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## Main Question or Discussion Point

Suppose there is an electic dipole that starts to oscillate with frequency ω at t=0, then how long does it take the electric dipole to emit a photon?

We know that the radiant power of such electric dipole calculated from quantum physics is [tex]P\left( \omega \right) = \frac{{{\omega ^4}}}{{3\pi {\varepsilon _0}{c^3}}}{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|^2}[/tex]. Does it mean that the time it take the electic dipole to emit a photon is [tex]\Delta t = \frac{{\hbar \omega }}{{P\left( \omega \right)}}[/tex]?([tex]{\boldsymbol{\hat d}}[/tex] is the dipole operator.)

We know that the radiant power of such electric dipole calculated from quantum physics is [tex]P\left( \omega \right) = \frac{{{\omega ^4}}}{{3\pi {\varepsilon _0}{c^3}}}{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|^2}[/tex]. Does it mean that the time it take the electic dipole to emit a photon is [tex]\Delta t = \frac{{\hbar \omega }}{{P\left( \omega \right)}}[/tex]?([tex]{\boldsymbol{\hat d}}[/tex] is the dipole operator.)

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