The sudden approximation

Problem:

Two infinite potential wells are extending from x = -a to x =  0 and from x = 0 to  x = a, respectively.  A particle is in its ground state in the right well.  At t = 0 the barrier at x = 0 is removed.  What is the probability of finding the particle in the first excited state of the new well?

Solution:

• Concepts:
  A particle in an infinite square well, the sudden approximation
• Reasoning:
  The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly.  The reaction time is so short that the transition amplitude <β|U(t₂,t₁)|α> is simply given by the overlap <β|α>.  The transition probability is |<β|α>|².  Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition.
• Details of the calculation:
  For t < 0, ψ_1i(x) = (2/a) ^½sin(πx/a),  E_1i = π²ħ²/(2ma²).
  For t > 0, the wave function of the first excited state is ψ_2f(x) = (1/a) ^½sin(πx/a),
  E_nf = n²π²ħ²/(8ma²).   E_2f = E_1i.  If we find the particle in state ψ_2f its energy has not changed.
  The wave function just after the wall is removed is ψ_1i(x).
  The probability of finding the particle in the state ψ_2f(x) of the new well is |<ψ_2f|ψ_1i |².
  |<ψ_2f|ψ_1i>|² = (2/a²)| ∫₀^asin²(πx/a)dx|² = ½.

Problem:

Before t = 0, the ground state wave function of a particle with mass m is
ψ(x,t) = N(sin(πx/L) + sin(2πx/L))exp(-iEt/ħ) for 0 < x < L, ψ(x,t) = 0 everywhere else. 
Set the potential energy of the particle to zero at x = L/2.
(a)  Find the ground state energy E and the potential energy function U(x) before t = 0.
(b)  At t = 0 the potential energy function U(x) suddenly changes to U(x) = 0 for 0 < x < L, U(x) = ∞ everywhere else.  For t > 0, find the probability of measuring the different energy eigenvalues of the particle in this infinite well for t > 0.
(c)  Find P(x,t) for t < 0 and t > 0.

Solution:

• Concepts:
  The Schroedinger equation, the sudden approximation
• Reasoning:
  The given wave function is a solution of the 1-dimensional, time-independent Schroedinger equation.
• Details of the calculation:
  (a)  Hψ(x) = Eψ(x),  (-ħ²/2m)(∂²/∂x²)Ψ(x) + U(x)Ψ(x) = EΨ(x).
  (∂²/∂x²)Ψ(x) = -N(π/L)²(sin(πx/L) + 4*sin(2πx/L))
  (ħ²/2m)(π/L)²(sin(πx/L) + 4*sin(2πx/L)) = (E - U(x))(sin(πx/L) + sin(2πx/L))
  Inserting U(x) = 0 at x = L/2 we find E = (π²ħ²/(2mL²).
  Inserting E we find
  U(x) = [-(ħ²/2m)(π/L)²(sin(πx/L) + 4*sin(2πx/L)) + (ħ²/2m)(π/L)²(sin(πx/L) + sin(2πx/L))]/(sin(πx/L) + sin(2πx/L))
  U(x) = -(3π²ħ²/(2mL²)sin(2πx/L)/(sin(πx/L) + sin(2πx/L)).
  (b)  The wave function ψ(x,0) Is a linear superposition of eigenstates of the new infinite-well Hamiltonian.
  Normalize:  ψ(x,0) = 2^-½[ψ₁(x,0) + ψ₂(x,0)], where 
  ψ₁(x,t) = (2/L)^½sin(πx/L))exp(-iE₁t/ħ)  and ψ₂(x,t) = (2/L)^½sin(2πx/L)exp(-iE₂t/ħ),
  with E₁ = π²ħ²/(2mL²)  and  E₂ = 4π²ħ²/(2mL²),
  are solutions to the one-dimensional infinite square-well problem.
  The probability of measuring E₁ is ½, the probability of measuring E₂ is ½, the probability of measuring any other eigenvalue is zero.
  (c)  For t < 0:  P(x,t) = |ψ(x,t)|² is the probability per unit length of finding the particle at position x.
  ψ(x,t) = (1/L)^½(sin(πx/L) + sin(2πx/L))exp(-iEt/ħ)
  |ψ(x,t)|² = (1/L)|sin(πx/L) + sin(2πx/L)|²
  = (1/L)[sin²(πx/L) + sin²(2πx/L) + sin(πx/L)sin(2πx/L)]
  P(x,t) = (1/L)[sin²(πx/L) + sin²(2πx/L) + 2sin(πx/L)sin(2πx/L)], independent of time.
  
  for t > 0:  P(x,t) = |ψ(x,t)|² = (1/L)|sin(πx/L)exp(-iE₁t/ħ) + sin(2πx/L)exp(-iE₂t/ħ)|²
  = (1/L)(sin(πx/L)exp(-iE₁t/ħ) + sin(2πx/L)exp(-iE₂t/ħ)
  *(sin(πx/L)exp(-iE₁t/ħ) + sin(2πx/L)exp(-iE₂t/ħ)
  = (1/L)[sin²(πx/L) + sin²(2πx/L) + sin(πx/L)sin(2πx/L)(exp(i(E₂-E₁)t/ħ) + exp(-i(E₂-E₁)t/ħ)]
  = (1/L)[sin²(πx/L) + sin²(2πx/L) + 2sin(πx/L)sin(2πx/L)cos((E₂-E₁)t/ħ)].
  P(x,t) = (1/L)[sin²(πx/L) + sin²(2πx/L) + 2sin(πx/L)sin(2πx/L)cos((E₂-E₁)t/ħ)].
  The probability per unit length of finding the particle at x is changing with time.

Problem:

A particle of mass m is in the ground state of a harmonic oscillator with spring constant k = mω².  
At t = 0, the spring constant changes suddenly to k' = 4k.  
Find the probability that the oscillator remains in its ground state.

∫_-∞^∞ exp(-x²/a²)dx = √(π) a

Solution:

• Concepts:
  The QM harmonic oscillator, the sudden approximation
• Reasoning:
  The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly.  The reaction time is so short that the transition amplitude <β|U(t₂,t₁)|α> is simply given by the overlap <β|α>.  The transition probability is |<β|α>|².  Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition.
• Details of the calculation:
  The normalized ground-state wave function of the 1D harmonic oscillator for t < 0 is
  Φ_0i(x) = (mω/(πħ))^¼exp(-½mωx²/ħ).
  The normalized ground-state wave function for t > 0 is
  Φ_0f(x) = (m2ω/(πħ))^¼exp(-mωx²/ħ).
  The probability that the oscillator remains in the ground state is
  P = |<Φ_0i|Φ_0f>|² =  |∫_-∞^∞dx Φ_0i*(x) Φ_0f(x)|² = 2√2/3.

Problem:

Consider a one-dimensional quantum particle with a Hamiltonian H = T + U(x),
T = -(ħ²/2m)(∂²/∂x²).  Suppose that m suddenly changes from m₀ to m₁ = m₀/λ at t = 0.
Assuming the particle was in the ground state at t < 0, find
(i) the probability it remains in the ground state at t > 0 and
(ii) the change in the energy expectation value <H>.  Consider two cases:
(a)  The infinite well, U(x) = 0 for 0 < x < L, and U(x) infinite otherwise.
(b)  The parabolic well,  U(x) = Cx²/2.

Hint:  for any a we have  ∫_-∞^∞dx exp(-ax²) = (π/a)^½.

Solution:

• Concepts:
  A particle in a well, the sudden approximation
• Reasoning:
  The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly.  The reaction time is so short that the transition amplitude <β|U(t₂,t₁)|α> is simply given by the overlap <β|α>.  The transition probability is |<β|α>|².  Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition.
• Details of the calculation:
  (a)  For t < 0, ψ_0i(x) = (2/L)^½sin(πx/L), E_0i = π²ħ²/(2m₀L²).
  For t > 0, the wave function of the ground state is ψ_0f(x) = (2/L)^½sin(πx/L),
  E_0f = λπ²ħ²/(2m₀L²).
  The probability that the particle remains in the ground state is 1.
  The change in energy is E_0f - E_0i = (λ - 1)π²ħ²/(2m₀L²).
  
  (b)  The normalized wave function of the ground state of a particle of mass m in the periodic well is
  ψ₀(x) = (mω/(πħ))^¼exp(-mωx²/(2ħ)), with ω² = C/m, or
  ψ₀(x) = (πb²)^-¼exp(-x²/(2b²)),  with b = (ħ²/(mC))^¼.
  The energy of the ground state is E₀ = ½ħω = ½ħ(C/m)^½.
  For t < 0, ψ_0i(x) = (πb_i ²)^-¼exp(-x²/(2b _i ²)),  E_0i = ½ħ(C/m₀)^½, with b_i = (ħ²/(m₀C))^¼.
  For t > 0, ψ_0f(x) = (πb_f ²)^-¼exp(-x²/(2b_f ²)),  E_0f = ½ħ(λC/m₀)^½, with b_f = (λħ²/(m₀C))^¼.
  The probability that the particle remains in the ground state is |< ψ_1f | ψ_0i >|².
  < ψ_1f | ψ_0i > = (πb_ib_f)^-½exp ∫_-∞^∞dx exp(-½x²(1/b_i² + 1/b_f²) = (2b_ib_f/(b_i² + b_f²))^½.
  The probability that the particle remains in the ground state is
  2b_ib_f/(b_i² + b_f²) = 2λ^¼/(1 + λ^½) < 1.
  The change in energy is E_0f - E_0i = ½(λ^½ - 1)ħ(C/m₀)^½.

Problem:

The ground state wave function for a hydrogen like atom is
Φ₁₀₀(r) = (1/√π)(Z/a₀)^3/2 exp(-Zr/a₀),
where  a₀ = ħ²/(μe²) and μ is the reduced mass, μ ~ m_e = mass of the electron. 
(a)  What is the ground state wave function of tritium?
(b)  What is the ground state wave function of ³He⁺?
(c)  An electron is in the ground state of tritium.  A nuclear reaction instantaneously changes the nucleus to ³He⁺.  Assume the beta particle and the neutrino are immediately removed from the system.  Calculate the probability that the electron remains in the ground state of ³He⁺.

Solution:

• Concepts:
  The hydrogenic atom, the sudden approximation
• Reasoning:
  Tritium and He⁺ are hydrogenic atoms.  Their wave functions and energy levels can be obtained from the wave functions and energy levels of the hydrogen atom using scaling rules.  We assume that right after the decay of tritium the wave function of the electron is the same as right before the decay.  It does not have time to change.  This is the sudden approximation.

• Details of the calculation:
  (a)  For a hydrogen atom we have
  [(∂²/∂r²) - l(l + 1)/r² + (2μ/ħ²)(e²/r + E_kl)]u_kl(r) = 0.
  
  Defining
  a₀ = ħ²/(μe²),  ρ = r/a₀,  λ_kl² = -2E_kla₀/e² = -E_kl/(μe²/(2ħ²)) = -E_kl/E_I,
  we write
  [(∂²/∂ρ²) - l(l + 1)/ρ² + (2/ρ) - λ_kl²)]u_kl(ρ) = 0.
  
  For a hydrogenic atom we have
  [(∂²/∂r²) - l(l + 1)/r² + (2μ'/ħ²)(Ze²/r + E_kl)]u_kl(r) = 0.
  
  Defining
  a₀'= ħ²/(μZe²),  ρ' = r/a₀',  λ'_kl² = -2E_kla₀'/(Ze²) = -E_kl/(μ'Z²e²/(2ħ²)) = -E_kl/E_I',
  we write
  [(∂²/∂ρ'²) - l(l + 1)/ρ'² + (2/ρ') - λ'_kl²)]u_kl(ρ') = 0.
  
  The same equations have the same solutions.  To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom a₀ by a₀' = ħ²/(μ'Ze²) = a₀(μ/μ')(1/Z), and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_I by E_I' = μ'Z²e⁴/(2ħ²) = E_I(μ'/μ)Z².
  (a)  Tritium: 
  The reduced mass is μ' = m_e3m_p/(m_e + 3 m_p) ≈ m_e
  assuming the neutron mass equals the proton mass.
  The nuclear charge is Z = 1.  Since a₀' ~ a₀ the ground state wave function of the tritium is the ground state wave function of the hydrogen atom.
  Φ₁₀₀(³H) = (1/√π)(1/a₀)^3/2 exp(-r/a₀).
  (b) ³He⁺: 
  The reduced mass is μ' = m_e3m_p/(m_e + 3 m_p) ≈ m_e
  assuming the neutron mass equals the proton mass.
  The nuclear charge is Z = 2.  Since  a₀' ~ a₀/2 the ground state wave function of the ³He⁺ is
  Φ₁₀₀(³He⁺) = (1/√π)(2/a₀)^3/2 exp(-2r/a₀).
  (c)  Sudden approximation:
  P = |<Φ₁₀₀(³H) |Φ₁₀₀(³He⁺) >|²
  = [(1/π)(√2/a₀)³∫d³r exp(-3r/a₀)]²
  = [(√2/a₀)³4∫₀^∞r²dr exp(-3r/a₀)]²
  = [(√2)³(4/27)∫₀^∞x²dx exp(-x)]²
  = [(√2)³(8/27)]² = 0.7.