Time-Dependent Perturbation Theory

Transcription

1 Time-Dependent Perturbation Theory Time-evolution operator as a product of elementary operators Let U(t 1, t ) be the time-evolution operator evolving the density matrix ˆρ(t ) into ˆρ(t 1 ) [see Eq. (22) in the Density Matrix chapter of Quantum Mechanics I]: ˆρ(t 1 ) = U(t 1, t )ˆρ(t )[U(t 1, t )] (t 1 > t ). (1) By its very definition, the evolution operator satisfies the identity where we assume U(t n, t n 1 )U(t n 1, t n 2 ) U(t 2, t 1 ) U(t 1, t ) U(t n, t ), (2) On the other hand, if the time step t n > t n 1 >... > t 1 > t. (3) t j t j 1 = ɛ (j = 1, 2, 3,..., n) (4) is small enough, then, up to small corrections in ɛ 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 iɛh tj + O(ɛ 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 ˆρ [equation (5) in the Density Matrix chapter of Quantum Mechanics I]. We thus arrive at the representation U(t n, t ) = (1 iɛh tn ) (1 iɛh t3 )(1 iɛh t2 )(1 iɛh t1 ) + O(ɛ). (6) Perturbative expansion for time-evolution and statistical operators 1

2 Let the Hamiltonian H t consist of two parts: H t = H + V t, (7) where H is time-independent and V t is a certain perturbation. Our goal is to expand U(t, t ) in powers of V t. Speaking practically, such an expansion becomes useful when V t is appropriately small allowing one to truncate the perturbative series to one or two first terms. Observing that 1 iɛh tj = (1 iɛh )(1 iɛv tj ) + O(ɛ 2 ), (8) we substitute the r.h.s. of (8) for (1 iɛh tj ) in (6) and then open the brackets with (1 iɛv tj ). This naturally introduces different powers of V tj along with the necessity to sum over the time moments t j. Taking the limit ɛ, we replace summation over t j with integration. Simultaneously, we replace the products of all the terms (1 iɛh ) with exponentials: V tb (1 iɛh ) (1 iɛh )V ta V tb e i(t b t a)h V ta. This allows us to take the limit of ɛ rendering our expansion exact. Finally, we decompose the exponentials to associate them with V t s: where we define e i(tc t b)h V tb e i(t b t a)h V ta e i(ta t d)h = This brings us to the result = e itch V (t b )V (t a )e it dh, V (t) = e i(t t )H V t e i(t t )H. (9) U(t, t ) = e i(t t )H [1 i dt 1 V (t 1 ) dt 1 dt 2 V (t 1 )V (t 2 ) +... t t t t 1 n 1 ] + ( i) n dt 1 dt 2 dt n V (t 1 )V (t 2 ) V (t n ) +..., (1) t t t Pay attention to the limits of integrating over t j s. Those limits reflect the fact that V tj s emerge from the expression (6) in the strict chronologic order. We can readily establish similar expression for the statistical operator e βh by treating β as an (imaginary) time interval. 1 e βh = (1 ɛh ) n + O(ɛ), ɛ = β/n. (11) 2

3 H = H + V. (12) (1 ɛh ) = (1 ɛh )(1 ɛv ) + O(ɛ 2 ). (13) [ β β τ1 e βh = e βh 1 dτ 1 V (τ 1 ) + dτ 1 dτ 2 V (τ 1 )V (τ 2 ) +... β τ1 τn 1 ] + ( 1) n dτ 1 dτ 2 dτ n V (τ 1 )V (τ 2 ) V (τ n ) +..., (14) V (τ) = e τh V e τh. (15) The chronologic order in (14) has the same origin as the chronologic order in (1). Linear Response: Kubo formula By linear response of an observable A to the perturbation V t one means the leading (in powers of V t ) effect of the perturbation on expectation value of A at a given moment of evolution. 1 The linear response is given by the Kubo formula: A = A(t) i dt 1 [A(t), V (t 1 )], t (16) where and A(t) = e i(t t )H A e i(t t )H, (17) (...) = Tr (...)ˆρ(t ). (18) Problem 11. Derive Kubo formula from (1). The similarity of the forms of Eqs. (9) and (17), as well as the deep meaning of this form will become clear in the next section, when we introduce the interaction picture. 1 In condensed matter physics, typical external perturbations used to probe the properties of various systems are periodic electric and magnetic fields, electromagnetic radiation, beams of particles (electrons, neutrons, etc.). 3

4 Time-dependent unitary transformations: Heisenberg and interaction pictures, rotating frames Let S t be some time-dependent unitary operator: S t S t = S t S t = 1. If for each observable A, we perform the unitary transformation Ã(t) = S t AS t A = S t Ã(t)S t, and do the same for the density matrix: ρ(t) = S t ρ(t)s t ρ(t) = S t ρ(t)s t, we will only change the form of the density matrix and the operators of all the observables, but not the physical properties of the system. This is because the measurement postulates of quantum mechanics are invariant with respect to any unitary transformation. Problem 12. Verify the above statement for both parts of the measuring postulate: (i) for the probability part and (ii) for the projection part. By performing this or that unitary transformation, we simply change representation of the very same physics. The change in the form of the density matrix and observables naturally implies a change in the form of the evolution operator. Let us see how the new evolution operator will look like. We have Hence Ũ(t, t ) ρ(t )Ũ (t, t ) = ρ(t) = S t ρ(t)s t = S t U(t, t )ρ(t )U (t, t )S t = S t U(t, t ) S t ρ(t ) S t U (t, t ) S t. Ũ(t, t ) = S t U(t, t ) S t. (19) We also want to relate the new and the old Hamiltonians. To this end we use the generic relation between the Hamiltonian and evolution operator (implied by the evolution equation for the density matrix): i t U(t, t ) = H t U(t, t ), i t Ũ(t, t ) = H t Ũ(t, t ). 4

5 In view of the unitarity of evolution operators, these relation imply [ ] [ ] H t = i t U(t, t ) U (t, t ), Ht = i t Ũ(t, t ) Ũ (t, t ). (2) Applying (2) to (19) brings us to the result We are especially interested in the cases when H t = S t H t S t + i S t t S t. (21) S t = e ih (t t ), (22) where H is a certain time-independent Hamiltonian. Here we have H t = e ih (t t ) H t e ih (t t ) H. (23) If H = H (since H does not depend on time, this implies that H is also time-independent), we are dealing with the Heisenberg picture, where H t = and the density matrix does not evolve. Problem 13. Show that for the second-quantized harmonic Hamiltonian (either bosonic or fermionic) H = s ɛ s â s â s, the creation and annihilation operators in the Heisenberg picture have the form (here we use t = ): â s (t) = â s e iɛst â s(t) = â s e iɛst. (24) Hint. The easiest way to proceed is to derive a certain differential equation for â s (t) and â s(t) by differentiating â s (t) = e iht â s e iht, â s(t) = e iht â s e iht with respect to time and then using the particular form of the Hamiltonian and the properties of creation and annihilation operators. 2 After that, one 2 Different for bosons and fermions! 5

6 simply verifies that (24) satisfy the obtained equations, with appropriate initial conditions. Yet another important class of pictures takes place when Here we have H t = H + V t, [H, H ] =. H t = H H + e ih (t t ) V t e ih (t t ). In particular, if H is time-independent, we can chose H = H to get the so-called interaction picture, in which only the perturbation term survives in the Hamiltonian H t V (t) = e ih (t t ) V t e ih (t t ) (interaction picture). (25) Comparing (25) to (9) and (17), one can guess that the interaction picture is especially convenient for considering perturbative effects. This is indeed the case as illustrated by the following problem. Problem 14. Derive Eq. (1) employing the interaction picture. In the interaction picture, find the evolution operator as a series of integrals by using expansion (6). Then return back to the original picture. In particular, make sure that Kubo formula can be derived in the interaction picture as well. Rotating frame. Suppose our Hilbert space consists of two subspaces, I and II, and the perturbation V has non-zero matrix elements V ab and V ba only if the state a belongs to the subspace I, while the state b belongs to the subspace II. In this case, the following transformation the so-called rotating frame might prove very useful. H = ω ˆP I (rotating frame), (26) where ˆP I is the projector on the subspace I and ω is a certain frequency/energy. In this case (below we set t = ), e ih t = ˆP II + e iωt ˆPI = 1 + ( e iωt 1 ) ˆPI, (27) 6

7 where ˆP II = 1 ˆP I is the projector on the subspace II. In the rotating frame, the matrix elements become: Ṽ ab = e iωt V ab, Ṽ ba = e iωt V ba. (28) Problem 15. Derive Eqs. (27) and (28). We conclude that in the rotating frame, all the energies in the subspace I get shifted by ω, while all the matrix elements from subspace I to subspace II acquire periodic phase factor e iωt, and all the matrix elements from subspace II to subspace I acquire periodic phase factor e iωt. This is extremely convenient for treating periodic perturbations. Periodic time dependence of matrix elements can be readily eliminated at the expense of simply shifting the energies of one of the two subspaces. The Golden Rule Suppose that at the initial time moment t =, the density matrix ˆρ() corresponds to either a pure eigenstate of the Hamiltonian H, or a statistical mixture of some but not all eigenstates of the Hamiltonian H. The actual Hamiltonian of the system is (7), so that ˆρ evolves in time. 3 Assuming that the perturbation V is small, we want to answer the following very important question. At the time moment t >, what is the probability, p Ω (t), to find the system in a certain span Ω of eigenstates of H orthogonal to the subspace of the initial state? As a typical example, think of and excited state of an atom or nucleus as the initial state ˆρ(). Then p Ω (t) is the probability to establish the decay of the excited atom/nucleus into a certain span Ω of the final states, provided corresponding measurement is performed at time t. 4 Correspondingly, Ω 3 Excluding the trivial case of [V t, ˆρ]. 4 Specifying the time of measurement matters, because, speaking generally, we cannot continuously observe quantum system without changing its properties (recall quantum Zeno effect). 7

8 is associated with certain areas in the momentum space of the products of decay. If Ω includes all possible momenta of the products of decay, then p Ω (t) is the probability for the excited state to decay by the time t. In accordance with the measurement axiom, p Ω (t) = Tr ˆP Ω ˆρ(t), ˆPΩ = f f, (29) f Ω where f is an eigenstate of H. The notation f comes from final to distinguish those states from the initial eigenstates of H contributing to ˆρ(): ˆρ() = w i i i. (3) i All the final states are orthogonal to all the initial states, and thus ˆP Ω ˆρ() = ˆρ() ˆP Ω =. (31) In view of Eq. (31), the leading term of the perturbative expansion of ˆρ(t) that yields non-vanishing contribution to p Ω (t) is the second-order term ˆρ (2) (t) = e ith dt V (t ) ˆρ() V (t) = e ith V t e ith. dt V (t ) e ith, Incidentally, note the convenience of the interaction picture where ˆρ (2) (t) = dt V (t ) ˆρ() dt V (t ) (interaction picture). The term ˆρ (2) is the minimal term in which ˆρ() is sandwiched between V s and the expression is thus protected from vanishing upon tracing with ˆP Ω. Note that ˆP Ω commutes with H, thus commuting with the exponentials e ±ith. This, in particular, means that it stays invariant when going to the interaction picture. We thus have where p Ω (t) Tr ˆP Ω ˆρ (2) (t) = dt dt V (t ) ˆP Ω V (t ), (32) (...) = Tr (...)ˆρ(). (33) 8

9 General expression (32) simplifies significantly when V t is either timeindependent, or a harmonic function of time (normally referred to as periodic perturbation). And the latter case actually reduces to the former one in the appropriate rotating-frame picture, where the frequency of the frame matches the frequency of the perturbation. Time-independent perturbation. Consider the case of time-independent perturbation. The first observation here is that V (t ) ˆP Ω V (t ) = V () ˆP Ω V (t t ), resulting in (we shift the integration variable t t t ) p Ω (t) = dt t t dt V () ˆP Ω V (t ). The next observation is that for p Ω (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 () ˆP Ω V (t ). In essence, this is the criterion of applicability of the perturbative treatment for p Ω (t). Vanishing V () ˆP Ω V (t ) at certain appropriately large t allows one at t t to extend the limits of integration over t to plus/minus infinity. This brings us to the result p Ω (t) = t W Ω, W Ω = dt V () ˆP Ω V (t ) (t t ), (34) known as Fermi s Golden Rule. According to the Golden Rule, the probability to find the system in one of the final states grows linearly with time. 5 The linearity of increasing p Ω with time is a crucial result on its own, allowing one to speak of the decay of the initial state and transitions to the final states, thus interpreting W Ω as the transition rate. In particular, extending Ω to all possible finite states, one gets the total transition/decay rate. With a natural requirement that V does not have non-zero matrix elements between any two initial states, 6 the total transition/decay rate acquires a very compact form W tot = 5 Within a wide range of times where t t while p Ω (t) 1. 6 Otherwise, those elements can be simply added to H. dt V () V (t ). (35) 9

10 An alternative and most frequently used form of the Golden Rule deals with the situation when ˆρ() is a pure state (called the initial state): In this case we have ˆρ() = i i. V () ˆP Ω V (t ) = i V () ˆP Ω V (t ) i = Taking into account f Ω i V () f f V (t ) i. H i = E i i, H f = E f f, the time-dependence can be explicitly factored: f V (t) i = e i(e f E i )t V fi, V fi f V i, (36) and the integration over time performed: dt e i(e f E i )t = 2πδ(E f E i ). This yields the following form of the Golden Rule (here we restore ) W Ω = 2π f Ω In particular, for the total rate we have W tot = 2π V fi 2 δ(e f E i ). (37) V fi 2 δ(e f E i ). (38) f The presence of the delta-function implies that the summation over the final states 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 ˆρ() is not a pure state. Using (3), we readily reduce the problem to the previous one, with the final answer W Ω = 2π i f Ω w i V fi 2 δ(e f E i ). (39) 1

11 It is very instructive to reveal the details of how the constant rate W Ω builds up in time. In particular, we want to clarify the conditions for the rate of the transitions into the subspace Ω to become constant. To this end, we use the above decomposition into matrix elements to perform explicit integration over t and t directly in (32), without making no further approximations. V (t ) ˆP Ω V (t ) = w i i V (t ) f f V (t ) i = i = i f Ω f Ω w i V fi 2 e i(e f E i )t e i(e i E f )t. Integrating over times, we find p Ω (t) = 2 w i V fi 2 f t (x), i x = E f E i ( = 1), (4) f Ω where 1 cos xt f t (x) = = 2 sin2 (xt/2). (41) x 2 x 2 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 (4) (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 allow one to make the replacement f t (x) = π t δ(x). (42) The replacement (42) is accurate if and only if the span of variation of E f E i is much larger than t. For example, if the initial state is pure and the subspace Ω deals with a rather narrow band of energies, E f E i ɛ, then the constant-rate regime for the subspace Ω will be achieved only at t ɛ 1. Furthermore, if the energetic width of the space Ω is much smaller than inverse t, then p Ω (t) is dramatically non-monotonic, in a profound contrast with the naive picture of transitions per unit time that is often used to interpret the Golden Rule. On the other hand, if the initial state ˆρ() is a mixed state, with the width of the energy distribution substantially exceeding the inverse decay time, then, the constant-rate approximation is immediately applicable, independently of the structure of the subspace Ω. 11

12 Based on the result (4) (41), we can quantify the energy uncertainty of the final state, E. We see that E is inversely proportional to the time of the perturbative evolution of the initial state (we restore here): E /t. Periodic perturbation The theory of the previous section is readily generalized to a harmonic time dependent perturbation of a special form obtained from time-independent one by the replacement (the matrix notation is especially convenient here) V if e iωt V if, V fi e iωt V fi = e iωt V if. (43) In this case, Eq. (36) gets modified to f V (t) i = e i(e f E i ω)t V fi, (44) bringing us to a simple but very important conclusion that the effect of the frequency ω reduces to the replacement E f E i E f E i ω ( = 1) (45) in all the final answers for p Ω. An alternative route towards the same conclusion as well as an instructive way of interpreting the latter is to utilize the rotating-frame picture (see Problem 16). In the case of two or more frequencies most often, the same frequency with two different signs, which corresponds to sinusoidal perturbation we can neglect the interference effects provided the frequencies are not too small. The simple reason for that is that the final states corresponding to different frequencies are also different, by conservation of energy. Problem 16. Use the rotating frame to generalize the Golden Rule to the perturbation of the form V t = Ae iωt + A e iωt, (46) 12

13 where A is a time-independent operator featuring the constraint Then show that a generic perturbation of the form A if =, f. (47) V t = Be iωt + B e iωt, (48) with B a generic time-independent operator, reduces to two periodic perturbations of the form (48) (47), with the opposite frequencies ±ω. Express matrix elements of those two perturbations in terms of B if and B fi. Creation of electron-hole pairs in a semiconductor by optical absorption The interaction of electronic subsystem of a semiconductor with light is perturbative. The effective Hamiltonian is as follows. H = H + V, H = E p (e) a pa p + E p (h) h ph p + cp f p,λ f p,λ, p p p,λ V = ) (ga k h p k f p,λ + g f p,λh p ka k. p,k,λ Here a p and h p are the annihilation operators of the electron and hole, respectively, for the momentum mode p; f p,λ is the annihilation operator of the photon in the mode with momentum p and polarization λ. The origin and momentum dependence of the coupling strength g are not important for our purposes. For simplicity, we will be treating g as momentum-independent constant. The other parameters of the Hamiltonian are the electron energy, the hole energy E p (h), and the velocity of light c. The system s volume is set equal to unity. 7 E (e) p Problem 17. Use the Golden Rule to calculate the pair-creation rate which, clearly, will simultaneously be the optical-absorption rate for the monochromatic radiation of the frequency ω. To this end, take the initial 7 This is a very convenient trick. Always use it! 13

14 condition in which there are no electrons and holes, and only one photon mode, k (ck = ω ) having a non-zero occupation number n. Use the following approximations (justified by the conditions in parentheses) E (e) p = 2 + p2 2m e E (h) p = 2 + p2 2m h ( p 2 2m e ( p 2 2m h ), ), where is the so-called insulating gap, and m e and m h are the effective masses of the electron and hole. Also, one can safely set the momentum (but not the frequency!) of the photon equal to zero. This is justified by the small parameters (of non-relativistic motion) p m e c, p m h c. The optical field can be treated classically. Strictly speaking this is justified only when the occupation numbers of relevant photon modes are much larger than unity. However, for the problem of optical absorption, the final answer is always essentially the same as for the classical electromagnetic field. Use the classical-field approximation for the electromagnetic field, 8 one first employs the Heisenberg picture for electromagnetic field and then replaces creation and annihilation operators with complex numbers. In accordance with the results (24), the effective Hamiltonian then reads (note that here f s are numbers, not operators). H = p E p (e) a pa p + p E (h) p h ph p, V = p,k,λ (gf p,λ e icpt a k h p k + g f p,λe icpt h p k a k ). Problem 18. Use the Golden Rule for the periodic perturbation (see Problem 16, or/and the recommended text) to find the absorption/pair-production 8 Or any other bosonic field for that matter. 14

15 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, a molecule, a nucleus, etc. undergoes a transition to a state with lower energy an emits a photon. The emission is called spontaneous if all the modes into which the photon is emitted are initially free of photons (have zero occupation numbers). As we will see, the rate of the emission (the decay rate of the excited 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 and the rate of spontaneous emission equal to the spontaneous-emission rate times n (s) ph is called stimulated emission. The (effective) Hamiltonian describing the emission (and not only the emission!) has the following form H = p H = H + V, p 2 2m a pa p + ) (E + p2 e 2m pe p + cp f p,λ f p,λ p p,λ V = ) (gf p,λa k p e k + g e k a k pf p,λ. p,k,λ Here a p and e p are the annihilation operators of the lower- and higher-energy state atom 9 for definiteness, we call our object atom in the momentum mode p, and f p,λ is the annihilation operator of the photon in the mode with momentum p and polarization λ. The origin and momentum dependence of the 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 spontaneous and stimulated emission of a single atom in the upper energy state. 9 Without loss of generality, we recon the energy of the atom from its lower-energy state with momentum zero. 15

16 For simplicity, assume that all the occupation numbers of the photons in the relevant modes are the same, being equal to n f. Note that, in view of the small non-relativistic parameter p/m c one can safely neglect the dispersion of the atom setting m =. Exponential decay Now we want to see how the perturbative treatment of the decay of an initial state i valid only at t Wtot 1 generalizes to times t Wtot 1. This can be controllably done with the model where there is no interaction between any two final states: V ff. (49) The model (49) yields a very good approximation to a realistic case, provided perturbation V is small enough. We confine ourselves with the case of pure initial state. The generalization to the case of a mixed state is trivial in view of the statistical-mixture interpretation of the density matrix. Let us solve our Schrödinger equation i t ψ(t) = (H + V ) ψ(t), (5) expanding the solution in the Fourier series of the eigenvectors of H : ψ(t) = C (t)e ite i i + f C f (t)e ite f 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 equations for the coefficients C i and C f (that are nothing but the amplitudes of the expansion of the wave function in the interaction representation). The righthand sides of those equations do not contain diagonal terms: Ċ = i f V if e iω fit C f, (52) where Ċ f = iv fi e iω fit C, (53) ω fi = E f E i. (54) 16

17 Equation (53) can be cast into the integral form: C f = iv fi dt e iω fit C (t ), (55) where the lower integration limit is chosen to satisfy the initial condition C f () =. Substituting (55) into (52) and making a substitution t = t t for the integration variable, we get an integro-differential equation for C (t): with the kernel Ċ = Q( t) C (t t) d t, (56) Q(t) = f V if 2 e iω fit V (t)v (). (57) Putting aside the specificity of the model expressed by Eq. (49), so far, we have not made any approximation. In particular, the structure V (t)v () that we saw before does not imply perturbative treatment. Now it is time to take essentially the same large-time limit that we discussed before in the context of perturbation theory. At t t [as before, by t we denote the characteristic time of vanishing Q(t) = V (t)v () ], we can extend the upper limit of integration in (56) to infinity. 1 As a result we get Ċ = Q( t) C (t t) d t (t t ). (58) Equation (58) is solved by the exponential substitution 11 C (t) = e iλt (59) reducing the integro-differential equation to the following equation for the (complex) Λ: Λ = iq Λ, (6) 1 Note that this is essentially the same limit that was bringing us to the constant-rate regime of perturbation theory. Also, speaking generally, there is a subtlety of taking this limit. We will address this subtlety later. 11 Because of the linearity of Eq. (58), the substitution comes with an arbitrary constant pre-factor. We set the latter equal to unity in view of the initial condition C () = 1. 17

18 where Q Λ = Q(t) e iλt dt. (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 E i. The quantity γ > describes the exponential decay of the magnitude of C. It is tempting to plug the sum of exponentials (57) and integrate over the time. However, given γ >, we cannot accomplish this seemingly straightforward procedure because of the divergence of the integrals at their upper limit. 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 original problem is not ill-defined without the regularization the answer can be expressed in the form of the (rather non-trivial) limit Q Λ Q iγ = i lim g [ f ] V if 2 E i E f + + i(g γ) a.c. from g>γ, (64) implying that the summation (integration) over f is performed strictly at g > γ, then the expression is analytically continued 12 to g γ, and only then the limit of g is taken. To better understand the mathematical procedure behind Eq. (64), introduce the spectral density η(ε) defined as η(ε) = f V if 2 δ(ε E f ). (65) According to the first equality in (57), the mathematical meaning of η(ε) is very transparent, the functions Q(t) and η(ε) are related to each other by the following Fourier transformation: Q(t) = e ie it dε η(ε) e iεt = dε η(ε+e i ) e iεt. (66) 12 We use the symbol [...] a.c. from g>γ to express the requirement of analytic continuation. 18

19 In terms of η, equation (64) reads [ ] η(ε) dε Q Λ Q iγ = i lim g E i + ε + i(g γ) a.c. from g>γ. (67) In this formula, it is absolutely crucial that the order of the limit and the integral are not switched. 13 Moreover, it is equally crucial that the integration is 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 condition of (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 majority of practically important cases, the procedure of replacing the upper limit of integration in (56) with infinity is ill-defined 14 unless we introduce a small but finite g. Let g be a positive quantity satisfying the following two inequalities: γ < g t 1. The two conditions guarantee that, on one hand, the integrals over time converge, while, on the other hand, the systematic error introduced by the regularization is negligibly small. Then, in view of the smallness of γ and g compared to t 1, the sum over f in Eq. (64) simplifies (the symbol p.v. stands for the principal-value integration): V if 2 E i E f + + i(g γ) f p.v. f V if 2 E i + E f iπ f and we arrive at the results ( is restored) = p.v. f V if 2 δ(e f E i ), V if 2 E i + E f, (69) 13 A nice piece from the internet:...and we d switch the order of the two operations without thinking about it. The professor would always say, I can interchange these two because I am a physicist and I am lazy. A mathematician would spend his whole life trying to prove this is permissible. 14 In the next section, we will clearly see why. 19

20 γ = π V if 2 δ(e f Ẽi), (7) f Ẽ i = E i +. (71) Equations (69) and (7) can be also written in terms of the spectral density: = p.v. η(ε) dε E i + ε, (72) γ = π η(ẽi). (73) As we see from (7) or (73), and will also see from other formulas, the only role played by is shifting the energy E i. That is why we introduced the shifted energy Ẽi. Note that, in the V limit, the value of scales as the first power of V, while γ is quadratic in V. Hence, it is meaningful to keep in the right-hand sides of (69) and (7), while neglecting γ. Furthermore, in many effective theories, the sum over f in (69) is ultraviolet divergent implying an ultraviolet cutoff. 15 In such a theory, neither E i nor are physically meaningful, since both are cutoff-dependent. However, the sum E i + makes perfect sense of the proper energy of the state i. Up to replacement E i Ẽi, the result (7) is nicely consistent with the Golden Rule. 16 Indeed, comparing (7) with (38), we see that 2γ = W tot. Hence, our result C 2 = e 2γt (t t ) (74) corresponds to that of the Golden Rule ( C 2 = 1 W tot t) within the range of times t t Wtot 1 where the Golden Rule applies. The condition (68) guarantees that at t t, we have C (t) 1, allowing us to safely substitute (59) into (55) and find C f (t) = V fi 1 e γt e i(e f Ẽi)t E f Ẽi + iγ, (75) C f (t) 2 = V fi e 2γt 2e γt cos(e f Ẽi)t (E f Ẽi) 2 + γ 2. (76) 15 An effective theory is not supposed to be accurate up to arbitrarily high energies. 16 Note also that Eq. (73) yields a compact representation of the Golden Rule in terms of the spectral density. 2

21 At t γ 1, the solution (76) corresponds to the perturbative result (4) (41). At t γ 1, the solution is dramatically different from (4) (41), saturating to the following time-independent limit: C f (t) 2 V fi 2 (E f Ẽi) 2 + γ 2 at t. (77) In this limit, it is the parameter γ not the inverse time that characterizes the energy uncertainty. The stochastic-decay interpretation and its limitations The exponential decay law (74) suggests the stochastic-decay interpretation that we already discussed in the context of the Golden Rule. Here one interprets W tot 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 having no decay events within the time interval [, t]. Problem 2. Derive the law (78) for the above-introduced process of stochastic decay. No matter how accurate, the decay-rate interpretation is fundamentally approximate, since, microscopically, quantum mechanics is stochastic only at the level of measurement; the evolution of the density matrix is absolutely deterministic. Therefore, it is very instructive to discuss the limitations of the stochastic-decay picture. One obvious limitation is that, strictly speaking, the law (74) does not apply at t t. This, however, is not a dramatic limitation because at t t we have C 1, so that the discrepancy occurs only at the level of small corrections. A really serious discrepancy with the stochastic-decay picture deals with the distribution of the final states 17 In the probability theory, such a process is known as a Poisson process a particular example of continuous-time Markov process. 21

22 over energies. The quantum mechanical analysis shows see Eqs. (4), (76) and (77) that the distribution gets narrower with time. 18 This effect is in a profound discrepancy with the stochastic-decay picture, where the properties of the decay processes cannot depend on time. Yet another limitation concerns very large times. The exponential behavior (59) was obtained by setting t = in (56). Let us see whether/when the latter procedure is consistent with (59). We have Q( t) C (t t) d t = Q( t) e iλ(t t) d t = e iλt Q( t) e i t e γ t d t, and realize that setting t = implies that Q(t) decays with time exponentially. Normally, that is not the case! The reason is that the spectrum of any realistic quantum system is bounded from below. Let E be the exact lower bound for E f. Then E is also the exact lower bound for the support of the spectral-density function η(ε) [see Eq. (65)], and we have Q(t) = e ie it E dε η(ε) e iεt = e i(e i E )t η(ε) = η(e + ε). dε η(ε) e iεt, Such an integral typically has a power-law rather than exponential decay: Q(t) ei(e i E )t t α (t ), (79) where the exponent α can be related (by dimensional analysis) to the exponent ν describing the asymptotic behavior η(ε) ε ν (ε ). Namely, α = ν + 1. (8) 18 And this is an excellent example of the energy-time uncertainty relation: E t. 22

23 Problem 21. A more controlled 19 and detailed (i.e., with the numeric prefactor) version of Eq. (79) can be derived by considering the integral ε ν e iεt e gε dε, with a certain real positive g. Perform corresponding derivation. In particular, make sure that the result the leading term does not depend on the choice of g in the limit of large t. Now we see that, depending on value of t t, there are two very different regimes. The first regime is when the leading contribution to the integral over t in (56) comes from t t. Here we can simply use the regularized kernel, Eq. (63), which does not change the value of the integral (up to small corrections in parameter γt ), while allowing us to set the limit of integration to infinity The second regime sets in at t t fail, such that the contribution to the integral over t in (56) associated t t becomes comparable to the contribution coming from t t. At t t fail. Now the whole physical picture changes, and, speaking generally, the decay becomes non-exponential. Let us estimate t fail. By dimensional analysis, based on (79), we find that the contributions from t t and t t become comparable at t t fail such that e γt fail (t /t fail ) α 1. When deriving this estimate, we also assumed based on the dimensional analysis that E t 1. A simple algebra brings us then to and a very instructive relation γt fail α ln(γt ), C (t fail ) 2 e 2γt fail (γt ) 2α = (γt ) 2(ν+1) 1, telling us that by the time the exponential decay law will become inapplicable, the probability to find the system in the initial state will be much 19 One can question the validity of the above-mentioned dimensional analysis because of the upper-limit divergence of the integral ε ν e iεt dε. 23

24 smaller than unity. Problem 22. As a very instructive example of how, in the absence of spectral-edge singularity, equation (67) works exactly as written and the theory is asymptotically accurate without the breakdown of the exponential law at very large times, consider a model where the spectral density has the Lorentzian form A η(ε) = (ε ε ) 2 + κ. 2 (a) Derive the system of two algebraic equations relating and γ to the parameters A, ε, 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, ε, and κ for the decay to be exponential? Resonant scattering Consider the same model as before see the Exponential Decay section but with a fundamentally different initial condition. Now the state i is not the initial state. Rather, it is an important intermediate state, while the initial state is one of the states f, call it state f. We will be solving the same equation (5), using essentially the same expansion of the solution 2 ψ(t) = C (t)e ite i + C (t)e ite f + f f C f (t)e ite f f, (81) but now C () = and C () = 1. For the rest of the coefficients we have C f () = and use the same trick of integrating corresponding equations: C f = iv f dt e iω ft C (t ). (82) Here we optimized the notation so that V f and ω f stand for what used to be V fi and ω fi, respectively. Below we will be also using V for V if and ω = E E. (83) 2 To avoid confusion and enhance readability, we replace E i with E and E f with E. 24

25 We thus have a system of two linear equations Ċ = Q( t) C (t t) d t iv e iωt C, (84) Ċ = iv e iωt C. (85) Since our system is linear, a general solution comes in the form of linear superposition of two particular solutions. One of the two is already known to us. It corresponds to the exponential decay of the state i. We thus need to find the second one. And it is this second solution that is most relevant to our case, since, in the limit of (macroscopically) small V, the first solution decays in time much faster than C (t), and by the time when the first solution vanishes completely, we still have C (t) 1. As before and for the same reason we set the upper limit of integration in (84) equal to infinity: Ċ = Q( t) C (t t) d t iv e iωt C, (86) after which the system (85) (86) is solved by the simple exponential substitution C (t) = Ae iλt, C (t) = Be i(λ+ω)t, leading to the following algebraic system of equations [ ] [ ] λ V A =, (87) V ω + λ + iq ω+λ B where Q ω+λ is defined by the formula (61) with Λ ω+λ. The characteristic equation of the system (87), has two solutions. 21 The first one, λ(ω + λ + iq ω+λ ) V 2 = λ 1 + ω = iq ω+λ1 (first solution), 21 Taking into account macroscopic smallness of the term V 2, we keep only the leading terms. 25

26 is known to us. It is identical to (6), with λ 1 + ω Λ As expected, the second solution yields macroscopically small λ: λ 2 = V 2 ω + iq ω+λ2 (second solution). (88) In view of the microscopical smallness of λ 2, we can 22 replace Q ω+λ2 with Q ω for which, in accordance with (64), we have [we use (83) to restore E ] Q ω = i lim g + f V if 2 E E f + ig. (89) Since now the regulator g is infinitesimally small, we can take the limit g + without any extra approximation. As a result, in the convenient parameterization Q ω = i + γ, we get 23 and the second solution becomes λ 2 = = p.v. f V if 2 E E f, (9) γ = π V if 2 δ(e f E ), (91) f V 2 E E + iγ ( = 1). (92) In view of macroscopic smallness of λ 2, really interesting is only its negative imaginary 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 W t, (93) W = 2Im λ 2 = 2γ V 2 (E E ) 2 + ( γ ) 2. (94) 22 And actually should, since we keep only the leading terms. 23 Here 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

27 This is the celebrated generic result for the rate of a resonant process (typically, resonant scattering). Problem 23. Derive the general relation: W = v grσ V, (95) between the (total) scattering rate in the system of the volume V, the (total) scattering cross-section σ and the group velocity v gr of the particle being scattered. Use the unit-volume version of relation (95): W = v gr σ (unit-volume normalization) (96) to convert Eq. (94) into the result for the (total) resonant-scattering crosssection in the following two characteristic cases: (a) Scattering of resonant light from an atom/molecule/micro-system (the so-called resonant fluorescence), (b) Resonant scattering of the α-particle from a nucleus (recall α-decay of an α-radioactive nucleus). In both cases, provide an explicit meaning of E, E,, and γ. Breit-Wigner formula Consider a single- or two-particle 24 scattering under the resonant conditions, when the result (94) applies. In this case, E = ɛ k, v gr = dɛ k dk. We will be assuming that the lifetime of the intermediate state is very large compared to the characteristic energy: γ ɛ k. (97) 24 Since a two-body problem reduces to a single-body one, our discussion will be about a single particle. 27

28 Under this condition, the resonance is very narrow and it makes sense to adopt yet another simplifying condition (of closeness to the resonance): In this case, we can neglect the k-dependence of γ : E E E. (98) γ γ = π f V if 2 δ(e f E ) ( = 1). (99) With the same degree of accuracy, is k-independent as well, and we can simply absorb it into E. Also, it is a bit more convenient 25 to work with the quantity ( is restored) Γ = 2 γ = 2π f V if 2 δ(e f E ), (1) called the decay width of the intermediate state. 26 For the resonant-scattering cross-section we thus have [see (96)]: σ = [ ] 1 dɛk Γ V 2 ( = 1). dk (ɛ k E ) 2 + (Γ/2) 2 And this is not the final result yet. In fact, we can write the answer in terms of Γ only! To this end we observe that, in our case of narrow resonance, so that V if 2 V 2, (11) Γ = 2π f V if 2 δ(e f E ) = 2π V 2 f δ(e f E ). (12) Then 2π f δ(e f E ) = 1 π δ(ɛ k1 E ) k 2 1dk 1 = 1 π [ dɛk dk ] 1 k 2, 25 Or, at least, conventional. 26 Comparing (12) to the Golden Rule, we see that Γ is equal to the decay rate of the intermediate state times. 28

29 where the momentum k is defined by so that We thus arrive at the relation ɛ k = E ɛ k, k k. V 2 = πγ k 2 dɛ k dk and the universal (dispersion-independent!) resonant cross-section: σ = = πγv gr k 2 (13) Breit-Wigner formula for the (Γ/2) 2 (ɛ k E ) 2 + (Γ/2) 2 σ max, (14) where (if is restored, k should be understood as the wave vector rather than momentum) σ max = 4π (s-channel). (15) k 2 is the maximal possible cross-section (in the s-channel, see below). Implicitly, the above consideration was actually dealing with the resonance in the l = channel only. For the derivation at l >, we should be more 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 = state, so that out of (2l + 1) intermediate states only the m = state is coupled to the incident plane wave. The problem is with a certain restriction on (11). Given the requirement m = for the intermediate state, Eq. (11) applies only if the final states are the eigenstates of the projection of the angular (as opposed to linear) momentum on the axis k, with m =. Denoting the set of such states as Ω m=, we have to modify (12) Γ = 2π V if 2 δ(e f E ) = 2π V 2 δ(e f E ). (16) f Ω m= f Ω m= Then, observing that, by the rotational symmetry, the result should be the same for any m, and, finally utilizing completeness of the m-states, we get f Ω m= δ(e f E ) 1 (2l + 1) l m= l f Ω m δ(e f E ) 29 1 (2l + 1) δ(e f E ). f

30 Hence, δ(e f E ) = f Ω m= and Eq. (13) modifies: 1 d 3 k 1 (2l + 1) (2π) δ(ɛ 3 k 1 E ), V 2 = (2l + 1) πγ k 2 dɛ k dk = (2l + 1) πγv gr k 2. (17) We arrive at (14) with σ max = 4π (2l + 1). (18) k2 3