Past exam of the laser science and quantum information processing course of the University of Oxford / 2019 / 3

Let $|g⟩,|e⟩$ be the atomic levels, ${\sigma }_{+}=|e⟩⟨g|$, and $a$ the annihilation operator of the single field mode. Choose the joint ground energy to be zero. The resonant Jaynes-Cummings model isThe first term counts field photons, the second the atomic excitation, and the interaction exchanges one photon for one atomic excitation. A phase redefinition makes $g$ real and positive. Resonance means $\omega ={\omega }_{eg}$; the total excitation $\mathcal{N}=a^{†}a+|e⟩⟨e|$ is conserved. Adding an overall constant gives the equivalent convention using ${\sigma }_z/2$ and a zero-point field term.

The model follows from the electric-dipole interaction and the rotating-wave approximation, retaining the energy-conserving terms while dropping $a{\sigma }_{-}$ and $a^{†}{\sigma }_{+}$. We assume a single quantized monochromatic mode and negligible cavity leakage and atomic relaxation over the time considered. With this convention, the resonant population Rabi angular frequency in the one-excitation subspace is $2g$.

Conservation of $\mathcal{N}$ confines evolution from $|e,0⟩$ to $|e,0⟩,|g,1⟩$. On this subspace,The interaction matrix squares to the identity, so its exponential separates into cosine and sine terms. The state isThe overall phase can be discarded or removed by the interaction picture. The relative phase $-i$ follows from Schrödinger evolution with the positive coupling in a. The atom and field are generally entangled; complete excitation transfer occurs at $gt=\pi /2$.

From b, the excited-state probability isFor a single short measurement interval $\Delta t$ beginning in $|e,0⟩$, requiring $P_e>0.9$ before the first departure from unity givesA zero-photon outcome resets the conditional state to $|e,0⟩$. After $n=T/\Delta t$ such tests, the probability of the all-zero-photon record isThus a 90 percent survival requirement over a specified total duration $T$ is stronger than the single-interval condition: approximately $\Delta t<-\ln (0.9)/(g^{2}T)$ in the short-interval limit. The limiting preservation as $\Delta t\to 0$ is the quantum Zeno effect.

The measurement instrument matters. If photons are absorbed and their first detection removes the excitation, this survival probability also equals the unconditional excited population. For ideal nondestructive photon-number measurements, excitation can transfer back after a one-photon result; the unconditional result after $n$ equal intervals is instead $P_e^{(n)}=[1+{\cos }^n(2g\Delta t)]/2$. It also tends to one for fixed $T$ as the intervals shrink. Neither interpretation guarantees indefinite 90 percent survival merely by imposing a 90 percent bound on each separate interval.

There is at most one field photon in b. Projection onto its one-photon sector givesBefore detector absorption, the normalized conditional joint state is $|g,1⟩$ up to an overall phase. A destructive photon count changes the field to its vacuum but leaves the atom in $|g⟩$. This conditional state is meaningful whenever the one-photon outcome has nonzero probability.

Discarding the photon-number result gives a classical mixture of the two outcome branches. The atomic state, in the basis $(|e⟩,|g⟩)$, isIt is also the reduced state before measuring the field; a nonselective operation on the field does not change the atomic reduced density matrix. The atom is not in a coherent superposition with these populations.

For an atomic energy gap $\hslash \omega$, the thermal population ratio is $p_e/p_g=e^{-\hslash \omega /(k_{B}T)}$. Matching it givesA finite positive temperature exists when $0<p_e<p_g$, with $T\to 0^{+}$ at $p_e=0$ and $T\to +\infty$ as $p_e\to 1/2$ from below. Equal populations correspond to zero inverse temperature. If $p_e>1/2$, the expression is a negative effective temperature for a bounded two-level system, not an equilibrium state at any ordinary positive temperature. For a concrete counterexample to an unrestricted positive-temperature interpretation, at $t=0$ the atom is certainly excited, whereas every positive-temperature Gibbs distribution has $p_e\le 1/2$. The fully excited endpoint is the formal limit $T\to 0^{-}$. Matching one density matrix does not establish genuine thermalization with a bath.