I Electron energy in atoms

[hokhani]

TL;DR Summary

Why electrons in atoms are in the atomic levels?

I don't know why the electrons in atoms are considered in the orbitals while they could be in sates which are superpositions of these orbitals? If electrons are in the superposition of these orbitals their energy expectation value is also constant, and the atom seems to be stable!

 

[pines-demon]

If electrons are in the superposition of these orbitals their energy expectation value is also constant

This is not true, unless I am reading it wrongly.

 

[hokhani]

This is not true, unless I am reading it wrongly.

If we have $\psi (0)= \sum_m c_m \psi_m$ at $t=0$, then $\psi(t)=\sum_m c_m e^{\frac{-iE_m t }{\hbar}} \psi_m$ so $\langle H \rangle=\sum_m |c_m|^2 E_m$ is constant.

 

[WernerQH]

A vector $\ket{\Psi}$ in Hilbert space, multiplied by any complex number(*), represents exactly the same state. That's why it is possible to call $e^{iEt} \ket{\Psi}$ "stationary" even though the phase factor may vary rapidly. Then all expectation values are time-independent. But this is no longer the case if you add two vectors with different energies. Then the expectation value of the position, say, can vary sinusoidally with the beat frequency $(E_2 - E_1)/h$. This is actually how Schrödinger wanted to explain how such an atom radiates.

(*) except zero, of course

 

[pines-demon]

If we have $\psi (0)= \sum_m c_m \psi_m$ at $t=0$, then $\psi(t)=\sum_m c_m e^{\frac{-iE_m t }{\hbar}} \psi_m$ so $\langle H \rangle=\sum_m |c_m|^2 E_m$ is constant.

Oh so you want a dynamic state (changing in time) but with stationary expectation value? But then the issue is that you never measure that energy, only statistically. I guess you could make it work but then everything seems more complicated, specially when studying spectra.

 

[JimWhoKnew]

TL;DR Summary: Why electrons in atoms are in the atomic levels?

I don't know why the electrons in atoms are considered in the orbitals while they could be in sates which are superpositions of these orbitals? If electrons are in the superposition of these orbitals their energy expectation value is also constant, and the atom seems to be stable!

I'm not sure I understand your question, but let me try nevertheless...

In the context of introductory QM textbooks, every state is a superposition of the energy eigenstates.

If $H$ is time independent, then in any state $$\frac{d}{dt}\langle H\rangle= \langle \left[H,H\right]\rangle+\langle \frac{\partial H}{\partial t}\rangle=0 \quad,$$in accord with your calculation in #3.

So if that was the whole story, the excited levels of the Hydrogen atom would have been stable too, as you've noted (no transitions). But in reality, there is more to it. Especially the interaction with radiation, which is usually totally omitted (for a reason) from the first discussion of the Hydrogen atom. When this interaction is included, along with other higher order corrections, ground states are more stable than the excited ones.

 

[PeroK]

TL;DR Summary: Why electrons in atoms are in the atomic levels?

I don't know why the electrons in atoms are considered in the orbitals while they could be in sates which are superpositions of these orbitals? If electrons are in the superposition of these orbitals their energy expectation value is also constant, and the atom seems to be stable!

Only the ground state is stable. An atom in an excited state will emit one or more photons and reduce to the ground state. The same would be true of an superposition of excited states.

The detection of a photon would be considered a measurement of the atom's energy and resolve the superposition into one of the excited states. The measurement of energy of excited state has the added consequence of reducing the atom to its ground state.

That said, the details of why an atom reduces to its ground state go beyond basic QM. You need the quantization of the EM field.

 

[hokhani]

I'm not sure I understand your question, but let me try nevertheless...

In the context of introductory QM textbooks, every state is a superposition of the energy eigenstates.

If $H$ is time independent, then in any state $$\frac{d}{dt}\langle H\rangle= \langle \left[H,H\right]\rangle+\langle \frac{\partial H}{\partial t}\rangle=0 \quad,$$in accord with your calculation in #3.

So if that was the whole story, the excited levels of the Hydrogen atom would have been stable too, as you've noted (no transitions). But in reality, there is more to it. Especially the interaction with radiation, which is usually totally omitted (for a reason) from the first discussion of the Hydrogen atom. When this interaction is included, along with other higher order corrections, ground states are more stable than the excited ones.

Thanks, the goal of raising this question was to know, step by step, how quantization manifests itself in practice while the system is not necessarily in an eigenstate and can be in a superposition of eigenstates. Also, the statistical measurement of an observable gives the expectation value corresponding to the superposition.
Could you (or others) please clarify?

 

[PeroK]

I wou
Thanks, the goal of raising this question was to know, step by step, how quantization manifests itself in practice while the system is not necessarily in an eigenstate and can be in a superposition of eigenstates. Also, the statistical measurement of an observable gives the expectation value corresponding to the superposition.
Could you (or others) please clarify?

The expected value is the average. You always get an energy corresponding to the transition between two eigenstates. The measurement of the energy of these eigenstates is indirect. It's inferred from the spectrum.

That's a common theme for many atomic phenomena. The theory is corroborated somewhat indirectly. As opposed to the phases of the Moon, which are directly observed.

PS or the period of a simple pendulum.