More Problems

Problem 1:

For a one-dimensional simple harmonic oscillator, consider a correlation function defined as C(t) = <x(t)x(0)>, where  x(t) = x cos(ωt) + p sin(ωt)/(mω), with x and p being the position and momentum operators.
Calculate C(t) explicitly for the ground state of a one-dimensional simple harmonic oscillator, i.e. C(t) = <0|x(t)x(0)|0>, where |n> refers to the eigenstate of a simple harmonic oscillator with energy (n + ½)ħω.

Hint: x and p can be expressed as x = (a + a^†)/(2α),  p = -i(a - a^†)/(2β), where α =√(mω/(2ħ)),  β =1/√(2mωħ).
a|n> = √(n) |n-1>,  a^†|n> = √(n+1) |n+1>.

Solution:

• Concepts:
  The harmonic oscillator, raising and lowering operators
• Reasoning:
  The mean value of any observable A is <ψ|A|ψ>.
• Details of the calculation:
  x(0) = x.  C(t) = x(t)x(0) = x²cos(ωt) + pxsin(ωt)/(mω).
  In terms of a and a^†
  x² = (ħ/(2mω))(aa + a^†a^† + a^†a  + aa^†),  px = -i(ħ/2)(aa - a^†a^† - a^†a  + aa^†).
  <0|x²|0> = (ħ/(2mω))(<0|aa^†)|0> = (ħ/(2mω)).
  <0|xp|0> = -i(ħ/2)(<0|aa^†)|0> = -iħ/2.
  <0|C(t)|0> = (ħ/(2mω))cos(ωt) - i(ħ/(2mω))sin(ωt) = (ħ/(2mω))exp(-iωt).

Problem 2:

Find the allowed energies of the "half" harmonic oscillator
U(x) =  ½mω²x²,  x > 0,
U(x) = ∞,  x ≤ 0.

Solution:

• Concepts:
  The eigenfunctions of the 1D harmonic oscillator
• Reasoning:
  The odd eigenfunctions of the harmonic oscillator with U(x) = ½ m ω²x² for all x are eigenfunctions of H for this potential.
• Details of the calculation:
  The eigenvalues therefore are E = (n + ½)ħω, n = odd
  or rewriting, E = (2n + 3/2)ħω, n = 0, 1, 2, ... .
  The ground state energy is (3/2)ħω = (3/2)(k/m)^½.

Problem 3

Consider the following non-pure state for a hydrogenic atom:
|ψ> = a₁|ψ₁₀₀> +  a₂|ψ₂₀₀> + a₃|ψ₂₁₀> + a₄|ψ_32-1> + a₅|ψ₄₃₂>.
a₁ = (3/10)^½,  a₂ = (1/10)^½,  a₃ = (2/10)^½,  a₄ = (1/10)^½,  a₅ = (3/10)^½.
i)  Show that |ψ> is normalized.  What property of the hydrogen wave functions must you exploit to show that?
ii)  Calculate the probability of observing a 1s state.
iii)  Calculate the probability of observing a state with n > 2.
iv)  What are the possible values you would get if you measured the quantity associated with L_z?  Give the probability of measuring each of these values.
v)  Calculate the expectation value of L_z.
vi)  What are the possible values you would get if you measured the quantity associated with L²?  Give the probability of measuring each of these values.
vii)  Calculate the expectation value of L².

Solution:

• Concepts:
  The hydrogen atom, postulates of QM
• 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) = ∑_i=0^gn|<u_nⁱ|ψ>|²,
  where {|u_nⁱ>} (i = 1, 2, ..., g_n) is an orthonormal basis in the eigensubspace E_n associated with the eigenvalue a_n.
• Details of the calculation:
  |ψ> = a₁|ψ₁₀₀> +  a₂|ψ₂₀₀> + a₃|ψ₂₁₀> + a₄|ψ_32-1> + a₅|ψ₄₃₂>.
  The |ψ_nlm> are normalized common eigenstates of H, L², and L_z.
  H|ψ_nlm> = E_n|ψ_nlm>,  L²|ψ_nlm> = l(l + 1)ħ²|ψ_nlm>,  L_z|ψ_nlm> = mħ|ψ_nlm>.
  <ψ_nlm|ψ_nlm> = 1.  <ψ_n'l'm'|ψ_nlm> = 0 unless n' = n, l' = l, m' = m.
  <ψ_nlm|ψ_nlm> stands for  ∫_{all space}|ψ_nlm (r,θ,φ)|²dV.
  i)  <ψ|ψ> = a₁² + a₂² + a₃² + a₄² + a₅² = 1.  The wave function is normalized because the hydrogen-atom wave functions are orthonormal.
  ii)  The probability of observing the ith state in the expansion is |a_i|².
  The probability of observing the 1s state is |a₁|² = 3/10.
  iii)  The 4^th and the 5^th state in the expansion have n > 2.
  P(n > 2) = probability of observing the 4^th state + probability of observing the 5^th state = 4/10.
  iv)  The values you can obtain for L_z are mħ, where m = 0, -1, or 2.
  P(m = 0 ) = a₁² + a₂² + a₃² = 6/10.  P(m = -1) = 1/10, P(m = 2) = 3/10.
  v)  <L_z> = ( 0 *6/10 + -1*1/10 + 2*3/10) mħ  = ½mħ.
  vi)  The values you can obtain for L² are l(l + 1)ħ², where l = 0, 1, 2, or 3.
  P(l = 0) = a₁² + a₂² = 4/10,  P(l = 1) = 2/10,  P(l = 2) = 1/10,  P(l = 3) = 3/10.
  vii)  <L²> =  (0 *4/10 + 2*2/10 + 6*1/10 + 12*3/10)ħ² = (23/5)ħ².

Problem 4:

Assume an antiproton is orbiting a bare ⁵⁶₂₆Fe nucleus.  The average radius of a nucleus with A nucleons is R = R₀A^1/3, where R₀ = 1.2*10^-15 m.
(a)  Treating this as a hydrogenic atom, what is the ground state energy of the atom?  Do you see any problems?
(b)  What is the most probable distance, r_{most_prob}, of the antiproton from the center of the nucleus in this hydrogenic approximation?  Compare this to the approximate radius of the nucleus.  Do you see any problems?
(c)  If you assumed a classical orbit with a radius of r_{most_prob} from part (b), what would be the effective charge "seen" by the antiproton?

Solution:

• Concepts:
  The hydrogenic atom, the ground-state energy of the hydrogen atom, the Bohr radius
• Reasoning:
  We use scaling rules, a₀ --> a₀(μ/μ')(1/Z), E_I --> E_I(μ'/μ)Z².
• Details of the calculation:
  (a)  E_ground = -Z²13.6 eV (μ'/μ). 
  μ'/μ = (1.67*10^-27 kg)/(9.1*10^-31 kg) = 1.84*10³.
  E = -26²*13.6 eV*1.84*10³ = -17 MeV.
  This is still a non-relativistic problem.
  (b)  r_{most_prob} = a₀μ/(μ'Z) = 5*29*10^-11 m/(1.84*10³*26) = 1.11*10^-15.
  The approximate radius of the nucleus is 4.59*10^-15 m.
  The antiproton is orbiting inside the nucleus.  The nucleus cannot be treated like a point charge.  The hydrogenic approximation is not valid for this system.
  (c)  Z_eff = 26(1.1/4.59)³ = 0.37.