Lectur
Lecture 13:
e 13: Time-
Time-dependent perturbati
dependent perturbation theory:
on theory:
photoelectric effect and Fermi’s golden rule (11/1/2005)
photoelectric effect and Fermi’s golden rule (11/1/2005)
Last time we discussed spontaneous and stimulated emission and stimulated absorption and their Last time we discussed spontaneous and stimulated emission and stimulated absorption and their relationships, using the insights from time-dependent perturbation theory. We also derived the rate relationships, using the insights from time-dependent perturbation theory. We also derived the rate of the emission of electric dipole radiation.
of the emission of electric dipole radiation. Addit
Additionaional l reareading if ding if yoyou u wish: wish: BranBransden and sden and JoacJoachainhain, , ch. ch. 11.711.7
At the beginning of our time-dependent adventures, we discussed the transition between two At the beginning of our time-dependent adventures, we discussed the transition between two states or a small number of states. But more typically, the number of states between which we find states or a small number of states. But more typically, the number of states between which we find trans
transitionitions s is infinite; the is infinite; the enerenergy of gy of the free particlthe free particles is es is concontintinuous, for uous, for exampexample. le. This requiThis requires someres some modificiations in our formulae. We will see what Fermi’s golden rule is and how these ideas are used modificiations in our formulae. We will see what Fermi’s golden rule is and how these ideas are used in a detailed analysis of the photoelectric effect.
in a detailed analysis of the photoelectric effect.
Approximating the frequency-dependence by a delta-function
Approximating the frequency-dependence by a delta-function
Consider the case of the harmonic perturbation Consider the case of the harmonic perturbation
H H
=
= V V cos( cos(ωtωt)), , H H
ni
ni = =
nn||
V V||
ii
cos(cos(ωtωt) =) = V V nini cos( cos(ωtωt))A week ago, we derived that the first-order probability for the initial state
A week ago, we derived that the first-order probability for the initial state
| |
ii
to change into one of to change into one of the final statesthe final states
| |
nn
is isP P ii→→nn = =
| |
V V nini||
22 ¯¯h h22 sin sin22[([(ωωnini−
−
ωω))t/t/2]2] ((ωωnini−
−
ωω))22 As timeAs time t t
→
→ ∞
∞
, , the second factor becomes strongly peaked aroundthe second factor becomes strongly peaked around ω ωnini = = ω ω. Using the result of the. Using the result of theintegral integral
∞∞ −∞ −∞ sin sin22xx x x22 dxdx = = π πcalculated by the residue techniques and the substitution
calculated by the residue techniques and the substitution xx = = ((ωω
−
−
ωωnini))t/t/2,2, dxdx = = ((t/t/2)2)dωdω, we may, we mayalso see that also see that
∞∞ −∞ −∞ sin sin22[([(ωωnini−
−
ωω))t/t/2]2] ((ωω−
−
ωωnini))22 dω dω = = πtπt 22 Because the integral becomes concentrated nearBecause the integral becomes concentrated near ωω == ωωnini for large tt, we can also generalize the for large , we can also generalize the
formula above to formula above to
∞∞ −∞ −∞ f f ((ωω))sinsin 2 2[([(ωω ni ni−
−
ωω))t/t/2]2] ((ωω−
−
ωωnini))22 dω dω = = πtπt 22 f f ((ωωnini)) The result only depends on the value ofThe result only depends on the value of f f ((ωω) at the right point which is allowed assuming that) at the right point which is allowed assuming that f f ch
changes slowly enough, anges slowly enough, namelnamely y if if ddf f dω dω
tt⇒
⇒
sin sin22[([(ωωnini−
−
ωω))t/t/2]2] ((ωω−
−
ωωnini))22 = = πtπt 22 δ δ ((ωω−
−
ωωnini)) Such a new form of the ratio allows us to writeSuch a new form of the ratio allows us to write P P ii→→nn = = π π 22
||
V V nini||
22 ¯¯h h22 tδ tδ ((ωω−
−
ωωnini))Does it make sense?
In other words, as time t goes to infinity, the transition is only allowed if the photons or other quanta of energy have the right value of the energy required by energy conservation. But the formula has two obvious problems
•
the probability can become infinite which seems unphysical (=physically meaningless)•
the probability does not describe “flopping” we studied previouslyHowever, both of these apparent problems evaporate if we the final states
|
n
form a continuum because:•
the total probabilities of well-defined outcomes will be obtained as integrals of this delta function and will be finite•
if the final states form a continuum, it is possible for the particles to escape to infinity and no flopping occursHow does it work quantitatively? Denote the number of states per unit energy as ρn(E n). To
find the probability of transition to any of these states, simply integrate the previous formula: P i→n (any) = π 2
|
V ni|
2 ¯h2 t
∞ −∞ ρn(E n)δ (ω−
ωni)dE nIn the integral we must change the variables of the delta-function to energy, gaining a factor of ¯h, and the integral equals ¯hρn(E n) where E n is the energy of the “correct” final state that preserves
the energy. We therefore have
P i→n(t) =
π 2
|
V ni
|
2¯h ρn(E n)t
Fermi’s golden rule
Let us make a final step and erase the factor of t, to find the rate of the transitions i.e. the number of transitions per unit time (if you multiply it by the number of “atoms” you have):
Rni =
π 2¯h
|
V ni|
2ρ n(E ).
This simple formula is called Fermi’s golden rule and it is useful for problems involving a contin-uum of states. Note that it resulted from the first-order perturbation theory; when we deal with a continuum, the reliability of perturbation theory also increases and most of the refinements we have seen for the two-level systems become unimportant.
More typically, you will see the rule written in the following form Rni = 2π
¯h
|
H
ni
|
2ρn(E )Photoelectric effect
Einstein received his 1921 Nobel prize for an explanation of this effect. Recall that the energy of outgoing electrons depends on the frequency, not intensity, of the incident light because the situation may be described as absorption of the photons whose energy is ¯hω by the electrons – and one must use the energy conservation.
Let us now study this effect in detail using the time-dependent perturbation theory. The “kick” of the photon only affects one electron. Let us choose the electron in the k-shell (klosest to the nucleus) of the atom. This electron sees the full charge Ze of the nucleus and the other electrons do not play much role. We are back to a Hydrogen-like problem. The initial state of our electron is
|
i
=
Z3
πa3
1/2exp(
−
Zr/a), a = 4π0¯h2
me2 .
Much like in the previous lecture, we are considering the transitions between two electronic states. However, in the present case, we choose the final state to be a free electron far away from the nucleus. It may be represented by a plane wave and these plane waves are normalized to a delta-function.
|
k
= e i k·r (2π)3/2,
k |
k
= δ (3)(k−
k )Despite this change towards the continuum, the arguments work much like before: the leading transitions are determined by the electric dipole effect interacting with the electric field of the electromagnetic wave:
|
V ni|
2 =|
E ·
P|
2 = . . . =|
ˆ·
P|
2E
02 =|
P|
2E
02 cos2θwhere θ is the angle between the matrix element of the electric dipole and the polarization vector ˆ. Thus, using our “first” Fermi’s golden rule, we have
Rni = π
2¯h
|
P|
2
E
02 cos2θ ρn(E n)What is the density of states? For a free particle, it is an easy question: ρ(E )dE = k2dk dΩ; E = p 2 2m = ¯h2k2 2m
⇒
dE = ¯h2k m dk⇒
ρ(E ) = mk2 ¯h2 dΩ. Substituting this to our rule, we haveRni = πmkn ¯h3
|
P|
2E
02 2 cos 2θ dΩCross sections
When we talk about the probabilities of transition in which the initial state contains two objects – for example, a beam of particles of one type directed to particles of other types – a useful concept is cross section. A cross section may be thought of as the area of a single target that one must hit in order for a particular reaction to take place.
Take this definition and imagine that a layer of thickness d of atoms whose total number density is N target/V is being bombed by N beam/t particles per unit time. The number of interactions per
unit time is then, following our definition of the cross section, equal to Rni = N t =
N beam t
σ
N target V
d⇒
σ = Rni (N /t)(N /V )dTo relate these formulae to our previous form of Rni, notice that the energy density may again be
written in two different ways:
energy density = 1 20
E
2
0 = N beam¯hω/V
We dropped an index on ω. Also realize that in the previous section, our target was only made of one atom
N target = 1
and the speed of light relates the thickness with the time in which the interactions occur: d
t = c With these substitutions, we obtain
σ = Rni
(1/2)0
E
02c/¯hωUsing our previous form of Rni, our cross section is therefore
σ = πmknω ¯h2c0 cos2θ
|
P|
2 dΩ⇔
dσ dΩ = πmknω ¯h2c0 cos2θ|
P|
2Switching to the momentum form
Up to some simple factors, we have expressed the cross section using the matrix element of the dipole moment. Now we want to calculate this matrix element of the electric dipole moment between our initial and final states:
P
= e
n|
r|
i
Actually, it will be helpful to express this matrix element in a form that superficially looks more complicated, by inserting the Hamiltonian as follows:
P
= e E n−
E i
n|
H ˆ0r−
r ˆH 0|
i
Because ˆ H 0 =−
¯h2 2m∇
2 + V (r) and [V (r), r] = 0, we have P
= e E n−
E i
−
¯h 2 2m
n|∇
2r−
r∇
2|
i
The commutator is easy to evaluate∇
2r−
r∇
2 = 2∇
⇒
P
=−
e¯hωm
n|
∇|
i
The last equation is the “momentum form” of
P
because p =−
i¯h∇
. This form, as we have said, is actually simpler for evaluation of the matrix elements. Finally, let us make the calculation of the matrix element for our case:
P
=−
e¯h ωm
d3r ψ∗ i(r)∇
ψn(r)
∗ =−
e¯h ωm
Z 3 πa3
1/2 1 (2π)3/2
e−Zr/a∇
e−i kn·r d3r = = +i e¯h ωm
Z 3 πa3
1/2 kn (2π)3/2
e−Zr/a e−i kn·r d3rThe integral is a standard Fourier transform:
e−Zr/a e−i kn·r d3r = 8π(Z/a) [(Z/a)2+ k2 n]2It can be calculated if you realize that the Laplacian has a pretty simple action on e−Zr/a
. Substi-tuting it to
P
, we obtain P
= ie¯h mωn 2√
2 π
Z a
5/2 kn [(Z/a)2+ k2 n]2Simplifying the cross section at high energies
Let us simplify this expression assuming that the energy of the incoming photon and the outgoing electron is still non-relativistic but actually much higher than the binding energy of the atom. Actually, we have already made this approximation when we pretended that the final state (the plane wave) was an energy eigenstate. For
kn2
Z a
2, ¯h2kn2 2m≈
¯hω⇒
k =
2mω ¯h Finally, dσ dΩ =
e2 4π0¯hc
32¯h mω
Z kna
5cos2θ = α32¯h mω
Z kna
5cos2θwhere we introduced α = 1/137.036..., the fine-structure constant that determines the squared charge of the electron in “natural units”. It is illuminating to express the cross section as a multiple of the squared Bohr radius
a = 4π0¯h
2
me2 =
¯h αmc in the following way:
dσ dΩ = 4
√
2α8Z 5a2
mc2 ¯hω
7/2 cos2θThe angle θ between the (electric) polarization direction ˆ and the direction of the outgoing electron is not observable. We should try to switch to the angle Θ between the incident photon and the outgoing electron. Draw a picture – one that should also imply
cos2θ = sin2Θcos2φ Averaging this over the angle φ gives
cos2θ = 1 2 sin 2Θ This leads us to dσ dΩ = 2
√
2α8Z 5a2
mc2 ¯hω
7/2 sin2ΘThe electron is usually emitted nearly Θ = 90 degrees to the photon, because of the transverse electric field. Note that all factors except for a2 are dimensionless and a2 has the correct dimension of an area. You could expect that σ is O(a2) but don’t forget that α8
≈
8×
10−18Graphs and errors in the periodic table
On the blackboard, I should be able to draw a graph showing how the cross section – more precisely, the absorption coefficient Nσ/ρ for platinum – depends on the wavelength λ in ˚A of the radiation that must be in the X-ray region to eject the inner electrons. You should notice that there are sharp edges. They correspond to ejecting of the electrons with energies
E = 13.6(Z
−
S n)2
n2 eV
where S n is an effective shielding by inner electrons. There is an edge for ejecting the k-shell
electrons; several edges for l-shell electrons, and so forth. In 1913, years before our derivations could have been made, H. G. J. Moseley plotted the energy of k- and l-shell edges and showed that
√
E∼
ZThis allowed him to determine Z of many elements he analyzed. Consequently, he could have corrected some mistakes in the periodic table of that time. After the errors are fixed, the graph looks very clean, and it should be drawn on the blackboard right now. Below, you find space to draw the two graphs, too.
