Consider a 2 by 2 matrix M with matrix elements m_{11},
m_{12}, m_{21}, and m_{22}.

(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^{†}= U^{†}U = I. - Details of the calculation:

(a) Let M = U be a Unitary matrix. Then UU^{†}= U^{†}U = I.

UU^{†}= I --> |m_{11}|^{2}+ |m_{12}|^{2}= 1, |m_{21}|^{2}+ |m_{22}|^{2}= 1, m_{11}m_{21}^{*}+ m_{12}m_{22}* = 0.

U^{†}U = I --> |m_{11}|^{2}+ |m_{21}|^{2}= 1, |m_{12}|^{2}+ |m_{22}|^{2}= 1, m_{22}m_{21}^{*}+ m_{12}m_{11}* = 0.

|m_{11}|^{2}+ |m_{12}|^{2}= |m_{11}|^{2}+ |m_{21}|^{2}= 1 implies |m_{21}| = |m_{21}|.

Since |m_{11}|^{2}+ |m_{12}|^{2}= 1 there exists an angle θ such that |m_{11}|= cosθ, |m_{12}| = |m_{21}| = sinθ.

|m_{11}|^{2}+ |m_{12}|^{2}= |m_{21}|^{2}+ |m_{22}|^{2}= 1 implies |m_{11}| = | m_{22}|, |m_{22}|= cosθ.

Let m_{ii}= |m_{ii}|exp(iφ_{ii}). Then m_{11}m_{21}^{*}+ m_{12}m_{22}* = 0 and m_{22}m_{21}^{*}+ m_{12}m_{11}* = 0.

implies φ_{11}- φ_{21}= φ_{12}- φ_{22}± π and φ_{22}- φ_{21}= φ_{12}- φ_{11}± π,

φ_{11}= -φ_{22}+ φ_{21}+ φ_{12}± π.

If we choose the φ_{11}= -φ_{22}, by factoring out a common factor exp(iΔ) from all matrix elements then

φ_{21}= -φ_{12}± π.

Therefore, except for some common factor exp(iΔ), we have

m_{11}= cosθ exp(iφ_{1}), m_{12}= sinθ exp(iφ_{2}),

m_{21}= -sinθ exp(-iφ_{2}), m_{22}= cosθ exp(-iφ_{1}).

(b) det(U) = exp(iΔ)(cos^{2}θ + sin^{2}θ) = exp(iΔ).

|det(U)| = 1.

An operator A has two normalized eigenstates ψ_{1} and ψ_{2},
with eigenvalues a_{1} and a_{2}, respectively. An operator
B, has two
normalized eigenstates, φ_{1} and φ_{2}, with eigenvalues b_{1}
and b_{2}, respectively. The eigenstates are related by**
**ψ

(a) Observable A is measured, and the value a

(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 a

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 a_{n}is given by

P(a_{n}) = Σ_{i=0}^{gn}|<u_{n}^{i}|ψ>|^{2}, where {|u_{n}^{i}>} (i=1,2,...,g_{n}) is an**orthonormal basis**in the eigensubspace E_{n}associated with the eigenvalue a_{n}.

If a measurement on a system in the state |ψ> gives the result a_{n}, then the state of the system immediately after the measurement is the normalized projection of |ψ> onto the eigensubspace associated with a_{n}. - Details of the calculation:

(a) After measuring a_{1}, the system is in the eigenstate ψ_{1}.

(b) If B is now measured, the probability of obtaining b_{1}is 9/25 and the probability of obtaining b_{2}is 16/25.

(c) There are two path. If b_{1}is measured the system is in the state φ_{1}= (3ψ_{1}+ 4ψ_{2})/5.

The probability of measuring a_{1}after measuring b_{1}is 9/25.

If b_{2}is measured the system is in the state φ_{2}= (4ψ_{1}- 3ψ_{2})/5.

The probability of measuring a_{1}after measuring b_{2}is 16/25.

The probability of getting a_{1}again after a measurement of B is (9/25)^{2}+ (16/25)^{2}= 0.5392.

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}) = (ml^{2}/6)(dθ/dt)^{2}), U = (mgl/2)cosθ, L = T - U.

∂L/∂(dθ/dt) = (ml^{2}/3)(dθ/dt), d/dt(∂L/∂(dθ/dt)) = (ml^{2}/3)d^{2}θ/dt^{2}, ∂L/∂θ = (mgl/2) sinθ.

The equation of motion is (ml^{2}/3)d^{2}θ/dt^{2}- (mgl/2)sinθ = 0.

For small θ we have d^{2}θ/dt^{2}= (3g/(2l))θ.

Most general solution: θ(t) = A exp(ct) + B exp(-ct), c = (3g/(2l))^{½}.

θ_{0}= A + B, ω_{0}= c(A - B), A = (cθ_{0}+ ω_{0})/(2c), B = (cθ_{0}- ω_{0})/(2c)

θ(t) = [(cθ_{0}+ ω_{0})/(2c)] exp(ct) + [(cθ_{0}- ω_{0})/(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: ∆θ_{0}∆ω_{0}≈ 4ħ/(ml^{2}) = B,

Different approaches lead to similar estimates.

θ(t) will be largest if θ_{0}and ω_{0}have the same sign, then we can safely ignore the decaying exponential.

exp(ct) = θ(t)2c/(cθ_{0}+ ω_{0}) = θ(t)2c/(cθ_{0}+ B/θ_{0}).

dexp(ct)/dθ_{0}= 0 --> θ_{0}= (B/c)^{½}.

Let l = 0.15 m, g = 10 m/s^{2}, m = 10^{-2}kg, and ħ = 10^{-34}Js.

Then c = 10/s and B = 1.78*10^{-30}/s.

θ_{0}= 4.22*10^{-16}, exp(ct) = 2.37*10^{14}, t = 3.3 s.

We can also argue in the following way.

The uncertainty principle restricts our ability to determine θ_{0}and ω_{0}simultaneously with arbitrary accuracy. The more precise we determine θ_{0}, the lower is the precision with which we can know ω_{0}. If θ_{0}is big, the pencil tips over quickly even if ω_{0}is very small, because the pencil starts far from the equilibrium position. If ω_{0}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 ∆θ_{0}and ∆ω_{0}make equal contributions.

(i) Let ω_{0}= 0, then A = B = θ_{0}/2. cosh(ct) = 0.1/θ_{0}

(ii) Let θ_{0}= 0, then A = -B = ω_{0}/(2c). sinh(ct) = 0.1/(ω_{0}/c)

Since cosh(ct) ≈ sinh(ct) for large values of ct, we want θ_{0}= ω_{0}/c = B/θ_{0}, or θ_{0}= (B/c)^{½}, t = 3.4 s.

(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*10^{8 }m/s)/(3.9*10^{-10}m) = 3180 eV.

(b) E_{max_electron}= hc/λ - 2.2 eV = 1.94 eV = 3.1*10^{-19}J.

p_{max_electron}= (2mE_{max_electron})^{½}= 7.52*10^{-25}kgm/s.

λ_{e_min}= h/p_{max_electron}= 8.8*10^{-10}m.

(c) E_{e}(max) = hf - Φ = hc/λ - Φ.

E_{e}(max) = 0 if λ > 500 mn, therefore Φ = hc/(500 nm).

With the unknown wavelength E_{e}(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.

Consider a two-state system governed by the Hamiltonian H with energy eigenstates |E_{1}> and |E_{2}>,
where H|E_{1}> = E_{1}|E_{1}> and H|E_{2}> = E_{2}
|E_{2}>. Consider also two other states,

|x> = (|E_{1}> + |E_{2}>)/√2, and |y> = (|E_{1}>
- |E_{2}>)/√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,t_{0})|ψ(t_{0})>.

If H does not explicitly depend on time, then U(t,t_{0}) = exp(-(i/ħ)H(t-t_{0})).

If |ψ(t_{0})> is an eigenfunction of H with eigenvalue E,

then |ψ(t)> = exp(-(i/ħ)E(t-t_{0}))|ψ(t_{0})>. - Details of the calculation:

|ψ(0)> = |x> = (|E_{1}> + |E_{2}>)/√2, |ψ(t)> = exp(-iHt/ħ)|ψ(0)>

= (exp(-iE_{1}t/ħ)|E_{1}> + exp(-iE_{2}t/ħ)|E_{2}>)/√2.

<y|ψ(t)> = (<E_{1}|ψ(t)> - <E_{2}|ψ(t)> /√2 = (exp(-iE_{1}t/ħ) - exp(-iE_{2}t/ħ))/2

= exp(-iE_{1}t/ħ) (1 - exp(-i(E_{2}- E_{1})t/ħ))/2.

Probability of finding the system in state |y>:

|<y|ψ(t)>|^{2}= (1 - exp(-i(E_{2}- E_{1})t/ħ))(1 - exp(+i(E_{2}- E_{1})t/ħ))/4

= ½(1 - cos((E_{2}- E_{1})t/ħ)).

The probability is maximized when cos((E_{2}- E_{1})t/ħ) = -1, or when t = nπħ/( E_{2}- E_{1}), n = 1, 3, 5, ... .