## Assignment 6, solutions

#### Problem 1:

Consider a 2 by 2 matrix M with matrix elements m11, m12, m21, and m22.
(a)  What relationships between these matrix elements must exist for the matrix M to be a unitary matrix, M = U?
(b)  Show that the determinant of M = U has magnitude 1.

Solution:

• Concepts:
Mathematical foundations of quantum mechanics
• Reasoning:
A matrix U is a unitary matrix if UU = UU = I.
• Details of the calculation:
(a)  Let M = U be a Unitary matrix.  Then UU = UU = I.

UU = I --> |m11|2 + |m12|2 = 1,  |m21|2 + |m22|2 = 1,  m11 m21* + m12 m22* = 0.
UU = I --> |m11|2 + |m21|2 = 1,  |m12|2 +  |m22|2 = 1,  m22 m21* + m12 m11* = 0.
|m11|2 + |m12|2 = |m11|2 + |m21|2 = 1 implies |m21| = |m21|.
Since |m11|2 + |m12|2 = 1 there exists an angle θ such that |m11|= cosθ,  |m12| = |m21| = sinθ.
|m11|2 + |m12|2 = |m21|2 + |m22|2 = 1 implies |m11| = | m22|,  |m22|= cosθ.
Let mii = |mii|exp(iφii).   Then m11 m21* + m12 m22* = 0 and m22 m21* + m12 m11* = 0.
implies φ11 - φ21 = φ12 - φ22 ± π and φ22 - φ21 = φ12 - φ11 ± π,
φ11 = -φ22 + φ21 + φ12 ± π.
If we choose the φ11 = -φ22, by factoring out a common factor exp(iΔ) from all matrix elements then
φ21 = -φ12 ± π.
Therefore, except for some common factor exp(iΔ), we have
m11 = cosθ exp(iφ1),  m12 = sinθ exp(iφ2),
m21 = -sinθ exp(-iφ2),  m22 = cosθ exp(-iφ1).
(b)  det(U) = exp(iΔ)(cos2θ + sin2θ) = exp(iΔ).
|det(U)| = 1.

#### Problem 2:

An operator A has two normalized eigenstates ψ1 and ψ2, with eigenvalues a1 and a2, respectively.  An operator B, has two normalized eigenstates, φ1 and φ2, with eigenvalues b1 and b2, respectively.   The eigenstates are related by
ψ1 = (3φ1 + 4φ2)/5,    ψ2 = (4φ1 - 3φ2)/5.
(a)  Observable A is measured, and the value a1 is obtained.  What is the state of the system immediately after this measurement?
(b)  If B is measured immediately afterwards, what are the possible results, and what are their probabilities?
(c)  If the result of the measurement of B is not recorded and right after the measurement of B, A is measured again, what is the probability of getting a1?

Solution:

• Concepts:
The postulates of Quantum Mechanics
• Reasoning:
When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue an is given by
P(an) = Σi=0gn|<uni|ψ>|2,  where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an.
If a measurement on a system in the state |ψ> gives the result an, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with an.
• Details of the calculation:
(a)  After measuring a1, the system is in the eigenstate ψ1.
(b)  If B is now measured, the probability of obtaining b1 is 9/25 and the probability of obtaining b2 is 16/25.
(c)  There are two path.  If b1 is measured the system is in the state φ1 = (3ψ1 + 4ψ2)/5.
The probability of measuring a1 after measuring b1 is 9/25.
If b2 is measured the system is in the state φ2 = (4ψ1 - 3ψ2)/5.
The probability of measuring a1 after measuring b2 is 16/25.
The probability of getting a1 again after a measurement of B is (9/25)2 + (16/25)2 = 0.5392.

#### Problem 3:

A pencil is placed with its point down on a flat surface.  Assume that the pencil point is infinitely "sharp" and the the surface is perfectly flat.  Ignore any extraneous effect such as vibration, air currents etc.  The uncertainty principle says that it is not possible to fix the pencil exactly vertically AND exactly at rest.  As a result, no matter how well it is positioned it will fall over in a finite time.  Using the uncertainty principle, estimate how long the pencil will take to fall if it is positioned as well as can be done.  Make reasonable estimates for the pencil mass, length, etc.

Solution:

• Concepts:
Lagrangian mechanics
• Reasoning:
We find the equation of motion for the pencil for small displacements.
• Details of the calculation:
Lagrangian mechanics
We find the equation of motion for the mass for small displacements.
(a)  T = ½I(dθ/dt)2) =  (ml2/6)(dθ/dt)2),  U = (mgl/2)cosθ,  L = T - U.
∂L/∂(dθ/dt) = (ml2/3)(dθ/dt),  d/dt(∂L/∂(dθ/dt)) = (ml2/3)d2θ/dt2, ∂L/∂θ = (mgl/2) sinθ.
The equation of motion is (ml2/3)d2θ/dt2 - (mgl/2)sinθ = 0.
For small θ we have d2θ/dt2 = (3g/(2l))θ.
Most general solution:  θ(t) = A exp(ct) + B exp(-ct),  c = (3g/(2l))½.
θ0 = A + B,  ω0 = c(A - B),  A = (cθ0 + ω0)/(2c),  B = (cθ0 - ω0)/(2c)
θ(t) = [(cθ0 + ω0)/(2c)] exp(ct) + [(cθ0 - ω0)/(2c)] exp(-ct).
exp(ct) = [θ(t) ± (θ(t)2 - 4AB)½]/(2A).
Let us find the time for the pencil to reach θ = 0.1 rad, so that the small angle approximation stays valid.  Once the pencil has  0.1 rad, it very quickly falls over completely.

Uncertainty principle:  ∆θ0∆ω0 ≈ 4ħ/(ml2) = B,
Different approaches lead to similar estimates.
θ(t) will be largest if θ0 and ω0 have the same sign, then we can safely ignore the decaying exponential.
exp(ct) = θ(t)2c/(cθ0 + ω0) = θ(t)2c/(cθ0 + B/θ0).
dexp(ct)/dθ0  = 0 --> θ0 = (B/c)½.
Let l = 0.15 m, g = 10 m/s2, m = 10-2 kg, and ħ = 10-34 Js.
Then c = 10/s and B = 1.78*10-30/s.
θ0 = 4.22*10-16, exp(ct) = 2.37*1014,  t = 3.3 s.

We can also argue in the following way.
The uncertainty principle restricts our ability to determine θ0 and ω0 simultaneously with arbitrary accuracy.  The more precise we determine θ0, the lower is the precision with which we can know ω0.  If θ0 is big, the pencil tips over quickly even if ω0 is very small, because the pencil starts far from the equilibrium position.  If ω0 is big, the pencil tips over quickly even if it starts from the equilibrium position, because it has a large initial velocity. The longest time to reach the ground must roughly be the time when ∆θ0 and ∆ω0 make equal contributions.
(i)  Let ω0 = 0, then A = B = θ0/2.  cosh(ct) = 0.1/θ0
(ii)  Let θ0 = 0, then A = -B = ω0/(2c).  sinh(ct) = 0.1/(ω0/c)
Since  cosh(ct) ≈ sinh(ct) for large values of ct, we want θ0 = ω0/c = B/θ0, or θ0 = (B/c)½, t = 3.4 s.

#### Problem 4:

(a)  What is the wavelength of a 10 eV electron and what is the energy of a photon with this same wavelength?
(b)  Light with a wavelength of 300 nm strikes a metal whose work function is 2.2 eV.  What is the shortest de Broglie wavelength for the electrons that are produced as photoelectrons?
(c)  A surface is irradiated with monochromatic light whose wavelength can be varied.  Above a wavelength of 500 nm, no photoelectrons are emitted from the surface.  With an unknown wavelength, a stopping potential of 3 V is necessary to eliminate the photoelectric current.  What is the unknown wavelength?

Solution:

• Concepts:
The de Broglie wavelength, the photoelectric effect
• Reasoning:
We are asked to compare the energy of an electron and a photon with the same wavelength.
Photoelectrons are only emitted if the photon energy is greater than the work function Φ of the surface.
• Details of the calculation:
(a)  For the electron:
p = h/λ,  λ = h/(2mE)½ = 2π*1.054*10-34 Js/(2*9.1*10-31 kg*10 eV *1.6*10-19 J/eV)½
= 3.9*10-10 m.
For the photon:
E = hc/λ = (2π*6.58*10-16 eV-s*3*108 m/s)/(3.9*10-10 m) = 3180 eV.
(b)  Emax_electron = hc/λ - 2.2 eV = 1.94 eV = 3.1*10-19 J.
pmax_electron = (2mEmax_electron)½  = 7.52*10-25 kgm/s.
λe_min = h/pmax_electron = 8.8*10-10 m.

(c)  Ee(max) = hf - Φ = hc/λ - Φ.
Ee(max) = 0 if λ > 500 mn, therefore Φ = hc/(500 nm).
With the unknown wavelength Ee(max) = 3 eV, so 3 eV = hc/λ - hc/(500 nm).
3 eV = (1240 eV/nm)/λ - (1240 eV/nm)/(500 nm).
λ = 226 nm.

#### Problem 5:

Consider a two-state system governed by the Hamiltonian H with energy eigenstates |E1> and |E2>,  where H|E1> = E1|E1> and H|E2> = E2 |E2>.  Consider also two other states,
|x> = (|E1> + |E2>)/√2,  and  |y> = (|E1> - |E2>)/√2.
At time t = 0 the system is in state |x>.  At what subsequent times is the probability of finding the system in state |y> the largest, and what is that probability?

Solution:

• Concepts:
The evolution operator, postulates of QM
• Reasoning:
The evolution operator is a unitary operator defined through |ψ(t)> = U(t,t0)|ψ(t0)>.
If H does not explicitly depend on time, then U(t,t0) = exp(-(i/ħ)H(t-t0)).
If |ψ(t0)> is an eigenfunction of H with eigenvalue E,
then |ψ(t)> =  exp(-(i/ħ)E(t-t0))|ψ(t0)>.
• Details of the calculation:
|ψ(0)>  = |x> = (|E1> + |E2>)/√2, |ψ(t)> = exp(-iHt/ħ)|ψ(0)>
= (exp(-iE1t/ħ)|E1> + exp(-iE2t/ħ)|E2>)/√2.
<y|ψ(t)> = (<E1|ψ(t)>  - <E2|ψ(t)> /√2 = (exp(-iE1t/ħ) - exp(-iE2t/ħ))/2
= exp(-iE1t/ħ) (1 - exp(-i(E2 - E1)t/ħ))/2.
Probability of finding the system in state |y>:
|<y|ψ(t)>|2 = (1 - exp(-i(E2 - E1)t/ħ))(1 - exp(+i(E2 - E1)t/ħ))/4
= ½(1 - cos((E2 - E1)t/ħ)).
The probability is maximized when cos((E2 - E1)t/ħ) = -1, or when t = nπħ/( E2 - E1), n = 1, 3, 5, ... .