When an atom emits a photon, for some time the photon is close to its source, the atom, because it takes some time for the photon to move away from the atom. So during that time the uncertainty in the position of the photon is small and according to Heisenberg's uncertainty relation, the uncertainty in the wave-vector of the photon is large.

My problem is that this means that the wave-length of the photon is not defined, and putting the numbers I saw that the spread is huge, much more than the width of the spectral line.

Formally, we consider an atom emitting a photon. Let's characterize the uncertainty in the position of the photon by $r_0$ = the size of the atom. From Heisenberg's uncertainty relation we have: $$r_0=\frac{1}{2\Delta k}$$ or (omitting small constants): $$r_0=\frac{\lambda(\lambda+\Delta\lambda)}{\Delta \lambda}$$ solving this for $\Delta\lambda$ we get: $$\Delta\lambda=\frac{\lambda^2}{r_0-\lambda}$$ therefore, if the size of the atom is small compare to the wavelength, we get: $$\Delta\lambda=\lambda$$

For example if the energy transition of the atom corresponds to $800$ nm, the uncertainty will be $800$ nm, sounds very wrong to me, what's my mistake?



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