# Rabi Oscillations: $\pi$-Pulse vs a single photon

I am puzzled by the following:

Assume an atom as a two-level-system. A $$\pi$$-Pulse acting on an atom in the ground state promotes this atom in the excited state. This is done by a continuous electromagnetic wave over a certain amount of time, such that $$\Omega t= \pi$$.

On the other hand, according to the photoelectric effect, the same two-level-system can be promoted from the groundstate to the excited state by the absorption of a single photon.

This absolutely blows my mind and I wanted to check if I am mistaken somewhere or if there are any other views on that issue. I mean: It sounds ridiculous to me that the effect of a single photon is the same as that of a light beam with ~$$10^{34}$$ photons. On top of that, it is a fundamentally different situation radiating an EM wave continuously on an atom or just shooting one resonant photon on it.

• A single-frequency photon does not have a temporal profile so the corresponding interaction time would be infinite, whereas the $\pi$ pulse requires finite temporal duration. Commented Aug 18, 2021 at 14:04
• @ZeroTheHero that's not so fair: a single photon inside of a cavity can excite an atom in that cavity, where the single photon is in a cavity mode with a well-defined frequency. Commented Aug 18, 2021 at 17:00

These both produce approximately the same result. The difference is the speed at which they produce them: the Rabi frequency in the interaction between a two-level system and a single mode of light scales as the square root of the intensity of the light. So a single photon $$\pi$$ pulse will be much slower than a pulse given by a laser beam with $$10^{34}$$ photons.

I say approximately because the intense beam will actually be slightly entangled with the atom and so the latter won't be a pure excited state. Still, that amount of entanglement drops precipitously with the intensity of the light, so for all practical purposes it generates a perfect $$\pi$$ pulse.

Things to look into are the Rabi model and the Jaynes-Cummings model.

• Thanks! One quick follow-up question on your answer if that's okay. Why wouldn't the atom be always excited then if we radiate so many photons on it? According to the "one-photon-situation" one photon excites the atom just like a continuous wave does, but, as you said, much slower. Hence, I would expect that if I radiate 10^34 photons on it, that each individual photon would excite the atom if that's not already the case. Therefore, the atom would be always excited instead of performing Rabi oscillations. I imagine it as a "photon onslaught" onto the atom if you know what I mean! Commented Aug 19, 2021 at 14:51
• In some sense, it is! What happens is that the atom radiates its excitation back into the field, then the field re-excites the atom. The more photons there are, the faster this transfer happens, so the probability of the atom being excited wiggles around really quickly ("Rabi flopping"). Then, if you treat the field quantum mechanically, eventally this flopping decays and you just get the atom being excited about 50% of the time (see the picture here) Commented Aug 19, 2021 at 16:22
• Remember that quantum mechanically, it's not a stream of equally spaced photons, it's more like shot noise (a Poisson distribution). The better way to put it is that the rate at which the atom gets excited is the same as the rate at which it decays (in this model), so increasing the number of photons increases both processes Commented Aug 19, 2021 at 16:25

Let us consider a TLS with states $$|e\rangle, |g\rangle$$ coupled to a single light mode $$|n\rangle$$, described by Hamiltonian $$H=\hbar\omega_e |e\rangle \langle e| + \hbar\Omega\left(|e\rangle\langle g|a + |g\rangle\langle e|a^\dagger\right) + \hbar \omega_0a^\dagger a$$ For a specific number of photons in a mode, $$n$$, the joint states of TLS and photon field are $$|e,n-1\rangle$$ and $$|g,n\rangle$$, and the Hamiltonian can be written as a 2-by-2 matrix in this sub-basis: $$\hat{H}=\begin{bmatrix} \hbar \omega_e + \hbar\omega_0(n-1)&\hbar\Omega\sqrt{n}\\\hbar\Omega\sqrt{n}&\hbar\omega_0 n\end{bmatrix}$$ The wave function can be written as $$|\psi(t)\rangle = c_g(t)|e,n-1\rangle + c_g(t)|g,n\rangle$$ The eigenvalues of this Hamilronian are $$\lambda_{\pm}=\hbar \left[\omega_0 n + \frac{\omega_e-\omega_0}{2}\pm\sqrt{\frac{(\omega_e-\omega_0)^2}{4}+n \Omega^2}\right],$$ that is the amplitudes $$c_{e,g}$$ oscillate with the Rabi frequency $$\omega_R=2\sqrt{\frac{(\omega_e-\omega_0)^2}{4}+n \Omega^2}=\sqrt{(\omega_e-\omega_0)^2+4n \Omega^2}$$ If the photon energy exactly matches the TLS level spacing, $$\omega_e\approx\omega_0$$, this frequency is simply $$2\Omega\sqrt{n}$$ As we see, the frequency is much higher for zillions of photons than for a single photon, that is the duration of $$\pi$$ pulse for a single photon is too long to wait.

On the other hand, according to the photoelectric effect, the same two-level-system can be promoted from the groundstate to the excited state by the absorption of a single photon.

A TLS can be promoted to an excited state by a single photon, but it does not have to. In photoelectric effect we are dealing with many atoms (which can be thought of as TLS) and many photons. Having huge amount of atoms increases the probability that some of them do absorb a photon, just as having many photons increases a probability that a specific atom absorbs a photon. Still, only a small fraction of atoms get excited, and they do so at random times. For a single atom and a single photon the probability that we will find the atom in an excited state is very small.

If we retake the picture with only one atom, after waiting for the same amount of time, the probability of finding the atom in the excited state depends on how many photons we have. If we take the length of a $$\pi$$ pulse for $$n$$ photons, $$t=\frac{\pi}{2\Omega\sqrt{n}}$$ but we actually have only one photon in the field, then the probability that the atom is excited is about $$1/\sqrt{n}$$ times smaller, than if we did have $$n$$ photons. In practice it is negligeable, when we talk about numbers like $$10^{34}$$.

Remark: A more correct analysis would require as consider interaction of TLS with a coherent state, rather than a mode with a known number of photons.