# Natural line width for doubly excited transitions

So a simple derivation for natural single electron emission line broadening is this:

$\Delta E \Delta t \approx \frac{h}{2\pi} \Leftrightarrow h\cdot\Delta f_{1/2}\cdot\tau \approx \frac{h}{2\pi} \Leftrightarrow \Delta f_{1/2}\approx \frac{1}{2\pi\tau}\Leftrightarrow \Delta \lambda_{1/2} \approx \frac{\lambda}{f} \cdot \frac{1}{2\pi\tau}$

(Where $\tau$ is the life time of the excited state)

This is a consequence of heisenbergs uncertainty principle. But how would it look for a doubly-excited transition say 2s2p-2d2d or something. If the two separate transitons have different life times $\tau_1$ and $\tau_2$ would the line width be something like:

$$\Delta \lambda_{1/2} \approx \frac{\lambda}{f} \cdot \frac{1}{2\pi} \cdot (\frac{1}{\tau_1} + \frac{1}{\tau_2})$$

If or if not can you also explain why it is what it is.

EDIT:

If the energy of the transition is described as:

$E_{tot}=E_1 + E_2$

Where $E_1$ and $E_2$ are the energies of the single electron transmission. Then the uncertainty of $E_{tot}$ is:

${\Delta E_{tot} }^2 ={\frac{\partial E}{\partial E_1}}^2{\Delta E_{1} }^2 +{\frac{\partial E}{\partial E_2}}^2{\Delta E_{2} }^2 \Rightarrow \Delta E_{tot}= \sqrt {{\Delta E_{1} }^2 + {\Delta E_{2} }^2}$

Which then by using same derivation as above gives: $$\Delta \lambda_{1/2} \approx \frac{\lambda}{f} \cdot \frac{1}{2\pi} \cdot \sqrt{{(\frac{1}{\tau_1})}^2 + {(\frac{1}{\tau_2})}^2}$$

• Hint: There is no energy time uncertainty relation in the "canonical" sense of the uncertainty relation. Hence you cannot derive the second equation from the first one. – Gonenc Mogol Mar 1 '18 at 0:41
• How come? Why is it wrong to assume that the uncertainty is a sum of the individual uncertainties of the individual transitions of the single electron emissions. – DrShellyCooper Mar 1 '18 at 8:27
• Because uncertainty being something like the standard deviation is not additive. – Gonenc Mogol Mar 1 '18 at 10:16
• Okay, is the reasoning in the edit correct then? – DrShellyCooper Mar 2 '18 at 0:11
• No! Because the "uncertainty" in the Heisenberg uncertainty principle is more like a misnomer and has nothing to do with the uncertainty of measurement. – Gonenc Mogol Mar 2 '18 at 12:14