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Time-Dependent Perturbation Theory Time-evolution operator as a product of elementary operators Let U (t 1 ,t 0 ) be the time-evolution operator evolving the density matrix ˆ ρ(t 0 ) into ˆ ρ(t 1 ) [see Eq. (22) in the Density Matrix chapter of Quantum Mechanics I]: ˆ ρ(t 1 )= U (t 1 ,t 0 ρ(t 0 )[U (t 1 ,t 0 )] (t 1 >t 0 ). (1) By its very definition, the evolution operator satisfies the identity U (t n ,t n-1 )U (t n-1 ,t n-2 ) ··· U (t 2 ,t 1 ) U (t 1 ,t 0 ) U (t n ,t 0 ), (2) where we assume t n >t n-1 >...>t 1 >t 0 . (3) On the other hand, if the time step t j - t j -1 = (j =1, 2, 3,...,n) (4) is small enough, then, up to small corrections in 2 , the evolution operator U (t j ,t j -1 ) can be related to the Hamiltonian as (we set ~ = 1) U (t j ,t j -1 )=1 - iH t j + O( 2 ), (5) where H t stands for the Hamiltonian at the time moment t. To check the validity of (5), substitute it into (1) and make sure that the result satisfies the equation of motion for ˆ ρ [equation (5) in the Density Matrix chapter of Quantum Mechanics I]. We thus arrive at the representation U (t n ,t 0 ) = (1 - iH tn ) ··· (1 - iH t 3 )(1 - iH t 2 )(1 - iH t 1 )+ O(). (6) Perturbative expansion for time-evolution and statistical operators 1

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Page 1: Time-Dependent Perturbation Theorypeople.umass.edu/bvs/T_perturb.pdf · Time-Dependent Perturbation Theory Time-evolution operator as a product of elementary operators Let U(t 1;t

Time-Dependent Perturbation Theory

Time-evolution operator as a product of elementary operators

Let U(t1, t0) be the time-evolution operator evolving the density matrixρ(t0) into ρ(t1) [see Eq. (22) in the Density Matrix chapter of QuantumMechanics I]:

ρ(t1) = U(t1, t0)ρ(t0)[U(t1, t0)]† (t1 > t0). (1)

By its very definition, the evolution operator satisfies the identity

U(tn, tn−1)U(tn−1, tn−2) · · · U(t2, t1)U(t1, t0) ≡ U(tn, t0), (2)

where we assumetn > tn−1 > . . . > t1 > t0. (3)

On the other hand, if the time step

tj − tj−1 = ε (j = 1, 2, 3, . . . , n) (4)

is small enough, then, up to small corrections in ε2, the evolution operatorU(tj, tj−1) can be related to the Hamiltonian as (we set ~ = 1)

U(tj, tj−1) = 1− iεHtj +O(ε2), (5)

where Ht stands for the Hamiltonian at the time moment t. To check thevalidity of (5), substitute it into (1) and make sure that the result satisfiesthe equation of motion for ρ [equation (5) in the Density Matrix chapter ofQuantum Mechanics I]. We thus arrive at the representation

U(tn, t0) = (1− iεHtn) · · · (1− iεHt3)(1− iεHt2)(1− iεHt1) + O(ε). (6)

Perturbative expansion for time-evolution and statistical operators

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Let the Hamiltonian Ht consist of two parts:

Ht = H0 + Vt, (7)

where H0 is time-independent and Vt is a certain perturbation. Our goal isto expand U(t, t0) in powers of Vt. Speaking practically, such an expansionbecomes useful when Vt is appropriately small allowing one to truncate theperturbative series to one or two first terms.

Observing that

1− iεHtj = (1− iεH0)(1− iεVtj) +O(ε2), (8)

we substitute the r.h.s. of (8) for (1−iεHtj) in (6) and then open the bracketswith (1− iεVtj). This naturally introduces different powers of Vtj along withthe necessity to sum over the time moments tj. Taking the limit ε → 0, wereplace summation over tj with integration. Simultaneously, we replace theproducts of all the terms (1− iεH0) with exponentials:

· · · Vtb(1− iεH0) · · · (1− iεH0)Vta · · · → · · · Vtb e−i(tb−ta)H0 Vta · · · .

This allows us to take the limit of ε → 0 rendering our expansion exact.Finally, we decompose the exponentials to associate them with Vt’s:

· · · e−i(tc−tb)H0 Vtb e−i(tb−ta)H0 Vta e

−i(ta−td)H0 · · · =

= · · · e−itcH0V (tb)V (ta)eitdH0 · · · ,

where we defineV (t) = ei(t−t0)H0 Vt e

−i(t−t0)H0 . (9)

This brings us to the result

U(t, t0) = e−i(t−t0)H0

[1− i

∫ t

t0

dt1V (t1)−∫ t

t0

dt1

∫ t1

t0

dt2 V (t1)V (t2) + . . .

+ (−i)n∫ t

t0

dt1

∫ t1

t0

dt2 · · ·∫ tn−1

t0

dtn V (t1)V (t2) · · ·V (tn) + . . .

], (10)

Pay attention to the limits of integrating over tj’s. Those limits reflect thefact that Vtj ’s emerge from the expression (6) in the strict chronologic order.

We can readily establish similar expression for the statistical operatore−βH

′by treating β as an (imaginary) time interval.

e−βH′

= (1− εH ′)n + O(ε), ε = β/n. (11)

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H ′ = H0 + V. (12)

(1− εH ′) = (1− εH0)(1− εV ) + O(ε2). (13)

e−βH′

= e−βH0

[1−

∫ β

0

dτ1V (τ1) +

∫ β

0

dτ1

∫ τ1

0

dτ2 V (τ1)V (τ2) + . . .

+ (−1)n∫ β

0

dτ1

∫ τ1

0

dτ2 · · ·∫ τn−1

0

dτn V (τ1)V (τ2) · · ·V (τn) + . . .

], (14)

V (τ) = eτH0 V e−τH0 . (15)

The chronologic order in (14) has the same origin as the chronologic order in(10).

Linear Response: Kubo formula

By linear response of an observable A to the perturbation Vt one means theleading (in powers of Vt) effect of the perturbation on expectation value of Aat a given moment of evolution.1 The linear response is given by the Kuboformula:

〈A 〉 = 〈A(t) 〉0 − i

∫ t

t0

dt1〈[A(t), V (t1)]〉0 , (16)

whereA(t) = ei(t−t0)H0 Ae−i(t−t0)H0 , (17)

and〈(. . .)〉0 = Tr (. . .)ρ(t0). (18)

Problem 11. Derive Kubo formula from (10).

The similarity of the forms of Eqs. (9) and (17), as well as the deep mean-ing of this form will become clear in the next section, when we introduce theinteraction picture.

1In condensed matter physics, typical external perturbations—used to probe the prop-erties of various systems—are periodic electric and magnetic fields, electromagnetic radi-ation, beams of particles (electrons, neutrons, etc.).

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Time-dependent unitary transformations: Heisenberg and interac-tion pictures, “rotating frames”

Let St be some time-dependent unitary operator:

S†tSt = StS†t = 1.

If for each observable A, we perform the unitary transformation

A(t) = S†tASt ⇔ A = StA(t)S†t ,

and do the same for the density matrix:

ρ(t) = S†t ρ(t)St ⇔ ρ(t) = Stρ(t)S†t ,

we will only change the form of the density matrix and the operators of allthe observables, but not the physical properties of the system. This is be-cause the measurement postulates of quantum mechanics are invariant withrespect to any unitary transformation.

Problem 12. Verify the above statement for both parts of the measuringpostulate: (i) for the probability part and (ii) for the projection part.

By performing this or that unitary transformation, we simply change repre-sentation of the very same physics.

The change in the form of the density matrix and observables naturallyimplies a change in the form of the evolution operator. Let us see how thenew evolution operator will look like. We have

U(t, t0)ρ(t0)U †(t, t0) = ρ(t) = S†t ρ(t)St = S†tU(t, t0)ρ(t0)U †(t, t0)St= S†t U(t, t0)St0 ρ(t0)S†t0 U

†(t, t0)St.Hence

U(t, t0) = S†t U(t, t0)St0 . (19)

We also want to relate the new and the old Hamiltonians. To this end we usethe generic relation between the Hamiltonian and evolution operator (impliedby the evolution equation for the density matrix):

i∂

∂tU(t, t0) = Ht U(t, t0), i

∂tU(t, t0) = Ht U(t, t0).

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In view of the unitarity of evolution operators, these relation imply

Ht = i

[∂

∂tU(t, t0)

]U †(t, t0), Ht = i

[∂

∂tU(t, t0)

]U †(t, t0). (20)

Applying (20) to (19) brings us to the result

Ht = S†tHtSt + i∂S†t∂tSt. (21)

We are especially interested in the cases when

St = e−iH∗(t−t0), (22)

where H∗ is a certain time-independent Hamiltonian. Here we have

Ht = eiH∗(t−t0) Ht e−iH∗(t−t0) − H∗. (23)

If H∗ = H (since H∗ does not depend on time, this implies that H is alsotime-independent), we are dealing with the Heisenberg picture, where Ht = 0and the density matrix does not evolve.

Problem 13. Show that for the second-quantized harmonic Hamiltonian(either bosonic or fermionic)

H =∑s

εs a†s as,

the creation and annihilation operators in the Heisenberg picture have theform (here we use t0 = 0):

as(t) = as e−iεst a†s(t) = a†s e

iεst. (24)

Hint. The easiest way to proceed is to derive a certain differential equationfor as(t) and a†s(t) by differentiating

as(t) = eiHt as e−iHt , a†s(t) = eiHt a†s e

−iHt

with respect to time and then using the particular form of the Hamiltonianand the properties of creation and annihilation operators.2 After that, one

2Different for bosons and fermions!

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simply verifies that (24) satisfy the obtained equations, with appropriate ini-tial conditions.

Yet another important class of pictures takes place when

Ht = H0 + Vt, [H0, H∗] = 0.

Here we have

Ht = H0 −H∗ + eiH∗(t−t0) Vt e−iH∗(t−t0).

In particular, if H0 is time-independent, we can chose H∗ = H0 to get theso-called interaction picture, in which only the perturbation term survives inthe Hamiltonian

Ht ≡ V (t) = eiH0(t−t0) Vt e−iH0(t−t0) (interaction picture). (25)

Comparing (25) to (9) and (17), one can guess that the interaction picture isespecially convenient for considering perturbative effects. This is indeed thecase as illustrated by the following problem.

Problem 14. Derive Eq. (10) employing the interaction picture. In the in-teraction picture, find the evolution operator as a series of integrals by usingexpansion (6). Then return back to the original picture. In particular, makesure that Kubo formula can be derived in the interaction picture as well.

Rotating frame. Suppose our Hilbert space consists of two subspaces, I andII, and the perturbation V has non-zero matrix elements Vab and Vba onlyif the state |a〉 belongs to the subspace I, while the state |b〉 belongs to thesubspace II. In this case, the following transformation—the so-called rotatingframe—might prove very useful.

H∗ = ωPI (rotating frame), (26)

where PI is the projector on the subspace I and ω is a certain frequency/energy.In this case (below we set t0 = 0),

e−iH∗t = PII + e−iωtPI = 1 +(e−iωt − 1

)PI , (27)

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where PII = 1 − PI is the projector on the subspace II. In the “rotatingframe,” the matrix elements become:

Vab = eiωt Vab, Vba = e−iωt Vba. (28)

Problem 15. Derive Eqs. (27) and (28).

We conclude that in the rotating frame, all the energies in the subspace Iget shifted by ω, while all the matrix elements from subspace I to subspaceII acquire periodic phase factor eiωt, and all the matrix elements from sub-space II to subspace I acquire periodic phase factor e−iωt. This is extremelyconvenient for treating periodic perturbations. Periodic time dependence ofmatrix elements can be readily eliminated at the expense of simply shiftingthe energies of one of the two subspaces.

The Golden Rule

Suppose that at the initial time moment t0 = 0, the density matrix ρ(0)corresponds to either a pure eigenstate of the Hamiltonian H0, or a statisti-cal mixture of some—but not all—eigenstates of the Hamiltonian H0. Theactual Hamiltonian of the system is (7), so that ρ evolves in time.3 Assum-ing that the perturbation V is small, we want to answer the following veryimportant question.

At the time moment t > 0, what is the probability, pΩ(t), to find the systemin a certain span Ω of eigenstates of H0 orthogonal to the subspace of theinitial state?

As a typical example, think of and excited state of an atom or nucleus asthe initial state ρ(0). Then pΩ(t) is the probability to establish the decay ofthe excited atom/nucleus into a certain span Ω of the final states, providedcorresponding measurement is performed at time t.4 Correspondingly, Ω

3Excluding the trivial case of [Vt, ρ] 6= 0.4Specifying the time of measurement matters, because, speaking generally, we cannot

continuously observe quantum system without changing its properties (recall quantumZeno effect).

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is associated with certain areas in the momentum space of the products ofdecay. If Ω includes all possible momenta of the products of decay, then pΩ(t)is the probability for the excited state to decay by the time t.

In accordance with the measurement axiom,

pΩ(t) = Tr PΩ ρ(t), PΩ =∑|f〉∈Ω

|f〉〈f |, (29)

where |f〉 is an eigenstate of H0. The notation f comes from “final” todistinguish those states from the “initial” eigenstates of H0 contributing toρ(0):

ρ(0) =∑i

wi |i〉〈i|. (30)

All the final states are orthogonal to all the initial states, and thus

PΩ ρ(0) = ρ(0) PΩ = 0. (31)

In view of Eq. (31), the leading term of the perturbative expansion of ρ(t)that yields non-vanishing contribution to pΩ(t) is the second-order term

ρ(2)(t) = e−itH0

∫ t

0

dt′ V (t′) ρ(0)

∫ t

0

dt′′ V (t′′) eitH0 ,

V (t) = eitH0 Vt e−itH0 .

Incidentally, note the convenience of the interaction picture where

ρ(2)(t) =

∫ t

0

dt′ V (t′) ρ(0)

∫ t

0

dt′′ V (t′′) (interaction picture).

The term ρ(2) is the minimal term in which ρ(0) is sandwiched between V ’sand the expression is thus protected from vanishing upon tracing with PΩ.Note that PΩ commutes with H0, thus commuting with the exponentialse±itH0 . This, in particular, means that it stays invariant when going to theinteraction picture. We thus have

pΩ(t) ≈ Tr PΩ ρ(2)(t) =

∫ t

0

dt′∫ t

0

dt′′ 〈V (t′′) PΩ V (t′)〉0 , (32)

where〈(. . .)〉0 = Tr (. . .)ρ(0). (33)

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General expression (32) simplifies significantly when Vt is either time-independent, or a harmonic function of time (normally referred to as peri-odic perturbation). And the latter case actually reduces to the former onein the appropriate rotating-frame picture, where the frequency of the framematches the frequency of the perturbation.

Time-independent perturbation. Consider the case of time-independent per-turbation. The first observation here is that

〈V (t′′) PΩ V (t′)〉0 = 〈V (0) PΩ V (t′ − t′′)〉0,

resulting in (we shift the integration variable t′ → t′ − t′′)

pΩ(t) =

∫ t

0

dt′′∫ t−t′′

−t′′dt′ 〈V (0) PΩ V (t′)〉0 .

The next observation is that for pΩ(t) to be appreciably different from zero,the time t has to be much larger than the typical time t∗ of variation (decay)of the function 〈V (0) PΩ V (t′)〉0. In essence, this is the criterion of applica-bility of the perturbative treatment for pΩ(t). Vanishing 〈V (0) PΩ V (t′)〉0 atcertain appropriately large t′ allows one—at t t′—to extend the limits ofintegration over t′ to plus/minus infinity. This brings us to the result

pΩ(t) = tWΩ, WΩ =

∫ ∞−∞

dt′ 〈V (0) PΩ V (t′)〉0 (t t∗), (34)

known as Fermi’s Golden Rule. According to the Golden Rule, the probabil-ity to find the system in one of the“final” states grows linearly with time.5

The linearity of increasing pΩ with time is a crucial result on its own, allow-ing one to speak of the decay of the initial state and transitions to the finalstates, thus interpreting WΩ as the transition rate. In particular, extendingΩ to all possible finite states, one gets the total transition/decay rate. Witha natural requirement that V does not have non-zero matrix elements be-tween any two initial states,6 the total transition/decay rate acquires a verycompact form

Wtot =

∫ ∞−∞

dt′ 〈V (0)V (t′)〉0 . (35)

5Within a wide range of times where t t∗ while pΩ(t) 1.6Otherwise, those elements can be simply added to H0.

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An alternative—and most frequently used—form of the Golden Rule dealswith the situation when ρ(0) is a pure state (called the initial state):

ρ(0) = |i〉〈i|.

In this case we have

〈V (0) PΩ V (t′)〉0 = 〈i|V (0)PΩV (t′)|i〉 =∑|f〉∈Ω

〈i|V (0)|f〉〈f |V (t′)|i〉.

Taking into account

H0|i〉 = Ei|i〉, H0|f〉 = Ef |f〉,

the time-dependence can be explicitly factored:

〈f |V (t)|i〉 = ei(Ef−Ei)t Vfi, Vfi ≡ 〈f |V |i〉, (36)

and the integration over time performed:∫ ∞−∞

dt ei(Ef−Ei)t = 2πδ(Ef − Ei).

This yields the following form of the Golden Rule (here we restore ~)

WΩ =2π

~∑|f〉∈Ω

|Vfi|2 δ(Ef − Ei). (37)

In particular, for the total rate we have

Wtot =2π

~∑f

|Vfi|2 δ(Ef − Ei). (38)

The presence of the delta-function implies that the summation over the finalstates is understood as an integral over the continuous quantum numbers—usually, momenta—specifying the final states.

It is absolutely straightforward to generalize (37) to the case when ρ(0) isnot a pure state. Using (30), we readily reduce the problem to the previousone, with the final answer

WΩ =2π

~∑i

∑|f〉∈Ω

wi |Vfi|2 δ(Ef − Ei). (39)

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It is very instructive to reveal the details of how the constant rate WΩ

builds up in time. In particular, we want to clarify the conditions for therate of the “transitions” into the subspace Ω to become constant. To thisend, we use the above decomposition into matrix elements to perform ex-plicit integration over t′ and t′′ directly in (32), without making no furtherapproximations.

〈V (t′′) PΩ V (t′)〉0 =∑i

∑|f〉∈Ω

wi 〈i|V (t′′)|f〉〈f |V (t′)|i〉 =

=∑i

∑|f〉∈Ω

wi |Vfi|2 ei(Ef−Ei)t′ei(Ei−Ef )t′′ .

Integrating over times, we find

pΩ(t) = 2∑i

∑|f〉∈Ω

wi |Vfi|2 ft(x), x = Ef − Ei (~ = 1), (40)

where

ft(x) =1− cosxt

x2= 2

sin2(xt/2)

x2. (41)

Recall the role played by a similar function in the theory of single-slit diffraction.—What are the counterparts of x and t in the theory of diffraction?

Comparing (40)–(41) with (39)—do not forget to either restore ~, or set~ = 1 in (39), we see that (39) is valid when t is appropriately large to allowone to make the replacement

ft(x) =→ π t δ(x). (42)

The replacement (42) is accurate if and only if the span of variation of |Ef −Ei| is much larger than t. For example, if the initial state is pure and thesubspace Ω deals with a rather narrow band of energies, |Ef −Ei| ≤ ε, thenthe constant-rate regime for the subspace Ω will be achieved only at t ε−1.Furthermore, if the energetic width of the space Ω is much smaller thaninverse t, then pΩ(t) is dramatically non-monotonic, in a profound contrastwith the naive picture of “transitions per unit time” that is often used tointerpret the Golden Rule. On the other hand, if the initial state ρ(0) is amixed state, with the width of the energy distribution substantially exceedingthe inverse decay time, then, the constant-rate approximation is immediatelyapplicable, independently of the structure of the subspace Ω.

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Based on the result (40)–(41), we can quantify the energy uncertainty ofthe final state, ∆E. We see that ∆E is inversely proportional to the time ofthe perturbative evolution of the initial state (we restore ~ here):

∆E ∼ ~/t.

Periodic perturbation

The theory of the previous section is readily generalized to a harmonictime dependent perturbation of a special form obtained from time-independentone by the replacement (the matrix notation is especially convenient here)

Vif → e−iωtVif , Vfi → eiωtVfi = eiωtV ∗if . (43)

In this case, Eq. (36) gets modified to

〈f |V (t)|i〉 = ei(Ef−Ei−ω)t Vfi, (44)

bringing us to a simple but very important conclusion that the effect of thefrequency ω reduces to the replacement

Ef − Ei → Ef − Ei − ω (~ = 1) (45)

in all the final answers for pΩ. An alternative route towards the sameconclusion—as well as an instructive way of interpreting the latter—is toutilize the rotating-frame picture (see Problem 16).

In the case of two or more frequencies—most often, the same frequencywith two different signs, which corresponds to sinusoidal perturbation—wecan neglect the interference effects provided the frequencies are not too small.The simple reason for that is that the final states corresponding to differentfrequencies are also different, by conservation of energy.

Problem 16. Use the rotating frame to generalize the Golden Rule to theperturbation of the form

Vt = Ae−iωt + A†eiωt, (46)

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where A is a time-independent operator featuring the constraint

Aif = 0, ∀f. (47)

Then show that a generic perturbation of the form

Vt = Be−iωt +B†eiωt, (48)

with B a generic time-independent operator, reduces to two periodic pertur-bations of the form (48)–(47), with the opposite frequencies ±ω. Expressmatrix elements of those two perturbations in terms of Bif and Bfi.

Creation of electron-hole pairs in a semiconductor by optical ab-sorption

The interaction of electronic subsystem of a semiconductor with light is per-turbative. The effective Hamiltonian is as follows.

H = H0 + V,

H0 =∑p

E(e)p a†pap +

∑p

E(h)p h†php +

∑p,λ

cp f †p,λfp,λ,

V =∑p,k,λ

(ga†kh

†p−kfp,λ + g∗f †p,λhp−kak

).

Here ap and hp are the annihilation operators of the electron and hole, re-spectively, for the momentum mode p; fp,λ is the annihilation operator of thephoton in the mode with momentum p and polarization λ. The origin andmomentum dependence of the coupling strength g are not important for ourpurposes. For simplicity, we will be treating g as momentum-independentconstant. The other parameters of the Hamiltonian are the electron energyE

(e)p , the hole energy E

(h)p , and the velocity of light c. The system’s volume

is set equal to unity.7

Problem 17. Use the Golden Rule to calculate the pair-creation rate—which, clearly, will simultaneously be the optical-absorption rate—for themonochromatic radiation of the frequency ω0. To this end, take the initial

7This is a very convenient trick. Always use it!

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condition in which there are no electrons and holes, and only one photonmode, k0 (ck0 = ω0) having a non-zero occupation number n0. Use thefollowing approximations (justified by the conditions in parentheses)

E(e)p =

2+

p2

2me

(p2

2me

),

E(h)p =

2+

p2

2mh

(p2

2mh

),

where ∆ is the so-called insulating gap, and me and mh are the effectivemasses of the electron and hole. Also, one can safely set the momentum (butnot the frequency !) of the photon equal to zero. This is justified by thesmall parameters (of non-relativistic motion)

p

me

c,p

mh

c.

The optical field can be treated classically. Strictly speaking this is justi-fied only when the occupation numbers of relevant photon modes are muchlarger than unity. However, for the problem of optical absorption, the finalanswer is always essentially the same as for the classical electromagnetic field.Use the classical-field approximation for the electromagnetic field,8 one firstemploys the Heisenberg picture for electromagnetic field and then replacescreation and annihilation operators with complex numbers. In accordancewith the results (24), the effective Hamiltonian then reads (note that heref ’s are numbers, not operators).

H0 =∑p

E(e)p a†pap +

∑p

E(h)p h†php,

V =∑p,k,λ

(gfp,λe

−icpta†kh†p−k + g∗f ∗p,λe

icpthp−kak

).

Problem 18. Use the Golden Rule for the periodic perturbation (see Prob-lem 16, or/and the recommended text) to find the absorption/pair-production

8Or any other bosonic field for that matter.

14

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rate for the above Hamiltonian. Compare the result with that of Problem 17.

Spontaneous and stimulated emission

The emission is a process in which an excited microscopic object—an atom, amolecule, a nucleus, etc.—undergoes a transition to a state with lower energyan emits a photon. The emission is called spontaneous if all the modes intowhich the photon is emitted are initially free of photons (have zero occupationnumbers). As we will see, the rate of the emission (the decay rate of theexcited state) into the modes with non-zero initial occupation numbers is

enhanced by the factor (1 +n(s)ph ), where n

(s)ph is the initial occupation number

of the s-th photon mode. The difference between the total emission rate andthe rate of spontaneous emission—equal to the spontaneous-emission ratetimes n

(s)ph —is called stimulated emission.

The (effective) Hamiltonian describing the emission (and not only theemission!) has the following form

H = H0 + V,

H0 =∑p

p2

2ma†pap +

∑p

(E∗ +

p2

2m

)e†pep +

∑p,λ

cp f †p,λfp,λ

V =∑p,k,λ

(gf †p,λa

†k−pek + g∗e†kak−pfp,λ

).

Here ap and ep are the annihilation operators of the lower- and higher-energystate atom9—for definiteness, we call our object “atom”—in the momentummode p, and fp,λ is the annihilation operator of the photon in the mode withmomentum p and polarization λ. The origin and momentum dependence ofthe coupling strength g are not important for our purposes. For simplicity,we will be treating g as momentum-independent constant.

Problem 19. Using the Golden Rule, calculate the total rates of sponta-neous and stimulated emission of a single atom in the upper energy state.

9Without loss of generality, we recon the energy of the atom from its lower-energy statewith momentum zero.

15

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For simplicity, assume that all the occupation numbers of the photons in therelevant modes are the same, being equal to nf . Note that, in view of thesmall non-relativistic parameter p/m c one can safely neglect the disper-sion of the atom setting m =∞.

Exponential decay

Now we want to see how the perturbative treatment of the decay of an initialstate |i〉—valid only at t W−1

tot —generalizes to times t & W−1tot . This can

be controllably done with the model where there is no interaction betweenany two “final” states:

Vff ′ ≡ 0. (49)

The model (49) yields a very good approximation to a realistic case, providedperturbation V is small enough. We confine ourselves with the case of pureinitial state. The generalization to the case of a mixed state is trivial in viewof the statistical-mixture interpretation of the density matrix.

Let us solve our Schrodinger equation

i∂

∂t|ψ(t)〉 = (H0 + V )|ψ(t)〉, (50)

expanding the solution in the Fourier series of the eigenvectors of H0:

|ψ(t)〉 = C0(t)e−itEi|i〉 +∑f

Cf (t)e−itEf |f〉. (51)

Pay attention to the convenient exponential factors. This trick—in essence,it is the same as working in the interaction picture—simplifies the equationsfor the coefficients Ci and Cf (that are nothing but the amplitudes of theexpansion of the wave function in the interaction representation). The right-hand sides of those equations do not contain diagonal terms:

C0 = −i∑f

Vif e−iωfitCf , (52)

Cf = −iVfi eiωfitC0, (53)

whereωfi = Ef − Ei. (54)

16

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Equation (53) can be cast into the integral form:

Cf = −iVfi∫ t

0

dt′ eiωfit′C0(t′), (55)

where the lower integration limit is chosen to satisfy the initial conditionCf (0) = 0. Substituting (55) into (52) and making a substitution t′ = t − tfor the integration variable, we get an integro-differential equation for C0(t):

C0 = −∫ t

0

Q(t)C0(t− t) dt , (56)

with the kernel

Q(t) =∑f

|Vif |2 e−iωfit ≡ 〈V (t)V (0)〉0. (57)

Putting aside the specificity of the model expressed by Eq. (49), so far, wehave not made any approximation. In particular, the structure 〈V (t)V (0)〉0—that we saw before—does not imply perturbative treatment. Now it is timeto take essentially the same large-time limit that we discussed before in thecontext of perturbation theory. At t t∗ [as before, by t∗ we denote thecharacteristic time of vanishing Q(t) = 〈V (t)V (0)〉0], we can extend theupper limit of integration in (56) to infinity.10 As a result we get

C0 = −∫ ∞

0

Q(t)C0(t− t) dt (t t∗). (58)

Equation (58) is solved by the exponential substitution11

C0(t) = e−iΛt (59)

reducing the integro-differential equation to the following equation for the(complex) Λ:

Λ = −iQΛ, (60)

10Note that this is essentially the same limit that was bringing us to the constant-rateregime of perturbation theory. Also, speaking generally, there is a subtlety of taking thislimit. We will address this subtlety later.

11Because of the linearity of Eq. (58), the substitution comes with an arbitrary constantpre-factor. We set the latter equal to unity in view of the initial condition C0(0) = 1.

17

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where

QΛ =

∫ ∞0

Q(t) eiΛtdt. (61)

It is convenient to introduce the real and imaginary parts of Λ:

Λ = ∆− iγ. (62)

The real part ∆ has the meaning of the shift of the energy Ei. The quantityγ > 0 describes the exponential decay of the magnitude of C0.

It is tempting to plug the sum of exponentials (57) and integrate over thetime. However, given γ > 0, we cannot accomplish this seemingly straight-forward procedure because of the divergence of the integrals at their upperlimit. What we can do is regularize the kernel Q as follows

Q(t) → Q(t)e−gt. (63)

observing that, at g > γ, the integrals converge and—provided the origi-nal problem is not ill-defined without the regularization—the answer can beexpressed in the form of the (rather non-trivial) limit

QΛ ≡ Q∆−iγ = i limg→0

[∑f

|Vif |2

Ei − Ef + ∆ + i(g − γ)

]a.c. from g>γ

, (64)

implying that the summation (integration) over f is performed strictly atg > γ, then the expression is analytically continued12 to g ≤ γ, and onlythen the limit of g → 0 is taken.

To better understand the mathematical procedure behind Eq. (64), in-troduce the spectral density η(ε) defined as

η(ε) =∑f

|Vif |2 δ(ε− Ef ). (65)

According to the first equality in (57), the mathematical meaning of η(ε) isvery transparent, the functions Q(t) and η(ε) are related to each other bythe following Fourier transformation:

Q(t) = eiEit

∫ ∞−∞

dε η(ε) e−iεt =

∫ ∞−∞

dε η(ε+Ei) e−iεt. (66)

12We use the symbol [. . .]a.c. from g>γ to express the requirement of analytic continuation.

18

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In terms of η, equation (64) reads

QΛ ≡ Q∆−iγ = i limg→0

[∫ ∞−∞

η(ε) dε

Ei + ∆− ε+ i(g − γ)

]a.c. from g>γ

. (67)

In this formula, it is absolutely crucial that the order of the limit and theintegral are not switched.13 Moreover, it is equally crucial that the integrationis performed at g > γ and the result is then analytically continued to g ≤ γ,and only then the limit is taken [cf. Problem 22, part (b)].

At this point we have to emphasize the central role played by the conditionof (semi-)perturbative regime (controlled by the smallness of V ):

γ t−1∗ . (68)

The importance of this condition comes from the fact that for the vast ma-jority of practically important cases, the procedure of replacing the upperlimit of integration in (56) with infinity is ill-defined14 unless we introduce asmall but finite g. Let g be a positive quantity satisfying the following twoinequalities:

γ < g t−1∗ .

The two conditions guarantee that, on one hand, the integrals over timeconverge, while, on the other hand, the systematic error introduced by theregularization is negligibly small. Then, in view of the smallness of γ andg compared to t−1

∗ , the sum over f in Eq. (64) simplifies (the symbol p.v.stands for the principal-value integration):∑

f

|Vif |2

Ei − Ef + ∆ + i(g − γ)≈

≈ p.v.∑f

|Vif |2

Ei + ∆− Ef− iπ

∑f

|Vif |2 δ(Ef − Ei −∆),

and we arrive at the results (~ is restored)

∆ = p.v.∑f

|Vif |2

Ei + ∆− Ef, (69)

13A nice piece from the internet: ‘...And we’d switch the order of the two operationswithout thinking about it. The professor would always say, “I can interchange these twobecause I am a physicist and I am lazy. A mathematician would spend his whole life tryingto prove this is permissible.” ’

14In the next section, we will clearly see why.

19

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γ =π

~∑f

|Vif |2 δ(Ef − Ei), (70)

Ei = Ei + ∆. (71)

Equations (69) and (70) can be also written in terms of the spectral density:

∆ = p.v.

∫ ∞−∞

η(ε) dε

Ei + ∆− ε, (72)

γ =π

~η(Ei). (73)

As we see from (70) or (73), and will also see from other formulas, theonly role played by ∆ is shifting the energy Ei. That is why we introducedthe shifted energy Ei. Note that, in the V → 0 limit, the value of ∆ scales asthe first power of V , while γ is quadratic in V . Hence, it is meaningful to keep∆ in the right-hand sides of (69) and (70), while neglecting γ. Furthermore,in many effective theories, the sum over f in (69) is ultraviolet divergentimplying an ultraviolet cutoff.15 In such a theory, neither Ei nor ∆ arephysically meaningful, since both are cutoff-dependent. However, the sumEi + ∆ makes perfect sense of the proper energy of the state |i〉.

Up to replacement Ei → Ei, the result (70) is nicely consistent with theGolden Rule.16 Indeed, comparing (70) with (38), we see that 2γ = Wtot.Hence, our result

|C0|2 = e−2γt (t t∗) (74)

corresponds to that of the Golden Rule (|C0|2 = 1−Wtott) within the rangeof times t∗ t W−1

tot where the Golden Rule applies.The condition (68) guarantees that at t . t∗, we have C0(t) ≈ 1, allowing

us to safely substitute (59) into (55) and find

Cf (t) = Vfi1− e−γtei(Ef−Ei)t

Ef − Ei + iγ, (75)

|Cf (t)|2 = |Vfi|21 + e−2γt − 2e−γt cos(Ef − Ei)t

(Ef − Ei)2 + γ2. (76)

15An effective theory is not supposed to be accurate up to arbitrarily high energies.16Note also that Eq. (73) yields a compact representation of the Golden Rule in terms

of the spectral density.

20

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At t γ−1, the solution (76) corresponds to the perturbative result (40)–(41). At t γ−1, the solution is dramatically different from (40)–(41),saturating to the following time-independent limit:

|Cf (t)|2 →|Vfi|2

(Ef − Ei)2 + γ2at t→∞. (77)

In this limit, it is the parameter γ—not the inverse time—that characterizesthe energy uncertainty.

The stochastic-decay interpretation and its limitations

The exponential decay law (74) suggests the stochastic-decay interpretationthat we already discussed in the context of the Golden Rule. Here oneinterprets Wtot dt = 2γ dt as an infinitesimal probability for the state |i〉 to“decay”—into one of the final states |f〉—during an infinitesimal time dt.Then, purely probabilistically,17 one readily obtains the (Poisson) law

p(t) = e−Wtott (78)

for the probability of not decaying by the time t, i.e., the probability of hav-ing no decay events within the time interval [0, t].

Problem 20. Derive the law (78) for the above-introduced process ofstochastic decay.

No matter how accurate, the decay-rate interpretation is fundamentallyapproximate, since, microscopically, quantum mechanics is stochastic only atthe level of measurement; the evolution of the density matrix is absolutelydeterministic. Therefore, it is very instructive to discuss the limitations of thestochastic-decay picture. One obvious limitation is that, strictly speaking,the law (74) does not apply at t . t∗. This, however, is not a dramaticlimitation because at t . t∗ we have |C0| ≈ 1, so that the discrepancy occursonly at the level of small corrections. A really serious discrepancy withthe stochastic-decay picture deals with the distribution of the final states

17In the probability theory, such a process is known as a Poisson process—a particularexample of continuous-time Markov process.

21

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over energies. The quantum mechanical analysis shows—see Eqs. (40), (76)and (77)—that the distribution gets narrower with time.18 This effect is in aprofound discrepancy with the stochastic-decay picture, where the propertiesof the decay processes cannot depend on time.

Yet another limitation concerns very large times. The exponential be-havior (59) was obtained by setting t =∞ in (56). Let us see whether/whenthe latter procedure is consistent with (59). We have∫ t

0

Q(t)C0(t− t) dt =

∫ t

0

Q(t) e−iΛ(t−t) dt = e−iΛt∫ t

0

Q(t) ei∆teγt dt,

and realize that setting t =∞ implies that Q(t) decays with time exponen-tially. Normally, that is not the case! The reason is that the spectrum of anyrealistic quantum system is bounded from below. Let E∗ be the exact lowerbound for Ef . Then E∗ is also the exact lower bound for the support of thespectral-density function η(ε) [see Eq. (65)], and we have

Q(t) = eiEit

∫ ∞E∗

dε η(ε) e−iεt = ei(Ei−E∗)t∫ ∞

0

dε η(ε) e−iεt,

η(ε) = η(E∗ + ε).

Such an integral typically has a power-law rather than exponential decay:

Q(t) ∝ ei(Ei−E∗)t

tα(t→∞), (79)

where the exponent α can be related (by dimensional analysis) to the expo-nent ν describing the asymptotic behavior

η(ε) ∝ εν (ε→ 0).

Namely,α = ν + 1. (80)

18And this is an excellent example of the energy-time uncertainty relation: ∆E∆t ∼ ~.

22

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Problem 21. A more controlled19 and detailed (i.e., with the numeric pre-factor) version of Eq. (79) can be derived by considering the integral∫ ∞

0

ενe−iεte−gεdε,

with a certain real positive g. Perform corresponding derivation. In partic-ular, make sure that the result—the leading term—does not depend on thechoice of g in the limit of large t.

Now we see that, depending on value of t t∗, there are two very differentregimes. The first regime is when the leading contribution to the integralover t in (56) comes from t ∼ t∗. Here we can simply use the regularizedkernel, Eq. (63), which does not change the value of the integral (up to smallcorrections in parameter γt∗), while allowing us to set the limit of integrationto infinity

The second regime sets in at t ∼ tfail, such that the contribution to theintegral over t in (56) associated t ∼ t becomes comparable to the contri-bution coming from t ∼ t∗. At t & tfail. Now the whole physical picturechanges, and, speaking generally, the decay becomes non-exponential.

Let us estimate tfail. By dimensional analysis, based on (79), we find thatthe contributions from t ∼ t∗ and t ∼ t become comparable at t ∼ tfail suchthat

eγtfail(t∗/tfail)α ∼ 1.

When deriving this estimate, we also assumed—based on the dimensionalanalysis—that E∗ ∼ t−1

∗ . A simple algebra brings us then to

γtfail ≈ −α ln(γt∗),

and a very instructive relation

|C0(tfail)|2 ∼ e−2γtfail ∼ (γt∗)2α = (γt∗)

2(ν+1) 1,

telling us that by the time the exponential decay law will become inappli-cable, the probability to find the system in the initial state will be much

19One can question the validity of the above-mentioned dimensional analysis because ofthe upper-limit divergence of the integral

∫ενe−iεtdε.

23

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smaller than unity.

Problem 22. As a very instructive example of how, in the absence ofspectral-edge singularity, equation (67) works exactly as written and thetheory is asymptotically accurate without the breakdown of the exponentiallaw at very large times, consider a model where the spectral density has theLorentzian form

η(ε) =A

(ε− ε0)2 + κ2.

(a) Derive the system of two algebraic equations relating ∆ and γ to theparameters A, ε0, and κ.(b) Make sure that doing the integral (67) at g < γ leads to a wrong answer.(c) Are there limitations on the values of the parameters A, ε0, and κ for thedecay to be exponential?

Resonant scattering

Consider the same model as before—see the Exponential Decay section—butwith a fundamentally different initial condition. Now the state |i〉 is not theinitial state. Rather, it is an important intermediate state, while the initialstate is one of the states |f〉, call it state |f∗〉. We will be solving the sameequation (50), using essentially the same expansion of the solution20

|ψ(t)〉 = C0(t)e−itE0|i〉 + C∗(t)e−itE∗|f∗〉 +

∑f 6=f∗

Cf (t)e−itEf |f〉, (81)

but now C0(0) = 0 and C∗(0) = 1. For the rest of the coefficients we haveCf (0) = 0 and use the same trick of integrating corresponding equations:

Cf = −iVf0

∫ t

0

dt′ eiωf0t′C0(t′). (82)

Here we optimized the notation so that Vf0 and ωf0 stand for what used tobe Vfi and ωfi, respectively. Below we will be also using V0∗ for Vif∗ and

ω = E∗ − E0. (83)

20To avoid confusion and enhance readability, we replace Ei with E0 and Ef∗ with E∗.

24

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We thus have a system of two linear equations

C0 = −∫ t

0

Q(t)C0(t− t) dt − iV0∗ e−iωtC∗ , (84)

C∗ = − iV ∗0∗ eiωtC0 . (85)

Since our system is linear, a general solution comes in the form of linearsuperposition of two particular solutions. One of the two is already knownto us. It corresponds to the exponential decay of the state |i〉. We thus needto find the second one. And it is this second solution that is most relevantto our case, since, in the limit of (macroscopically) small V , the first solutiondecays in time much faster than C∗(t), and by the time when the first solutionvanishes completely, we still have C∗(t) ≈ 1.

As before—and for the same reason—we set the upper limit of integrationin (84) equal to infinity:

C0 = −∫ ∞

0

Q(t)C0(t− t) dt − iV0∗ e−iωtC∗ , (86)

after which the system (85)–(86) is solved by the simple exponential substi-tution

C∗(t) = Ae−iλt, C0(t) = Be−i(λ+ω)t,

leading to the following algebraic system of equations[λ −V ∗0∗−V0∗ ω + λ+ iQω+λ

] [AB

]= 0, (87)

where Qω+λ is defined by the formula (61) with Λ→ ω+λ. The characteristicequation of the system (87),

λ(ω + λ+ iQω+λ)− |V0∗|2 = 0

has two solutions.21 The first one,

λ1 + ω = −iQω+λ1 (first solution),

21Taking into account macroscopic smallness of the term |V0∗|2, we keep only the leadingterms.

25

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is known to us. It is identical to (60), with λ1 + ω ≡ Λ As expected, thesecond solution yields macroscopically small λ:

λ2 =|V0∗|2

ω + iQω+λ2

(second solution). (88)

In view of the microscopical smallness of λ2, we can22 replace Qω+λ2 with Qω

for which, in accordance with (64), we have [we use (83) to restore E∗]

Qω = i limg→+0

∑f

|Vif |2

E∗ − Ef + ig. (89)

Since now the regulator g is infinitesimally small, we can take the limitg → +0 without any extra approximation. As a result, in the convenientparameterization

Qω = i∆∗ + γ∗,

we get23

∆∗ = p.v.∑f

|Vif |2

E∗ − Ef, (90)

γ∗ =π

~∑f

|Vif |2 δ(Ef − E∗), (91)

and the second solution becomes

λ2 =|V0∗|2

E∗ − E0 −∆∗ + iγ∗(~ = 1). (92)

In view of macroscopic smallness of λ2, really interesting is only its negativeimaginary part describing the decay rate of the survival probability |C∗(t)|2.We thus leave in Eq. (92) only the imaginary parts in both sides and find (~is restored by dimensional analysis)

|C∗(t)|2 = e−Wt, (93)

W = −2Imλ2 =2γ∗|V0∗|2

(E∗ − E0 −∆∗)2 + (~γ∗)2. (94)

22And actually should, since we keep only the leading terms.23Here we restore ~ in such a way that the physical meaning of γ∗ is most close to that

of γ in the exponential-decay section).

26

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This is the celebrated generic result for the rate of a resonant process (typi-cally, resonant scattering).

Problem 23. Derive the general relation:

W =vgrσ

V, (95)

between the (total) scattering rate in the system of the volume V , the (total)scattering cross-section σ and the group velocity vgr of the particle beingscattered.

Use the unit-volume version of relation (95):

W = vgr σ (unit-volume normalization) (96)

to convert Eq. (94) into the result for the (total) resonant-scattering cross-section in the following two characteristic cases:(a) Scattering of resonant light from an atom/molecule/micro-system (theso-called resonant fluorescence),(b) Resonant scattering of the α-particle from a nucleus (recall α-decay ofan α-radioactive nucleus).In both cases, provide an explicit meaning of E∗, E0, ∆∗, and γ∗.

Breit-Wigner formula

Consider a single- or two-particle24 scattering under the resonant conditions,when the result (94) applies. In this case,

E∗ = εk,

vgr =dεkdk

.

We will be assuming that the lifetime of the intermediate state is very largecompared to the characteristic energy:

γ∗ εk. (97)

24Since a two-body problem reduces to a single-body one, our discussion will be abouta single particle.

27

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Under this condition, the resonance is very narrow and it makes sense toadopt yet another simplifying condition (of closeness to the resonance):

|E∗ − E0| E0. (98)

In this case, we can neglect the k-dependence of γ∗:

γ∗ ≈ γ0 = π∑f

|Vif |2 δ(Ef − E0) (~ = 1). (99)

With the same degree of accuracy, ∆∗ is k-independent as well, and we cansimply absorb it into E0.

Also, it is a bit more convenient25 to work with the quantity (~ is restored)

Γ = 2~γ0 = 2π∑f

|Vif |2 δ(Ef − E0), (100)

called the decay width of the intermediate state.26 For the resonant-scatteringcross-section we thus have [see (96)]:

σ =

[dεkdk

]−1Γ |V0∗|2

(εk − E0)2 + (Γ/2)2(~ = 1).

And this is not the final result yet. In fact, we can write the answer in termsof Γ only! To this end we observe that, in our case of narrow resonance,

|Vif |2 ≈ |V0∗|2, (101)

so that

Γ = 2π∑f

|Vif |2 δ(Ef − E0) = 2π|V0∗|2∑f

δ(Ef − E0). (102)

Then

2π∑f

δ(Ef − E0) =1

π

∫ ∞0

δ(εk1 − E0) k21dk1 =

1

π

[dεk0dk0

]−1

k20,

25Or, at least, conventional.26Comparing (102) to the Golden Rule, we see that Γ is equal to the decay rate of the

intermediate state times ~.

28

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where the momentum k0 is defined by

εk0 = E0 ≈ εk,

so thatk0 ≈ k.

We thus arrive at the relation

|V0∗|2 =πΓ

k2

dεk0dk0

=πΓvgr

k2(103)

and the universal (dispersion-independent!) Breit-Wigner formula for theresonant cross-section:

σ =(Γ/2)2

(εk − E0)2 + (Γ/2)2σmax, (104)

where (if ~ is restored, k should be understood as the wave vector ratherthan momentum)

σmax =4π

k2(s-channel). (105)

is the maximal possible cross-section (in the s-channel, see below).Implicitly, the above consideration was actually dealing with the reso-

nance in the l = 0 channel only. For the derivation at l > 0, we should bemore cautious. First, we have to recall that now the intermediate state is(2l + 1)-fold degenerate. All by itself, this does not yet create a problem,since in the incident plane wave, there is only the m = 0 state, so that out of(2l + 1) intermediate states only the m = 0 state is coupled to the incidentplane wave. The problem is with a certain restriction on (101). Given therequirement m = 0 for the intermediate state, Eq. (101) applies only if thefinal states are the eigenstates of the projection of the angular (as opposedto linear) momentum on the axis k, with m = 0. Denoting the set of suchstates as Ωm=0, we have to modify (102)

Γ = 2π∑

f∈Ωm=0

|Vif |2 δ(Ef − E0) = 2π|V0∗|2∑

f∈Ωm=0

δ(Ef − E0). (106)

Then, observing that, by the rotational symmetry, the result should be thesame for any m, and, finally utilizing completeness of the m-states, we get∑f∈Ωm=0

δ(Ef −E0) ≡ 1

(2l + 1)

l∑m=−l

∑f∈Ωm

δ(Ef −E0) ≡ 1

(2l + 1)

∑f

δ(Ef −E0).

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Hence, ∑f∈Ωm=0

δ(Ef − E0) =1

(2l + 1)

∫d3k1

(2π)3δ(εk1 − E0),

and Eq. (103) modifies:

|V0∗|2 = (2l + 1)πΓ

k2

dεk0dk0

= (2l + 1)πΓvgr

k2. (107)

We arrive at (104) with

σmax =4π

k2(2l + 1). (108)

30