**
Problem 1:**

Consider a composite system made of two spin ½ particles. For t < 0 the
Hamiltonian does not depend on time and can be taken to be zero. For t > 0
the Hamiltonian is given by H = (4∆/ħ^{2})**S**_{1}∙**S**_{2}, where ∆ is a constant. Suppose that the system is in the state |+ -> for t
≤ 0. Find, as a function of time, the probability for being in each of the
states |+ +>, |+ ->, |- +>, and |- ->,

(a) by solving the problem exactly, using |Ψ(t)> = U(t, t_{0})Ψ(t_{0})>
and

(b) by solving the problem assuming the validity of first-order
time-dependent perturbation theory with H as a perturbation which is switched on
a t = 0.

(b) Under what conditions does the perturbation calculation disagree with
the exact solution and why?

Solution:

Concepts: The state space of two spin ½ particles, time-dependent perturbation theory. | |

Reasoning:S_{1}∙S_{2}
= ½(S^{2} – S_{1}^{2} – S_{2}^{2}). The
eigenstates of S_{1}∙S_{2} are the singlet state
and the triplet states, {|S, M_{s}>}, S = 0, 1. These are
common eigenfunctions of S^{2} and S_{z}. | |

Details of the calculation: (a) The common eigenfunctions of S ^{2} and S_{z}
are|1,1> = |++>, |1,0> = ½ ^{½}(|+-> + |-+>),|1,-1> =|-->, |0, 0> = ½ ^{½}(|+-> - |-+>).Therefore |+-> = ½ ^{½}(|0,0> + |1,0>),
|-+> = ½^{½}(|0,0> - |1,0>).For the singlet state we have S_{1}∙S_{2} |0,0> = -(3/4)ħ^{2}, H |0,0> = -3∆|0,0>.For the triplet states we have S_{1}∙S_{2} |1, M_{s}>
= (1/4)ħ^{2}, H |1, M_{s}> = ∆|1, M_{s}>.|ψ(0)> = |+-> = ½ ^{½}(|0,0> + |1,0>), |ψ(t)> = U(t, 0)|+ -> = 2 ^{-½}(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>).P(t) of being in the state |+ +> = |1 1> = 0. P(t) of being in the state |- -> = |1 -1> = 0. P(t) of being in the state |+ -> = |<+ -| ψ(t)>| ^{2}.<+ -|ψ(t)> = ½(<0,0| + <1,0|)(exp(i3∆*t/ħ)|0,0> + exp(-i∆*t/ħ)|1,0>) = ½(exp(i3∆*t/ ħ) + exp(-i∆*t/ ħ)) = ½ exp(i∆*t/ ħ) (exp(i2∆*t/ ħ) + exp(-i2∆*t/ħ)) = exp(i∆*t/ħ) cos(2∆*t/ħ). P(t) of being in the state |+ -> = cos ^{2}(2∆*t/ħ).For 2∆*t/ħ << 1, P(t) ~ 1. P(t) of being in the state |- +> = sin ^{2}(2∆*t/ħ), since the total probability of being in one of the 4 basis states must be
equal to 1.For 2∆*t/ħ << 1, P(t) ~ (2∆*t/ħ) ^{2}.(b) Time-dependent perturbation theory P _{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}.The initial and the final state must be different for this formula to be valid. For i = f, P(t) ~ 1. W _{fi} = <f|H|i>. Here W_{fi} = <f|H |+ ->.
W _{fi} = 0 for |f> = |+ +> and |f> = |- ->, so P_{if}(t) = 0 for
these states.For |f> = |- +> we have W _{fi} = 2^{-½}<- +|H| (|00> + |10>)
= 2 ^{-½}[-3∆<- +| 0, 0> + ∆<- +| 1, 0>] = ½(3∆
+ ∆) = 2∆.ω _{fi} = (E_{f} – E_{i})/ω_{fi} = 0 for the
unperturbed states.P _{if}(t) = (1/ħ^{2})|∫_{0}^{t}(2∆)dt'|^{2}
= 4∆^{2}t^{2}/ħ^{2}.(c) The time-dependent perturbation theory calculation result agrees with the exact result as long as x = 2∆*t/ħ << 1, so that cos(x) = 1 and sin(x) = x are very good approximations. For large x (or t) the probability of finding the particle in the state |- +> oscillates between 0 and 1. |

**Problem 2:**

A hydrogen atom with Hamiltonian H_{0}(r) is placed in a
time-dependent electric field **E** = E(t) **k**. The perturbed
Hamiltonian is H(**r**,t) = H_{0}(r) + H’(**r**,t).

(a) Show that H’(**r**,t) = q_{e}E(t) r cos(θ).

(b) Assuming the electron is
initially in the ground state, and recalling that the first excited state of
hydrogen is quadruply degenerate, to which state of the quadruply degenerate
first excited states is a dipole transition from the ground state possible?
Prove this.

(c) If the electron is in the
ground state at t = 0, find the probability (to first order in perturbation theory) that at time
t the
electron will have made the transition to the state determined in (b), as a
function of E(t).

Solution:

Concepts: Time dependent perturbation theory | |

Reasoning: We consider the interactions of the atomic electrons with the electric field perturbation to the atomic Hamiltonian. | |

Details of the calculation: (a) The uniform electric field changes the energy of the proton and the electron, and therefore perturbs the Hamiltonian. The energy of the proton in the field is -∫ _{0}^{zp}q_{e}E(t)dz
= -q_{e}E(t)z_{p}.The energy of the electron in the field is ∫ _{0}^{ze}q_{e}
E(t)dz = q_{e}E(t)z_{e}.The potential energy of both particles is q _{e}E(t)(z_{e} - z_{p})
= q_{e}E(t)z = q_{e}E(t) r cos(θ).(b) For a dipole transition to be allowed, we need <Φ _{f}|H’(r,t)|Φ_{i}> to be nonzero.Φ _{nlm}(r,θ,φ) = R_{nl}(r)Y_{lm}(θ,φ).The ground state wave function is Φ _{100}(r,θ,φ) = π^{-½} a_{0}^{-3/2
}exp(-r/a_{0}).The wave functions of the quadruply degenerate first excited states are Φ _{200}(r,θ,φ) = (4π)^{-½} (2a_{0})^{-3/2} (2 -
r/a_{0}) exp(-r/(2a_{0})),Φ _{211}(r,θ,φ) = (8π)^{-½} (2a_{0})^{-3/2} (r/a_{0})
exp(-r/(2a_{0})) sinθ e^{iφ},Φ _{210}(r,θ,φ) = (4π)^{-½} (2a_{0})^{-3/2} (r/a_{0})
exp(-r/(2a_{0})) cosθ,Φ _{21-1}(r,θ,φ) = (8π)^{-½} (2a_{0})^{-3/2} (r/a_{0})
exp(-r/(2a_{0})) sinθ e^{-iφ}.<Φ _{100}| H’(r,t)|Φ_{2lm}>
∝ ∫_{0}^{∞}r^{3}dr
R_{00}^{*}(r)R_{2l}(r) ∫_{0}^{π}sinθ
dθ∫_{0}^{2π}dφ Y_{00}^{*}(θ,φ) cosθ Y_{lm}(θ,φ)∝ ∫ _{0}^{π}sinθ
dθ∫_{0}^{2π}dφ Y_{00}^{*}(θ,φ)^{
}Y_{10}(θ,φ)
Y_{lm}(θ,φ).We can invoke the properties if the spherical harmonics. Or we can show explicitly <Φ _{100}|H’(r,t)|Φ_{200}>
∝ ∫_{0}^{π}sinθ
cosθ dθ = 0, <Φ_{100}|H’(r,t)|Φ_{211}>
∝ ∫_{0}^{2π}dφ
e^{iφ} = 0,<Φ _{100}|H’(r,t)|Φ_{21-1}>
∝ ∫_{0}^{2π}dφ
e^{-iφ} = 0, <Φ_{100}|H’(r,t)|Φ_{210}>
∝ ∫_{0}^{π}sinθ
dθ cos^{2}θ ≠ 0.The only transition from the ground state to an n = 2 state that is allowed is the transition to the n = 2, l = 1, m = 0 state, Φ _{210}(r,θ,φ).(c) P _{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2},with ω _{fi} = (E_{f} - E_{i})/ħ and W_{fi}(t) =
<Φ_{f}|H’(r,t)|Φ_{i}>, with H’(r,t) = q_{e}E(t)
r cos(θ).Let Φ _{i} = Φ_{100} and Φ_{f} = Φ_{210}. Then ω_{fi}
= ω_{21} = (E_{2} – E_{1})/ħ = (10.2 eV)/ħ and
W _{fi}(t) = <Φ_{210}|H’(r,t)|Φ_{000}> = [q_{e}E(t)
/(2^{3/2}a_{0}^{4})] ∫_{0}^{∞}r^{4}dr
exp(-3r/(2a_{0}))∫_{0}^{π}sinθ dθ cos^{2}θ= q _{e}E(t) 2^{½
}2^{7} a_{0}/3^{5}.Therefore the probability that at time t the electron will have made the transition is P _{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}
= 0.555 q_{e}^{2}a_{0}^{2}|∫_{0}^{t}exp(iω_{21}t')E(t')dt'|^{2}. |

**Problem 3:**

Conside a 2-dimensional system containing a gas of
electrons completely free to move in the x direction but confined by a
square-well potential of infinite depth and total width w in the y
direction. In such a system, the electrons can often be approximated as
non-interacting, provided that the mass of the electron is replaced by an
effective mass. Assume for this problem that the effective mass of the
electrons is about 1/10 the mass of free electrons.

(a) This system contains states that involve quantum-mechanical motion in both
the x and y directions; describe qualitatively the nature of the
absorption spectrum that you expect.

(b) Write or derive a formula for the discrete levels expected for quantized
motion in the y direction in terms of the width w.

(c) How small does the width w have to be before the transition energy between
the first two discrete levels found in b is larger than the average energy
available from thermal excitation at room temperature?

Solution:

Concepts: Separation of variable, the infinite square well | |

Reasoning: This is a two-dimensional problem, with confinement in one dimensions. | |

Details of the calculation: (a) H = (-ħ ^{2}/(2m))[∂^{2}/∂x^{2}
+ ∂^{2}/∂y^{2}]
+ U(y), H ψ(x,y) = E ψ(x,y).Separation of variables is possible. ψ(x,y) = Χ(x) Φ(y). E = E _{x} + E_{y}. (-ħ ^{2}/(2m))∂^{2}Χ(x)/∂x^{2}
= E_{x}Χ(x).E _{x} is a continuous eigenvalue. E_{x} =
ħk^{2}/(2m).(-ħ ^{2}/(2m))∂^{2}Φ(y)/∂y^{2}
+ U(y) Φ(y) = E_{y}Φ(y).E _{y} is quantized. The E_{yn} are the eigenenergies of the
infinite square well. E = ħk ^{2}/(2m)
+ E_{yn}, k = (2m(E – E_{yn})/ħ^{2})^{½}.The density of states dN/dE dk/dE = [m/(2ħ ^{2}(E – E_{yn}))]^{½}
will have spikes when E = E_{yn}, and the absorption spectrum, for
example, should reflect those spikes. (Fermi's golden rule.)(b) E _{yn} = n^{2}π^{2}ħ^{2}/(2mw^{2})(c) 3π ^{2}ħ^{2}/(2mw^{2})
> kT, w < 20 nm. |

**Problem 4:**

In a one-dimensional structure a particle of mass m is in the ground state of a
potential energy function U(x) = ½ kx^{2}. A phase transition occurs,
and the effective spring constant suddenly doubles. What is the probability that
the particle will be found in an exited state after the phase transition? Give a
numerical answer.

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(-½mω^{2}x^{2}/ħ), ω =
(k/m)^{½}.The normalized ground-state wave function for t > 0 is Φ _{0f}(x) = (m√(2)ω/(πħ))^{¼}exp(-½m√(2)ω^{2}x^{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^{5/4}/(1 + √2) = 0.985.The probability that the particle will be found in an exited state after the phase transition is 1 - 0.985 = 0.015. |

**Problem 5:**

In one dimension, consider a spinless particle trapped in a delta-function potential U(x) = -Cδ(x), C > 0.

At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.

Assume ω >> mC^{2}/(2ħ^{3}) so that the particle can be ejected
from the trap. Use perturbation theory to find the transition rate.

You can assume that the free particles will be in a box of size L, L =
very large.

Solution:

Concepts: Fermi's golden rule, the delta-function potential | |

Reasoning: Only one bound state exist in the delta-function potential. This is the initial state. The final state of the particle is a continuum state. | |

Details of the calculation: (a) Let H _{0} = p^{2}/(2m) - Cδ(x).Normalized eigenfunction of H _{0}: Φ(x) = ρ^{½}
exp(-ρ|x|), ρ ^{2} = -2mE_{i}/ħ^{2}, E_{i} = -mC^{2}/(2ħ^{2}),
ρ = mC/ħ^{2}.To find the transition rate we use Fermi's golden rule. If W(t) = Wcos(ωt), then the transition probability per unit time is given by w(i,E) = (π/(2ħ))ρ(E)|W _{Ei}|^{2}δ_{E-Ei,ħω}, where
W_{Ei} = <Φ_{E}|W|Φ_{i}>.Since ħω >> -mC ^{2}/(2ħ^{2}), we can assume that the ejected
particle is a nearly free particle
with energy E = ħω + E_{i} and k^{2} = 2mE/ħ^{2}.We assume that the particle is confined to a region of width L >> a and use periodic boundary conditions. Then Φ _{k}(x) = L^{-½}exp(ikx), with k = 2πn/L, n = 0,
±1, ±2, ... is the final state wave function.The number of states with wave vectors whose magnitudes lie between k and k + dk is dN = 2 dk/(2π/L) = Ldk/π. (The particle can move towards the left or to the right.) dN/dk = L/π. The density of states is dN/dE = (dN/dk)(dk/dE). With E = ħ ^{2}k^{2}/(2m) we have ρ(E) = dN/dE = Lm/(πkħ^{2}).W _{Ei} = (ρ/L)^{½}W[∫_{0}^{∞}dx e^{ikx}e^{-ρx}
+ ∫_{-∞}^{0}dx e^{ikx}e^{ρx}]= (ρ/L) ^{½}W 2ρ/(ρ^{2} + k^{2}).w(i,E) = (π/(2ħ))(Lm/(πkħ ^{2}))(ρ/L)W^{2} 4ρ^{2}/(ρ^{2}
+ k^{2})^{2
}= (2mW^{2}/ħ^{3}) ρ^{3}/[k(ρ^{2} + k^{2})^{2}].(ρ ^{2} + k^{2})^{2} = 4m^{2}ω^{2}/ħ^{2},
ρ^{3}/[(ρ^{2} + k^{2})^{2}] = mC^{3}/(4ħ^{4}ω^{2}), This is the transition probability per unit time. |