## Assignment 7, solutions

Problem 1:

Calculate the uncertainty product <Δx><Δp> for the ground state of the infinite square-well potential.   How can we tell if the ground-state wave function is or is not a minimum uncertainty state?

Solution:

 Concepts: Fundamental assumptions of QM, the eigenstates of the 1D infinite well Reasoning: The eigenfunctions of the infinite well are ψn(x) = (2/L)½sin(nπx/L), En = n2π2ħ2/(2mL2) for a well of width L. For any operator A we have ∆A = (<(A - )2>)½ = ( - 2)½. Details of the calculation: ∆x =  ( - 2)½.  = L/2 from symmetry. = (2/L)∫0L dx  x2 sin2(πx/L) = (2L2/π3)∫0π dy  y2 sin2(y) = (2L2/π3)(π3/6 - π/4) = L2/3 - L2/(2π2).∆x = (L2/3 - L2/(2π2) - L2/4)½ = (L/2)(⅓ - 2/π2)½. ∆p =  ( -

2)½.

= 0 for any stationary state. =  -(2ħ2/L)∫0L dx  sin(πx/L) (∂2/∂x2) sin(πx/L)= (2ħ2π/L2)∫0π dy  sin2(y) = (ħ2π2/L2).∆p = (ħπ/L).∆x∆p = (ħ/2)π(⅓ - 2/π2)½ =1.136*(ħ/2).For a minimum uncertainty state, we should have ∆x∆p = (ħ/2) and since 1.136 > 1 we see that the ground state is NOT a minimum uncertainty state.

Problem 2:

Consider a spinless particle of mass m moving in one dimension in the presence of a delta-function potential well, U(x) = -λ δ(x),  λ > 0.
(a)  Evaluate the transmission coefficient T(E) as a function of the incident energy E > 0.
(b)  Find the energy of the bound state for this potential well.
(c)  Comment on the pole in the expression for T(E) from part (a) in light of your result
from part (b).

Solution:

 Concepts: Piecewise constant potentials Reasoning: This is a "square potential" problem.  We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions. U(x) = 0 everywhere except at x = 0. Details of the calculation: (a)  E > 0.  Φ1(x) = A1 exp(ikx) + A1'exp(-ikx) for x < 0.  k2 = 2mE/ħ2.Φ2(x) = A2 exp(ikx)  for x > 0.Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 + A1' = A2.∂Φ/∂x has a finite discontinuity at x = 0.∂2Φ(x)/∂x2 + (2m(E - U)/ħ2)Φ(x) = 0.Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.∂Φ2ε(x1 + ε)/∂x - ∂Φ1ε(x1 - ε)/∂x = -(2m/ħ2)∫x1-εx1+ε (E + λ δ(x)) Φ(x) dx= -(2mλ/ħ2)Φ(0).iA1k – iA1'k = iA2k + A22mλ/ħ2,  A1 – A1' = A2 + A22mλ/(ikħ2)  = [1 + 2mλ/(ikħ2)]A2.Eliminate A1':  2A1 = [2 + 2mλ/(ikħ2) ]A2,  A2/A1 = 1/[1 + mλ/(ikħ2)] = ikħ2/(ikħ2 + mλ).T(E) = (k|A2|2)/(k|A1|2) = ħ4k2/(ħ4k2 + m2λ2) = E/(E + mλ2/(2ħ2)). (b)  E < 0.Φ1(x) = A1 exp(ρx) + A1'exp(-ρx) for x < 0.  ρ2 = -2mE/ħ2.Φ2(x) = A2 exp(ρx) + A2'exp(-ρx) for x > 0.Φ is finite at infinity.  A1' = A2 = 0.  Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 = A2'.∂Φ2/∂x|x=0 = ∂Φ1/∂x|x=0 - (2mλ/ħ2)Φ(0).-ρA2' + (2mλ/ħ2)A2' = ρA2',  ρ = mλ/ħ2.m2λ2/ħ4 = -2mE/ħ2,  E = -mλ2/(2ħ2).Only one bound state exists. (c)  The expression for T(E) has a pole when E = -mλ2/(2ħ2) = energy of the bound state.T --> ∞, but E is negative, we have no incident wave.

Problem 3:

Use what you know about the eigenfunctions of the harmonic oscillator in coordinate space to find the momentum-space eigenfunctions of a harmonic oscillator

Solution:

 Concepts: The eigenfunctions of the harmonic oscillator Reasoning: The same equations have the same solutions. Details of the calculation: H = p2/(2m) + ½mω2x2. In coordinate space (∂2/∂x2)Φ(x) + (2m/ħ2)(E - ½mω2x2 )Φ(x) = 0 is the eigenvalue equation for H. The eigenvalues are En = (n + ½)ħω, n = 0, 1, 2, ...  .  With y = βx and β = (mω/ħ)½ the eigenvalue equation becomes (∂2/∂y2)Φ(y) + (γ2E - y2 )Φ(y) = 0, with γ = (2/ωħ)½. and the normalized eigenfunctions are Φn(x) = (n!2n)-½(β½/π¼)Hn(βx)exp(-(βx)2/2), Φn(y) = (n!2n)-½(1/π¼)Hn(y)exp(-y2/2). (Note: ∫-∞∞|Φn(y)|2dy = ∫-∞∞|Φn(y)|2d(βx) =  ∫-∞∞|Φn(x)|2dx =1.) Write out the eigenvalue equation for H in momentum space.This is accomplished by replacing x with iħd/dp.  [p2/(2m) - ½mω2ħ2d2/dp2]Φ(p) = EΦ(p).(d2/dp2)Φ(p) + (2/mω2ħ2)(E - p2/(2m))Φ(p) = 0.With y = αp and α = (mωħ)-½ the eigenvalue equation becomes(∂2/∂y2)Φ(y) + (γ2E - y2 )Φ(y) = 0,and the normalized eigenfunctions are Φn(y) = (n!2n)-½(1/π¼)Hn(y)exp(-y2/2). Φn(p) = (n!2n)-½(α½/π¼)Hn(αp)exp(-(αp)2/2).The eigenvalues are En = (n + ½)ħω, independent of the representation.

Problem 4:

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) d2Φ/dt2 + I/C = d2Φ/dt2 + Φ/(LC) = 0.With C = m and k = 1/L we have d2Φ/dt2 + (k/m)Φ = 0.  This is the equation of motion of a simple harmonic oscillator with Φ being a generalized coordinate.(b)  LLag = ½C(dΦ/dt)2 - ½Φ2/L. PΦ = ∂LLag/∂(dΦ/dt) = C(dΦ/dt). H = PΦ2/(2C) + ½Φ2/L.(c)  Let PΦ and Φ become non-commuting operators, [Φ, PΦ] = iħ.Then H becomes the Hamiltonian operator  -ħ2/(2C) ∂2/∂Φ2 + ½Φ2/L,and Hψ(Φ) = Eψ(Φ).The eigenvalues of H = (n + ½)ħω, with ω = 1/(LC)½.  The ground state energy is ½ħω/(LC)½.(d)  We are assuming zero resistance.  If kT << ħω/(LC) the probability of thermal excitation and de-excitation is small.  But in an excited state, the system radiates.

Problem 5:

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> = E0,  <n|H| n±1> = -A, <n|H| n±2> = B,  <n|H|m> = 0 for all other m, with
E0 = (7/8)ħ2/(ma2),  A = ½ħ2/(ma2),  and B = (1/16)ħ2/(ma2).

(a) Assume the eigenstates of H are of the form |Φ> = ∑n b(xn)|n>.
Write down the coupled linear equations for the b(xn).  Note that b(xn±1) = b(xn ± a).
(b)  Try solutions of the form b(xn) = exp(ikxn) and show that E as a function of k is given by
E(k) = [ħ2/(ma2)][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)   = ∑n b(xn) .E b(xm) = E0 b(xm) – A b(xm+1) – A b(xm-1) +  B b(xm+2) + B b(xm-2)= E0 b(xm) – A b(xm + a) – A b(xm - a) + B b(xm + 2a) + B b(xm – 2a). (b)  Eexp(ikxn) = E0expik(xn) – Aexp(ik(xn + a)) – Aexp(ik(xn - a)) + Bexp(ik(xn + 2a))  + Bexp(ik(xn - 2a)).E = E0 – Aexp(ika) – Aexp(-ika))  + Bexp(i2ka)  + Bexp(i2ka) = E0 – 2Acos(ka) + 2Bcos(2ka).E(k) = [ħ2/(ma2)][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 E0 – 2(A - B) to E0 + 2(A – B). (c)  An expansion of E(k) about k = 0 yields E(k) = E0 - 2(A - B) + Ak2a2 - 4Bk2a2 = Emin + Ak2a2 - 4Bk2a2.Let E(k) – Emin = Ak2a2 - 4Bk2a2 = ½m*vg2.vg = dω/dk|k0 = (1/ħ)dE/dk|k0 = (2Aa/ħ)sin(k0a) - (4Ba/ħ)sin(2k0a).An expansion of vg(k) about k = 0 yields (2Aka2/ħ) - (8Bka2/ħ) = (ħk/m)½.(ħ2k2/4m))  = ½ m*(( ħk/(2m))2,  m* = 2m.We can also use m* =  ħ2(d2E/dk2)-1.   dE/dk = (ħ2k/(2m))  d2E/dk2 = (ħ2/(2m)),  m* = 2m.And expansion about k = π/a yields E(k) = Emax - Ak2a2 + 4Bk2a2 = Emax - (ħ2k2/(4m)).m* = ħ2(d2E/dk2)-1 = -2m. Note dvg/dt = Fext/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(Ta).In coordinate representation U(Ta)f(x) = f(x - a) for any function f(x).All non-degenerate eigenfunctions of H are eigenfunctions of the translation operator U(Ta).Wave functions of the form ψ(x) = eikx v(x) are eigenfunctions of U(Ta) if v(x + a) = v(x), i.e. if v(x) is a periodic function with period a.U(Taψ(x) = U(Ta)eikx v(x) = eik(x-a) v(x - a) = e-ikaeikx v(x) = e-ikaψ(x).  The eigenvalue is e-ikaψ(x).We can therefore find a complete set of eigenfunctions of H of this form.  These functions are called Bloch functions.In coordinate representation |Φ> = ∑n exp(ikxn)|n> becomes Φ(x) = ∑n exp(ikxn)Φn(x).If we assume that the Φn(x) = Φ0(x - na), then Φ(x) is a Bloch function.Φ(x) = ∑nexp(ikna)Φn(x) = eikx∑nexp(-ik(x - na))Φ0(x - na) = eikx v(x).  v(x) = ∑nexp(-ik(x - na))Φ0(x - na),  n = -∞ to +∞.  v(x + a) = ∑nexp(-ik(x - (n - 1)a))Φ0(x - (n - 1)a) = ∑mexp(-ik(x - ma))Φ0(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.