Tmp Lecture 25

<framebox> <attributes> <width>None</width> <pos>None</pos> </attributes> Lecture XXV </framebox>

Two-phase absorption, Fano profiles

Let us assume large one-photon detuning, $\Delta \gg \Gamma$, weak probe $\omega _{1}$ and strong control field $\omega _{2}$ (we also define the two-photon detuning $\delta =\Delta _{1}-\Delta _{2}$).

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In this limit analytic expressions for the absorption cross section for beam $\omega _{1}$ and the refractive index seen by beam $\omega _{1}$ exist, e.g. [| Muller et al., PRA 56, 2385 (1997)]

refractive index: $n(d)=1+{\frac {3}{8\pi }}{\tilde {n}}\lambda ^{3}f(\Delta )$ ${\tilde {n}}$ atomic density $f(\delta ){\frac {r}{2}}\delta {\frac {A}{A^{2}+B^{2}}},A=\omega _{2}^{2}-\delta \Delta _{2},B=\delta r$ for zero ground-state linewidth $T(d)=T_{o}g(d)$ $T_{o}={\frac {3}{2\pi }}\lambda ^{2}$, $g(\Delta )={\frac {?_{2}}{2}}{\frac {dB}{A^{2}+B^{2}}}$

The absorption cross section is for $\Delta _{2}>0$

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This is like a ground state coupling to one narrow and one wide excited state, except that there is EIT in between because both states decay to the same continuum.

? r (BI)

|g> - ? $r_{se}$

$\delta =\Delta _{2}$

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one-photon absorption is two-photon scattering process

$r=-{\frac {\omega _{2}^{2}}{\Delta _{2}}}$

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two-photon absorption is (at least) four-photon scattering process

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For the EIT condition $\delta =0$, there is no coupling to the excited state, and the refractive index is zero. In the vicinity of EIT, there is steep dispersion, resulting in a strong alteration of the group velocity of light$\rightarrow$ slowing and stopping light.

Slow light, adiabatic changes of velocity of light

The group velocity of light in the presence of linear dispersion ${\frac {dn}{d\omega }}$ is given by (Harris and Han, PRC <underline> <attributes> </attributes> 82 </underline>, 4611 (1999))

$V_{g}={\frac {c}{1+\omega {\frac {dn}{d\omega }}}}$ for light of frequency $\omega$

A strong linear dispersion with positive slope near EIT then corresponds to very slow light

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$n=1\rightarrow$ electric field ???$\rightarrow$ power per area

${\frac {P}{A}}={\frac {1}{2}}\epsilon _{o}C|E|^{2}$ unchanged

Pulse is compressed in medium $\rightarrow$ energy density is increased, light is partly in the form of an atomic excitation : polarization

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For sufficiently small $\omega _{2}$, velocity of light may be very small...(L.V. Hau, S.E. Harris, Z. Dutton, and C.H. Behrooti, Nature <underline> <attributes> </attributes> 397 </underline>, 594 (1999)), as observed in a BEC. What happens if experiment performed in room temperature vapor when ???? ? The same if setup is Doppler free (co-propagating probe and control fields $\omega _{1}$ and $\omega _{2}$). If we change control field adiabatically while the pulse is inside the medium, we can coherently stop light, i.e. convert it into an atomic excitation or spin wave. With the reverse process we can then convert the stored spin-wave back into the original light field. The adiabatic conversion is made possible by the finite splitting between bright and dark states. In principle, all coherence properties and other (qm) features of the light are maintained, and it is possible to store non-classical states of light by mapping photon properties one-to-one onto quantized spin waves. More about these quanta called "dark-state polaritons" once we have introduced Dicke states. Is it possible to make use of EIT for, e.g. atom detection without absorption? Answer: no improvement for such linear processes. However: improvement for non-linear processes is possible.

Superradiance

Assume that two identical atoms, one in its ground and the other in its excited state, are placed within a distance $d\gg \lambda _{eg}$ of each other. What happens?

For a simgle atom we have for the emission rate R(t) at time t an emission probability $p(t)=\int _{o}^{t}R(t')dt'$ to have emitted a photon by time t:

$R_{1}(t)=re^{-rt}$

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$p_{1}(t)=\int _{o}^{t}dt'R(t')=1-e^{-rt}$ for a single atom

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what about two atoms?

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It turns out that the correct answer is

$R_{2}(t)=re^{-2pt}$

$p_{2}(t)=\int _{o}^{t}R_{2}(t')dt'={\frac {1}{2}}(1-e^{-2rt})$

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The photon is emitted with the same initial rate, but has only probability ${\frac {1}{2}}$ of being emitted at all! How can we understand this? The interaction Hamiltonian is

$V=-{\vec {d}}_{1}\cdot {\vec {E}}({\vec {\gamma }},t)-{\vec {d}}_{2}\cdot {\vec {E}}({\vec {\gamma }},t)=-{\vec {D}}\cdot {\vec {E}}({\vec {\gamma }},t)$ class.

$V=\hbar g(\sigma _{1}^{+}+\sigma _{1}^{-})(a+a^{+})g(\sigma _{2}^{+}+\sigma _{2}^{-})(a+a^{+})QED$ $=\hbar g((\sigma _{1}^{+}+\sigma _{2}^{+}+\sigma _{1}^{-}+\sigma _{2}^{-})(a+a^{+})$ $=\hbar g(\Sigma ^{+}+\Sigma ^{-})(a+a^{+})$

with $\Sigma ^{\pm }{\dot {\Sigma }}_{i=1}^{z}\sigma _{i}^{\pm },{\vec {D}}=\Sigma _{i=1}^{z}{\vec {d}}_{i}$