Calculate the probability current density for of a freely moving particle whose quantum state is described by a de Broglie plane wave.

Solution:

- Concepts:

The Schroedinger equation, ρ(x,t) = |ψ(x,t)|^{2}. - Reasoning:

Using the Schroedinger equation we can find an expression for ∂ρ(**r**,t)/∂t in terms of space derivatives. - Details of the calculation:

Let the plane wave propagate in the positive x-direction. Then we can work in one dimension.

The probability per unit length of finding a particle described by the normalized wave function ψ(x,t) is given by dP(x,t)/dx = |ψ(x,t)|^{2}.

The total probability of finding the particle at time t anywhere in space is ∫_{all space}|ψ(x,t)|^{2}dx = <ψ|ψ> = 1.

Local conservation of a classical quantity is usually expressed through the equation**∇∙j**= -(∂/∂t)ρ. Here ρ(**r**,t) is the volume density and**j**(**r**,t) is the current density.

In one dimension this becomes ∂j(x,t)/∂x = = -(∂/∂t)ρ(x,t).

Assuming local conservation of probability we write

∂ρ/∂t = ∂ψ^{*}ψ/∂t = ψ^{*}∂ψ/∂t + ψ∂ψ^{*}/∂t

= (1/(iħ))[ψ^{*}((-ħ^{2}/(2m))∂^{2}ψ/∂x^{2}+ Vψ) - ((-ħ^{2}/(2m))∂^{2}ψ/∂x^{2}+ Vψ)^{*}ψ]

= (-ħ/(2im)[ψ^{*}∂^{2}ψ/∂x^{2}- (∂^{2}ψ/∂x^{2})^{*}ψ], since V is real.

[We use iħ∂ψ/∂t = Hψ, ∂ψ/∂t = (1/(iħ))Hψ, ∂ψ^{*}/∂t = (-1/(iħ))(Hψ)^{*}.]

∂ρ/∂t = (-ħ/(2im)(∂**/**∂x)[ψ^{*}∂ψ/∂x - ψ∂ψ^{*}/∂x] = -∂j(x,t)/∂x**.**j(x,t) = (-ħ/(2im)[ψ^{*}∂ψ/∂x - ψ∂ψ^{*}/∂x] = (-ħ/(m) Im(ψ^{*}∂ψ/∂x).

We interpret ρ(x,t) as the probability density and j(x,t) as the probability current density.

For the plane wave wave function of a free particle ψ(x,t) = Aexp(ikx - ωt) we have

(ħ/i)∂ψ(x,t)/∂x = ħkψ(x,t) = p_{x}ψ(x,t).

ψ*∂ψ/∂x = ik|ψ(x,t)|^{2}, ψ∂ψ*/∂x = ik|ψ(x,t)|^{2},

j(x,t) = (ħk/m)|ψ(x,t)|^{2}= (ħk/m)|A|^{2}.

Since the eigenfunctions of the operator p_{x}are not square integrable, we cannot normalize the wave function in the traditional way.

Find <x> and ∆x for the nth stationary state of a free particle in one dimension restricted to the interval 0 < x < a. Show that as n --> ∞ these become the classical values.

Solution:

- Concepts:

The eigenfunctions of the infinite square well, the postulates of quantum mechanics - Reasoning:

The expression for the mean value of an observable x in the normalized state |ψ> is

<x> = <ψ|x|ψ>. The root mean square deviation ∆x is given by

∆x = (<x^{2}> - <x>^{2})^{½}. Here ψ_{n}(x) = (2/a)^{½}sin(nπx/a), E_{n}= n^{2}π^{2}ħ^{2}/(2ma^{2}). - Details of the calculation:

<x>_{n}= (2/a)∫_{0}^{a}x sin^{2}(nπx/a)dx = a/2.

<x^{2}>_{n}= (2/a)∫_{0}^{a}x^{2}sin^{2}(nπx/a)dx = (2/a)(a/(nπ))^{3}∫_{0}^{nπ}x^{2}sin^{2}(x)dx

= (2a^{2}/(nπ)^{3})(nπ)^{3}/6 - (2a^{2}/(nπ)^{3})nπ/4 = a^{2}/3 - ½a^{2}/(nπ)^{2}.

∆x_{n}= (a^{2}/12 - ½a^{2}/(nπ)^{2})^{½}--> a/12^{½}as n --> ∞.

Classically, the particle is moving back and forth in the well. It is equally likely to be found at any position.

The average position x_{avg}is a/2.

For x^{2}_{avg}we have x^{2}_{avg}= a^{2}/3. (∫_{0}^{a}P(x) x^{2}dx with P(x) = 1/a.)

If we define ∆x = (x^{2}_{avg}- x_{avg}^{2})^{½}we have ∆x = a/12^{½}.

Consider a system of N particles with only 3 possible energy levels separated by ε. Let the ground state energy be zero. The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at temperature T. Ignore interactions between particles and assume Boltzmann statistics apply.

(a) What is the partition function for a single particle in the system?

(b) What is the average energy per particle?

(c) What is the probability that the 2ε level is occupied in the
high-temperature limit k_{B}T >> ε?

Explain your answer on physical grounds.

(d) What is the average energy per particle in the high-temperature limit k_{B}T
>> ε?

(e) At what energy is the ground state 1.1 times as likely to be occupied
as the 2ε level?

(f) Find the heat capacity C_{V} of the system, analyze the low-T
(k_{B}T >> ε) and high-T (k_{B}T >> ε) limit, and sketch CV as a
function of T.

Solution:

- Concepts:

Boltzmann statistics - Reasoning:

In thermodynamic equilibrium P(E_{i}) ∝ g_{i}exp(-E_{i}/(kT).

Here the degeneracy g_{i}is 1 for all i. - Details of the calculation:

(a) P(E) = Cexp(-E/(k_{B}T)).

Σ_{j}P(E_{j}) = CΣ_{j}exp(-E_{j}/(k_{B}T)) = 1, C = 1/Σ_{j}exp(-E_{j}/(k_{B}T)) = 1/Z.

Let β = 1/(k_{B}T)).

Z = 1 + exp(-βε) + exp(-2βε).

(b) <ε> = (ε exp(-βε) + 2ε exp(-2βε))/Z

= ε(exp(-βε) + 2exp(-2βε))/(1 + exp(-βε) + exp(-2βε)).

(c) P = exp(-2βε)/(1 + exp(-βε) + exp(-2βε)) ~ (1 - 2βε)/( 1 + 1 - βε + 1 - 2βε) = 1/3.

In the high energy limit the probability is the same for all three levels.

(d) <ε> = 0/3 + ε/3 + 2ε/3 = ε.

(e) exp(-2βε) = 1/1.1, 2βε ln(1.1), T = 2ε/(k_{B}*ln(1.1)).

(f) C_{V}= dU/dT = Nd<ε>/dT = Nd<ε>/dβ)(dβ/dT).

C_{V}= (Nε^{2}/(k_{B}T^{2}))[ exp(-βε) + 4 exp(-2βε) + exp(-3βε)]/[1 + exp(-βε) + exp(-2βε)]^{2}.^{ }Low temperature limit: C_{V}= (Nε^{2}/(k_{B}T^{2})) exp(-ε/(k_{B}T)). (C_{V}--> 0 as T --> 0)

High temperature limit: C_{V}= (2/3)(Nε^{2}/(k_{B}T^{2})). ( C_{V}--> 0 as T --> ∞)

A river of width D flows on the northern hemisphere at a geographical
latitude φ toward the north with a certain flow speed v_{0}. By
which amount is the right bank higher that the left one?

First apply the equation of motion in a non-inertial frame to the problem at hand, and
then use D = 2 km, v_{0}
= 5 km/h, and φ = 45^{o} to find the super-elevation of the river for the
given parameters.

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Assume an observer in the rotating frame at mid latitudes φ = 90^{o}- θ.

The equations of motion in a rotating coordinate system contain fictitious forces.

md**v**/dt =**F**_{inertial}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**. - Details of the calculation:

(a) Consider a volume element dV of the water with mass m = ρdV.

For this volume element we have

-m**Ω**× (**Ω**×**r**) = centrifugal force.

(**Ω**×**r**) = ΩRsinθ**j**.**Ω**× (**Ω**×**r**) = -Ω^{2}Rsin^{2}θ**k**- Ω^{2}Rsinθcosθ**i**.

-2m**Ω**×**v**= Coriolis force.**Ω**×**v**= -Ωvcosθ**j.**For each volume element dV of the water we want dv

_{y}/dt = 0.We therefore need 0 = F

_{y inertial}+ 2ρdVΩvcosθ.

F_{y inertial}= -2ρdVΩvcosθ.

F_{y inertial}is a force on ρdV due to a horizontal pressure gradient, if the banks have different heights.

[P(y + dy) - P(y)] dA = -2ρdAdyΩvcosθ.

dP = -2ρdyΩvcosθ, ρgdh = -2ρdyΩvcosθ, dh/dy = -2Ωvcosθ/g.

(b) Ω = 2π/(24*3600 s), |dh/dy| = 1.46*10^{-5}.

Δh = 2000*1.46*10^{-5}m ~ 3 cm.

The right bank is 3 cm higher that the left one.

Two equal mass particles are connected by a spring and are executing
simple harmonic motion when they are placed in a horizontal, frictionless open pipe
rotating at a constant angular velocity **Ω** about a vertical axis
through its midpoint. The midpoint of the spring is initially at the midpoint of the pipe,
but neither the particles nor the spring are attached to the pipe.

(a) Find the maximum angular velocity **Ω** for which the particles stay in the pipe.

(b) Assume (a) is satisfied and the particles stay in the pipe. Find their
positions as a function of time.

Solution:

- Concepts:

Motion in a non-inertial frame - Reasoning:

For a uniformly rotating frame the equations of motion for a particle with mass m is

md**v**/dt = F_{inertial}- 2m**Ω**×**v**- m**Ω**× (**Ω**×**r**). - Details of the
calculation:

(a) Chose coordinate in the rotating frame.

Let x_{0}be the equilibrium length of the spring. Let x_{1}and x_{2}be the coordinate of the particles in the rotating frame.

The equations of motions for the particles constrained to move in the pipe are

md^{2}x_{1}/dt^{2}= -k(x_{1}- x_{2}- x_{0}) + mΩ^{2}x_{1},

md^{2}x_{2}/dt^{2}= k(x_{1}- x_{2}- x_{0}) + mΩ^{2}x_{2}.

Introduce new coordinates.

y_{1}= (x_{1}+ x_{2})/2, y_{2}= (x_{1}- x_{2})/2. Then

d^{2}y_{1}/dt^{2}= Ω^{2}y_{1}, d^{2}y_{2}/dt^{2}= Ω^{2}y_{2}- (2k/m)y_{2}+ (2k/m)x_{0}.

For the initial conditions x_{1}= -x_{2}, , dx_{1}/dt = -dx_{2}/dt, we have that y_{1}(t) = 0, for all t, the midpoint of the spring stays at the midpoint of the pipe. (unstable equilibrium)

Since x_{1}= -x_{2}, the equation describing the motion of x_{1}= y_{2}is

d^{2}y_{2}/dt^{2}= -(2k/m - Ω^{2})y_{2}+ (2k/m)x_{0}.

If 2k/m > Ω^{2}, then y_{2}= C cos(ωt + φ) + y_{20},

with ω^{2}= 2k/m - Ω^{2}and y_{20}= (2k/m)x_{0}/(2k/m - Ω^{2}) = x_{0}/(1 - ½mΩ^{2}/k).

The particle can stay in the pipe if Ω < (2k/m)^{½},**Ω**= ±Ω**k**.

We also need y_{20}+ C = x_{0}/(1 - ½mΩ^{2}/k) + C < L, where L is the length of the pipe.

(b) x_{1}= - x_{2}= ½C cos(ωt + φ) + ½y_{20}. The constants C and φ are determined by the initial conditions.