__Formal__

A system makes transitions between eigenstates of H_{0} under
the action of the time dependent Hamiltonian H_{0 }+ εU_{0}sinωt,
εU_{0}
<< H_{0}. Assume that at t = 0 the system is in the state |Φ_{i}>. Find an expression for the probability of transition from |Φ_{i}>
to |Φ_{f}>, where |Φ_{i}>
and |Φ_{f}> are eigenstates of H_{0}
with eigenvalues E_{i} and E_{f}. Show that this
probability is small unless E_{f} - E_{i} ~ ±ħω.

[This shows that a charged particle in an oscillating electric field with
angular frequency ω will exchange energy with the
field only in multiples of E = ħω.]

Solution:

- Concepts:

Time-dependent perturbation theory - Reasoning:

For small perturbations time-dependent perturbation theory can be used to calculate transition probabilities. - Details of the calculation:

The probability of finding the system in the state |Φ_{f}> (f ≠ i) at time t is

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}|W(t)|Φ_{i}>, with W(t) = εU_{0}sinωt = Wsinωt and W = εU_{0}.

We have to consider a sinusoidal perturbation starting at t = 0.

∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt' = (W_{fi}/2i)∫_{0}^{t}[exp(i(ω_{fi}+ω)t' - exp(i(ω_{fi}-ω)t']dt'

= (W_{fi}/2)[(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω)) - (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω))].

P_{if}(t) =[|W_{fi}|^{2}/(4ħ^{2})]|(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω)) - (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω))|^{2},

with W_{fi}= <Φ_{f}|εU_{0}|Φ_{i}>.

The expression

(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω)) - (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω))

has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ω_{fi}= ±ω, or E_{f}= E_{i}± ħω. The first order effect of a perturbation that varies sinusoidally with time is to receive from or transfer to the system a quantum of energy hω.

If the system is initially in the ground state, then E_{f }> E_{i}, and only the second term needs to be considered. Then

P_{if}(t) = [|W_{fi}|^{2}/(4ħ^{2})] sin^{2}((ω_{fi}- ω)t/2)/((ω_{fi}- ω)/2)^{2}.

Let β = ω_{fi }- ω, and plot sin^{2}(βt/2)/(β/2)^{2}= t^{2}* sinc(βt/2) versus β.

If β = 0, i.e. ω_{fi}= ω, then sin^{2}(βt/2)/(β/2)^{2}= t^{2}. Therefore, if ω_{fi }= ω, then the probability of finding the system in the state |Φ_{f}>, P_{if}= |W_{fi}|^{2}t^{2}/(4ħ^{2}), increases with time proportional to t^{2}.

For a first order approximation to be valid, we need t << ħ/|W_{fi}|.

On the other hand, to justify neglecting the first term in the above formula, we need 2ω_{fi }>> Δω. 2ω_{fi}is the difference in the positions of the peaks due to the first term and the second term in the above formula, Δω is the width of the peaks, Δω » 4π/t. We therefore need t >> 1/ω_{fi}= 1/ω.

Combining these two conditions we obtain 1/ω_{fi}<< ħ/|W_{fi}|, or ħω_{fi}= (E_{f}- E_{i}) >> |W_{fi}|.

A system makes transitions between eigenstates of H_{0} under
the action of the time dependent Hamiltonian H_{0 }+
Wcosωt, W << H_{0}. Assume that at t =
0 the system is in the state |ψ_{i}>.

(a) Find an expression for the probability of transition from |ψ_{i}>
to |ψ_{f}>, where |ψ_{i}>
and |ψ_{f}> are eigenstates of H_{0}
with eigenvalues E_{i} and E_{f}.

(b) Find an expression for the probability of transition from |ψ_{i}>
to |ψ_{f}>,
for a constant perturbation W.

Solution:

- Concepts:

Time-dependent perturbation theory - Reasoning:

For small perturbations time-dependent perturbation theory can be used to calculate transition probabilities. - Details of the calculation:

(a) The probability of finding the system in the state |ψ_{f}> (f ≠ i) at time t is

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) = <ψ_{f}|W(t)|ψ_{i}>, with W(t) = Wcosωt.

We have to consider a sinusoidal perturbation starting at t = 0.

∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt' = (W_{fi}/2)∫_{0}^{t}[exp(i(ω_{fi}+ω)t' + exp(i(ω_{fi}-ω)t']dt'

= (W_{fi}/2)[(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω) + (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω)].

P_{if}(t) =[|W_{fi}|^{2}/(4ħ^{2})]|(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω) + (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω)|^{2},

with W_{fi}= <ψ_{f}|W|ψ_{i}>.

The expression

(1 - exp(i(ω_{fi}+ω)t)/(ω_{fi}+ω) + (1 - exp(i(ω_{fi}-ω)t)/(ω_{fi}-ω)

has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ω_{fi}= ±ω, or E_{f}= E_{i}± ħω. The first order effect of a perturbation that varies sinusoidally with time is to receive from or transfer to the system a quantum of energy hω.

If the system is initially in the ground state, then E_{f }> E_{i}, and only the second term needs to be considered. Then

P_{if}(t) = [|W_{fi}|^{2}/(4ħ^{2})] sin^{2}((ω_{fi}- ω)t/2)/((ω_{fi}- ω)/2)^{2}.

(b) For a constant perturbation W we have

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}dt'|^{2}= (1/ħ^{2})|W_{fi}∫_{0}^{t}exp(iω_{fi}t')dt'|^{2}

(1/ħ^{2})|W_{fi}(1 - exp(i(ω_{fi})t)/ω_{fi}|^{2}= [|W_{fi}|^{2}/ħ^{2}]sin^{2}(ω_{fi}t/2)/(ω_{fi}/2)^{2}.

__Infinite well__

Consider a one-dimensional, infinitely deep well. Let U(x) = 0, for
0 < x < a, and U(x) = ∞ everywhere else. The eigenstates of H_{0} = p^{2}/(2m) +
U(x) are Φ_{n}(x) = (2/a)^{½}sin(nπx/a)
with eigenvalues E_{n} = n^{2}π^{2}ħ^{2}/(2ma^{2}).

Assume that at t = 0 you put a coin on the bottom of the well.

W(x) = W_{0}, a/4 < x < 3a/4 for t > 0, W(x) = 0 everywhere
else, H = H_{0 }+ W.

If at t = 0 the system is in the state Φ_{3}(x),
what is the probability of finding it in Φ_{1}(x)
at time t?

Solution:

- Concepts:

The eigenfunctions of the infinite square well, time-dependent perturbation theory - Reasoning:

W is a small correction to H_{0}. The problem governed by H_{0}is already solved.

W = 0 for t < 0. The probability P(E_{k},t) of finding the system in the eigenstate |k> of H_{0}at time t, i.e . the probability of measuring the eigenvalue E_{k}, is

P(E_{k},t) = (1/ħ^{2})|∫_{0}^{t}exp((i/ħ)(E_{k}-E_{3})t')<Φ_{k}|W|Φ_{3}>dt'|^{2}.

This is the result of first order time dependent perturbation theory. P(E_{k},t) is the transition probability.

[Note: We are calculating the probability of finding the system in the ground state of the unperturbed Hamiltonian H_{0}, not of the perturbed Hamiltonian H. We are calculating the probability that we find the system in the ground state after we take the coin out at time t.] - Details of the calculation:

P(E_{1},t) = (1/ħ^{2})|∫_{0}^{t}exp((i/ħ)(E_{1}-E_{3})t')<Φ_{1}|W|Φ_{3}>dt'|^{2}

= |(<Φ_{1}|W|Φ_{3}>/ħ)∫_{0}^{t}exp(iω_{13}t')dt'|^{2}

[∫_{0}^{t}exp(iω_{13}t')dt' = (-i/ω_{13})(exp(iω_{13}t) - 1)

= (-i/ω_{13})exp(iω_{13}t/2)(exp(iω_{13}t/2) - exp(-iω_{13}t/2))

= (2/ω_{13})exp(iω_{13}t/2)sin(ω_{13}t/2)]

P(E_{1},t) = (|<Φ_{1}|W|Φ_{3}>|^{2}/(ħω_{13})^{2})4 sin^{2}(ω_{13}t/2).

ω_{13}= 8π^{2}ħ/(2ma^{2}).

<Φ_{1}|W|Φ_{3}> = ∫_{a/4}^{3a/4}dx Φ_{1}(x)W_{0}Φ_{3}(x)

= (2W_{0}/a)∫_{a/4}^{3a/4}dx sin(πx/a) sin(3πx/a)

= (2W_{0}/π)∫_{π/4}^{3π/4}dx sin(x) sin(3x)

= -W_{0}/π.

P(E_{1},t) = (2W_{0}/(ħπω_{13}))^{2}sin^{2}(ω_{13}t/2) = (m^{2}a^{4}W_{0}^{2}/(4ħ^{4}π^{6}))sin^{2}(2π^{2}ħt/(ma^{2})).

P(E_{1},t) oscillates with time t.

For t > 0 the Hamiltonian is H. At t = 0 the system is not in an eigenstate of H but in a eigenstate of H_{0}, an operator that does not commute with H. The probability of finding the system in an eigenstate of an operator that does not commute with the Hamiltonian changes with time.

Consider a particle in the ground state of a two-dimensional infinite
square well. The width of the well is L. The ground state is not
degenerate, but the excited energy levels are degenerate.

A perturbation H' = A x sinωt with A =
h^{2}/(100 m L^{3}) is turned on at t = 0. Find the
probability that at time t the particle will in the first excited state, for
each of the degenerate states.

Solution:

- Concepts:

The eigenfunctions of the two-dimensional infinite square well, time-dependent perturbation theory - Reasoning:

We have a sinusoidal perturbation starting at t = 0. For small perturbations time-dependent perturbation theory can be used to calculate transition probabilities. - Details of the calculation:

The energy eigenfunctions of a particle in a two-dimensional, infinite square well are

Φ_{nx,ny}(x,y) = (2/L)sin(n_{x}πx/L)sin(n_{y}πy/L)

with eigenvalues E_{nx,ny}= (n_{x}^{2}+ n_{y}^{2})π^{2}ħ^{2}/(2mL^{2}), n_{x}, n_{y}= 1, 2, 3, ... .

The ground state wave function is Φ_{1,1}(x,y) = (2/L)sin(πx/L)sin(πy/L).

The first excited state wave functions are

Φ_{1,2}(x,y) = (2/L)sin(πx/L)sin(2πy/L), Φ_{2,1}(x,y) = (2/L)sin(2πx/L)sin(πy/L).

We have to consider a sinusoidal perturbation H' = A x sinωt starting at t = 0.

(A*L < E_{11}, E_{12 }= E_{21}, first order time-dependent perturbation theory is valid.)

From first order time dependent perturbation theory we have

P_{if}(t) = [|W_{fi}|^{2}/(4ħ^{2})] sin^{2}((ω_{fi}- ω)t/2)/((ω_{fi}- ω)/2)^{2}.

(See problem 1.)

Here ω_{fi}= (E_{f}- E_{i})/ħ, and W(t) = Wsinωt, W = A*x.

For transition from the ground state to Φ_{1,2}and Φ_{2,1}ω_{fi}= 3π^{2}ħ^{2}/(2mL^{2}).

For transitions to the Φ_{2,1}state we have

W_{fi}= (4A/L^{2})∫_{0}^{L}sin(πx/L) sin(2πx/L) x dx ∫_{0}^{L}sin^{2}(πy/L) dy

= (2A/L)∫_{0}^{L}½(cos(πx/L) - cos(3πx/L)) x dx

= (A/L)(-2L^{2}/π^{2}+ 2L^{2}/(9π^{2})) = -16AL/(9π^{2}).

P_{if}= 12.64 ħ^{2}/(10^{4}m^{2}L^{4})sin^{2}((ω_{fi}- ω)t/2)/((ω_{fi}- ω)/2)^{2}.

For transitions to the Φ_{1,2}state we have

W_{fi}= (4A/L^{2})∫_{0}^{L}sin(πy/L) sin(2πy/L) dy ∫_{0}^{L}sin^{2}(πx/L) x dx = 0

P_{if}= 0.

A particle of mass m and electric charge e, moving in one dimension, is
confined to an interval of length a and is subject to a uniform electric field
E.

Initially it is in the eigenstate of kinetic energy with eigenvalue E_{k}
= k^{2}π^{2}ħ^{2}/(2ma^{2}), where k
is an integer.

Find, to first order in E^{2}, the probability that after a time t its
kinetic energy will be found to be E_{l} where k ≠
l.

Solution:

- Concepts:

Time-dependent perturbation theory - Reasoning:

For small perturbations time-dependent perturbation theory can be used to calculate transition probabilities. Here the perturbation is due to the uniform electric field. - Details of the calculation:

H = H_{0}+ W. H_{0}= T + U, U = 0, 0 < x < a, U = infinite elsewhere.

Eigenfunctions of H_{0}: Φ_{k}(x) = (2/a)^{½}sin(kπx/a)

with eigenvalues E_{k}= k^{2}π^{2}ħ^{2}/(2ma^{2}).

Assume the electric field points into the x-direction.

The potential energy of a particle in this field is -eEx, with the zero of the potential energy at x = 0.

W = -eEx.

First order time-dependent perturbation theory yields

P_{kl}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{lk}t')W_{lk}(t')dt'|^{2},

with with ω_{lk}= (E_{l}- E_{k})/ħ = (l^{2}-k^{2})π^{2}ħ/(2ma^{2}) and W_{lk}(t) = -eE<Φ_{l}|x|Φ_{k}>.

W_{lk}(t) = -(2eE/a)∫_{0}^{a }x sin(lπx/a) sin(kπx/a) dx

= (2eE/a)∫_{0}^{a }x ½[cos((k+l)πx/a) - cos((k-l)πx/a)] dx

= (eE/a)∫_{0}^{a }x [cos((k+l)πx/a) dx - (eE/a)∫_{0}^{a }x [cos((k-l)πx/a) dx

= (eE/a)(a^{2}/((k+l)^{2}π^{2})∫_{0}^{(k+l)π }x cos(x) dx - (eE/a)(a^{2}/((k-l)^{2}π^{2})∫_{0}^{(k-l)π }x cos(x) dx.

∫^{ }x cos(x) dx = cos(x) + xsin(x).

W_{lk}(t) = (eE/(aπ^{2}))[(cos((k+l)π) -1)/(k+l)^{2}- (cos((k-l)π) -1)/(k-l)^{2}].

If k+l = even then (cos((k+l)π) = (cos((k-l)π) = 1, W_{lk}(t) = 0.

If k+l = odd then (cos((k+l)π) = (cos((k-l)π) = -1,

W_{lk}(t) = -2(eE/(aπ^{2}))[((k-l)^{2}- (k+l)^{2})/((k-l)^{2}(k+l)^{2})] = W_{lk}.

P_{kl}(t) = (1/ħ^{2})W_{lk}^{2}|∫_{0}^{t}exp(iω_{lk}t')dt'|^{2}= (1/ħ^{2})W_{lk}^{2}(4ħ^{2}/(E_{l}- E_{k})^{2})sin^{2}[(E_{l}- E_{k})t/2ħ)

= (4W_{lk}^{2}/(E_{l}- E_{k})^{2})sin^{2}[(E_{l}- E_{k})t/2ħ).

__Harmonic oscillator__

A one-dimensional harmonic oscillator is in its ground state for t < 0.

For t > 0 it is subjected to a time-dependent but spatially uniform force in
the x-direction,

F = F_{0}exp(-t/τ).

(a) Using time-dependent perturbation theory to first order, obtain the
probability of finding the oscillator in its first excited state for t > 0.
Show that as t approaches infinity (with τ finite), the limit of your expression
is independent of time. Is this result reasonable, or surprising?

(b) Can the oscillator be found in higher excited states (for t > 0 )?

Solution:

- Concepts:

The harmonic oscillator, time dependent perturbation theory - Reasoning:

We are asked to find the transition probability from the ground state to an excited state for a perturbed harmonic oscillator. - Details of the calculation:

(a) Time dependent perturbation theory:

P_{n0}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{n0}t')W_{n0}(t')dt'|^{2 }W_{n0}= <n|W(t)|0>, where W(t) = -F_{0}xe^{-t/τ}, and {|n>} are the normalized eigenstates of the harmonic oscillator Hamiltonian.

<n|W(t)|0> = -F_{0}e^{-t/τ}<n|x|0> = CF_{0}e^{-t/τ}<n|(a + a^{T})|0> = CF_{0}e^{-t/τ}δ_{1n}.

Here C = -(ħ/(2mω))^{½}.

For the harmonic oscillator with the potential energy function ½mω^{2}x^{2}we have

E_{n}= (n + ½)ħω. Therefore ω_{n0}= nω.

P_{10}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{10}t')W_{n0}(t')dt'|^{2}= (CF_{0}/ħ)^{2}|∫_{0}^{t}exp(iωt')e^{-t’/τ}dt'|^{2}_{. }∫_{0}^{t}exp((iω-1/τ)t’)dt' = [τ/(iωτ-1)][exp((iω-1/τ)t) - 1]

|∫_{0}^{t}exp(iωt') e^{-t’/τ}dt'|^{2}= [τ^{2}/(ω^{2}τ^{2}+ 1)][1 + exp(-2t/τ) + 2exp(-t/τ)cos(ωt)]

P_{10}(t) = (CF_{0}/ħ)^{2}[τ^{2}/(ω^{2}τ^{2}+ 1)][1 + exp(-2t/τ) - 2exp(-t/τ)cos(ωt)]

As t -->infinity, |∫_{0}^{t}exp(iωt') e^{-t’/τ}dt'|^{2}= [τ^{2}/(ω^{2}τ^{2}+ 1)].

Since the perturbation decreases exponentially with time, we expect the probability that a transition has occurred to no longer change as t --> infinity.

(b) In first order time-dependent perturbation the probability of finding the oscillator in a higher excited state is zero since <n|x|0> = 0 for n > 1.

Consider a one-dimensional oscillator in its ground state.

The unperturbed Hamiltonian is H_{0 }= p^{2}/(2m) + ½mω_{0}^{2}x^{2}.

At t = 0 the Hamiltonian becomes H = H_{0 }+ H_{1}, where H_{1
}= ½mω_{1}^{2}x^{2}cosft, ω_{1 }<< ω_{0}.

Calculate the transition probability to the second excited state.

Can there be a transition to any other excited state?

Note:

The stationary states of H_{0} = p^{2}/(2m) + mω_{0}^{2}x^{2}/2
are u_{n}(x) = N_{n}H_{n}(αx)exp(-α^{2}x^{2}/2)

with N_{n} = [α/(π^{½}2^{n}n!)]^{½}, α = (mω_{0}/ħ)^{½}
and H_{n}(αx) a Hermite polynomial.

Recursion relation: H_{n+1}(αx) = 2αxH_{n}(αx) - 2nH_{n-1}(αx).

H_{0}(αx) = 1, H_{1}(αx) = 2αx, H_{2}(αx)
= 4(αx)^{2}
- 2.

Solution:

- Concepts:

Time-dependent perturbation theory, the 1D harmonic oscillator - Reasoning:

H = H_{0}+ H_{1}, H_{1}= W is a small, periodic perturbation. We use time-dependent perturbation theory to calculate the probability of transitions between stationary states of H_{0}. - Details of the calculation:

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) = <ψ_{f}|W(t)|ψ_{i}>, with W(t) = Wcosft.

We have a periodic perturbation W(t) = Wcosft starting at t = 0.

∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt' = (W_{fi}/2)∫_{0}^{t}[exp(i(ω_{fi }+ f)t') + exp(i(ω_{fi}- f)t')]dt'

= (iW_{fi}/2)[(1 - exp(i(ω_{fi }+ f)t)/(ω_{fi }+ f)) + (1 - exp(i(ω_{fi }- f)t))/(ω_{fi }- f))].

P_{if}(t) = [|W_{fi}|^{2}/(4ħ^{2})]|(1 - exp(i(ω_{fi }+ f)t))/(ω_{fi }+ f) + (1 - exp(i(ω_{fi }- f)t))/(ω_{fi}- f))|^{2},

with W_{fi}= <ψ_{f}|W|ψ_{i}>.

The expression

(1 - exp(i(ω_{fi }+ f)t))/(ω_{fi }+ f)) + (1 - exp(i(ω_{fi }- f)t))/(ω_{fi }- f))

has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ω_{fi}= ±f, or E_{f}= E_{i}± ħf. If the system is initially in the ground state, then E_{f }> E_{i}, and only the second term needs to be considered. Then

P_{if}(t) = [|W_{fi}|^{2}/(4ħ^{2})] sin^{2}((ω_{fi}- f)t/2)/((ω_{fi}- f)/2)^{2}.

P_{02}(t) = [(½mω_{1}^{2})^{2}/(4ħ^{2})]|<0|x^{2}|2>|^{2}sin^{2}((2ω_{0}- f)t/2)/((2ω_{0}- f)/2)^{2}, since ω_{fi}= 2ω_{0}.

P_{02}(t) is the probability that at time t we will find the oscillator in its second excited state.

<0|x^{2}|2> = N_{0}N_{2}∫_{-∞}^{∞}dx H_{0}(αx) x^{2}H_{2}(αx) exp(-α^{2}x^{2})

= N_{0}N_{2}∫_{-∞}^{∞}dx xH_{0}(αx) xH_{2}(αx) exp(-α^{2}x^{2})

= N_{0}N_{2}(½α)∫_{-∞}^{∞}dx H_{1}(αx) xH_{2}(αx) exp(-α^{2}x^{2})

= N_{0}N_{2}(1/α^{2})∫_{-∞}^{∞}dx H_{1}^{2}(αx) exp(-α^{2}x^{2})

= [N_{0}N_{2}/(N_{1}^{2}α^{2})] = ħ/(2^{½}mω_{0}).

P_{02}(t) = [(½mω_{1}^{2})^{2}/(4ħ^{2})](ħ^{2}/(2m^{2}ω_{0}^{2})) sin^{2}((2ω_{0}- f)t/2)/((2ω_{0}- f)/2)^{2 }= [ω_{1}^{4}/(32ω_{0}^{2})]sin^{2}((2ω_{0}- f)t/2)/((2ω_{0}- f)/2)^{2}.

In first-order, time-dependent perturbation theory no other transitions are allowed.

__Rotator__

Consider a symmetrical top, spinning about is symmetry axis with non-zero
angular momentum **J**. The moment of inertia of the top is I and its
Hamiltonian is given by H = J^{2}/2I. Assume the top is perturbed
by an external magnetic field **B**, with interaction H' = -**μ∙B**,
where **μ** = G**J**μ_{B}/ħ
is the magnetic moment of the top, G > 0, and μ_{B}
is the Bohr magneton. Assume that the top has an integer total angular
momentum quantum number.

(a) What is the energy of the top when B = 0?

(b) What are the ground state energy and the energy of the first excited state
of the spinning top when **B** = B_{0}**k**, B_{0} > 0?

(c) Assume that **B** becomes time dependent,

**B**(t) = B_{0}**k**, t < 0, **B**(t) = B_{0}**k**
+ **i** ΔBexp(-λt), t >
0.

where ΔB << B_{0} and λ > 0.

If the top is in its first excited state for t < 0, find the probability that
the top is in the ground state of part (b) for t > 0.

Solution:

- Concepts:

The eigenstates of the angular momentum operator, time-dependent perturbation theory - Reasoning:

The eigenstates of the unperturbed Hamiltonian are the common eigenstates of J^{2}and J_{z}. A small perturbation is introduced and we are asked to calculate the probability or a transition between two eigenstates of the unperturbed Hamiltonian.

- Details of the calculation:

(a) H = J^{2}/2I, E^{j}= ħ^{2}j(j+1)/2I, where j is the total angular momentum quantum number.

(b) H = J^{2}/2I - (Gμ_{B}/ħ)**J∙B**= J^{2}/2I - ( Gμ_{B}/ħ)B_{0}J_{z}.

The eigenstates of H are denoted by |j,m> (m = -j, -j+1, ..., j-1, j).

H|jm> = (ħ^{2}j(j+1)/2I - (Gμ_{B}/ħ)B_{0}mħ)|j,m>.

E^{jm}= (ħ^{2}j(j+1)/2I -(Gμ_{B}/ħ)B_{0}mħ).

The ground state for a given j has m = j, the first excited state has m = j - 1.

E_{0}= ħ^{2}j(j+1)/2I - Gμ_{B}B_{0}j, E_{1}= ħ^{2}j(j+1)/2I - Gμ_{B}B_{0}(j-1).

(c) Time dependent perturbation theory yields

P_{if}(t) = (1/ħ^{2})|∫_{0}^{t}exp(iω_{fi}t')W_{fi}(t')dt'|^{2}, when H = H_{0}+ W(t),

with ω_{fi}= (E_{f}- E_{i})/ħ and W_{fi}(t) = <Φ_{f}|W(t)|Φ_{i}>.

W(t) = -(Gμ_{B}/ħ)ΔBJ_{x}e^{-λt}, |Φ_{i}> = |j,j-1>, |Φ_{f}> = |j,j>

W_{fi}(t') = -(Gμ_{B}/ħ)ΔB<j,j|J_{x}|j,j-1>e^{-λt'}.

2J_{x}= J_{+}+ J_{-}, J_{±}|k,j,m> = [j(j+1)-m(m±1)]^{½}ħ|k,j,m±1>.

<j,j|J_{x}|j,j-1> = [j(j+1)-(j-1)j]^{½}ħ/2 = (ħ/2)[2j]^{½}.

W_{fi}(t') = -(Gμ_{B}/2)ΔB[2j]^{½}e^{-λt'}.

ω_{fi}= -Gμ_{B}B_{0}/ħ.

P_{if}(t) = (Gμ_{B}ΔB/2)^{2}2j|∫_{0}^{t}exp(-iGμ_{B}B_{0}t'/ħ)e^{-λt'}dt'|^{2}.

Let λ' = (iGμ_{B}B_{0}/ħ + λ).

∫_{0}^{t}exp(-λ't')^{'}dt' = (1 - e^{-λ't})/λ', P_{if}(t) = (Gμ_{B}ΔB/2)^{2}2j|(1 - e^{-λ't})/λ'|^{2}.

__Atoms__

Assume an atom is interacting with a monochromatic (visible light) EM plane
wave **
E**(

Assume the wave functions Φ

Solution:

- Concepts:

Time dependent perturbation theory - Reasoning:

We consider the interactions of the atomic electrons with the EM wave a perturbation to the atomic Hamiltonian. - Details of the calculation:

The energy of an electron in the field of the plane wave to first order is W_{DE}= e**r∙E**.

If we define the dipole moment**p**= -e**r**, we have W_{DE}= -**p∙E**.

Here W_{DE}= eE_{0}zcos(ωt), since we can neglect the spatial variations of visible light the plane wave over the dimensions of an atom.

W_{DE}= eE_{0}zcos(ωt) is a perturbation to the atomic Hamiltonian H_{0}whose eigenstates are the Φ_{nlm}(**r**).

Time dependent perturbation theory yields

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}|W_{DE}(t)|Φ_{i}>, with W_{DE}(t) = eE_{0}zcos(ωt).

P_{if}(t) = 0 if <Φ_{f}|W_{DE}(t)|Φ_{i}> = 0. This yields the selection rules.

<Φ_{f}|W_{DE}(t)|Φ_{i}> = eE_{0}cos(ωt)<Φ_{f}|z|Φ_{i}>.

Let Φ_{i}(**r**) = R_{ni li}(r)Y_{li mi}(θ,φ), Φ_{f}(**r**) = R_{nf lf}(r)Y_{lf mf}(θ,φ)

With z = r cosθ ∝ r Y_{10}(θ,φ) we have

<Φ_{f}|W_{DE}(t)|Φ_{i}> ∝ ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ Y^{*}_{lf mf}(θ,φ) Y_{10}(θ,φ) Y_{li mi}(θ,φ).

The integrant is a product of three spherical harmonics.

The integral is zero unless

(i) m_{1}+ m_{2}+ m_{3}= 0,

(ii) |l_{1}- l_{2}| ≤ l_{3}≤ (l_{1 }+ l_{2}),

(iii) l_{1}+ l_{2}- l_{3}= even.

Y_{lm}*(θ,φ) = (-1)^{m}Y_{l-m}(θ,φ).

We therefore have that <Φ_{f}|W_{DE}(t)|Φ_{i}> = 0 unless m_{f}= m_{i}, l_{f }= l_{i}± 1. If we choose another direction for the polarization of**E**, i.e.**E**= E_{0}**i**or E_{0}**i**, then we find m_{f }= m_{i}± 1, l_{f }= l_{i}± 1.

The**dipole transition selection rules**therefore are

Δl = ±1, Δm = 0, ±1.

These selection rules result as a**consequence of the properties of the spherical harmonics**.

The electron in a hydrogen
atom is in a 3d state. Neglect the fine structure.

(a) To what state or states (i.e. 1s etc.) can it go by radiating a photon in
an allowed transition?

(b) What is the degeneracy of the electron (include spin, but ignore spin-orbit
interaction) in a 3d state?

Solution:

- Concepts:

Selection rules Δl = 1, degeneracy of the energy level of the hydrogen atom - Reasoning:

Optical transitions are only "allowed" if Δl = 1. - Details of the calculation:

(a) Selection rules: Δl = 1.

An electron can only go to a 2p state in an allowed transition involving photon emission.

(b) l = 2 m = -2, -1, 0, 1, 2, s = + ½, - ½

Degeneracy = 10.