Physics Asked on August 28, 2021
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?
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".
Correct answer by Andrew on August 28, 2021
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