2
$\begingroup$

Naively I assumed that an electron would sit in an excited state (in the shell of an atom) having some stochastic lifetime, so nothing changes and suddenly the electron would jump back into the ground state (emitting a photon), and the process of this jump is instantaneous...

However, from what I learned in this thread Do electrons really perform instantaneous quantum leaps? and this Does the new finding on "reversing a quantum jump mid-flight" rule out any interpretations of QM?, this view is wrong.

My current understanding: Quantum mechanically, both the ground state $|g\rangle$ as well as the excited state $|e\rangle$ are eigenvectors of the Hamilton operator $H$ of the atom, meaning the corresponding eigenvalues are observables $E_g$ and $E_e$ (which can be measured or observed). As $|g\rangle$ and $|e\rangle$ are base vectors, we can write linear combinations $|\Psi\rangle$ (not corresponding to an obervable eigenvalue of $H$, i.e. if we measure, the state would collapse into an eigenvector). However, temporal evolution of quantum states is deterministic, and as soon as the electron is in the excited state $|e\rangle$, the transition or quantum jump starts and the electron's state can be described as a linear combination $|\Psi\rangle = a(t) |e\rangle + b(t) |g\rangle$, which develops continuously into $|g\rangle$ when $a(t$) changes with time from 1 to 0 while $b(t$) changes from 0 to 1. And during this, the electromagnetic wave (the photon) is emitted meaning the longer it takes (the more stable the excited level is) the longer the emitted wave train (more coherent light is emitted).

Question: Is that correct, meaning is the lifetime of an excited state the transition time (or quantum jump time)? If so, where is the randomness of the lifetime, i.e. what determines $a(t)$ and $b(t)$ in the equation above?

$\endgroup$
1
$\begingroup$

Lifetime is the time that characterizes the decay of an excited state to the ground state. E.g., the probability to find system in the excited state may be given by $$ P_e(t)=e^{-\frac{t}{\tau}}, $$ where $\tau$ is the lifetime.

Transition time generalizes this concept to any two states (i.e., not necessarily the ground state among them) and to any direction of the transition (i.e., not necessarily towards the ground state).

The quantum jump
Thes times say nothing (or vrey little) about the instant at which the transition (the quantum jump) actually happens, and about its duration.

See here, here, and here for more background.

$\endgroup$
4
  • $\begingroup$ So this means, quantum jump is a moment in time distinguishing between two real pysical states or eigenstates of H (electron is in upper state, and in the next moment in the lower state). The lifetime is just the timespan during which this event becomes more and more probable - and the probability is calculated with help of Psi (linear combination in my question above)? $\endgroup$ Aug 27 '21 at 12:22
  • 1
    $\begingroup$ @CharlesTucker3 If one wants to consider transitions, one cannot take only the Hamiltonian evolution of the two states, but need to consider the joint evolution of the atom and the field modes. (Fermi golden rule avoids explicitly doing it, but makes a lot of handwaving arguments in this diretcion.) I added also this link to my answer (I couldn't find it earlier): physics.stackexchange.com/a/623662/247642 $\endgroup$ Aug 27 '21 at 12:35
  • $\begingroup$ Quantum jump is the moment when the transition actually happens in a particular atom. Lifetime and transition time have probabilistics or ensemble meaning. $\endgroup$ Aug 27 '21 at 12:37
  • 1
    $\begingroup$ Many thanks, your answer in the last link (and there under "update") helps! It is especially interesting that the emitted photon is in a superposition of modes until measured (guess that can take some time e.g. when emitted by a star...;-)! $\endgroup$ Aug 27 '21 at 13:37

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.