A two-state system has a
Hamiltonian H = E_{0}

1 | 2i | ||

-3i | 6 |

in the {|1>, |2>} basis.

Find the eigenvalues and normalized eigenvectors of this Hamiltonian.

Solution:

- Concepts:

The eigenvalues and eigenvectors of a Hermitian operator - Reasoning:

We are given the matrix of the Hermitian operator H in some basis. To find the eigenvalues E we set the determinant of the matrix (H - EI) equal to zero and solve for E. To find the corresponding eigenvectors {|Ψ>}, we substitute each eigenvalue E back into the equation (H - EI)|Ψ> and solve for the expansion coefficients of |Ψ> in the given basis. - Details of the calculation:

The eigenvalues are λE_{0}, where

1 - λ 2i -2i 6 - λ = 0.

(1 - λ)(6 - λ) - 4 = 0, λ^{2}- 7λ + 2 = 0, λ_{1}= (7 + (41)^{½})/2, λ_{2}= (7 - (41)^{½})/2.

The normalized eigenvector corresponding to λ_{1}is |1'> = a|1> + b|2>,

where a(1 - λ_{1}) + i2b = 0, and |a|^{2}+ |b|^{2}= 1.

-a(1 - (7 + (41)^{½})/2)/(2i) = b, b = -i2.85a,

|a|^{2}(1 + 2.85^{2}) = 1, |a|^{2}= 0.11, |b|^{2}= 0.89, |1'> = 0.33|1> - i0.94|2>.

The normalized eigenvector corresponding to λ_{2}is |2'> = c|1> + d|2>,

where c(1 - λ_{2}) + i2c = 0, and |c|^{2}+ |d|^{2}= 1.

-c(1 - (7 - (41)^{½})/2)/(2i) = d, d = i0.35a,

|c|^{2}(1 + 0.35^{2}) = 1, |c|^{2}= 0.89, |d|^{2}= 0.11, |1'> = 0.94|1> + i0.33|2>.

A cylindrical solenoid L = 50 cm long with a radius of r = 3 mm has N = 500
tightly-wound turns of low-resistance wire uniformly distributed along its
length. The solenoid is connected, using low-resistance wires, in series with a
R_{S} = 20 ohm resistor, a V = 9-volt battery, and a switch, which is
initially open. Around the middle of the solenoid is a two-turn
rectangular loop l = 3 cm by w = 2 cm made of resistive wire having a total
resistance of R_{L} = 150 ohms. The plane of the loop is
perpendicular to the axis of the solenoid and the centers of the loop and
solenoid coincide. The switch is closed at t = 0.

Showing all your work, develop an expression, in terms of the above symbolic
quantities, for the current as a function of time in the rectangular loop, and
evaluate that current at t = 1 microsecond after the switch is closed.

Solution:

- Concepts:

Ampere's law, self inductance, induced emf - Reasoning:

Since the length L of the solenoid is much larger than its radius r, Ampere's law (and the right-hand rule) can be used to find the magnetic field**B**of the solenoid, except near its ends.

If the axis of the solenoid is the z-axis and the current I in the solenoid flows in the**φ**/φ direction, then**B**= (N/L)μ_{0}I**k**inside the solenoid and zero outside the solenoid near the center of the solenoid. - Details of the calculation:

The self inductance of the solenoid is L_{S}= Φ/I = Nπr^{2}B/I = μ_{0}πr^{2}N^{2}/L.

If we ignore the mutual inductance, then V - L_{S}dI/dt - R_{S}I = 0, dI/dt = V/L_{S}- R_{S}I/L_{S}.

I(t) = (V/R_{S})(1 - exp(-R_{S}t/L_{S})) is the current in the solenoid as a function of time.

I(t) = 0.45*(1 - exp(-1.13*10^{6}/s*t)) A.

Flux through rectangular loop: Φ_{loop }= ∫_{A}**B**·**n**dA_{loop}= N_{loop}πr^{2}(N/L)μ_{0}I, if we ignore the self inductance of the loop.

The magnetic field produced by the coil is zero except inside the coil, so only the area of the coil contributes to the integral.

|emf_{loop}| = dΦ_{loop}/dt = N_{loop}πr^{2}(N/L)μ_{0}dI/dt

= N_{loop}πr^{2}(N/L)μ_{0}(V/L_{S})exp(-R_{S}t/L_{S}) = 3.6*10^{-2}*exp(-1.13*10^{6}/s * t) V.

I_{loop}= |emf_{loop}|/R_{L}= 2.4*10^{-4}*exp(-1.13*10^{6}/s * t) A.

I_{loop}(10^{-6}s) = 7.8*10^{-5}A.

I_{loop}is very small, so we are justified in neglecting the magnetic field produced by the loop in the calculations.

A particle of mass m and energy E > 0 whose wave function is a
one-dimensional plane wave moving in the +x direction is incident on a
one-dimensional square potential U(x) = -|U_{0}|, 0 ≤ x ≤ a, U(x)
= 0 everywhere else.

There are certain discrete energies for which the probability density of the
transmitted wave is equal to the probability density of the incident wave.
Find an expression for the values of these resonant energies.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential and are asked to find the resonant energies. - Details of the calculation:

The most general solutions in regions 1, 2, and 3 are

Φ_{1}(x) = A exp(ikx) + B exp(-ikx),

Φ_{2}(x) = C exp(ik'x) + D exp(-ik'x),

Φ_{3}(x) = F exp(ikx).

Here k^{2}= (2m/ħ^{2})E and k'^{2}= (2m/ħ^{2})(E + |U_{0}|).

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A + B = C_{ }+ D , ikA_{ }- ikB = ik'C_{ }- ik'D.

x = a: C exp(ik'a) + D exp(-ik'a) = F exp(ika),

C exp(ik'a) - D exp(-ik'a) = (k/k')F exp(ika).

We want to find k for|A|^{2}= |F|^{2}and B = 0.

F = A exp(iφ), where φ is a possible phase shift.

A = C_{ }+ D , ikA = ik'C_{ }- ik'D.

C exp(ik'a) + D exp(-ik'a) = A exp(iφ) exp(ika),

C exp(ik'a) - D exp(-ik'a) = (k/k')A exp(iφ)exp(ika).

(i) C + D = (k'/k)(C - D), (k'/k - 1)C = (k'/k + 1)D.

(ii) C exp(ik'a) + D exp(-ik'a) = (k'/k)(C exp(ik'a) - D exp(-ik'a))

exp(ik'a) (k'/k - 1)C = exp(-ik'a) (k'/k + 1)D = exp(-ik'a)(k'/k - 1)C.

exp(ik'a) = exp(-ik'a), k'a = nπ, k'^{2}= (2m/ħ^{2})(E + |U_{0}|) = n^{2}π^{2}/a^{2},

E = (ħ^{2}/(2m))n^{2}π^{2}/a^{2}- |U_{0}|, n big enough so that E > 0.

Refer to the figure. One end of a conducting rod rotates with angular
velocity ω in a circle of radius a making contact with a horizontal, conducting
ring of the same radius. The other end of the rod is fixed. Stationary
conducting wires connect the fixed end of the rod (A) and a fixed point on the
ring (C) to either end of a resistance R. A uniform vertical magnetic field
**B** passes through the ring.

(a) Find the current I flowing through the resistor and the rate at which
heat is generated in the resistor.

(b) What is the sign of the current, if positive I corresponds to flow in the
direction of the arrow in the figure?

(c) What torque must be applied to the rod to maintain its rotation at the
constant angular rate ω?

What is the rate at which mechanical work must be done?

Solution:

- Concepts:

Motional emf - Reasoning:

The conducting rod is moving in a plane perpendicular to**B**. - Details of the calculation:

(a) Speed of the rod as a function of the distance r from the origin: v(r) = ωr.

Force on an electron: F_{e}= q_{e}vB = q_{e}Bωr. (towards point A)

Work done per unit charge: emf = Bω∫_{0}^{a}rdr = Bωa^{2}/2.

Assume that the resistance of the conducting rod and the wires is negligible.

I = emf/R = Bωa^{2}/(2R) is the current flowing through the resistor.

P_{e}= I^{2}R = B^{2}ω^{2}a^{4}/(4R) is the rate heat is generated.

(b) The sign of the current is positive.

(c) Force on a section dr of the current carrying rod: dF = IdrB (direction clockwise in the figure).

An external force of equal magnitude and opposite direction is needed to maintain the constant angular speed.

dτ = rdF = r IBdr, τ = IB∫_{0}^{a}rdr = IBa^{2}/2 = B^{2}ωa^{4}/(4R).

The rate at which mechanical work is done is P_{m}= τω = B^{2}ω^{2}a^{4}/(4R) = P_{e}.

A particle of mass m moves in a one-dimensional potential given by

V(x) = -W for |x| < a, V(x) = 0 for |x| ≥ a.

Demonstrate that this potential has at least one even bound state.

Solution:

- Concepts:

This is a "square potential" problem. We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

We are given a piecewise constant potential and are asked to show a bound-state solutions exists. - Details of the calculation:

We have a symmetric potential. The wave function must be symmetric or anti-symmetric. The ground state wave function has no nodes except at infinity, so it must be symmetric. We therefore only have to consider the region x ≥ 0.

Let -W_{ }< E < 0. The wave function must remain finite at x = ±∞. Solutions to HΦ(x) = EΦ(x) in regions 1 (x < a) and region 2 (x > a) therefore are

Φ_{1}(x) = Acos(kx), Φ_{2}(x) = Be^{-ρx},

with k^{2}= (2m/ħ^{2})(E + W) and ρ^{2}= (2m/ħ^{2})(-E).

Φ and ∂Φ/∂x are continuous at x = a. This implies

A cos(ka) = B e^{-ρa}, -kA sin(ka) = -ρB e^{-ρa}, tan(ka) = ρ/k.

We can try a graphical solution.

Define k_{0}^{2}= 2mW/ħ^{2}= k^{2}+ ρ^{2}.

Then 1/cos^{2}(ka) = 1 + tan^{2}(ka) = (k^{2}+ ρ^{2})/k^{2}= k_{0}^{2}/k^{2},

or |cos(ka)| = k/k_{0}for all k for which tan(ka) ≥ 0.

Note: We changed from a tangent to a cosine function for easier graphing.

In regions (1), (2), (3,) ... tan(ka) ≥ 0. Three solutions exist for the given k_{0}in the graph. As we increase k_{0}more solutions become possible. For every k_{0}at least one solution is possible. (The two curves always cross in region 1.) There exists at least one bound state.