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Sandbox Physics
Cavity and waveguide quantum electrodynamics

L3 · Chapter 4

Why two photons cannot always be scattered separately

Sandbox Physics ·

An excited two-level atom cannot absorb a second identical excitation into a higher atomic state. Two nearby photons therefore share the response of the emitter. Revealing their correlation requires paired position outcomes: even a finely resolved photon density retains only part of that information.

Equal density can hide different pairs

Consider left and right output regions. One state always places both photons on the same side, choosing left or right equally. Another always places one on each side. Both give the same average photon number on either side. Their density profiles cannot distinguish them.

A joint map records one position on each axis. Weight in its same-side and opposite-side regions reveals the pair structure. Photons carry no permanent first and second identities, so exchanging the coordinates leaves the amplitude unchanged.

Reading: Nysteen et al.

Include both ways to distribute two excitations

Number chain sites from left to right. The operator aj†a_j^\dagger adds a photon at site jj, and σ+\sigma_+ excites the atom. With two total excitations, there are either two field photons or one photon and an excited atom. A symmetric matrix Φjk\Phi_{jk} and a vector eje_j describe these branches.

The factor 1/21/\sqrt2 handles identical-particle counting. A pair on distinct sites occurs twice in the full matrix, while double occupation of one site carries its bosonic square-root factor. The resulting state norm is the sum of all squared matrix entries and the excited-atom branch probability.

∣Ψ⟩=12∑j,kΦjkaj†ak†∣g,0⟩+∑jejσ+aj†∣g,0⟩,Φjk=Φkj|\Psi\rangle=\frac{1}{\sqrt2}\sum_{j,k}\Phi_{jk}a_j^\dagger a_k^\dagger|g,0\rangle+\sum_j e_j\sigma_+a_j^\dagger|g,0\rangle,\qquad\Phi_{jk}=\Phi_{kj}
There is no state with two atomic excitations. Removing that option is the source of saturation.
∑j,k∣Φjk∣2+∑j∣ej∣2=1,Pe=∑j∣ej∣2\sum_{j,k}|\Phi_{jk}|^2+\sum_j|e_j|^2=1,\qquad P_e=\sum_j|e_j|^2
Atomic excitation probability cannot exceed one, while the field can hold two photons. State norm and excitation number are separate checks.

Reading: Nysteen et al.

Evolve the two amplitude tables together

Let hjℓh_{j\ell} be the bare single-photon hopping matrix: minus the coupling for neighbours and zero otherwise. The atom connects at site cc, has detuning Δ\Delta and coupling gg. Insert the state into the Schrödinger equation and equate coefficients of the two branches to obtain the equations below.

The first equation moves either photon coordinate and adds emission from the excited atom. The second propagates the remaining photon and allows absorption. Both branches must evolve together.

The square-root-two factor can be checked on one site. A normalized two-photon state is (a†)2∣0⟩/2(a^\dagger)^2|0\rangle/\sqrt2; annihilating one photon gives 2∣1⟩\sqrt2|1\rangle. This ladder factor couples the two branches, while the atom itself has no next excitation level.

iΦ˙jk=∑ℓ(hjℓΦℓk+hkℓΦjℓ)+g2(δjcek+δkcej),ie˙j=∑ℓhjℓeℓ+Δej+2gΦcji\dot\Phi_{jk}=\sum_\ell(h_{j\ell}\Phi_{\ell k}+h_{k\ell}\Phi_{j\ell})+\frac g{\sqrt2}(\delta_{jc}e_k+\delta_{kc}e_j),\qquad i\dot e_j=\sum_\ell h_{j\ell}e_\ell+\Delta e_j+\sqrt2g\Phi_{cj}
A Kronecker delta is one for equal indices and zero otherwise, confining atom–field exchange to the coupling site. Frequency and time units are consistent.

Reading: Sandbox Physics FR11

Construct a genuine linear comparison

Replace the atom with a harmonic oscillator, keeping its frequency, coupling and waveguide. A second oscillator excitation is allowed. Propagate each incoming mode using the same single-excitation equations, then symmetrize their product.

Let normalized incident modes be uu and vv, with overlap s=⟨u∣v⟩=∑juj∗vjs=\langle u|v\rangle=\sum_j u_j^*v_j, summed over the complete chain-and-oscillator single-excitation space. Their overlap changes the normalization factor. Keep the oscillator double-occupation branch as part of the reference; removing it would change the linear model.

Worked example: for identical modes the overlap is one. The numerator becomes twice their product and the denominator is two, leaving the product itself. For orthogonal modes the denominator is square-root two. These two limits independently check the bosonic normalization.

Φjklin=ujvk+vjuk2(1+∣s∣2)\Phi_{jk}^{\rm lin}=\frac{u_jv_k+v_ju_k}{\sqrt{2(1+|s|^2)}}
The complete single-particle space includes chain and oscillator. Only its field–field component is shown here.

Ask what a difference map subtracts

The spatial map compares ∣Φjk∣2|\Phi_{jk}|^2 with ∣Φjklin∣2|\Phi_{jk}^{\rm lin}|^2. Neither is independently renormalized. Their integrals retain the probability that both excitations are in the field. An excited-atom branch makes that weight smaller than one.

Writing the amplitude difference as C=Φ−ΦlinC=\Phi-\Phi^{\rm lin} shows that the probability difference also contains interference with the reference. A bright patch is not directly a count of correlated photons, and a dark patch is not a negative probability.

A connected scattering amplitude is defined between incoming and asymptotic outgoing states by subtracting the symmetrized product of single-photon scattering. A transient spatial probability difference and a density-subtracted correlation answer other questions.

∣Φ∣2−∣Φlin∣2=2Re⁡ ⁣[(Φlin)∗C]+∣C∣2|\Phi|^2-|\Phi^{\rm lin}|^2=2\operatorname{Re}\!\left[(\Phi^{\rm lin})^*C\right]+|C|^2
The probability difference combines a reference interference term with the squared amplitude change.

Reading: Shen and Fan

Test whether saturation explains the change

First decouple the emitter: the two calculations must agree. Then separate the pulses enough for the atom largely to recover between them; independent single-photon scattering becomes a useful approximation. Finally overlap the pulses near the atom and compare excitation, pair structure and departure from the linear reference.

Photon density is nj=2∑k∣Φjk∣2+∣ej∣2n_j=2\sum_k|\Phi_{jk}|^2+|e_j|^2. Its sum plus atomic excitation probability must be two. This conservation law follows from the model. To interpret a completed scattering event, choose a time after the pulses leave the atom and before a reflected boundary wave returns.

∑jnj+Pe=2,Gjk(2)=⟨aj†ak†akaj⟩=2∣Φjk∣2\sum_j n_j+P_e=2,\qquad G^{(2)}_{jk}=\langle a_j^\dagger a_k^\dagger a_ka_j\rangle=2|\Phi_{jk}|^2
The normally ordered pair quantity has a known factor of two relative to the ordered-position probability map, including on its diagonal.

Try the prediction

Bring two pulses together at the atom

  1. Predict whether coincident or separated arrivals should depart more from the linear reference.
  2. Change only pulse separation. Compare atom excitation, field density and the two-dimensional difference map.
  3. Choose a time after the pulse has left the emitter. Export the record and check norm, excitation number and boundary probability before interpreting the map.

Exercises and explanations

  1. Why can same-side photon pairs and one photon per side have identical single-photon densities?

    Read the explanation

    Density retains average occupation. Two photons on either side with equal probability average to one per side, just as one photon on each side does. Pairing information is lost.

  2. The position map integrates to 0.80.8 and atomic excitation probability is 0.20.2. Is twenty percent of the probability missing?

    Read the explanation

    No. The branch probabilities add to one. Mean field photon number is 1.8 and the atom contributes 0.2, giving two total excitations.

  3. Why does normalizing each pair map separately change the comparison?

    Read the explanation

    It conditions on both excitations being in the field and removes differences in branch probability. Conditional shapes can also be compared, but that is a different question from the unconditioned scattering map.

Sources and calculation scope

  1. Shen and Fan, Strongly correlated two-photon transport, Phys. Rev. Lett. 98, 153003 (2007)

    Fixed author version v2, 13 March 2007. Used for the asymptotic two-photon scattering distinction and correlated component. The finite tight-binding wavepacket model has its own bandwidth and boundaries; it is not the paper’s infinite linear-dispersion S matrix.

  2. Nysteen et al., Scattering of two photons on a quantum emitter, New J. Phys. 17, 023030 (2015)

    Section 2 and the symmetric two-excitation wavefunction establish normalization and photon-density conventions. Author preprint arXiv:1409.1256v1 was also read. Published article is CC BY 3.0; no figure or dataset is copied here.

  3. Sandbox Physics FR11, finite two-excitation scattering model

    State basis, ordered pair-map normalization, harmonic comparison, convergence and independent single-mode checks. Exported spatial data remain separate from continuous detector records.

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