Time-dependent perturbation theory

Up to now, every Hamiltonian in this chapter has been independent of time. That makes for an excessively quiet world: as section 2.5 told us, a stationary state only spins its phase in place, its probability distribution frozen solid. An electron that moves into $\ket{2p}$ ought to live there forever.

Reality objects at once: put sodium vapour in white light and it precisely devours the two yellow lines near 589 nm — not a bit more, not a bit less; and a sodium lamp, conversely, glows with the very same pair of yellow lines. The atom jumps back and forth between two levels, and it is fussy about the light’s frequency to a punishing degree.

Where is the contradiction? In “the Hamiltonian does not depend on time”. Light is an electromagnetic wave; when it strikes the atom, the electron feels an electric field oscillating in time, and the Hamiltonian picks up a time-dependent piece:

$$\hat H(t)=\hat H_0+\hat V(t), \qquad \hat V(t)=e\mathcal E z\cos\omega t\tag{7.5.1}$$

$\hat H_0$ is the isolated atom (whose levels $E_n$ and eigenstates $\ket{n}$ we have already solved), and $\hat V(t)$ is the light field’s shoving. The moment the Hamiltonian depends on time, the very concept of an “energy eigenstate” loses its footing — there are no stationary states to solve for, and every static method of the last four sections is out of a job. We must return to the equation of motion itself.

Moving the problem into the coefficients

The idea: the eigenstates of $\hat H_0$ still form a complete basis at every instant, so any state can be expanded. We cannot solve the time-dependent Schrödinger equation outright, but we can squeeze all the time dependence into the expansion coefficients:

$$\ket{\Psi(t)}=\sum_n c_n(t)\,\ee^{-\ii E_nt/\hbar}\,\ket{n}\tag{7.5.2}$$

Note that we write the “boring” phase $\ee^{-\ii E_nt/\hbar}$ out explicitly — with no perturbation it is the whole of the evolution, and the $c_n$ are constants. This bookkeeping is precisely the interaction picture of section 3.9: let $\hat H_0$‘s evolution spin its own phases, while $c_n(t)$ records only the changes caused by the perturbation. $|c_n(t)|^2$ is then the probability of finding the system in $\ket{n}$ at time $t$.

The coefficient equations and the first-order transition amplitudebasic~10 minexpand

Step 1: substitute into the Schrödinger equation.

Insert the expansion into $\ii\hbar\,\partial_t\ket{\Psi}=(\hat H_0+\hat V)\ket{\Psi}$. Differentiating the left side in $t$ produces two kinds of terms: $\dot c_n$ terms and $E_nc_n$ terms; the latter cancel term by term against the $\hat H_0$ contribution on the right — exactly the payoff for writing the phase factor explicitly. What remains is

$$\ii\hbar\sum_n\dot c_n\,\ee^{-\ii E_nt/\hbar}\ket{n} =\sum_n c_n\,\ee^{-\ii E_nt/\hbar}\,\hat V(t)\ket{n}\tag{7.5.3}$$

Step 2: project.

Multiply from the left by $\bra{f}$ and use orthogonality:

$$\dot c_f=-\frac{\ii}{\hbar}\sum_n c_n(t)\,V_{fn}(t)\,\ee^{\ii\omega_{fn}t}, \qquad V_{fn}\equiv\bra{f}\hat V(t)\ket{n},\quad \omega_{fn}\equiv\frac{E_f-E_n}{\hbar}\tag{7.5.4}$$

Everything so far is still exact — we have merely rewritten one partial differential equation as infinitely many coupled ordinary ones. The trouble has not vanished, only changed address: the rate of change of each coefficient depends on all the others.

Step 3: the first-order approximation.

Suppose at $t=0$ the system is in the initial state $\ket{i}$: $c_i(0)=1$, all others zero. For a weak perturbation and not too long a time, the coefficients have not had time to run far: in the sum on the right keep $c_i\approx1$ and set the rest to zero — replace the evolved values by the initial values, the same spirit as time-independent perturbation theory’s “average with the old wavefunction”. Hence

$$c_f(t)\approx-\frac{\ii}{\hbar}\int_0^t V_{fi}(t')\,\ee^{\ii\omega_{fi}t'}\,\dd t' \qquad(f\neq i)\tag{7.5.5}$$

The first-order transition amplitude: the perturbation matrix element, dressed with a phase factor, integrated over history. This one line is the master plan for this section and the next.

How to read it: the integral is exactly the Fourier transform of $V_{fi}(t')$ evaluated at the frequency $\omega_{fi}$. In other words, the system uses the transition frequency as its radio tuning dial and listens to only that one station in the perturbation’s spectrum. If the perturbation contains a component at frequency $\omega_{fi}$, the transition happens; if not, the system barely stirs. The atom’s fussiness about colour — laid bare in a single line.

Periodic perturbations and resonance

Insert the light field: $\hat V(t)=\hat V\cos\omega t=\frac{\hat V}{2}(\ee^{\ii\omega t}+\ee^{-\ii\omega t})$. The integral is elementary:

$$c_f(t)=-\frac{\ii V_{fi}}{2\hbar} \left[\frac{\ee^{\ii(\omega_{fi}+\omega)t}-1}{\ii(\omega_{fi}+\omega)} +\frac{\ee^{\ii(\omega_{fi}-\omega)t}-1}{\ii(\omega_{fi}-\omega)}\right]\tag{7.5.6}$$

The two denominators become small at $\omega=-\omega_{fi}$ and $\omega=+\omega_{fi}$ respectively. Take $E_f$ above $E_i$ (absorption); then near $\omega\approx\omega_{fi}$ the second term stands alone, and we drop the first (the rotating-wave approximation: the non-resonant term oscillates rapidly and averages to zero). Take the modulus squared:

$$P_{i\to f}(t)=\frac{|V_{fi}|^2}{\hbar^2}\, \frac{\sin^2\!\left[(\omega_{fi}-\omega)t/2\right]}{(\omega_{fi}-\omega)^2}\tag{7.5.7}$$

Plug in numbers: sodium’s $3s\to3p$ transition sits at 589 nm, $\omega_{fi}=2\pi c/\lambda\approx3.2\times10^{15}\ \text{rad/s}$ (photon energy 2.1 eV). If the atom interacts with the light for $1\ \mu\text{s}$ (flying through a laser beam, say), the main-peak width is $2\pi/t\approx6\times10^6\ \text{rad/s}$ — a relative width of order $10^{-9}$. Atoms really are punishing listeners; atomic clocks push this punishing taste to the limit: the longer the interaction time, the more precisely the frequency is recognised.

∎Key formulas

• Coefficient equations (exact)

  $$\dot c_f=-\frac{\ii}{\hbar}\sum_n c_n V_{fn}(t)\,\ee^{\ii\omega_{fn}t}$$

  Interaction picture: phases go to H₀, changes are logged in the coefficients

• First-order transition amplitude

  $$c_f(t)=-\frac{\ii}{\hbar}\int_0^t V_{fi}(t')\,\ee^{\ii\omega_{fi}t'}\dd t'$$

  The perturbation spectrum sampled at the transition frequency ω_fi

• Periodic perturbation

  $$P_{i\to f}=\frac{|V_{fi}|^2}{\hbar^2}\frac{\sin^2[(\omega_{fi}-\omega)t/2]}{(\omega_{fi}-\omega)^2}$$

  Resonance peak width ~2π/t: the longer the illumination, the stricter the selection

• Rabi oscillations

  $$P=\sin^2\!\left(\tfrac{\Omega t}{2}\right),\quad\Omega=|V_{fi}|/\hbar$$

  The exact solution at resonance; first order is its opening at Ωt≪1

?Self-check4 questions

1. 1.

   The best reading of the first-order transition amplitude c_f(t) ∝ ∫V_fi(t′)e^(iω_fi t′)dt′ is:

   • The Fourier spectrum of the perturbation evaluated at the transition frequency ω_fi — a transition needs that frequency component to be present
   • The time average of the perturbation
   • The perturbation energy divided by the level spacing
   • The overlap integral of the initial and final wavefunctions

   Submit
2. 2.

   As the periodic perturbation acts for a longer time t, the resonance peak of the transition probability versus frequency will:

   • Get narrower and narrower (width about 2π/t), with ever better frequency selectivity
   • Get wider and wider, because more disturbance has accumulated
   • Keep the same width, with only the peak height growing
   • Split into two peaks

   Submit
3. 3.

   At resonance, first-order theory gives P ∝ t², which will eventually exceed 1. The correct handling is:

   • Accept that the first-order result is valid only while P ≪ 1; the long-time behaviour at resonance is described by Rabi oscillations, P = sin²(Ωt/2)
   • Clip the probability at 1
   • Conclude that quantum mechanics fails in strong fields
   • Adding the second-order correction fixes it

   Submit
4. 4.

   The sodium 3s→3p transition has wavelength 589 nm. Find the transition angular frequency ω_fi, in units of 10¹⁵ rad/s. (c = 3.0×10⁸ m/s)

   ×10¹⁵ rad/s20% relative tolerance

   Submit

What comes next

For a single discrete final state, the first-order formula predicts either oscillation or $t^2$ growth — yet what experiments actually measure is different behaviour altogether: excited atoms decay at a constant rate, their number falling off as $\ee^{-t/\tau}$, with each 2p hydrogen atom living 1.6 nanoseconds on average. Where does a constant rate come from? The key is that when an atom emits light, the photon may fly in any direction with any polarisation — the final state is not one state but a whole continuum. Sum the sinc² peak over that continuum, and the oscillations vanish as if by magic, leaving one clean rate formula: Fermi’s golden rule.