A one-dimensional potential barrier or square well problem is defined by the Hamiltonian H = (P^{2}/2m)
+ U(x),> with U(x) = ζU_{0}Θ(ℓ/2 - |x|).
Here Θ(z) = 0 for z < 0 and Θ(z) = 1 for z > 0, ζ
= +1 for potential barriers and ζ = -1 for square wells.

(a) Calculate the transmission
coefficient for E = 1 eV incident electrons facing a potential barrier of U_{0}
= 2 eV and ℓ = 1 Å. What is the probability
that a 1 eV protons will tunnel through the barrier?

(b) Infer from (a) the general expression for the transmission
coefficient T, and draw T as a function of ℓ
for E = 2.25 eV electrons.

(c) Calculate eigenfunctions and eigenvalues of the square well in the
limit U_{0} --> ∞.

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. We have a potential that has a value of U_{0}(barrier) or -U_{0}(well) from x = -ℓ/2 to x = ℓ/2 and is zero otherwise. We are asked to find the transmission coefficient T. - Details of the calculation:

Let x' = x + ℓ/2. Divide space into 3 regions.

region 1: x' < 0, region 2: 0 < x' < ℓ, region 3: x' > ℓ.

(a) potential barrier, E < U_{0}.

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

Φ_{1}(x') = A_{1}exp(ik_{1}x') + A_{1}'exp(-ik_{1}x'),

Φ_{2}(x') = A_{2}exp(ρ_{2}x') + A_{2}'exp(-ρ_{2}x'),

Φ_{3}(x') = A_{3}exp(ik_{1}x').

Here k_{1}^{2}= (2m/ħ^{2})E and ρ_{2}^{2}= (2m/ħ^{2})(U_{0}- E).

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

x' = 0: A_{1}+ A_{1}' = A_{2 }+ A_{2}' , ik_{1}A_{1 }- ik_{1}A_{1}' = ρ_{2}A_{2 }- ρ_{2}A_{2}'.

x' = ℓ: A_{2}exp(ρ_{2}ℓ) + A_{2}'exp(-ρ_{2}ℓ) = A_{3}exp(ik_{1}ℓ),

ρ_{2}A_{2}exp(ik_{2}ℓ) - ρ_{2}A_{2}'exp(-ρ_{2}ℓ) = ik_{i}A_{3}exp(ik_{1}ℓ).

We need to solve these equations for A_{3}in terms of A_{1}.

(i) Solve for A_{2}and A_{2}' in terms of A_{3}.

A_{2}= ½exp((ik_{1 }- ρ_{2})ℓ)(1 + ik_{1}/ρ_{2})A_{3}= C A_{3}.

A_{2}' = ½exp((ik_{1 }+ ρ_{2})ℓ)(1 - ik_{1}/ρ_{2})A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}.

2A_{1}= A_{2}+ A_{2}' + (ρ_{2}/ik_{1})(A_{2 }- A_{2}') = (C + C' + (ρ_{2}/ik_{1})(C - C'))A_{3}.

A_{1}= ([(k_{1}^{2}- ρ_{2}^{2}))/(2ik_{1}ρ_{2})]sinhρ_{2}ℓ + coshρ_{2}ℓ)exp(ik_{1}a)A_{3}.

T = |A_{3}/A_{1}|^{2}= 4k_{1}^{2}ρ_{2}^{2}/[(k_{1}^{2}- ρ_{2}^{2})^{2}sinh^{2}ρ_{2}ℓ + 4k_{1}^{2}ρ_{2}^{2}cosh^{2}ρ_{2}ℓ]

= 4k_{1}^{2}ρ_{2}^{2}/[(k_{1}^{2}+ ρ_{2}^{2})^{2}sinh^{2}ρ_{2}ℓ + 4k_{1}^{2}ρ_{2}^{2}].

[cosh^{2}x - sinh^{2}x = 1.]

T = 4E(U_{0}- E)/[U_{0}^{2}sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{1/2}ℓ] + 4E(U_{0}- E)].

If U_{0}= 2 eV, E = 1 e V, and ℓ = 1 Å and the particle is an electron then

T = 1/(1 + sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{1/2}ℓ] = 1/(1 + sinh^{2}(0.54)) = 0.76.

If U_{0}= 2 eV, E = 1 eV, and ℓ = 1 Å and the particle is an proton then

T = 3.3*10^{-20}.

(b) If we have a barrier of height U_{0}and 0 < E < U_{0}the general expression for T is

T = 4E(U_{0}- E)/[U_{0}^{2}sinh^{2}[(2m(U_{0}- E)/ħ^{2})^{1/2}ℓ] + 4E(U_{0}- E)].

If we have a barrier of height U_{0}and E > U_{0}, or if we have a well (U_{0}< 0) and E > 0 the general expression for T is

T = 4E(E - U_{0})/[U_{0}^{2}sin^{2}(2m(E - U_{0})/ħ^{2})^{1/2}ℓ] + 4E(E - U_{0})].

We can consider this second expression the general expression for T valid for all one-dimensional potential barrier or square well scattering problems, if we use the fact that sin^{2}(ix) = -sinh^{2}(x).

For U_{0}= 2 eV, E = 2.25 eV we have

T = 2.25/[4sin^{2}(0.5 eVm/ħ^{2})^{1/2}ℓ] + 2.25].

(c) We have a symmetric infinite square well.

H = -(ħ^{2}/2m)(∂^{2}/∂x^{2}) + U(x), U(x)_{ }= 0 for |x| < ℓ and U(x)_{ }= ∞ for |x| > ℓ.

(∂^{2}/∂x^{2})Φ(x) + (2m/ħ^{2})EΦ(x) = 0 in the region |x| < ℓ, Φ(x) = 0 for |x| = ℓ.

Φ(x) = Aexp(ikx) + Bexp(-ikx), with E = ħ^{2}k^{2}/(2m).

Aexp(ikℓ) + Bexp(-ikℓ) = 0, Aexp(-ikℓ) + Bexp(ikℓ) = 0.

Two solutions:

A = B: coskℓ = 0, kℓ = nπ/2, n = odd.

A = -B: sinkℓ = 0, kℓ = nπ/2, n = even.

E_{n}= n^{2}π^{2}ħ^{2}/(8mℓ^{2}), Φ_{n}(x) = Ncos(nπx/(2ℓ)), n = odd, Φ_{n}(x) = Nsin(nπx/(2ℓ)), n = even.

Normalization: N^{2}= 1/ℓ.

A one-dimensional potential well is given in the form of a delta function at x =
0,
U(x) = Cδ(x), C < 0. A stream of non-relativistic particles of
mass m and energy E approaches the origin from one side.

(a) Derive an expression
for the reflectance R(E).

(b) Can you express R(E) in terms of sin^{2}(δ), where δ is the
phase shift of the transmitted wave?

Solution:

- Concepts:

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

V(x) = 0 everywhere except at x = 0. - Details of the calculation:

(a) Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx) for x < 0. k^{2}= 2mE/ħ^{2}.

Φ_{2}(x) = A_{2}exp(ikx) for x > 0.

Φ is continuous at x = 0. Φ_{1}(0) = Φ_{2}(0). A_{1 }+ A_{1}' = A_{2}.

∂^{2}Φ(x)/∂x^{2}+ (2m(E - V(x))/ħ^{2})Φ(x) = 0.

Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.

∂Φ_{ε}(x_{1 }+ ε)/∂x + ∂Φ_{ε}(x_{1 }- ε)/∂x = (2m/ħ^{2})∫_{x1-ε}^{x1+ε }(Cδ(x) - E) Φ(x) dx

= (2mC/ħ^{2})Φ(0).

If V does not remain finite at the step, then ∂Φ/∂x has a finite discontinuity at the step.

iA_{1}k - iA_{1}'k = iA_{2}k - A_{2}2mC/ħ^{2}, A_{1}- A_{1}' = A_{2}- A_{2}2mC/(ikħ^{2}) = [1 - 2mC/(ikħ^{2})]A_{2}.

Eliminate A_{1}':

2A_{1}= [2 - 2mC/(ikħ^{2})]A_{2}, A_{1}= [1 + imC/(kħ^{2})]A_{2},

T(E) = (k|A_{2}|^{2})/(k|A_{1}|^{2}) = ħ^{4}k^{2}/(ħ^{4}k^{2}+ m^{2}C^{2}) = E/(E + mC^{2}/(2ħ^{2})).

R(E) = 1 - T(E) = [mC^{2}/(2ħ^{2})]/(E + mC^{2}/(2ħ^{2})) = m^{2}C^{2}/(ħ^{4}k^{2}+ m^{2}C^{2}).

(b) A_{2}= A_{1}/ [1 + imC/(kħ^{2})] = A_{1}(1 - imC/(kħ^{2}))/[1 + m^{2}C^{2}/(k^{2}ħ^{4})] = |A_{2}|exp(iδ).

sinδ = [-mC/(kħ^{2})]/[1 + m^{2}C^{2}/(k^{2}ħ^{4})]^{1/2}.

sin^{2}δ = [m^{2}C^{2}/(k^{2}ħ^{4})]/[1 + m^{2}C^{2}/(k^{2}ħ^{4})] = R(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}= 2mU_{0}/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.

Alternative solution:

- The variational method

We can also use the variational method to show that the upper bound for the ground-state energy is negative.

Use the ground state wave function of the 1D harmonic oscillator as the trial function.

Φ(x) = (mω/(πħ))^{¼}exp(-½mωx^{2}/ħ) = (b/π)^{¼}exp(-½bx^{2}). Let b be the adjustable parameter.

<H> = <T> + <V>, <T>_{harm. osc}= ½<E>_{harm. osc}= ¼ħω = ¼ħ^{2}b/m.

<V> = -W(b/π)^{½}∫_{-a}^{a}exp(-bx^{2})dx --> -2aW(b/π)^{½}for small values of b.

For small values of b <H> = <T> + <V> --> A b - B b^{½}, with A and B positive constants.

<T> approaches zero faster than |<V>|, so for some values of b <H> will be negative.

For b = 0, <H> = 0, so <H> has a negative minimum for some value of the adjustable parameter b.

This minimum is the upper bound for the ground-state energy. A bound ground state exists.

Consider a simple, small, but macroscopic LC circuit made from conventional superconducting material and kept at a temperature below the critical temperature (~1 K). The circuit has no resistance.

(a) Write down a second order differential equation describing the time
evolution of the magnetic flux Φ = LI in the inductor.

(Consider this "the equation of motion" of the circuit.) Relabel C = m, k =
1/L, and compare this equation with the equation of motion of a simple harmonic
oscillator.

(b) Write down a Lagrangian for the LC circuit. (Lagrange's equation then is
the "equation of motion" of the circuit.) Find the generalized momentum
corresponding to the generalized coordinate in the Lagrangian, and write down
the Hamiltonian for the LC circuit.

(c) Assume that the circuit is cooled down to a
temperature of near zero K (~1 mK). At such a low temperature, only the
lowest allowed energy states are accessible to the system and observable
quantum-mechanical effects can appear. Quantize the system and find the
ground state energy of the system,

(d) Qualitatively reason why excited states of such a system would be unstable
even at very low temperature.

Solution:

- Concepts:

LC circuits, the Lagrangian and Hamiltonian of a system, quantization of a classical Hamiltonian - Reasoning:

The same equations have the same solutions. We explore the mathematical analogy between an LC circuit and a harmonic oscillator. - Details of the calculation:

(a) Φ = LI. emf = -dΦ/dt = Q/C. (Kirchhoff's rule)

d^{2}Φ/dt^{2}+ I/C = d^{2}Φ/dt^{2}+ Φ/(LC) = 0.

With C = m and k = 1/L we have d^{2}Φ/dt^{2}+ (k/m)Φ = 0.

This is the equation of motion of a simple harmonic oscillator with Φ being a generalized coordinate.

(b) L_{Lag}= ½C(dΦ/dt)^{2}- ½Φ^{2}/L.

We can obtain the equation of motion from this Lagrangian.

P_{Φ}= ∂L_{Lag}/∂(dΦ/dt) = C(dΦ/dt).

H = P_{Φ}^{2}/(2C) + ½Φ^{2}/L.

(c) Let P_{Φ}and Φ^{ }become non-commuting operators, [Φ, P_{Φ}] = iħ.

Then H becomes the Hamiltonian operator -ħ^{2}/(2C) ∂^{2}/∂Φ^{2}+ ½Φ^{2}/L,

and Hψ(Φ) = Eψ(Φ).

The eigenvalues of H = (n + ½)ħω, with ω = 1/(LC)^{½}. The ground state energy is ½ħ/(LC)^{½}.

(d) We are assuming zero resistance. If kT << ħ/(LC)^{½}the probability of thermal excitation and de-excitation is small. But in an excited state, the system radiates.

Consider a one-dimensional crystal with primitive lattice translation a.

Let {|n>} be a set orthonormal electron states, n = -∞ to +∞.

Assume that in the subspace spanned by {|n>} the matrix elements of the electron Hamiltonian are
given by

<n|H|n> = E_{0}, <n|H|_{ }n±1> = -A, <n|H|_{ }n±2> =
B, <n|H|m> = 0 for all other m, with

E_{0} = (7/8)ħ^{2}/(ma^{2}), A = ½ħ^{2}/(ma^{2}),
and B = (1/16)ħ^{2}/(ma^{2}).

(a) Assume the eigenstates of H are of the form |Φ> = ∑_{n} b(x_{n})|n>.

Write down the coupled linear equations for the b(x_{n}). Note that b(x_{n±1})
= b(x_{n} ± a).

(b) Try solutions of the form b(x_{n}) = exp(ikx_{n}) and show
that E as a function of k is given by

E(k) = [ħ^{2}/(ma^{2})][7/8 - cos(ka) + (1/8) cos(2ka)].

(c) Determine the effective mass at the bottom of the energy band and at the
top of the band from a quadratic expansion of E in the departure of k from these
points.

Solution:

- Concepts:

Electrons in a periodic structure - Reasoning:

We must solve coupled eigenvalue equations. - Details of the calculation:

(a) <m|H|Φ> = ∑_{n}b(x_{n}) <m|H|n> .

E b(x_{m}) = E_{0 }b(x_{m}) - A b(x_{m+1}) - A b(x_{m-1}) + B b(x_{m+2}) + B b(x_{m-2})

= E_{0}b(x_{m}) - A b(x_{m}+ a) - A b(x_{m}- a) + B b(x_{m}+ 2a) + B b(x_{m}- 2a).

(b) Eexp(ikx_{n}) = E_{0}expik(x_{n}) - Aexp(ik(x_{n}+ a)) - Aexp(ik(x_{n}- a))

+ Bexp(ik(x_{n}+ 2a)) + Bexp(ik(x_{n}- 2a)).

E = E_{0}- Aexp(ika) - Aexp(-ika)) + Bexp(i2ka) + Bexp(i2ka) = E_{0}- 2Acos(ka) + 2Bcos(2ka).

Plug in A and B.

E(k) = [ħ^{2}/(ma^{2})][7/8 - cos(ka) + (1/8) cos(2ka)].

We only need to consider -π/a ≤ k < π/a because k values differing by 2π/a give the same coefficients.

The possible eigenvalues of H lie in a band from E_{0}- 2(A - B) to E_{0}+ 2(A - B).

(c) An expansion of E(k') about k' = 0 for k' = 0 + k yields

E(k) = E_{0}- 2(A - B) + Ak^{2}a^{2}- 4Bk^{2}a^{2}= E_{min}+ Ak^{2}a^{2}- 4Bk^{2}a^{2}.

Let E(k) - E_{min}= Ak^{2}a^{2}- 4Bk^{2}a^{2}= (ħ^{2}k^{2}/4m)) = ½m*v_{g}^{2}.

Definition of group velocity:

v_{g}= dω/dk|_{k0}= (1/ħ)dE/dk|_{k0}= (2Aa/ħ)sin(k_{0}a) - (4Ba/ħ)sin(2k_{0}a).

An expansion of v_{g}(k') about k' = 0 for k' = 0 + k yields (2Aka^{2}/ħ) - (8Bka^{2}/ħ) = ½(ħk/m).

(ħ^{2}k^{2}/4m)) = ½ m*(( ħk/(2m))^{2}, m* = 2m.

We can also use m* = ħ^{2}(d^{2}E/dk^{2})^{-1}. dE/dk = (ħ^{2}k/(2m)) d^{2}E/dk^{2}= (ħ^{2}/(2m)), m* = 2m.

And expansion about k' = π/a for k' = π/a - k yields

E(k) = E_{0}+ 2(A - B) - Ak^{2}a^{2}+ 4Bk^{2}a^{2}= E_{max}- (ħ^{2}k^{2}/(4m)).

m* = ħ^{2}(d^{2}E/dk^{2})^{-1}= -2m.

Note dv_{g}/dt = F_{ext}/m*. If an external force is applied, the electron in a crystal is subject to this force and internal forces from the lattice. The net force may point opposite to the external force.

Remarks:

The potential energy function for the one-dimensional crystal is periodic, U(x + a) = U(x).

Therefore the Hamiltonian H commutes with the translation operator U(T_{a}).

In coordinate representation U(T_{a})f(x) = f(x - a) for any function f(x).

Wave functions of the form ψ(x) = e^{ikx }v(x) are eigenfunctions of U(T_{a}) if v(x + a) = v(x), i.e. if v(x) is a periodic function with period a.

U(T_{a})ψ(x) = U(T_{a})e^{ikx }v(x) = e^{ik(x-a) }v(x - a) = e^{-ika}e^{ikx }v(x) = e^{-ika}ψ(x).

The eigenvalue is e^{-ika}ψ(x).

All eigenfunctions of U(T_{a}) are of this form.

since H and U(T_{a}) commute, we can therefore find a complete set of eigenfunctions of H of this form. These functions are called Bloch functions.

For the current problem, |Φ> = ∑_{n}exp(ikx_{n})|n> becomes Φ(x) = ∑_{n}exp(ikx_{n})Φ_{n}(x) in coordinate representation.

If we assume that the Φ_{n}(x) = Φ_{0}(x - na), then Φ(x) is a Bloch function.

Φ(x) = ∑_{n}exp(ikna)Φ_{n}(x) = e^{ikx}∑_{n}exp(-ik(x - na))Φ_{0}(x - na) = e^{ikx }v(x).

v(x) = ∑_{n}exp(-ik(x - na))Φ_{0}(x - na), n = -∞ to +∞.

v(x + a) = ∑_{n}exp(-ik(x - (n - 1)a))Φ_{0}(x - (n - 1)a) = ∑_{m}exp(-ik(x - ma))Φ_{0}(x - ma) = v(x).

The set {|n>} of orthonormal electron states, n = -∞ to +∞ is not complete. But if we restrict ourselves to this subspace, we may be able to approximately solve for the possible E(k) in one energy band of the crystal.