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I am reading Mukhanov and Winitzki's book Introduction to Quantum Effects in Gravity and in second paragraph of Sec. 1.4.2 they say that : The spontaneous emission by a hydrogen atom is the transition between the electron states $ 2p \rightarrow 1s$ with the production of a photon. This effect can be explained only by an interaction of electrons with vacuum fluctuations of the electromagnetic field. Without these fluctuations, the hydrogen atom would have remained forever in the stable $2p$ state.

But isn't the ground state and the most stable state of Hydrogen atom $1s$ ? If so, why should we even consider that the electron is in $2p$ state? The only reason for me to consider that would be the energy-time uncertainty relation which would imply that the electron cannot be in a single energy state otherwise the uncertainty relation would be violated. But then the authors say that without fluctuations the hydrogen atom would have remained forever in the stable $2p$ state which would violate uncertainty relation.

Am I missing something or the argument is flawed?

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Let's say the hydrogen atom started in the $2p$ state. Suppose we keep it in a vacuum so it doesn't interact with any other particles. Then the $2p$ state is an eigenstate of the Hamiltonian, and therefore the energy does not change and the only change in time of the state is to pick up a phase $e^{i \omega t}$, where $\hbar \omega=E_{2p}$, and $E_{2p}$ is the energy of the $2p$ state.

What Mukhanov and Winitzki are pointing out is that even if we put the hydrogen atom in a completely isolated vacuum with no other particles, we cannot remove the quantum vacuum. So we can't really think of the hydrogen atom as an isolated system, the system is really "hydrogen + electromagnetic field." Then state "no photons, hydrogen in $2p$ state" is not an energy eigenstate of the combined system. The system will evolve so that its state becomes "emitted photon + hydrogen in ground state".

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  • $\begingroup$ Can you please confirm that this needs to assume the hydrogen atom to start in the $2p$ state? Also, disregarding the fluctuations for the moment, is the uncertainty principle okay with hydrogen being forever in the $2p$ state? $\endgroup$ Jul 23, 2020 at 14:08
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    $\begingroup$ No, it's just a simpler to explain. We can start with the hydrogen in the ground state and an external field with a frequency tuned to the energy difference between the ground state and excited state. In general this will produce a superposition of the ground state and $2p$ state. If we are careful we can arrange that the external field is applied for exactly the right time that the amplitude for being in the ground state is zero and the hydrogen atom is in the $2p$ state. Then, ignoring the effect of the quantum vacuum, the hydrogen atom is in a stationary state and won't evolve. $\endgroup$
    – Andrew
    Jul 23, 2020 at 14:18
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    $\begingroup$ There's no contradiction with the uncertainty principle. In an energy eigenstate, the uncertainty in the energy is zero. This is compensated by the fact that the system remains in this state for an infinite time. $\endgroup$
    – Andrew
    Jul 23, 2020 at 14:19
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    $\begingroup$ There's a small subtlety in that for "cartoon hydrogen" there is a degeneracy between the different angular momentum states with $n=2$. But I don't think this really matters for your question, and we could deal with it if necessary. $\endgroup$
    – Andrew
    Jul 23, 2020 at 14:20

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