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 left 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(t_{2},t_{1})|a> 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)^{ ½}sin(πx/a), E_{1i}= π^{2}ħ^{2}/(2ma^{2}).

For t > 0, the wave function of the first excited state is ψ_{2f}(x) = (1/a)^{ ½}sin(πx/a),

E_{nf}= n^{2}π^{2}ħ^{2}/(8ma^{2}). 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/a^{2})| ∫_{0}^{a}sin^{2}(πx/a)dx|^{2}= ½.

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), (-ħ^{2}/2m)(∂^{2}/∂x^{2})Ψ(x) + U(x)Ψ(x) = EΨ(x).

(∂^{2}/∂x^{2})Ψ(x) = -N(π/L)^{2}(sin(πx/L) + 4*sin(2πx/L))

(ħ^{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 = (π^{2}ħ^{2}/(2mL^{2}).

Inserting E we find

U(x) = [-(ħ^{2}/2m)(π/L)^{2}(sin(πx/L) + 4*sin(2πx/L)) + (ħ^{2}/2m)(π/L)^{2}(sin(πx/L) + sin(2πx/L))]/(sin(πx/L) + sin(2πx/L))

U(x) = -(3π^{2}ħ^{2}/(2mL^{2})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}[ψ_{1}(x,0) + ψ_{2}(x,0)], where

ψ_{1}(x,t) = (2/L)^{½}sin(πx/L))exp(-iE_{1}t/ħ) and ψ_{2}(x,t) = (2/L)^{½}sin(2πx/L)exp(-iE_{2}t/ħ),

with E_{1 }= π^{2}ħ^{2}/(2mL^{2}) and E_{2 }= 4π^{2}ħ^{2}/(2mL^{2}),

are solutions to the one-dimensional infinite square-well problem.

The probability of measuring E_{1}is ½, the probability of measuring E_{2}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)[sin^{2}(πx/L) + sin^{2}(2πx/L) + sin(πx/L)sin(2πx/L)]

P(x,t) = (1/L)[sin^{2}(πx/L) + sin^{2}(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(-iE_{1}t/ħ) + sin(2πx/L)exp(-iE_{2}t/ħ)|^{2 }= (1/L)(sin(πx/L)exp(-iE_{1}t/ħ) + sin(2πx/L)exp(-iE_{2}t/ħ)

*(sin(πx/L)exp(-iE_{1}t/ħ) + sin(2πx/L)exp(-iE_{2}t/ħ)

= (1/L)[sin^{2}(πx/L) + sin^{2}(2πx/L) + sin(πx/L)sin(2πx/L)(exp(i(E_{2}-E_{1})t/ħ) + exp(-i(E_{2}-E_{1})t/ħ)]

= (1/L)[sin^{2}(πx/L) + sin^{2}(2πx/L) + 2sin(πx/L)sin(2πx/L)cos((E_{2}-E_{1})t/ħ)].

P(x,t) = (1/L)[sin^{2}(πx/L) + sin^{2}(2πx/L) + 2sin(πx/L)sin(2πx/L)cos((E_{2}-E_{1})t/ħ)].

The probability per unit length of finding the particle at x is changing with time.

A particle of mass m is in the ground state of a harmonic
oscillator with spring constant
k = mω^{2}.

At t = 0, the
spring constant changes suddenly to k' = 4k.

Find the probability that the oscillator remains in its ground state.

∫_{-∞}^{∞} exp(-x^{2}/a^{2})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(t_{2},t_{1})|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))^{¼}exp(-½mωx^{2}/ħ).

The normalized ground-state wave function for t > 0 is

Φ_{0f}(x) = (m2ω/(πħ))^{¼}exp(-mωx^{2}/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.

Consider a one-dimensional quantum particle with a Hamiltonian H = T + U(x),

T = -(ħ^{2}/2m)(∂^{2}/∂x^{2}). Suppose that m suddenly
changes from m_{0} to m_{1} = m_{0}/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) = Cx^{2}/2.

Hint: for any a we have ∫_{-∞}^{∞}dx exp(-ax^{2}) =
(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(t_{2},t_{1})|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)^{½}sin(πx/L), E_{0i}= π^{2}ħ^{2}/(2m_{0}L^{2}).

For t > 0, the wave function of the ground state is ψ_{0f}(x) = (2/L)^{½}sin(πx/L),

E_{0f}= λπ^{2}ħ^{2}/(2m_{0}L^{2}).

The probability that the particle remains in the ground state is 1.

The change in energy is E_{0f}- E_{0i}= (λ - 1)π^{2}ħ^{2}/(2m_{0}L^{2}).

(b) The normalized wave function of the ground state of a particle of mass m in the periodic well is

ψ_{0}(x) = (mω/(πħ))^{¼}exp(-mωx^{2}/(2ħ)), with ω^{2}= C/m, or

ψ_{0}(x) = (πb^{2})^{-¼}exp(-x^{2}/(2b^{2})), with b = (ħ^{2}/(mC))^{¼}.

The energy of the ground state is E_{0}= ½ħω = ½ħ(C/m)^{½}.

For t < 0, ψ_{0i}(x) = (πb_{i}^{ 2})^{-¼}exp(-x^{2}/(2b_{ i}^{ 2})), E_{0i}= ½ħ(C/m_{0})^{½}, with b_{i}= (ħ^{2}/(m_{0}C))^{¼}.

For t > 0, ψ_{0f}(x) = (πb_{f}^{ 2})^{-¼}exp(-x^{2}/(2b_{f}^{ 2})), E_{0f}= ½ħ(λC/m_{0})^{½}, with b_{f}= (λħ^{2}/(m_{0}C))^{¼}.

The probability that the particle remains in the ground state is |< ψ_{1f}| ψ_{0i}>|^{2}.

< ψ_{1f}| ψ_{0i}> = (πb_{i}b_{f})^{-½}exp ∫_{-∞}^{∞}dx exp(-½x^{2}(1/b_{i}^{2}+ 1/b_{f}^{2}) = (2b_{i}b_{f}/(b_{i}^{2}+ b_{f}^{2}))^{½}.

The probability that the particle remains in the ground state is

2b_{i}b_{f}/(b_{i}^{2}+ b_{f}^{2}) = 2λ^{¼}/(1 + λ^{½}) < 1.

The change in energy is E_{0f}- E_{0i}= ½(λ^{½}- 1)ħ(C/m_{0})^{½}.

The ground state wave function for a hydrogen like atom is

Φ_{100}(r) = (1/√π)(Z/a_{0})^{3/2
}exp(-Zr/a_{0}),

where a_{0} = ħ^{2}/(μe^{2})
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
^{3}He^{+}?

(c) An electron is in the ground state of tritium. A nuclear reaction
instantaneously changes the nucleus to ^{3}He^{+}. 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 ^{3}He^{+}.

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}/∂r^{2}) - l(l + 1)/r^{2}+ (2μ/ħ^{2})(e^{2}/r + E_{kl})]u_{kl}(r) = 0.

Defining

a_{0}= ħ^{2}/(μe^{2}), ρ = r/a_{0}, λ_{kl}^{2}= -2E_{kl}a_{0}/e^{2}= -E_{kl}/(μe^{2}/(2ħ^{2})) = -E_{kl}/E_{I},

we write

[(∂^{2}/∂ρ^{2}) - l(l + 1)/ρ^{2}+ (2/ρ) - λ_{kl}^{2})]u_{kl}(ρ) = 0.

For a hydrogenic atom we have

[(∂^{2}/∂r^{2}) - l(l + 1)/r^{2}+ (2μ'/ħ^{2})(Ze^{2}/r + E_{kl})]u_{kl}(r) = 0.

Defining

a_{0}'= ħ^{2}/(μZe^{2}), ρ' = r/a_{0}', λ'_{kl}^{2}= -2E_{kl}a_{0}'/(Ze^{2}) = -E_{kl}/(μ'Z^{2}e^{2}/(2ħ^{2})) = -E_{kl}/E_{I}',

we write

[(∂^{2}/∂ρ'^{2}) - l(l + 1)/ρ'^{2}+ (2/ρ') - λ'_{kl}^{2})]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 a_{0}by a_{0}' = ħ^{2}/(μ'Ze^{2}) = a_{0}(μ/μ')(1/Z), and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_{I}by E_{I}' = μ'Z^{2}e^{4}/(2ħ^{2}) = E_{I}(μ'/μ)Z^{2}.

(a) Tritium:

The reduced mass is μ' = m_{e}3m_{p}/(m_{e}+ 3 m_{p}) ≈ m_{e}

assuming the neutron mass equals the proton mass.

The nuclear charge is Z = 1. Since a_{0}' ~ a_{0}the ground state wave function of the tritium is the ground state wave function of the hydrogen atom.

Φ_{100}(^{3}H) = (1/√π)(1/a_{0})^{3/2 }exp(-r/a_{0}).

(b)^{3}He^{+}:

The reduced mass is μ' = m_{e}3m_{p}/(m_{e}+ 3 m_{p}) ≈ m_{e}

assuming the neutron mass equals the proton mass.

The nuclear charge is Z = 2. Since a_{0}' ~ a_{0}/2 the ground state wave function of the^{3}He^{+}is

Φ_{100}(^{3}He^{+}) = (1/√π)(2/a_{0})^{3/2 }exp(-2r/a_{0}).

(c) Sudden approximation:

P = |<Φ_{100}(^{3}H) |Φ_{100}(^{3}He^{+}) >|^{2 }= [(1/π)(√2/a_{0})^{3}∫d^{3}r exp(-3r/a_{0})]^{2}

= [(√2/a_{0})^{3}4∫_{0}^{∞}r^{2}dr exp(-3r/a_{0})]^{2 }= [(√2)^{3}(4/27)∫_{0}^{∞}x^{2}dx exp(-x)]^{2 }= [(√2)^{3}(8/27)]^{2}= 0.7.