3
$\begingroup$

During a mock Cambridge interview, one of the questions was about how small a computer could theoretically be.The way I approached it was in terms of what something must be to be considered a computer.

I eventually made some progress, but I made a point that I am quite doubtful about. I said that one can use the photoelectric effect as a timer.

When you shine light on an atom and promote an electron to a higher energy level, there is a time delay before it returns to ground state (or at least there is a time delay between light being absorbed and light being re-radiated), which ought to be constant. I know this point may not really relate to the interview question.

Nevertheless, is what I said true? Is there a (measurable) time delay between the absorption of a photon and the re-radiation of a photon?

$\endgroup$
  • 1
    $\begingroup$ Let's put it this way, if you had said that in my interview I would have failed you. There is no such time delay. Baring more complicated scenarios the decay of a quantum state has a probability that is constant. The state can decay today or a million years from now and there is no way to make a reliable timer out of a single decay. You could take a large number of these decays and trigger a circuit when half of them are done, but if you gave me that as a design for a timing circuit in an engineering interview I would have failed you just as well. $\endgroup$ – CuriousOne Jun 27 '15 at 17:31
9
$\begingroup$

Yes excited states have a non-zero lifetime. Electronically excited states of atoms have lifetimes of a few nanoseconds, though the lifetime of other excited states can be as long as 10 million years.

The decay probability can be calculated using Fermi's golden rule. The lifetime is then an average lifetime derived from the decay probability.

The lifetime can be measured directly for long lifetimes, or for short lifetimes by measuring the broadening of the peak in the emission spectrum. If the lifetime is $\tau$ then the uncertainty principle tells us that the energy difference between the excited and ground states is uncertain by around:

$$ \Delta E \approx \frac{\hbar/2}{\tau} $$

This results in a broadening of the emission peak by a frequency of:

$$ \Delta \nu \approx \frac{\Delta E}{h} $$

This is known as lifetime broadening.

$\endgroup$
  • $\begingroup$ Is it worth mentioning the $\nu$^3 law due to einstein A coefficient and thus the difficulty of manufacturing UV + higher energy lasers due to short lifetimes... the few ns is ok for visible/near UV - which in the context of this question is probably ok, but will vary with frequency of course. $\endgroup$ – tom Sep 10 '15 at 10:26
  • 1
    $\begingroup$ It's important to note that there are also long-lifetime electronic excitations. This includes metastable states that can only decay by quadrupole radiation or higher (which are still electronic excitations but can have >1s lifetimes), and also Rydberg states which normally have very long lifetimes, (both at high angular momentum because you need a slow cascade, but even for low-$\ell$ states). $\endgroup$ – Emilio Pisanty Dec 8 '16 at 20:38
  • $\begingroup$ It's also worth linking to the NIST Atomic Spectra Database, which includes transition rates (the $A_{ki}$) for a wide range of electronic transitions. $\endgroup$ – Emilio Pisanty Dec 8 '16 at 20:39
0
$\begingroup$

The characteristic time of interaction - energy of interaction relation between two systems is usually written as $\delta E\cdot\delta t\sim\hbar/2$ (do NOT mix with the uncertainty principle). So the characteristic time would be about $\delta t\sim\hbar/(2\delta E)$, where for $\delta E$ we can take the difference of energies between two states.

$\endgroup$
  • 1
    $\begingroup$ This is incorrect. In this setting $\delta E$ is not the energy difference between the states; instead, it is the width of each line due to homogeneous broadening (i.e. the intrinsic uncertainty in the transition energy due to the fact that it takes a finite time). $\endgroup$ – Emilio Pisanty Dec 8 '16 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.