### 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, ψ1i(x) = (2/a) ½sin(πx/a),  E1i = π2ħ2/(2ma2).
For t > 0, the wave function of the first excited state is ψ2f(x) = (1/a) ½sin(πx/a),
Enf = n2π2ħ2/(8ma2).   E2f = E1i.  If we find the particle in state ψ2f its energy has not changed.
The wave function just after the wall is removed is ψ1i(x).
The probability of finding the particle in the state ψ2f(x) of the new well is |<ψ2f1i |2.
|<ψ2f1i>|2 = (2/a2)| ∫0asin2(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))
2/2m)(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) = 21(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
Φ0i(x) = (mω/(πħ))¼exp(-½mωx2/ħ).
The normalized ground-state wave function for t > 0 is
Φ0f(x) = (m2ω/(πħ))¼exp(-mωx2/ħ).
The probability that the oscillator remains in the ground state is
P = |<Φ0i0f>|2 =  |∫-∞dx Φ0i*(x) Φ0f(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, ψ0i(x) = (2/L)½sin(πx/L), E0i = π2ħ2/(2m0L2).
For t > 0, the wave function of the ground state is ψ0f(x) = (2/L)½sin(πx/L),
E0f = λπ2ħ2/(2m0L2).
The probability that the particle remains in the ground state is 1.
The change in energy is E0f - E0i = (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)exp(-x2/(2b2)),  with b = (ħ2/(mC))¼.
The energy of the ground state is E0 = ½ħω = ½ħ(C/m)½.
For t < 0, ψ0i(x) = (πbi 2)exp(-x2/(2b i 2)),  E0i = ½ħ(C/m0)½, with bi = (ħ2/(m0C))¼.
For t > 0, ψ0f(x) = (πbf 2)exp(-x2/(2bf 2)),  E0f = ½ħ(λC/m0)½, with bf = (λħ2/(m0C))¼.
The probability that the particle remains in the ground state is |< ψ1f | ψ0i >|2.
< ψ1f | ψ0i > = (πbibf)exp ∫-∞dx exp(-½x2(1/bi2 + 1/bf2) = (2bibf/(bi2 + bf2))½.
The probability that the particle remains in the ground state is
2bibf/(bi2 + bf2) = 2λ¼/(1 + λ½) < 1.
The change in energy is E0f - E0i = ½(λ½ - 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 ~ me = 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 + Ekl)]ukl(r) = 0.

Defining
a0 = ħ2/(μe2),  ρ = r/a0,  λkl2 = -2Ekla0/e2 = -Ekl/(μe2/(2ħ2)) = -Ekl/EI,
we write
[(∂2/∂ρ2) - l(l + 1)/ρ2 + (2/ρ) - λkl2)]ukl(ρ) = 0.

For a hydrogenic atom we have
[(∂2/∂r2) - l(l + 1)/r2 + (2μ'/ħ2)(Ze2/r + Ekl)]ukl(r) = 0.

Defining
a0'= ħ2/(μZe2),  ρ' = r/a0',  λ'kl2 = -2Ekla0'/(Ze2) = -Ekl/(μ'Z2e2/(2ħ2)) = -Ekl/EI',
we write
[(∂2/∂ρ'2) - l(l + 1)/ρ'2 + (2/ρ') - λ'kl2)]ukl(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 EI by EI' = μ'Z2e4/(2ħ2) = EI(μ'/μ)Z2.
(a)  Tritium:
The reduced mass is μ' = me3mp/(me + 3 mp) ≈ me
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' = me3mp/(me + 3 mp) ≈ me
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∫0r2dr exp(-3r/a0)]2
= [(√2)3(4/27)∫0x2dx exp(-x)]2
= [(√2)3(8/27)]2 = 0.7.