Lecture 13
Short Description
Time-dependent perturbation theory - 4...
Description
Lecture Lecture 13: Time-dependent Time-dependent perturbation perturbation theory: photoelectric effect and Fermi’s golden rule (11/1/2005) 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 of the emission of electric dipole radiation.
Additiona Additionall reading reading if you you wish: Bran Bransden sden and Joachain Joachain,, ch. 11.7 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 transition transitionss is infinite; the energy energy of the free particles particles is contin continuous, uous, for example. example. This requires requires some 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.
Approximating the frequency-dependence by a delta-function Consider the case of the harmonic perturbation
cos(ωt)), H = V V cos(ωt
cos(ωt)) = V = V ni cos(ωt)) H ni = n V i cos(ωt ni cos(ωt
| |
A week ago, we derived that the first-order probability for the initial state i to change into one of the final states n is 2 2 [(ωni ω )t/2] V ni t/2] ni sin [(ω P i n = 2 (ωni ω )2 ¯ h
|
|
| |
→
As time t time t integral
− −
→ ∞, the second factor becomes strongly peaked around ω around ω
sin2 x dx = dx = π π x2 calculated by the residue techniques and the substitution x = (ω also see that sin2 [(ω [(ωni ω )t/2] t/2] πt = dω = dω (ω ωni)2 2
= ω.. ni = ω
Using the result of the
t/2, ni )t/2,
dx = dx = (t/2) t/2)dω dω,, we may
∞
−∞
∞
−
−∞
−ω
−
Because the integral becomes concentrated near ω = ωni for large t, we can also generalize the formula above to sin2 [(ω [(ωni ω )t/2] t/2] πt = f ( f (ω ) dω = dω f ( f (ωni) (ω ωni )2 2 ∞
−
−∞
−
The result only depends on the value of f ( f (ω ) at the right point which is allowed assuming that f changes changes slowly enough, namely namely if sin2 [(ω [(ωni ω )t/2] t/2] πt = δ (ω (ω ωni )2 2
df dω
t ⇒
−
−
Such a new form of the ratio allows us to write P i
n =
→
2 π V ni ni tδ (ω 2 h ¯2
| | 1
ni )
−ω
ni )
−ω
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 previously However, 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 occurs
How 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: π V ni 2 P i n (any) = t 2 h ¯2
| |
→
∞
ρn (E n )δ (ω
−∞
ni )dE n
−ω
In 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 π V ni 2 P i n (t) = ρn (E n )t 2 h ¯ →
| |
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 V ni 2 ρn (E ). 2¯h
| |
This simple formula is called Fermi’s golden rule and it is useful for problems involving a continuum 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 ni 2 ρn (E ) ¯ h
| |
in which the general time-dependence was recovered, gaining (surprisingly) a factor of four.
2
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 Z e 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 3
|i =
1/2
Z
4π0 h ¯2 a = . me2
exp( Zr/a),
−
πa3
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. ·
eik r k = , (2π)3/2
(3)
k |k = δ
|
(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 | = |E · P| ni
2
2
= . . . = ˆ 2 02 = 2 02 cos 2 θ
| · P| E |P| E
where 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 π 2 2 2 Rni = 0 cos θ ρn (E n ) 2¯h What is the density of states? For a free particle, it is an easy question:
|P| E
2
ρ(E )dE = k dk dΩ;
¯ 2k2 p2 h = E = 2m 2m
⇒
¯ 2k h dE = dk m
⇒
ρ(E ) =
mk2 dΩ. ¯2 h
Substituting this to our rule, we have πmkn 2 02 cos2 θ dΩ Rni = 3 2 ¯ h
|P| E
Cross 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 N N beam N target = Rni = σ d t t V
3
⇒
σ =
Rni (N beam /t)(N target /V )d
To relate these formulae to our previous form of Rni , notice that the energy density may again be written in two different ways: 1 energy density = 0 02 = N beamh ¯ ω/V 2 We dropped an index on ω. Also realize that in the previous section, our target was only made of one atom N target = 1
E
and the speed of light relates the thickness with the time in which the interactions occur: d = c t With these substitutions, we obtain Rni (1/2)0 02 c/¯hω
σ =
E
Using our previous form of R ni , our cross section is therefore σ =
πmkn ω cos2 θ 2 dΩ 2 ¯ c0 h
|P|
dσ πmkn ω = 2 cos2 θ 2 dΩ ¯ c0 h
⇔
|P|
Switching 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: = e n r i
P | |
Actually, it will be helpful to express this matrix element in a form that superficially looks more complicated, by inserting the Hamiltonian as follows: e ˆ 0r rH ˆ 0 i = n H E n E i
P
Because ˆ 0 = H
¯2 h 2m
− ∇
we have
− |
2
+ V (r)
P
E n The commutator is easy to evaluate
|
and
[V (r), r] = 0,
2
e
=
−
− E
i
h¯
2
2
− 2m n|∇ r − r∇ |i
= − e¯h n|∇| i ⇒ P ωm because p = −i¯h∇. This form, as we have said, The last equation is the “momentum form” of P 2
∇ r − r∇
2
= 2
∇
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
d r ψ (r)∇ ψ (r) 3
∗
i
n
∗
=
e¯h Z 3 ωm πa3
1/2
− e¯h Z 3
= +i
ωm 4
πa3
1 (2π)3/2 1/2 kn
e e
(2π)3/2
Zr/a
−
i k ·r 3
−
∇e
Zr/a
−
d r =
i k ·r 3
−
e
n
n
dr
The integral is a standard Fourier transform:
e
Zr/a
−
i k ·r 3
−
e
n
d r =
8π(Z/a) [(Z/a)2 + kn2 ]2
It can be calculated if you realize that the Laplacian has a pretty simple action on e tuting it to , we obtain 5/2 kn = ie¯h 2 2 Z [(Z/a)2 + kn2 ]2 mωn π a
Zr/a
−
P
√
P
. Substi-
Simplifying 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 2
Z k , 2 n
a
Finally, dσ = dΩ
¯ 2 kn2 h 2m
≈ h¯ ω
32¯h Z e2 4π0 h ¯ c mω kn a
5
⇒
k =
2mω ¯h
5
32¯h Z cos θ = α cos θ 2
2
mω kn a
where 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 4π0 h ¯2 ¯ h = a = me2 αmc in the following way: 2 7/2 dσ 8 5 2 mc = 4 2α Z a cos2 θ ¯ω dΩ h
√
The 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 cos2 θ = sin2 Θcos2 φ Averaging this over the angle φ gives 1 cos2 θ = sin 2 Θ 2 This leads us to
√
dσ mc2 = 2 2α8 Z 5 a2 ¯ω dΩ h
7/2
sin2 Θ
The electron is usually emitted nearly Θ = 90 degrees to the photon, because of the transverse electric field. Note that all factors except for a 2 are dimensionless and a 2 has the correct dimension of an area. You could expect that σ is O(a2 ) but don’t forget that α 8 8 10 18.
≈ ×
5
−
Graphs 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
n2
2
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
∼ Z
This 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.
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