Energy needed for electron's change in energy level in an atom

[Nathi ORea]

TL;DR Summary

'An electron can jump to a higher energy level if it absorb's energy which "exactly" matches the difference between final and initial energy level' What is mean't be "exactly"?

Textbooks always give an explanation for why electrons move between energy levels in atoms with an explanation something like this:

'An electron can jump to a higher energy level if it absorb's energy which exactly matches the difference between final and initial energy level'.

My question, whenever I see this, is what does 'exactly' mean? Like how exact does the photon have to be to make the jump. I would think even the slightest under or overshoot would mean its wave function would not be stable at that level.

Also I would think that it would be impossible to ever get a photon at that 'exact' energy because you would be expecting a photon at basically an infinitely small frequency band. There must be some wiggle room?

Thanks in advance

 

[Drakkith]

There must be some wiggle room?

I believe the uncertainty principle causes the energy levels to have some small amount of uncertainty in them. This would manifest as a slight 'fuzziness' in the spectral lines in even an otherwise perfect spectrographic device.

 

[Nathi ORea]

Thanks Drakkith. That makes a bit of sense.

I appreciate the reply.

 

[PeroK]

Summary:: 'An electron can jump to a higher energy level if it absorb's energy which "exactly" matches the difference between final and initial energy level' What is mean't be "exactly"?

Textbooks always give an explanation for why electrons move between energy levels in atoms with an explanation something like this:

'An electron can jump to a higher energy level if it absorb's energy which exactly matches the difference between final and initial energy level'.

My question, whenever I see this, is what does 'exactly' mean? Like how exact does the photon have to be to make the jump. I would think even the slightest under or overshoot would mean its wave function would not be stable at that level.

Also I would think that it would be impossible to ever get a photon at that 'exact' energy because you would be expecting a photon at basically an infinitely small frequency band. There must be some wiggle room?

Thanks in advance

Good question.

The absorption of a photon by an atom is - like everything in QM - governed by statistical rules. The closer a photon's energy is to the difference between energy levels the more likely it is to be absorbed. The distribution of probabilities is very tightly centred around the mean and drops off very quickly as the photon's energy varies from the difference in the atomic energy levels.

This leads to an example of the energy-time uncertainty principle: $\Delta E \Delta t \ge \frac \hbar 2$. In this case $\Delta E$ is the difference between the absorbed energy and the energy level and $\Delta t$ is the expected time that the atom remains in the excited state. If an atom absorbs a photon of very close to the required energy, then it may remain in the excited state for a relatively long time before emitting a photon and dropping to a lower energy state. And, if the atom absorbs a photon further from the required energy, then it will remain in the exited state for a relatively short time.

In other words, the stabiity of the excited state depends on how close the energy is to the precise energy level. In theory, if an atom absorbed a photon of precisely the required it would be stable (if $\Delta E \approx 0$ then $\Delta t \rightarrow \infty$).

 

[Nathi ORea]

Gee Perok. That is so well explained. Even a serious physics novice like me can understand.

Can I ask, when the electron absorbs a photon at a certain distance from the required energy level, does it go to a slightly different orbital level? Like... I have wondered too why absorption/ emission spectra have width to the lines. Perhaps this is just an illusion.

 

[PeterDonis]

the stabiity of the excited state depends on how close the energy is to the precise energy level. In theory, if an atom absorbed a photon of precisely the required it would be stable (if $\Delta E \approx 0$ then $\Delta t \rightarrow \infty$).

I don't think this is correct.

The expected lifetime of a given excited state is the same regardless of how that state is prepared; the quantum states that we call the energy levels of an atom are determined by the properties of the joint Hamiltonian of the atom and the electromagnetic field, including the interaction between the two.

It is true that the finite expected lifetime of an excited state corresponds to an uncertainty in its energy, and therefore causes line broadening. But this is not due to any difference in the state depending on how it was prepared (including any properties of the photon that was originally absorbed to prepare the state). It is simply due to the fact that the emission of a photon from the excited state is a process subject to quantum uncertainty.

 

[PeterDonis]

what does 'exactly' mean?

It means that the textbook that used that word was being sloppy. (Btw, you haven't referenced any specific textbooks; can you actually find one that uses the word "exactly" in this context?)

There must be some wiggle room?

Yes, there is, in the sense that all spectral lines have finite width. The reasons why are complicated; the Wikipedia article on spectral lines has a section on line broadening and shift that will at least give some idea of the complexities:

https://en.wikipedia.org/wiki/Spectral_line#Line_broadening_and_shift

It is true that energy-time uncertainty is one of the factors involved (it's the reason for what the article calls "natural broadening"), but it is by no means the only one, and in general its effect is likely to be small compared to the effects of other factors.

when the electron absorbs a photon at a certain distance from the required energy level, does it go to a slightly different orbital level?

No. See my response to @PeroK in post #6.

 

[PeroK]

In other words, the stabiity of the excited state depends on how close the energy is to the precise energy level. In theory, if an atom absorbed a photon of precisely the required it would be stable (if $\Delta E \approx 0$ then $\Delta t \rightarrow \infty$).

I don't think this is correct.

You're right, it's really the other way round: as $\Delta t \rightarrow \infty$, so $\Delta E \rightarrow 0$. And, yes, the excited state is unstable.

The wikipedia page simply quotes the energy-time uncertainty, but that goes no way to explaining how an atom can absorb a photon of anything but the precise energy in the first place. Ultimately, the explanation must use QED or QFT. The best I could find is this:

http://www-star.st-and.ac.uk/~kw25/teaching/nebulae/lecture08_linewidths.pdf

But, it still doesn't explain the width of the absorption lines. As you say, you need the EM field for that.

 

[andrew s 1905]

As I understand it the life time of excited states ( and hence line width) depends also on conservation laws which require different interactions with the EM field to occur. Dipole v quadrupole etc.

Regards Andrew

 

[vanhees71]

The "natural linewidth" of the of the

You're right, it's really the other way round: as $\Delta t \rightarrow \infty$, so $\Delta E \rightarrow 0$. And, yes, the excited state is unstable.

The wikipedia page simply quotes the energy-time uncertainty, but that goes no way to explaining how an atom can absorb a photon of anything but the precise energy in the first place. Ultimately, the explanation must use QED or QFT. The best I could find is this:

http://www-star.st-and.ac.uk/~kw25/teaching/nebulae/lecture08_linewidths.pdf

But, it still doesn't explain the width of the absorption lines. As you say, you need the EM field for that.

You are right. The semiclassical theory, where the em. field is treated classically, cannot explain the finite lifetime of the excited states of the atom, because in this approximation there is no spontaneous emission, i.e., in this approximation the energy eigenstates of the atom are exact energy eigenstates and thus have infinite lifetime and zero energy uncertainty.

If you now quantize the em. field, then you always have and uncertainty in the electromagnetic field, i.e., in the em. vacuum the expectation values of the em. field vanish but not their standard deviations. Now if you have an atom in an excited state of the semiclassical approximation you have to consider the additional terms in the Hamiltonian when you include the em. field as a quantum dynamical degree of freedom (in this way working within full QED). Taking into account these additional terms lead among other things to spontaneous emission, i.e., none of the excited states are true energy eigenstates anymore and thus are in fact instable, due to the fluctuations of the em. field. Thus all the energy levels of the semiclassical approximation are unstable in complete QED and thus these energy values have a finite uncertainty resulting in the "natural line width" of the spectral lines of the atom.

The fact that there is spontaneous emission is the most obvious phenomenon making it necessary from an empirical point of view to also quantize the em. field and not only the "particles" (here the electrons). It's indeed a bit misleading to quote the photoelectric effect and the Compton effect to motivate the "photon picture", because the leading-order QED results (tree-level Feynman diagrams) give indeed the same results when solving the Dirac equation of the electron in semiclassical approximation with the em. wave field treated classically.

 

[PeterDonis]

You're right, it's really the other way round: as $\Delta t \rightarrow \infty$, so $\Delta E \rightarrow 0$.

As far as the energy-time uncertainty is concerned, it's neither "way round": $\Delta t$ and $\Delta E$ are combined in an uncertainty relation, but neither one causes the other.

However, my issue was not the relationship between $\Delta t$ and $\Delta E$, but your implication that $\Delta E$ refers to how close the energy of the photon that was absorbed to create the excited state is to the "exact" energy of the state. That's not correct. $\Delta E$ is an inherent property of the excited state itself, and so is $\Delta t$ (the expected lifetime of the state). You can't vary those things by varying the preparation procedure. That was my point.

(To be clear, by $\Delta E$ here I am referring to the "natural" line broadening, in the terminology of the Wikipedia article I linked to. The expected lifetime $\Delta t$ and the associated uncertainty in energy $\Delta E$ for this are inherent properties of the state in the same way that the half-life is an inherent property of a radioactive atom.)

 

[Nathi ORea]

It means that the textbook that used that word was being sloppy. (Btw, you haven't referenced any specific textbooks; can you actually find one that uses the word "exactly" in this context?)

This is the passage from Australian Curriculum Pearson Physics 12, NSW ed (high school textbook). This is what made me finally ask this question, but I was just paraphrasing the sort of explanation I feel like I usually see.

Please keep in mind that I would never read anything with heavy maths.

 

[Nathi ORea]

Thank you so much for the answers guys, and I really appreciate the links (PeterDonis and Perok). I'll keep these on hand for future reference.

Some of it is going over my head atm, but I'll have another go reading it later.

Obviously you need a good mathematical knowledge to understand this stuff, and it is something I'd like to teach myself sometime in the future, but certainly not near there yet.

Appreciate it all

Nathan.

 

[Twigg]

To give you a little sense of scale for what "exactly" might mean, for a rubidium atom transitioning from the 5s orbital to 5p, a photon with a frequency of roughly 380 THz will have a "wiggle room" of about 6 MHz to excite the atom. This doesn't mean that a photon 7MHz away can't excite the transition. Literally, it means that photons that're 6MHz away from the transition will be absorbed at 1/2 the rate that photons that're exactly on the transition will be absorbed. This "wiggle room" will vary wildly from transition to transition, element to element, or molecule to molecule. It's just to give you an idea for a low-lying transition in a hydrogen-like atom like what Bohr would've had in mind.

This is for the idealest of ideal cases, in reality you won't find such a low uncertainty in a room temperature rubidium vapor, as mentioned by others here.

 

[PeterDonis]

This is the passage from Australian Curriculum Pearson Physics 12, NSW ed (high school textbook)

This passage is describing "Bohr's main ideas". That's not necessarily the same as our best current understanding of QM today. In Bohr's "old quantum theory" model there was no broadening of spectral lines.

 

[vanhees71]

As far as the energy-time uncertainty is concerned, it's neither "way round": $\Delta t$ and $\Delta E$ are combined in an uncertainty relation, but neither one causes the other.

However, my issue was not the relationship between $\Delta t$ and $\Delta E$, but your implication that $\Delta E$ refers to how close the energy of the photon that was absorbed to create the excited state is to the "exact" energy of the state. That's not correct. $\Delta E$ is an inherent property of the excited state itself, and so is $\Delta t$ (the expected lifetime of the state). You can't vary those things by varying the preparation procedure. That was my point.

(To be clear, by $\Delta E$ here I am referring to the "natural" line broadening, in the terminology of the Wikipedia article I linked to. The expected lifetime $\Delta t$ and the associated uncertainty in energy $\Delta E$ for this are inherent properties of the state in the same way that the half-life is an inherent property of a radioactive atom.)

The energy-time uncertainty relation is more subtle, because time is not an observable but a parameter. In the context here, i.e., the lifetime of an unstable state (resonance) and the energy uncertainty it turns out that it means that the width of the resonance is inversely proportional to the lifetime.

This is easily seen from the fact that for a narrow resonance you can approximate the transition-matrix element by
$$f(E)=-\frac{\Gamma/2}{(E-E_0)+\mathrm{i} \Gamma}/2.$$
This translates in the time domain to
$$\psi(t)=\int_{\mathbb{R}} \mathrm{d} E \exp(-\mathrm{i} E t) f(E) \propto \exp(-\Gamma t/2 -\mathrm{i} E_0 t).$$
Thus $|\psi(t)|^2\propto \exp(-\Gamma t)$. So the (mean) lifetime is $\tau=1/\Gamma$, i.e., $\tau \Gamma=1$. It is in some sense mathematically related to the uncertainty relation for observables as it has to do with the fact that a distribution narrow in $E$ is broad in $t$ and vice versa.

Another point of view of the energy-time uncertainty relation is to define the "time uncertainty" as the time over which the expectation value of an observable changes by the amount of its standard deviation. This is most simply derived in the Heisenberg picture of time evolution, where the operators of observables carry the entire time evolution and the states are constant with time. So we consider an arbitrary observable $A$ with its self-adjoint operator $\hat{A}(t)$ in the Heisenberg picture. Then the operator that represents its time derivative is just
$$\dot{\hat{A}}=\frac{1}{\mathrm{i}} [\hat{A},\hat{H}].$$
Now the usual Heisenberg-Robertson uncertainty relation tells us that
$$\Delta E \Delta A \geq \frac{1}{2} \left |\frac{1}{\mathrm{i}} \langle [\hat{A},\hat{H}] \right|=\frac{1}{2} |\langle \dot{A} \rangle|.$$
Dividing by $|\langle \dot{A} \rangle|$ you get
$$\Delta E \Delta t \geq 1/2,$$
where
$$\Delta t=\frac{\Delta A}{|\langle \dot{A} \rangle|},$$
which is an estimate for the time the system needs that the observable changes by an amount larger than the uncertainty of the observable.

For a detailed discussion, see Messiah, Quantum Mechanics.