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intermediate · Physics · Time-Dependent Perturbation & Fermi's Golden Rule

Harmonic Perturbations

Most time-dependent perturbations of physical interest oscillate: an atom feels an electromagnetic wave ∝cos⁡(ωt)\propto \cos(\omega t), a spin feels an oscillating magnetic field, a qubit is driven by a microwave tone. We now specialize the first-order machinery to a harmonic (sinusoidal) perturbation.

The harmonic perturbation

Take

H^′(t)=V^cos⁡(ωt)=V^2(e iωt+e−iωt),\hat H'(t) = \hat V\cos(\omega t) = \frac{\hat V}{2}\left(e^{\,i\omega t} + e^{-i\omega t}\right),

where V^\hat V is a time-independent operator and ω\omega is the drive frequency. Splitting the cosine into two exponentials is the key step — each exponential will combine with the intrinsic phase eiωfit′e^{i\omega_{fi}t'} of the transition.

First-order amplitude

Insert this into the first-order formula cf(1)(t)=−iℏ∫0tHfi′(t′)eiωfit′ dt′c_f^{(1)}(t) = -\frac{i}{\hbar}\int_0^t H'_{fi}(t')e^{i\omega_{fi}t'}\,dt' with Hfi′(t′)=Vficos⁡(ωt′)H'_{fi}(t') = V_{fi}\cos(\omega t'):

cf(1)(t)=−i Vfi2ℏ∫0t(e i(ωfi+ω)t′+e i(ωfi−ω)t′) dt′.c_f^{(1)}(t) = -\frac{i\,V_{fi}}{2\hbar}\int_0^t \Big(e^{\,i(\omega_{fi}+\omega)t'} + e^{\,i(\omega_{fi}-\omega)t'}\Big)\,dt' .

Carrying out the integrals,

cf(1)(t)=−Vfi2ℏ[e i(ωfi+ω)t−1ωfi+ω+e i(ωfi−ω)t−1ωfi−ω].c_f^{(1)}(t) = -\frac{V_{fi}}{2\hbar}\left[ \frac{e^{\,i(\omega_{fi}+\omega)t}-1}{\omega_{fi}+\omega} + \frac{e^{\,i(\omega_{fi}-\omega)t}-1}{\omega_{fi}-\omega} \right].

The rotating-wave approximation

The two terms have denominators ωfi+ω\omega_{fi}+\omega and ωfi−ω\omega_{fi}-\omega. If the drive is tuned near the transition, ω≈ωfi>0\omega \approx \omega_{fi} > 0, then ωfi−ω\omega_{fi}-\omega is small while ωfi+ω≈2ωfi\omega_{fi}+\omega \approx 2\omega_{fi} is large. The second (co-rotating) term has a tiny denominator and dominates; the first (counter-rotating) term is suppressed by ∼(ωfi−ω)/(ωfi+ω)\sim (\omega_{fi}-\omega)/(\omega_{fi}+\omega). Dropping the small term is the rotating-wave approximation (RWA):

cf(1)(t)≈−Vfi2ℏ e i(ωfi−ω)t−1ωfi−ω.c_f^{(1)}(t) \approx -\frac{V_{fi}}{2\hbar}\,\frac{e^{\,i(\omega_{fi}-\omega)t}-1}{\omega_{fi}-\omega} .

Defining the detuning Δ≡ωfi−ω\Delta \equiv \omega_{fi}-\omega and squaring:

  Pi→f(t)=∣Vfi∣24ℏ2 sin⁡2(Δt/2)(Δ/2)2  \boxed{\;P_{i\to f}(t) = \frac{|V_{fi}|^2}{4\hbar^2}\,\frac{\sin^2(\Delta t/2)}{(\Delta/2)^2}\;}

This is the same sinc-squared shape as before, but now centred on Δ=0\Delta=0, i.e. on ω=ωfi\omega = \omega_{fi}.

Absorption and stimulated emission

The two exponentials are not just a mathematical convenience — they are two physical processes. The co-rotating term peaks at ω=ωfi\omega=\omega_{fi}, i.e. Ef=Ei+ℏωE_f = E_i + \hbar\omega: the system absorbs a quantum of energy ℏω\hbar\omega and climbs to a higher state. The counter-rotating term peaks at ω=−ωfi\omega=-\omega_{fi}, i.e. Ef=Ei−ℏωE_f = E_i - \hbar\omega: the system emits a quantum and drops to a lower state (stimulated emission). The drive frequency selects which process is resonant.

Try it

For a harmonic drive with Vfi=0.4V_{fi}=0.4, ωfi=5.0\omega_{fi}=5.0, ω=4.0\omega=4.0, t=2.0t=2.0, and ℏ=1\hbar=1, compute the detuning Δ=ωfi−ω\Delta=\omega_{fi}-\omega and the RWA transition probability Pi→f(t)=(∣Vfi∣2/4ℏ2) sin⁡2(Δt/2)/(Δ/2)2P_{i\to f}(t) = (|V_{fi}|^2/4\hbar^2)\,\sin^2(\Delta t/2)/(\Delta/2)^2. return the number.

Run your code to see the quantum state.

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