + EyBy) = ExBx + EyBy.
(EB)2 is invariant under a Lorentz transformation.
If EB is zero in one frame, it is zero in every frame.

(b)  E2 - c2B2 = ExEx + EyEy + EzEz - c2(BxBx + ByBy + BzBz).
E'zE'z - c2B'zB'z = EzEz - c2BzBz.
E'xE'x = γ2(Ex2 - 2vExBy + v2By2).
E'yE'y = γ2(Ey2 + 2vEyBx + v2Bx2).
B'xB'x = γ2(Bx2 + 2(v/c2)EyBx + (v2/c4)Ey2).
B'yB'y = γ2(By2 - 2(v/c2)ExBy + (v2/c4)Ex2).
E'xE'x + Ey'E'y - c2(B'xB'x + B'yB'y)
= γ2(Ex2 + v2By2 + Ey2 + v2Bx2 - c2Bx2 - (v2/c2)Ey2 - c2By2 - (v2/c2)Ex2)
= γ2(1 - v2/c2)(Ex2 + Ey2) - γ2(c2 - v2)(Bx2 + By2) = ExEx + EyEy - c2(BxBx + ByBy).
E2 - c2B2 is invariant under a Lorentz transformation.
If |E| > c|B| in one inertial frame then |E| > c|B| in every inertial frame.