Spectral lines and superposition of states

[Yuras]

TL;DR Summary

Any linear combination of atom's states is possible, so atom should be able to absorb any amount of energy. Though, as we all know, it doesn't. Why?

Let's say atom has two energy levels, $E_1$ and $E_2$. If atom is in the first state $|E_1\rangle$, then it's able to absorb a photon with energy $E_2-E_1$, while transitioning to the second state $|E_2\rangle$. In atom's spectrum we can see an absorption line at the corresponding frequency. So far so good.
But atom doesn't need to be in energy eigenstates, does it? So one can imaging atom transitioning from $|E_1\rangle$ to e.g. $\frac{1}{\sqrt 2}(|E_1\rangle + |E_2\rangle)$. Energy, required for the transition, is $\frac{1}{2}(E_2-E_1)$, why don't we see an absorption line at the corresponding frequency?

 

[malawi_glenn]

It is like saying that the daily average number of eggs in my fridge over the course of a year is 13.8 but how come I always have eggs corresponding to a positive integer.

That the atom is not in an energy eigenstate means that we do now know in which state it is.
By interacting with light, one energy eigenstate is selected.

 

[PeroK]

TL;DR Summary: Any linear combination of atom's states is possible, so atom should be able to absorb any amount of energy. Though, as we all know, it doesn't. Why?

Let's say atom has two energy levels, $E_1$ and $E_2$. If atom is in the first state $|E_1\rangle$, then it's able to absorb a photon with energy $E_2-E_1$, while transitioning to the second state $|E_2\rangle$. In atom's spectrum we can see an absorption line at the corresponding frequency. So far so good.
But atom doesn't need to be in energy eigenstates, does it? So one can imaging atom transitioning from $|E_1\rangle$ to e.g. $\frac{1}{\sqrt 2}(|E_1\rangle + |E_2\rangle)$. Energy, required for the transition, is $\frac{1}{2}(E_2-E_1)$, why don't we see an absorption line at the corresponding frequency?

In introductory QM the absorption and emission of light is taken as an initial and final measurement of the atom's energy, which forces both an an initial and final energy eigenstate. The possible emission and absorption wavelengths correspond to differences between energy eigenstates. Although, this cannot be the whole story. For example, there is no calculation of the expected time that an excited energy state survives.

In particular, the initial or final energy measurement cannot correspond to a superposition of energy eigenstates. You cannot measure the expectation value of a superposed energy eigenstates.

In more advanced treatments, you need to consider the full interaction with the quantized EM field. This is where the calculations of the expected lifetime of an excited state and the probabilities for each energy transition are understood. This is the theory of QED (Quantum Electrodynamics). And, in fact, the spectral lines themselves (if you look very closely) are not infinitely precise, but have a Lorentzian shape.

PS I think this is a good question.

 

[Yuras]

That the atom is not in an energy eigenstate means that we do now know in which state it is.

Hmm, indeed. Thanks!

By interacting with light, one energy eigenstate is selected.

Why is it energy eigenstate though? Why not e.g. momentum eigenstate or position eigenstate?

 

[malawi_glenn]

Why is it energy eigenstate though?

How would energy be conserved otherwise

 

[PeroK]

How would energy be conserved otherwise

I don't understand this question. If we have a superposition of energy eigenstates and we measure the energy, then we do not get the expected value. How can that conserve energy? The answer must be that the atom is entangled with another object in superposition of energy eigenststes. And, conservation of energy is only seen when we measure the energy of both entangled objects.

 

[LittleSchwinger]

Hmm, indeed. Thanks!

Why is it energy eigenstate though? Why not e.g. momentum eigenstate or position eigenstate?

Roughly speaking when a photon interacts with an atom the frequency of the photon becomes correlated with the energy levels of the atom. So if the atom was in a state:
$\ket{\psi} = \sum_{i} c_{i}\ket{E_{i}}$
then after the photon interacts with the atom, the state of the atom + photon is:
$\ket{\phi} = \sum_{i} c_{i}\ket{E_{i}}\otimes\ket{\nu_{i}}$
Where $\nu_{i}$ refers to some frequency. The photon frequency is connected to energy level due to the details of electron-photon interactions.

When we measure the photons frequency then, via this correlation, we learn the atom's energy level and thus have done an energy measurement on the atom.

I'm passing over a lot of details here and presenting a very very idealized case.

 

[Yuras]

Roughly speaking when a photon interacts with an atom the frequency of the photon becomes correlated with the energy levels of the atom. So if the atom was in a state:
$\ket{\psi} = \sum_{i} c_{i}\ket{E_{i}}$
then after the photon interacts with the atom, the state of the atom + photon is:
$\ket{\phi} = \sum_{i} c_{i}\ket{E_{i}}\otimes\ket{\nu_{i}}$
Where $\nu_{i}$ refers to some frequency. The photon frequency is connected to energy level due to the details of electron-photon interactions.

Thank you!

Let me rephrase it to make sure I understand it correctly. When photon interacts with the atom, we get $\ket{E_0}\otimes\ket{photon\space is\space not\space absorbed} + \ket{E_1}\otimes\ket{photon\space is\space absorbed}$. And energy eigenstates are selected later, when we pass light to a spectrometer, which measures whether photon was or was not absorbed. (I assume by "selected" @malawi_glenn meant that state collapsed as a result of a measurement.)

If the above is correct, then what happens when we don't use spectrometer? E.g. we pass photons through a gas and measure it's temperature. If temperature increases, then we deduce that photons were absorbed. Since temperature is related to internal energy, I guess we are measuring energy of atoms here. And I'm not able to come up with a setup where you don't actually measure energy, which is suspicious. I mean, is there a reason why whatever you do, you still measure energy?

 

[PeterDonis]

is there a reason why whatever you do, you still measure energy?

You don't have to measure energy; there are other observables you could measure. But none of them will tell you whether the photons were absorbed or not. Only an energy measurement will tell you that.

 

[Hyperfine]

... E.g. we pass photons through a gas and measure it's temperature. If temperature increases, then we deduce that photons were absorbed. Since temperature is related to internal energy, I guess we are measuring energy of atoms here. And I'm not able to come up with a setup where you don't actually measure energy, which is suspicious. I mean, is there a reason why whatever you do, you still measure energy?

Temperature is a macroscopic property of the ensemble. The energy eigenstates of the atom are microscopic properties of the individual atom.

As PeterDonis pointed out, you do not need to measure energy, but that is the way to probe the absorbtion or emission of radiation.

 

[Yuras]

Temperature is a macroscopic property of the ensemble. The energy eigenstates of the atom are microscopic properties of the individual atom.

That's true of course. But I don't quite see what's your point. I mean, any measurement involves some kind of macroscopic property, doesn't it? E.g. pointer angle, state of digital display, height of mercury etc.
Do you mean that change of temperature of the gas is not related to energy of individual atoms?

 

[Hyperfine]

...
Do you mean that change of temperature of the gas is not related to energy of individual atoms?

What property defines the temperature of an ensemble of atoms? How is that property related to the internal energy levels of a single distinct atom? How do you know that a change in temperature, even if observable, is the direct result of the photoexcitation event under question?

I think you have selected one example of the general "measurement problem" for discussion. I also think the responses in posts #3 and #7 are good.