44. Natural Line Width, Long-Time Behavior and the Lamb Shift

Introduction¶

Let us return to the Kramers-Heisenberg formula. For simplicity we consider the case of elastic scattering, $A+\gamma \to A+\gamma'$, in which the state $A$ is the ground state. The cross is

where

which are (46) and (47). The first term under the sum, which corresponds to the Feynman diagram in Figure 1, is resonant when $\omega=\omega_{IA}$ for some intermediate state $I$, that is, the denominator goes to zero at this frequency and the cross nominally becomes infinite. Note that there is a sum over intermediate states $I\ne A$, so states $I$ must be excited states, and when $\omega=\omega_{IA}$ it means that the incident photon has the right energy to lift the system into the excited state $I$. Only one state $I$ (or a collection of degenerate states) can be resonant for a given incident photon frequency $\omega$; when $\omega$ is near $\omega_{IA}$ for some intermediate state $I$, then that one term in the sum in the Kramers-Heisenberg formula dominates and we can approximate the amplitude by neglecting the others.

The resonant divergence of the scattering cross is nonphysical, so a number of questions are raised. First, what is wrong with our derivation of the Kramers-Heisenberg formula, that we obtain non-physical divergences, and how do we correct them? Next, how close to $\omega=\omega_{IA}$ can we come before the Kramers-Heisenberg formula (47) breaks down? And finally, since we know the cross cannot really be infinite at resonance $\omega=\omega_{IA}$, how high does it go? Recall that for non-resonant frequencies, that is, $\omega$ that are not close to $\omega_{IA}$ for any intermediate state $I$, the Kramers-Heisenberg cross is of order $r_e^2 = \alpha^4 a_0^2$. That is, it is of order $\alpha^4\sim 10^{-9}$ in atomic units, which is very small by atomic standards. It is of interest to see by what factor this small cross is enhanced at resonance.

To see some of the qualitative features of the divergences in the Kramers-Heisenberg formula, consider the Feynman diagram of Figure 1. A question that arises is whether energy is conserved at the vertices of the diagram, for example, at the lower vertex (circled), where the energy of the initial state is $E_A+\hbar\omega$ and that of the intermediate state is $E_I$. The answer depends on the photon energy $\hbar\omega$ but for most values of this parameter energy is not conserved. There is no reason why it should be, since the intermediate states originate in a resolution of the identity, which is a sum over all states, regardless of their energy. Nevertheless, the Feynman diagram suggests that somehow the initial state is transformed into the intermediate state at a certain time, which later transforms into the final state. In quantum mechanics energy is only precisely defined in processes that take an infinite amount of time, while for a process that takes time $\Delta t$ the energy is uncertain by an amount $\Delta E \sim \hbar/\Delta t$. In fact, intermediate states in Feynman diagrams only survive for a limited amount of time, or at least this is an interpretation that can be given to the machinery of the Dyson series developed in 34. Time-Dependent Perturbation Theory. Such a state is called virtual. For the lower vertex of the Feynman diagram in question, note that $\Delta E= E_I-E_A-\hbar\omega$, so that $\Delta E$ is proportional to the resonant denominator we see in the Kramers-Heisenberg formula. The whole amplitude is proportional to $1/\Delta E$, that is, it is proportional to the amount of time that the intermediate state is allowed to survive.

If $\omega$ is not near any resonant frequency $\omega_{IA}$ for any intermediate state $I$, then $\Delta E$ is of order unity in atomic units, so any given intermediate state $I$ survives only for a time of order unity before disappearing again. This means that the contribution of the state $I$ to the transition amplitude, which is computed in the Dyson series as an integral over time, can only accumulate to a small value before the virtual state disappears. The time integral in question is the one over $t''$ in (38); the integrand has the phase factor $e^{i\omega_{ki}t''}$, which in the context of the Kramers-Heisenberg formula is

If $\omega$ is not resonant, then the contribution to the transition amplitude just oscillates with a frequency $\Delta E/\hbar$ and remains small.

If however we let $\omega$ approach a resonant frequency $\omega_{IA}$ for some intermediate state $I$, then $\Delta E$ gets smaller, the virtual state lives for a longer time, and the transition amplitude accumulates. This is reflected by the fact that the denominator gets smaller. As $\Delta E\to 0$, it would appear that the virtual state could live for an arbitrarily long time, which would mean that it is becoming a real state.

But the intermediate state $I$ in question must be an excited state, since we are assuming that $A$ is the ground state, so it cannot live forever, rather it will eventually undergo the spontaneous emission of a photon and drop into a lower state. This will become a significant process when $\Delta E$ is comparable to $\Gamma=\hbar/\tau$, where $\tau$ is the lifetime of the excited state $I$. $\Gamma$ is the usual symbol used for the decay rate of an excited state, measured in energy units. Thus we can expect that the predictions of the Dyson series and the Kramers-Heisenberg formula, as derived in 43. Scattering of Radiation by Matter, will break down when

As explained in 42. Emission and Absorption of Radiation, for E1 transitions $\Gamma$ is of order $\alpha^3 \sim 10^{-6}$ in atomic units, so the fraction of the $\omega$ axis over which the breakdown occurs is very small. For M1 and higher order transitions $\Gamma$ is even smaller. In other applications (nuclear physics, particle physics) $\Gamma$ may be relatively much larger (that is, broad resonances exist).

Time-dependent perturbation theory, as we have presented it in 34. Time-Dependent Perturbation Theory, is based on the method of successive approximations, in which a lower order approximation is substituted back into the Schr"odinger equation in the interaction picture and integrated to get the next approximation. It does not take into account the depletion of probability from the initial state due to the perturbation, so its validity is limited to short times, that is, $t\ll \hbar/\Gamma$. To cure the divergences in the Kramers-Heisenberg formula we need a theory capable of handling the long-time behavior of the system, that is, one that is valid for times $t>\approx \hbar/\Gamma$.

Decay of an Excited State; the Decay Rate $\Gamma$¶

To analyze the long-time behavior of quantum systems, we will take a model problem, that of the decay of an atom in an excited state $B$. We outline the problem in this , which contains a slight rephrasing of some of the material given in Spontaneous Emission. For the rest of this analysis we set $\hbar=1$ except that we restore $\hbar$’s selectively when it makes final formulas look more familiar.

We assume the atom is initially in an excited state $B$ with no photons in the field, and that it subsequently makes a transition to a lower state $A$ with the emission of a photon. There may be more than one lower energy state to which a transition can be made, $A'$, $A''$, etc., as illustrated in Figure 2. As in 42. Emission and Absorption of Radiation{} we let the initial state be $\ket{B0}=\ket{B}\ket{0}$, where $\ket{0}$ is the vacuum state of the field, and we let the final state be $\ket{A\lambda}=\ket{A}a_\lambda^\dagger\ket{0}$, that is, the state in which the atom is in state $\ket{A}$ and there is one photon in the field of mode $\lambda=(\kvec,\mu)$. We take the Hamiltonian to be

where $K_0$, $K_1$ and $K_2$ are given by (5) and where we set $K'=K_1+K_2$ as it is convenient to have a symbol for the entire perturbation.

According to first-order time-dependent perturbation theory, the transition amplitude in the interaction picture to find the system in state $\ket{A\lambda}$ at time $t$ is

where $\omega_{ni}$ is interpreted as

See (40) and (12). We note that $K_1$ can only change the number of photons by $\pm1$ while $K_2$ can change them only by 0 or $\pm2$, so $K'$ in (6) can be replaced by $K_1$.

Squaring the transition amplitude to get a probability we have

where $\Delta_t$ is the usual symbol we have used for an approximate $\delta$-function, as defined by (51). Dividing this by $t$ and summing over final states gives us the transition rate, according to first-order time-dependent perturbation theory. If we want the total rate of decay, we must sum not only over all photon states $\lambda$ but all atomic states $A$. This gives

where we have taken $t$ long enough to replace the function $\Delta_t$ by a genuine $\delta$-function. There is no harm in summing over all states $A$, including those higher in energy than $B$, since for those the $\delta$-function will give zero.

A $\delta$-function is only meaningful under an integral, so when interpreting formulas such as (9) it must be understood that the sum on photon modes $\lambda$ includes a sum on wavevectors $\kvec$ of the photon, which in the infinite volume limit can be written as an integral over the frequency $\omega$ of the photon,

The volume factor $V$ will cancel when we compute the matrix element.

As noted in 42. Emission and Absorption of Radiation, this calculation of the decay rate really gives the slope of the decay curve at $t=0$. See Fig.~\figr\radnmatt.1, in which the decay rate is denoted $A_E$ (the Einstein $A$-coefficient) rather than $\Gamma$. (The notation $A_E$ is appropriate for the decay rate to a specific final state, while $\Gamma$ is the decay rate to all possible final states.) First-order time-dependent perturbation theory is giving the probability of remaining in the initial state $\ket{B0}$ as

which is correct for $t\ll 1/\Gamma$ but absurd for longer times. For longer times we might expect an exponential decay law $P_{B0}(t) = e^{-\Gamma t}$ but that is not proven by first-order time-dependent perturbation theory. We do have the first term in the Taylor expansion of the exponential,

but we haven’t proven that the higher order terms are the right ones.

To access the long-time behavior of the system it won’t help to go to higher order in time-dependent perturbation theory. These higher order terms will generate powers of $t$, that is, a Taylor series expansion of the probability in time about $t=0$. This is a poor way to understand the long-time behavior.

An Approach Based on Laplace Transforms and Green’s Functions¶

Let us return to the formalism of time-dependent perturbation theory in the general notation used in 34. Time-Dependent Perturbation Theory. We write the Hamiltonian as $H=H_0+H_1$, and we let the initial state be

where $\ket{i}$ is an eigenstate of $H_0$,

The state of the system at a later time is

where we have expanded $\ket{\psi(t)}$ as a linear combination of eigenstates of $H_0$, $H_0\ket{n} = E_n\ket{n}$, for simplicity assumed to be discrete. The expansion coefficients $a_n(t)$ are the transition amplitudes in the Schr"odinger picture,

In this analysis we will work in the Schr"odinger picture, but for reference the transition amplitudes in the interaction picture, which we denote by $c_n(t)$ as in 34. Time-Dependent Perturbation Theory, are related to those of the Schr"odinger picture by

See (30). The two amplitudes differ by a phase factor, so the transition probability, which is the square of the amplitudes, is the same in both pictures. Differentiating (16) with respect to time we obtain an evolution equation for the transition amplitudes in the Schr"odinger picture,

or,

This is really just a version of the Schr"odinger equation and it is exact.

We will solve (19) by means of Laplace transforms. A Laplace transform is essentially a one-sided Fourier transform, that is, the range of integration only goes from 0 to $\infty$ instead of $-\infty$ to $+\infty$. (In comparison to the usual definition of the Laplace transform there is also the question of a deformation of the contour of integration in the complex plane, which we gloss over, the main point being that we will be using a one-sided Fourier transform.) The one-sided transform is necessary in our applications since we are interested in decay processes that go to zero as $t\to\infty$ but which diverge as $t\to-\infty$ (and in any case we are only interested in $t\ge 0$). The variable conjugate to $t$ is an energy-like variable, so we want to consider transforms like

The essence of this integral is

However, as discussed in The Outgoing Energy-Dependent Green’s Operator, this integral does not converge at the upper limit. To fix this we push $E$ into the upper half complex energy plane, making the replacement $E\to z=E+i\epsilon$, where $z$ is a complex energy. Then the integral (21) gives the Green’s operator, to within constant factors. To get these right, we define the “Laplace transform” for our purposes as the integral operator

for $\Im z>0$. Then we have

where $G(z)$ is the Green’s operator as defined by (58) (with $\hbar=1$), and where we denote the Laplace transform of $a_n(t)$ by ${\tilde a}_n(z)$.

The advantage of Laplace transforms is that they convert differential equations into algebraic equations, which are easier to solve, and they also deal nicely with initial conditions and transients. This is like the method of complex impedances when analyzing an electrical circuit, which is much easier than solving coupled differential equations for the currents and charges in the circuit elements.

Applying the Laplace transform to the left-hand side of the Schr"odinger equation (19) and integrating by parts we obtain

Note that $a_n(0)=\delta_{ni}$. Applying the Laplace transform also to the right-hand side of (19) we obtain

This is an algebraic equation for the Laplace transforms of the transition amplitudes that is equivalent to the Schr"odinger equation and is still exact.

We can obtain the same result by noting that

so that

Now sandwiching this by $\bra{n}$ on the left and $\ket{i}$ on the right, we obtain

which is the same as (25).

Solving for the Transition Amplitudes¶

We now apply these results to the decay problem described in \secrnlw-2, making the identifications $\ket{i}\to\ket{B0}$, $E_i = E_B$, $H_0=K_0$, and $H_1=K'=K_1+K_2$. Also, we make the approximation of working on the subspace of the ket space spanned by $\ket{B0}$ and $\ket{A\lambda}$, ignoring all states with two or more photons. This means that we interpret the resolution of the identity as

We can enlarge this space to get a better approximation, at the expense of solving a more complicated system of equations.

We specialize (25) to the case $n=i$, which for our decay problem means that the final state is the same as the initial state $\ket{B0}$. The corresponding amplitude is $a_{B0}(t)$, the amplitude to remain in the initial state. Making the replacement (29) for this case we obtain

Taking into account the numbers of photons that $K_1$ and $K_2$ can create or destroy, we can write this as

Similarly, taking the case $n\ne i$ and identifying $\ket{n}$ with $\ket{A\lambda}$, so that $E_n=E_A+\omega$, we can write (25) as

where we have restricted the resolution of the identity as in (29).

Equations~(31) and (32) are algebraic equations for the Laplace transforms of the transition amplitudes, ${\tilde a}_{B0}(z)$ and ${\tilde a}_{A\lambda}(z)$. The various terms can be assigned an order in the perturbation $K_1+K_2$, where we count $K_2$ as of order $K_1^2$. The amplitude ${\tilde a}_{B0}(z)$ is of order unity in the perturbation, since it is non-zero even when $K_1=K_2=0$, as shown by (31). Then since the matrix element $\matrixelement{A\lambda}{K_1}{B0}$ is of order $K_1$, we see from (32) that the amplitude ${\tilde a}_{A\lambda}(z)$ is also of order $K_1$. This also follows from the fact that transitions to the states $A\lambda$ are induced by $K_1$. From this it follows that the final term in (32) is of order $K_1^3$. As for (31), its terms are of at most second order in $K_1$.

We will work to second order in the perturbation, neglecting the final, third order term in (32) (the final sum). This allows us to solve for the amplitude ${\tilde a}_{A\lambda}(z)$ in terms of ${\tilde a}_{B0}(z)$,

We substitute this back into (31), which allows us to solve for the amplitude ${\tilde a}_{B0}(z)$,

where we define

The symbol $D$ stands for “denominator.” Finally, we can substitute this back into (33) to obtain

Equations~(34) and (36) are explicit solutions for the Laplace transforms of the transition amplitudes $a_{B0}(t)$ and $a_{A\lambda}(t)$. We must now carry out the inverse Laplace transform to get the amplitudes themselves.

Inverting the Laplace Transform¶

Let us revert to a general notation, denoting transition amplitudes by $a_n(t)$ and their Laplace transforms by ${\tilde a}_n(z)$. To carry out the inverse Laplace transform we consider the integral

where $C_+$ is a contour that runs just above the real axis in the complex energy plane, as illustrated in Figure 3. The essence of this integral is

If $t<0$ then the integrand is exponentially damped in the upper half plane, and the contour can be closed by a semicircle at infinity which does not contribute to the integral, as illustrated in Figure 4. This contour encloses no singularities since $G(z)$ is analytic in the upper half plane, so the value of the integral is zero.

For positive times $t>0$ the integrand diverges exponentially in the upper half-plane so we cannot close the contour there. Nor can we close it in the lower half-plane, since that would involve crossing the real energy axis where $G(z)$ has singularities (poles and branch cuts). To evaluate the integral (38) for positive times we consider a different integral,

where $C_-$ runs just below the real energy axis, as illustrated in Figure 5. For positive times $t>0$ the integrand damps exponentially in the lower half-plane, so the contour can be closed as in Figure 6. But since $G(z)$ is also analytic in the lower half-plane, the integral (39) vanishes for $t>0$.

To return to the integral (38) for positive times, we can subtract the integral (39) without changing the value since the latter is zero. That is, for $t>0$ we have

where as the contours $C_+$ and $C_-$ are pushed toward the real axis we get an integral running over the real axis itself, in which the integrand contains the difference in $G(z)$ just above and just below the real axis. But this is the discontinuity in the Green’s operator across the real axis, which is $-2\pi i\,\delta(E-H)$. See (73) and (77). Thus the integral (40) becomes

Altogether, we have

and the inverse Laplace transform of the transition amplitude is

Transition Amplitude $a_{B0}(t)$ in the Unperturbed System¶

Now we apply the inverse Laplace transform to the amplitudes ${\tilde a}_{B0}(z)$ and ${\tilde a}_{A\lambda}(z)$, given by (34) and (36). We start with the amplitude to remain in the initial state $\ket{B0}$,

Before doing this in full detail, let us neglect the terms of $D(z)$ seen in (35) that involve the perturbing Hamiltonians $K_1$ and $K_2$, in order to see the evolution due to the unperturbed system. We do this to gain some experience and confidence with the formalism.

In that case $D(z)$ becomes simply $z-E_B$, and we have

The integrand has a pole on the real axis at $z=E_B$ and no other singularities. For $t>0$ the contour $C_+$ can be closed with a semicircle in the lower half-plane and can then be contracted into a small circle running in the clockwise (negative) direction around the pole. See Figs.~Figure 7 and Figure 8. Then by the residue theorem we have

which is exactly the time evolution we expect for the amplitude to remain in the initial state when there is no perturbation. We see that the time dependence of the amplitude depends on the location of the poles of the integrand of the inverse Laplace transform.

Denominator $D(z)$ on the Real Axis¶

To handle the full expression (35) for $D(z)$ let us first set $z=E+i\epsilon$ for a point on the contour $C_+$,

where we have changed the sign of the denominator of the last term and compensated with the overall sign of that term. We remember that $\sum_\lambda$ actually implies the integral (10) over $\omega$.

We want to let $\epsilon\to0$ in this expression to get $D(z)$ on the real axis, but the final term is tricky. It looks like we could just set $\epsilon=0$ in this term and we would obtain a quantity that was purely real, but this is incorrect. In fact, if we just set $\epsilon=0$ we get an integral that has the mathematical form

where $f(x)$ is a smooth function and where the interval $[a,b]$ straddles the singularity at $x_0$. (The variable $x$ in the integral (48) is like $\omega$ in the integral (47).) The problem is that the integral (48) is not defined. The part of the integral from $x=a$ to $x=x_0$ diverges with one sign and the part from $x=x_0$ to $x=b$ diverges with the opposite sign. Thus the integral has the form $\infty-\infty$, and is not defined.

We cannot just set $\epsilon=0$ in an expression like (47), instead we must take the limit $\epsilon\to0$. That is, we have something of the form

Assuming $f$ is real, the real and imaginary parts of this integrand are obtained from

As for the imaginary part, as $\epsilon\to0$ we get a representation of the $\delta$-function,

so the imaginary part of the final sum in (47) is nonzero as $\epsilon\to0$ and is given by

A bare formula like (51) is not meaningful if interpreted as the limit of a function, since the so-called “$\delta$-function” is not a real function. It does not make sense to say that the $\delta$-function is zero everywhere except at one point where it is infinite, and that its integral is unity. Instead a formula like (49) must be understood as being used under an integral sign, in which the left-hand side is integrated against a smooth function $f(x)$ and the limit is taken. The “$\delta$-function” is an example of what is called a distribution, something that can be regarded as the limit of a family of functions used under an integral.

There are other distributions besides the $\delta$-function, in fact, the real part of (50) is another example which is denoted

This is called the “principal value” of $1/(x-x_0)$. This equation is a “bare” formula like (51). Its real meaning is

Without the $P$ the left-hand side of this equation would not be defined, as explained above.

Now we can take the limit $\epsilon\to0$ to push $D(z)$ down onto the real axis. We write

where

The Zero of $D(z)$ and the Complex Energy of State $B$¶

The zero of $D(z)$ is the pole of the integrand of the inverse Laplace transform (44) and is important for evaluating that integral. It also turns out to be a kind of complex energy of the state $B$, once the perturbation is turned on. In the unperturbed system $D(z)$ has a zero on the real axis at $z=E_B$ but when the perturbation is turned on $D(E)$ is nonzero everywhere on the real axis because the imaginary part is positive there. Thus the zero must have moved to some complex location nearby. We denote the zero by $z_B$, so that $D(z_B)=0$. We write $z_B=E_B+\delta z$ so that $\delta z$ is the shift in the zero of $D(z)$ due to the perturbation. To find $\delta z$ we write

Assuming $\delta z$ is small we neglect it in the last two terms which are already small, finding

or,

where we have rearranged signs in the third term. We see that the zero of $D(z)$ has moved off the real axis into the lower half plane. As for its imaginary part, a comparison with (9) shows that it is $-\Gamma_B/2$, where $\Gamma_B/\hbar$ is the decay rate of state $B$.

The real part of $\delta z$ bears a close relationship to the energy shifts of bound state perturbation theory, as explained in 23. Bound-State Perturbation Theory. The state $\ket{B0}$ is a bound (normalizable) eigenstate of $K_0$ with energy $E_B$. It is an eigenstate of the unperturbed system, that is, the atom with the interactions with the electromagnetic field switched off. Of course, in reality we cannot switch off those interactions; do they lead to shifts in atomic energy levels?

To answer this question we consider using bound state perturbation theory to compute the shift in level $E_B$ due to the perturbation $K'=K_1+K_2$. According to (26), there is a first order shift

where $K_1$ does not contribute. Also, according to (27) the nominal second order shift is

If we replace $K'$ by $K_1$ to get the lowest order contribution, then the only states $k$ that we need sum over are the single photon states $\ket{A\lambda}$, and we obtain

This is almost the third term on the right in (59), missing only the principal value sign.

Actually, without the principal value $\Delta E^{(2)}_B$ in (62) is not defined, nor is it justified to use bound state perturbation theory to find the energy shift in the state $\ket{B0}$ due to the interaction with the electromagnetic field. Bound state perturbation theory applies when we have discrete energy levels that are separated from one another, and in this system the state $\ket{B0}$ is a discrete state imbedded in the continuum. Nevertheless, our search for the zero of $D(z)$ has led to expressions that are obviously related to the energy shifts of bound state perturbation theory, except that the answers are well defined and the energy shift includes an imaginary part. Let us therefore simply define $\Delta E^{(2)}_B$ to be the third term on the right in (59) (that is, by (62) but with a principal value sign), so that

where

Then ${\tilde E}_B$ is the real part of the energy of atomic state $B$, corrected for interactions with the field, while the imaginary part is $-\Gamma_B/2$.

The Transition Amplitude $a_{B0}(t)$ and the Exponential Decay Law¶

We return to the integral (44) for the transition amplitude $a_{B0}(t)$. The integrand has a pole at $z=z_B$, located near and below the energy $E_B$ of the unperturbed state $\ket{B0}$, as illustrated in Figure 9. The contour can be closed with a semicircle in the lower half plane and can then be contracted into a small circle around the pole at $z_B$ in the lower half plane.

Near $z=z_B$ we can expand $D(z)$,

But the first term vanishes and by (35) we have $D'(z) = 1 + O(K_1^2)$. Therefore to lowest order,

when $z$ is near $z_B$. The contour in Figure 10 can be contracted arbitrarily close to the pole at $z_B$ so the integral (44) can be written

where the circle is traversed in a clockwise (negative) direction. We can also write this as

This shows that the amplitude to remain in the initial state evolves with a real phase factor indicating an energy ${\tilde E}_B$, that is, one corrected for the interactions with the field, but with an exponential decay factor. The probability to remain in the initial state is the square of the amplitude,

which shows the exponential decay law.

We see that the exponential decay law is not a rigorous consequence of the Schr"odinger equation, but rather it follows from a series of nontrivial approximations. In fact, it is known that there are deviations from the exponential decay law for long times.

The unperturbed system has a genuine bound state at $E=E_B$ but when the perturbation is turned on there is only a continuous spectrum in the neighborhood of $E=E_B$. The discrete state has “dissolved into the continuum,” and become a resonance. A similar mechanism explains the resonances associated with the doubly excited states in helium, as discussed in The Doubly Excited States.

This derivation has been based on a model of an atom, but it applies in its essence to the decay of any excited state of a system, including radioactive decays of nuclei and decays of unstable particles. The derivation looks complicted, but that is because the exponential law is not exact. The usual “simple” derivations of the exponential decay law rely on a type of statistical assumption of the Markov type, which means that the transition probability for a system depends only on the present state and not the previous history. This is what we usually assume about radioactive decay. For example, if a nucleus has not decayed after ten times its average lifetime, the probability that it will decay in the next second is the same as it always was. The nucleus does not try to “hurry up and decay” because it has failed to do so for so long. At least this is what we usually assume. But studying the decay by solving the Schr"odinger equation, we find that the exponential decay law is only approximate. Various aspects of this situation are discussed in an appendix to the first edition of Sakurai’s book, Modern Quantum Mechanics.

The Transition Amplitude $a_{A\lambda}(t)$¶

We can also get the amplitude to make a transition from the initial state $\ket{B0}$ to some specific final state $\ket{A\lambda}$. According to (36) this is

The matrix element is independent of $z$ and can be taken out of the integral. As for the rest of the integrand, it has poles at $z=E_A+\omega$ and at $z=z_B$, as illustrated in Figure 11. The figure is drawn for the case $E_A < E_B$, as required if there is to be a transition $\ket{B0} \to \ket{A\lambda}$.

Evaluation of the integral by the residue theorem gives two terms, one for each pole,

This is valid for any final state $\ket{A\lambda}$ but the amplitude will be small unless energy is approximately conserved, which means that $E_A+\omega$ is near to $E_B$, and therefore near to $z_B$. If this is the case then (66) gives

where we define

Then the two denominators in (71) are the same to within a sign and the result can be written

Amplitude and Probability as $t\to\infty$¶

The long-time behavior of the system is of interest but as $t\to\infty$ the amplitude $a_{A\lambda}(t)$ oscillates and does not approach a limit due to the phase factor $e^{-i(E_A+\omega)t}$. This factor is due to the evolution under the unperturbed Hamiltonian $H_0$, which is all that matters at long times because the photon will have left the atom behind and is no longer interacting with it. A similar situation holds in scattering theory, in which a wave packet directed at a target evolves according to the free particle Hamiltonian at large negative and large positive times. See the discussion in The $S$-Matrix.