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I know that the probability of the electron tunnelling out of the orbital is non-zero, but my question is strictly wrt the energy-time uncertainty at zero Kelvin. Is that probability non-zero as well?

Satyajit Sen
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  • Do you mean at zero temperature? Obviously at any finite temperature there is a probability of finding the electron in any energy level or even that they hydrogen is ionised. – ProfRob Nov 01 '23 at 13:25
  • @ProfRob, yes you are right. I had forgotten to add that detail in. Thanks for pointing it out, I have edited the question appropriately. – Satyajit Sen Nov 01 '23 at 16:41
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    What "energy-time uncertainty" do you mean? When we write $\Delta E \Delta t$ those $\Delta E$ and $\Delta t$ may not mean what you think they mean, see https://physics.stackexchange.com/q/53802/50583 – ACuriousMind Nov 01 '23 at 16:50

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The probability of finding the 1s orbital empty under those conditions is zero by any possible measurement. The "borrowing" of some amount of energy $\Delta E$ to reach an excited state can only occur with the subsequent return of that energy on a timescale $\Delta t$ with $\Delta E \Delta t \geq \hbar/2$. In other words, the probability of exciting the electron without decay by the same process is 0.

Otherwise, we may e.g. consider a process where an electron is excited by the uncertainty principle, we measure this state, and it generates a photon through radiative decay. We end up in the same state, but have gained a photon, clearly violating energy conservation. Consider similarly the tunneling example you mention yourself. What is the probability of the electron tunnelling out of potential well, up to a higher energy? 0. Tunneling can only occur through an intermittent virtual state with higher energy, if the overall system then returns to the same amount of energy it had before. The electron tunnels through a potential wall, not to the top of it.

The lifetime of such virtual states, as they occur e.g. in Raman scattering and non-linear optics, can be estimated using the uncertainty principle, which is it's primary application in atomic physics.

KarimAED
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what's the probability of finding the 1s orbital empty in an electrically neutral hydrogen atom at zero Kelvin?

We are working at fixed particle number (one electron) and fixed temperature (zero Kelvin), so the appropriate ensemble is the canonical ensemble.

In the canonical ensemble, the probability assigned to a given microstate is: $$ p_i = \frac{e^{-\beta E_i}}{Z}\;, $$ where $Z=\sum_i e^{-\beta E_i}$ and $\beta = \frac{1}{k_B T}$, where $T$ is temperature.

Since we are considering a single electron in the field of a classical proton (classical coulomb potential), the lowest energy state is the usual 1s hydrogen state. Therefore, as $T\to 0$, we have $\beta \to \infty$, and we have: $$ p_{\text{1s}} = 1 $$ $$ p_{\text{all other energies}} = 0 $$

hft
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