### Time-dependent perturbation theory

Formal

#### Problem:

A system makes transitions between eigenstates of H0 under the action of the time dependent Hamiltonian H0 + εU0sinωt, εU0 << H0.  Assume that at t = 0 the system is in the state |Φi>.  Find an expression for the probability of transition from |Φi> to |Φf>, where |Φi> and |Φf> are eigenstates of H0 with eigenvalues Ei and Ef.  Show that this probability is small unless Ef - Ei ~ ±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 |Φf> (f ≠ i) at time t is
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <Φf|W(t)|Φi>, with W(t) = εU0sinωt = Wsinωt and W = εU0.
We have to consider a sinusoidal perturbation starting at t = 0.
0t exp(iωfit')Wfi(t')dt' = (Wfi/2i)∫0t [exp(i(ωfi+ω)t' - exp(i(ωfi-ω)t']dt'
= (Wfi/2)[(1 - exp(i(ωfi+ω)t)/(ωfi+ω)) - (1 - exp(i(ωfi-ω)t)/(ωfi-ω))].
Pif(t) =[|Wfi|2/(4ħ2)]|(1 - exp(i(ωfi+ω)t)/(ωfi+ω)) - (1 - exp(i(ωfi-ω)t)/(ωfi-ω))|2,
with Wfi = <Φf|εU0i>.
The expression
(1 - exp(i(ωfi+ω)t)/(ωfi+ω)) - (1 - exp(i(ωfi-ω)t)/(ωfi-ω))
has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e.  if ωfi = ±ω, or Ef = Ei ± 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 Ef > Ei, and only the second term needs to be considered.  Then

Pif(t) = [|Wfi|2/(4ħ2)] sin2((ωfi - ω)t/2)/((ωfi - ω)/2)2.

Let β = ωfi - ω, and plot sin2(βt/2)/(β/2)2 = t2 * sinc(βt/2) versus β.

If β = 0, i.e.  ωfi = ω, then sin2(βt/2)/(β/2)2 = t2.  Therefore, if ωfi = ω, then the probability of finding the system in the state |Φf>,  Pif = |Wfi|2t2/(4ħ2), increases with time proportional to t2.
For a first order approximation to be valid, we need t << ħ/|Wfi|.
On the other hand, to justify neglecting the first term in the above formula, we need 2ωfi >> Δω.  2ωfi 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/ωfi = 1/ω.
Combining these two conditions we obtain 1/ωfi << ħ/|Wfi|, or ħωfi = (Ef - Ei) >> |Wfi|.

#### Problem:

A system makes transitions between eigenstates of H0 under the action of the time dependent Hamiltonian H0 + Wcosωt, W << H0.  Assume that at t = 0 the system is in the state |ψi>.
(a)  Find an expression for the probability of transition from |ψi> to |ψf>, where |ψi> and |ψf> are eigenstates of H0 with eigenvalues Ei and Ef.
(b)  Find an expression for the probability of transition from |ψi> to |ψf>, 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 |ψf> (f ≠ i)  at time t is
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <ψf|W(t)|ψi>, with W(t) = Wcosωt.
We have to consider a sinusoidal perturbation starting at t = 0.
0t exp(iωfit')Wfi(t')dt' = (Wfi/2)∫0t [exp(i(ωfi+ω)t' + exp(i(ωfi-ω)t']dt'
= (Wfi/2)[(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)].
Pif(t) =[|Wfi|2/(4ħ2)]|(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)|2,
with Wfi = <ψf|W|ψi>.
The expression
(1 - exp(i(ωfi+ω)t)/(ωfi+ω) + (1 - exp(i(ωfi-ω)t)/(ωfi-ω)
has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ωfi = ±ω, or Ef = Ei ± 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 Ef > Ei, and only the second term needs to be considered.  Then
Pif(t) = [|Wfi|2/(4ħ2)] sin2((ωfi - ω)t/2)/((ωfi - ω)/2)2.
(b) For a constant perturbation W we have
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfidt'|2 = (1/ħ2)|Wfi0texp(iωfit')dt'|2
(1/ħ2)|Wfi(1 - exp(i(ωfi)t)/ωfi|2 = [|Wfi|22]sin2fit/2)/(ωfi/2)2.

Solution:

• Concepts:
The eigenfunctions of the infinite square well, time-dependent perturbation theory
• Reasoning:
W is a small correction to H0.  The problem governed by H0 is already solved.
W = 0  for  t < 0.  The probability P(Ek,t) of finding the system in the eigenstate |k> of H0 at time t, i.e . the probability of measuring the eigenvalue Ek, is
P(Ek,t) = (1/ħ2)|∫0texp((i/ħ)(Ek-E3)t')<Φk|W|Φ3>dt'|2.
P(Ek,t) is the transition probability.
[Note: We are calculating the probability of finding the system in the ground state of the unperturbed Hamiltonian H0, 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(E1,t) = (1/ħ2)|∫0texp((i/ħ)(E1-E3)t')<Φ1|W|Φ3>dt'|2
= |(<Φ1|W|Φ3>/ħ)∫0texp(iω13t')dt'|2
[∫0texp(iω13t')dt' = (-i/ω13)(exp(iω13t) - 1)
= (-i/ω13)exp(iω13t/2)(exp(iω13t/2) - exp(-iω13t/2))
= (2/ω13)exp(iω13t/2)sin(ω13t/2)]
P(E1,t) = (|<Φ1|W|Φ3>|2/(ħω13)2) 4 sin213t/2).
ω13 = 8π2ħ/(2ma2).
1|W|Φ3> = (2W0/a)∫a/43a/4dx sin(πx/a)sin(3πx/a)
= (2W0/π)∫π/43π/4 dx' sin(x')sin(3x')
= (2W0/π)[(sin(2x')/4 - sin(4x')/6))|π/43π/4  = -W0/π.

P(E1,t) = (2W0/(ħπω13))2sin213t/2) = (m2a4W02/(4ħ4π6))sin2(2π2ħt/(ma2)).
P(E1,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 H0, 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.

Infinite well

#### Problem:

Consider a one-dimensional, infinitely deep well.  Let U(x) = 0, for 0 < x < a, and U(x) = ∞ everywhere else.  The eigenstates of H0 = p2/(2m) + U(x) are Φn(x) = (2/a)½sin(nπx/a) with eigenvalues En = n2π2ħ2/(2ma2).
Assume that at t = 0 you put a coin on the bottom of the well.
W(x) = W0, a/4 < x < 3a/4 for t > 0, W(x) = 0 everywhere else, H = H0 + W.
If at t = 0 the system is in the state Φ3(x), what is the probability of finding it in Φ1(x) at time t?

Solution:

• Concepts:
The eigenfunctions of the infinite square well, time-dependent perturbation theory
• Reasoning:
W is a small correction to H0.  The problem governed by H0 is already solved.
W = 0  for  t < 0.  The probability P(Ek,t) of finding the system in the eigenstate |k> of H0 at time t, i.e .  the probability of measuring the eigenvalue Ek, is
P(Ek,t) = (1/ħ2)|∫0texp((i/ħ)(Ek-E3)t')<Φk|W|Φ3>dt'|2.
This is the result of first order time dependent perturbation theory.  P(Ek,t) is the transition probability.
[Note: We are calculating the probability of finding the system in the ground state of the unperturbed Hamiltonian H0, 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(E1,t) = (1/ħ2)|∫0texp((i/ħ)(E1-E3)t')<Φ1|W|Φ3>dt'|2
= |(<Φ1|W|Φ3>/ħ)∫0texp(iω13t')dt'|2
[∫0texp(iω13t')dt' = (-i/ω13)(exp(iω13t) - 1)
= (-i/ω13)exp(iω13t/2)(exp(iω13t/2) - exp(-iω13t/2))
= (2/ω13)exp(iω13t/2)sin(ω13t/2)]
P(E1,t) = (|<Φ1|W|Φ3>|2/(ħω13)2)4 sin213t/2).
ω13 = 8π2ħ/(2ma2).
1|W|Φ3> = ∫a/43a/4 dx Φ1(x)W0Φ3(x)
= (2W0/a)∫a/43a/4 dx sin(πx/a) sin(3πx/a)
= (2W0/π)∫π/43π/4 dx sin(x) sin(3x)
= -W0/π.
P(E1,t) = (2W0/(ħπω13))2sin213t/2) = (m2a4W02/(4ħ4π6))sin2(2π2ħt/(ma2)).
P(E1,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 H0, 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.

#### Problem:

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 = h2/(100 m L3) 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
Φnx,ny(x,y) = (2/L)sin(nxπx/L)sin(nyπy/L)
with eigenvalues Enx,ny = (nx2 + ny22ħ2/(2mL2),  nx, ny = 1, 2, 3, ...  .
The ground state wave function is Φ1,1(x,y) = (2/L)sin(πx/L)sin(πy/L).
The first excited state wave functions are
Φ1,2(x,y) = (2/L)sin(πx/L)sin(2πy/L),  Φ2,1(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 < E11, E12 = E21, first order time-dependent perturbation theory is valid.)
From first order time dependent perturbation theory we have
Pif(t) = [|Wfi|2/(4ħ2)] sin2((ωfi - ω)t/2)/((ωfi - ω)/2)2.
(See problem 1.)
Here ωfi = (Ef - Ei)/ħ, and W(t) = Wsinωt,  W = A*x.
For transition from the ground state to Φ1,2 and Φ2,1 ωfi = 3π2ħ2/(2mL2).
For transitions to the Φ2,1 state we have
Wfi = (4A/L2)∫0Lsin(πx/L) sin(2πx/L) x dx ∫0Lsin2(πy/L) dy
= (2A/L)∫0L½(cos(πx/L) - cos(3πx/L)) x dx
= (A/L)(-2L22 + 2L2/(9π2)) = -16AL/(9π2).
Pif = 12.64 ħ2/(104m2L4)sin2((ωfi - ω)t/2)/((ωfi - ω)/2)2.
For transitions to the Φ1,2 state we have
Wfi = (4A/L2)∫0Lsin(πy/L) sin(2πy/L) dy ∫0Lsin2(πx/L) x dx = 0
Pif = 0.

#### Problem:

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 Ek = k2π2ħ2/(2ma2), where k is an integer.
Find, to first order in E2, the probability that after a time t its kinetic energy will be found to be El 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 = H0 + W.  H0 = T + U,  U = 0,  0 < x < a,  U = infinite elsewhere.
Eigenfunctions of H0:   Φk(x) = (2/a)½sin(kπx/a)
with eigenvalues Ek = k2π2ħ2/(2ma2).
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
Pkl(t) = (1/ħ2)|∫0texp(iωlkt')Wlk(t')dt'|2,
with with  ωlk = (El - Ek)/ħ = (l2 -k22ħ/(2ma2)  and Wlk(t) = -eE<Φl|x|Φk>.
Wlk(t) = -(2eE/a)∫0a x sin(lπx/a) sin(kπx/a) dx
= (2eE/a)∫0a x ½[cos((k+l)πx/a)  - cos((k-l)πx/a)] dx
= (eE/a)∫0a x [cos((k+l)πx/a) dx - (eE/a)∫0a x [cos((k-l)πx/a) dx
= (eE/a)(a2/((k+l)2π2)∫0(k+l)π x cos(x) dx - (eE/a)(a2/((k-l)2π2)∫0(k-l)π x cos(x) dx.
x cos(x) dx = cos(x) + xsin(x).
Wlk(t) = (eE/(aπ2))[(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, Wlk(t)  = 0.
If k+l = odd then (cos((k+l)π) = (cos((k-l)π) = -1,
Wlk(t)  = -2(eE/(aπ2))[((k-l)2 - (k+l)2)/((k-l)2(k+l)2)] = Wlk.
Pkl(t) = (1/ħ2)Wlk2|∫0texp(iωlkt')dt'|2 = (1/ħ2)Wlk2(4ħ2/(El - Ek)2)sin2[(El - Ek)t/2ħ)
= (4Wlk2/(El - Ek)2)sin2[(El - Ek)t/2ħ).

Harmonic oscillator

#### Problem:

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 = F0exp(-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:
Pn0(t) = (1/ħ2)|∫0texp(iωn0t')Wn0(t')dt'|2
Wn0 = <n|W(t)|0>, where W(t) = -F0xe-t/τ, and {|n>} are the normalized eigenstates of the harmonic oscillator Hamiltonian.
<n|W(t)|0> = -F0e-t/τ<n|x|0> = CF0e-t/τ<n|(a + aT)|0> = CF0e-t/τδ1n.
Here C = -(ħ/(2mω))½.
For the harmonic oscillator with the potential energy function ½mω2x2 we have
En = (n + ½)ħω.  Therefore ωn0 = nω.
P10(t) = (1/ħ2)|∫0texp(iω10t')Wn0(t')dt'|2 = (CF0/ħ)2|∫0texp(iωt')e-t’/τdt'|2.
0texp((iω-1/τ)t’)dt' = [τ/(iωτ-1)][exp((iω-1/τ)t) - 1]
|∫0texp(iωt') e-t’/τdt'|2 = [τ2/(ω2τ2 + 1)][1 + exp(-2t/τ) + 2exp(-t/τ)cos(ωt)]
P10(t) = (CF0/ħ)22/(ω2τ2 + 1)][1 + exp(-2t/τ) - 2exp(-t/τ)cos(ωt)]
As t -->infinity, |∫0texp(iωt') e-t’/τdt'|2 = [τ2/(ω2τ2 + 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.

#### Problem:

Consider a one-dimensional oscillator in its ground state.
The unperturbed Hamiltonian is H0 = p2/(2m) + ½mω02x2.
At t = 0 the Hamiltonian becomes H = H0 + H1, where H1 = ½mω12x2cosft, ω1 << ω0.
Calculate the transition probability to the second excited state.
Can there be a transition to any other excited state?

Note:
The stationary states of H0 = p2/(2m) + mω02x2/2 are un(x) = NnHn(αx)exp(-α2x2/2)
with Nn = [α/(π½2nn!)]½, α = (mω0/ħ)½  and Hn(αx) a Hermite polynomial.
Recursion relation: Hn+1(αx) = 2αxHn(αx) - 2nHn-1(αx).
H0(αx) = 1,   H1(αx) = 2αx,   H2(αx) = 4(αx)2 - 2.

Solution:

• Concepts:
Time-dependent perturbation theory, the 1D harmonic oscillator
• Reasoning:
H = H0 + H1, H1 = W is a small, periodic perturbation.  We use time-dependent perturbation theory to calculate the probability of transitions between stationary states of H0.
• Details of the calculation:
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <ψf|W(t)|ψi>, with W(t) = Wcosft.
We have a periodic perturbation W(t) = Wcosft starting at t = 0.
0t exp(iωfit')Wfi(t')dt' = (Wfi/2)∫0t [exp(i(ωfi + f)t') + exp(i(ωfi-  f)t')]dt'
= (iWfi/2)[(1 - exp(i(ωfi + f)t)/(ωfi + f)) + (1 - exp(i(ωfi - f)t))/(ωfi - f))].
Pif(t) = [|Wfi|2/(4ħ2)]|(1 - exp(i(ωfi + f)t))/(ωfi + f) + (1 - exp(i(ωfi - f)t))/(ωfi-  f))|2,
with Wfi = <ψf|W|ψi>.
The expression
(1 - exp(i(ωfi + f)t))/(ωfi + f)) + (1 - exp(i(ωfi - f)t))/(ωfi - f))
has an appreciable amplitude only if the denominator of one of the two terms is approximately zero, i.e. if ωfi = ±f, or Ef = Ei ± ħf.  If the system is initially in the ground state, then Ef > Ei, and only the second term needs to be considered.  Then
Pif(t) = [|Wfi|2/(4ħ2)] sin2((ωfi - f)t/2)/((ωfi - f)/2)2.
P02(t) = [(½mω12)2/(4ħ2)]|<0|x2|2>|2 sin2((2ω0 - f)t/2)/((2ω0 - f)/2)2,  since ωfi = 2ω0.
P02(t)  is the probability that at time t we will find the oscillator in its second excited state.
<0|x2|2> = N0N2-∞dx H0(αx) x2 H2(αx) exp(-α2x2)
= N0N2-∞dx xH0(αx) xH2(αx) exp(-α2x2)
= N0N2(½α)∫-∞dx H1(αx) xH2(αx) exp(-α2x2)
= N0N2(1/α2)∫-∞dx H12(αx) exp(-α2x2)
= [N0N2/(N12α2)] = ħ/(2½0).
P02(t) = [(½mω12)2/(4ħ2)](ħ2/(2m2ω02)) sin2((2ω0 - f)t/2)/((2ω0 - f)/2)2
=  [ω14/(32ω02)]sin2((2ω0 - f)t/2)/((2ω0 - f)/2)2.
In first-order, time-dependent perturbation theory no other transitions are allowed.

Rotator

#### Problem:

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 = J2/2I.  Assume the top is perturbed by an external magnetic field B, with interaction H' = -μ∙B, where μ = GJμB/ħ is the magnetic moment of the top, G > 0, and μB 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 = B0k, B0 > 0?
(c)  Assume that B becomes time dependent,
B(t) = B0k, t < 0,  B(t) = B0k + i ΔBexp(-λt), t > 0.
where ΔB << B0 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 J2 and Jz.  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 = J2/2I,  Ej = ħ2j(j+1)/2I, where j is the total angular momentum quantum number.
(b)  H = J2/2I - (GμB/ħ)J∙B  = J2/2I - ( GμB/ħ)B0Jz.
The eigenstates of H are denoted by |j,m> (m = -j, -j+1, ..., j-1, j).
H|jm> = (ħ2j(j+1)/2I - (GμB/ħ)B0mħ)|j,m>.
Ejm = (ħ2j(j+1)/2I -(GμB/ħ)B0mħ).
The ground state for a given j has m = j, the first excited state has m = j - 1.
E0 = ħ2j(j+1)/2I - GμBB0j,  E1 = ħ2j(j+1)/2I - GμBB0(j-1).
(c)  Time dependent perturbation theory yields
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2, when H = H0 + W(t),
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <Φf|W(t)|Φi>.
W(t) = -(GμB/ħ)ΔBJxe-λt, |Φi> = |j,j-1>, |Φf> = |j,j>
Wfi(t') = -(GμB/ħ)ΔB<j,j|Jx|j,j-1>e-λt'.
2Jx = J+ + J-,  J±|k,j,m> = [j(j+1)-m(m±1)]½ħ|k,j,m±1>.
<j,j|Jx|j,j-1> = [j(j+1)-(j-1)j]½ħ/2 = (ħ/2)[2j]½.
Wfi(t') = -(GμB/2)ΔB[2j]½e-λt'.
ωfi = -GμBB0/ħ.
Pif(t) = (GμBΔB/2)22j|∫0texp(-iGμBB0t'/ħ)e-λt'dt'|2.
Let λ' = (iGμBB0/ħ + λ).
0texp(-λ't')'dt' = (1 - e-λ't)/λ',  Pif(t) = (GμBΔB/2)22j|(1 - e-λ't)/λ'|2.

Atoms

#### Problem:

Assume an atom is interacting with a monochromatic (visible light) EM plane wave
E
(r,t) = E0k cos(ky - ωt).
Assume the wave functions Φnlm(r) of the atomic electrons are characterized by the quantum numbers n, l, and m.  Derive the selection rules for electric dipole transitions.

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 WDE = er∙E.
If we define the dipole moment p = -er , we have WDE = -p∙E.
Here WDE = eE0zcos(ωt), since we can neglect the spatial variations of visible light the plane wave over the dimensions of an atom.
WDE = eE0zcos(ωt) is a perturbation to the atomic Hamiltonian H0 whose eigenstates are the Φnlm(r).
Time dependent perturbation theory yields
Pif(t) = (1/ħ2)|∫0texp(iωfit')Wfi(t')dt'|2,
with  ωfi = (Ef - Ei)/ħ and Wfi(t) = <Φf|WDE(t)|Φi>, with WDE(t) = eE0zcos(ωt).
Pif(t) = 0 if <Φf|WDE(t)|Φi> = 0.  This yields the selection rules.

f|WDE(t)|Φi> = eE0cos(ωt)<Φf|z|Φi>.
Let Φi(r) = Rni li(r)Yli mi(θ,φ),  Φf(r) = Rnf lf(r)Ylf mf(θ,φ)

With z = r cosθ ∝ r Y10(θ,φ) we have
f|WDE(t)|Φi> ∝ ∫0πsinθ dθ∫0dφ Y*lf mf(θ,φ) Y10(θ,φ) Yli mi(θ,φ).
The integrant is a product of three spherical harmonics.
The integral is zero unless
(i) m1 + m2 + m3 = 0,
(ii) |l1 - l2| ≤ l3 ≤ (l1 + l2),
(iii) l1 + l2 - l3 = even.
Ylm*(θ,φ) = (-1)mYl-m(θ,φ).
We therefore have that <Φf|WDE(t)|Φi> = 0 unless  mf = mi, lf = li ± 1.  If we choose another direction for the polarization of E, i.e.  E = E0i or E0i, then we find  mf = mi ± 1, lf = li ± 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.

#### Problem:

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.