Orbitals and selection rules in a synchrotron

[snorkack]

What orbitals are occupied in a synchrotron, and which selection rules limit charged particles in a synchrotron?
A Schrödinger hydrogen atom is not actually the Bohr atom, but the Bohr theory gives correct energies (but not transition probabilities). Why?
Because in a hydrogen-like atom, 2s (n=2, l=0) and 2p (n=2, l=1) orbitals are degenerate with each other. Therefore they are also degenerate with an orbit where n=2, l=2, which is the Bohr orbit which is impossible and unphysical because of the wave nature of electron (the electron must wave to the sides of the orbit as well, not only along the orbit as on a Bohr orbit).

The first 3 l values are designated s, p, d. Then on come orbitals which have letters of alphabet starting from f - 1s, 2p, 3d, 4f, 5g, 6h, 7i, 8j, 9k, 10l, 11m, 12n, 13o... but then? p is taken. What is the designation of orbital with n=14, l=13?

Note that although l=n is unphysical and forbidden by wave nature of electron, l=n-1 is legal... and for large n and l, it should look much like a classical circular orbit.
Hydrogen-like atoms are subject to a number of selection rules. One simple one is Δl=+-1.
Which means that an atom with electron in 13o has only one legal emission. Since it can only have transition to l values 11 or 13, and there is no allowed orbital with n<13, l=13, the only option is n=12, l=11.
Even in a near-circular orbit that does not precisely have l=n-1, the options are limited. For an electron at 13n, transitions can still be only to m orbitals, and only 2 are available (11m and 12m). Still a line spectrum, not continuum.

Electron in the electrostatic field of a distant proton follows an orbit (circle or ellipse) and, given free lower states (many if the proton is distant), emits to go to such where selection rules allow (which is NOT always many!)

An electron in the magnetostatic field of a synchrotron also follows an orbit (circle or helix), and given free lower states (many because the synchrotron is macroscopic) emits to go where selection rules allow.

But what ARE selection rules for an electron in a synchrotron?

 

[EigenState137]

Electron in the electrostatic field of a distant proton follows an orbit (circle or ellipse) and, given free lower states (many if the proton is distant), emits to go to such where selection rules allow (which is NOT always many!)

An electron in the magnetostatic field of a synchrotron also follows an orbit (circle or helix), and given free lower states (many because the synchrotron is macroscopic) emits to go where selection rules allow.

But what ARE selection rules for an electron in a synchrotron?

Greetings,

The orbits are meaningful only in the case of bound states. Free-free transitions (bremsstrahlung) involve Coulombic interactions between free electrons and heavier charged particles that serve to decelerate the free electron with concomitant emission of a photon. The emission spectrum is a continuum. There are no selection rules based on the structure of an atom because no bound states are involved..

Synchroton radiation is generated when freely moving, relativistic charged particles are accelerated in a magnetic field. The emission spectrum is again continuous ranging from the X-ray to radiofrequency domains. Again, there are no selection rules based on some assumed atomic structure because no bound states are involved..ES

 

[snorkack]

The orbits are meaningful only in the case of bound states.

Synchroton radiation is generated when freely moving, relativistic charged particles are accelerated in a magnetic field. The emission spectrum is again continuous ranging from the X-ray to radiofrequency domains. Again, there are no selection rules based on some assumed atomic structure because no bound states are involved..

But an electron on a closed, circular orbit in the uniform magnetic field of a cyclotron is bound to the cyclotron. Therefore it is in a bound state?

 

[WernerQH]

Yes, you can view it as a bound state, but the quantum numbers are enormous. The letters of the alphabet won't suffice for the l-values. :-)

Selection rules are related to the dipole approximation, which is applicable in the context of atomic transitions. In a typical optical transition the size of the orbitals is much smaller than the wavelength of the emitted radiation, and one simplifies the calculation by taking only the matrix elements of the dipole moment between the two electronic states. But this is an approximation. If the matrix element of the dipole moment vanishes (for example if the initial and final states have the same parity), then the line is called "forbidden", but this does not mean that a transition is impossible. It could proceed via quadrupole emission, with a much reduced intensity (by a factor $(a / \lambda)^2$).

In a cyclotron, and even more so in a synchrotron, this factor is no longer small. The dipole approximation and selection rules cease to be useful. An electron can jump over many Landau levels and emit higher harmonics of the cyclotron frequency.

 

[EigenState137]

But an electron on a closed, circular orbit in the uniform magnetic field of a cyclotron is bound to the cyclotron. Therefore it is in a bound state?

Greetings,

From the lengthy introduction to your original post I assumed you were talking about bound atomic states characterized by good quantum numbers.

As @WernerQH mentioned, Landau quantization applies for uniform magnetic fields. However, recall that synchroton emission is observed in astrophysical objects in which the uniformity of the magnetic fields is not obvious.ES

 

[snorkack]

Selection rules are related to the dipole approximation, which is applicable in the context of atomic transitions. In a typical optical transition the size of the orbitals is much smaller than the wavelength of the emitted radiation, and one simplifies the calculation by taking only the matrix elements of the dipole moment between the two electronic states. But this is an approximation. If the matrix element of the dipole moment vanishes (for example if the initial and final states have the same parity), then the line is called "forbidden", but this does not mean that a transition is impossible. It could proceed via quadrupole emission, with a much reduced intensity (by a factor $(a / \lambda)^2$).

Does quadrupole radiation allow all transitions, or only those for Δl=2?

In a cyclotron, and even more so in a synchrotron, this factor is no longer small. The dipole approximation and selection rules cease to be useful. An electron can jump over many Landau levels and emit higher harmonics of the cyclotron frequency.

Is the emission of harmonics of cyclotron radiation caused by size of cyclotron, or by imperfections in the uniformity of cyclotron magnetic field?

 

[Twigg]

Does quadrupole radiation allow all transitions, or only those for Δl=2?

As a rule of thumb, dipole transition means 1 photon absorbed, quadrupole means 2 photons absorbed, and so on. Each photon has angular momentum of 1 (in natural units where $\hbar=1$). So you can't have a quadrupole transition with $\Delta l = 3$. There's a summary going up to octapole on wikipedia. If you want to go past that, I suggest you learn the identities that govern integrals over a product 3 Wigner matrices. You could also directly get the results with integrals of 3 spherical harmonics, it's just a pain in the butt. Well, both ways are a hassle. I have a feeling that @EigenState137 is probably fluent with that math and thinks I'm a wimp

Edit: crossed out false claims. Whoops!

 

[EigenState137]

As a rule of thumb, dipole transition means 1 photon absorbed, quadrupole means 2 photons absorbed, and so on. Each photon has angular momentum of 1 (in natural units where $\hbar=1$). So you can't have a quadrupole transition with $\Delta l = 3$. There's a summary going up to octapole on wikipedia. If you want to go past that, I suggest you learn the identities that govern integrals over a product 3 Wigner matrices. You could also directly get the results with integrals of 3 spherical harmonics, it's just a pain in the butt. Well, both ways are a hassle. I have a feeling that @EigenState137 is probably fluent with that math and thinks I'm a wimp

Greetings,

One need be careful to distinguish the characterization of the incident radiation field in terms of a multipole expansion and a physical excitation that involves more than one photon.

The selection rules described in the Wikipedia page mentioned above are all for one-photon processes, and the designation as electric quadrupole (E2) or magnetic dipole (M1) refer to the multipole expansion of the radiation field.

In the case of multiphoton excitation, one simply adds the allowed changes in angular momentum over the number of photons involved. For example, for two-photon excitation $\Delta J= 0,\pm 1,\pm 2$.

Wimp?! I never said that, and do not think it.ES

 

[EigenState137]

Is the emission of harmonics of cyclotron radiation caused by size of cyclotron, or by imperfections in the uniformity of cyclotron magnetic field?

Greetings,

The harmonics, and line broadening, are indeed associated with inhomogeneities in the magnetic field. See for example Cyclotron radiationES

 

[WernerQH]

Is the emission of harmonics of cyclotron radiation caused by size of cyclotron, or by imperfections in the uniformity of cyclotron magnetic field?

It has nothing to do with the size of the cyclotron or synchrotron. And it occurs also in a perfectly homogeneous magnetic field. But for low energy electrons the gyroradius is small compared to the wavelength $\frac {mc} {eB}$ corresponding to the gyrofrequency. Then the system is very much like a rotating dipole, radiating practically only at the fundamental frequency. In the classical picture the radiation field varies sinusoidally.

But this is vastly different for relativistic electrons. Relativistic electrons radiate strongly in the forward direction, and the radiation field will have strong spikes whenever the "beam" sweeps over the "observer". If you Fourier transform that you will see a frequency comb at multiples of the gyrofrequency. The quantum calculations confirm this because of the correspondence principle. Of course, in astrophysical plasmas you will never observe these lines, because the longitudinal motion of relativistic electrons produces huge Doppler shifts, and the Doppler-broadened lines overlap to a broad continuum. (And the magnetic fields are rarely homogeneous.)

 

[snorkack]

Yes, you can view it as a bound state, but the quantum numbers are enormous. The letters of the alphabet won't suffice for the l-values. :-)

Selection rules are related to the dipole approximation, which is applicable in the context of atomic transitions. In a typical optical transition the size of the orbitals is much smaller than the wavelength of the emitted radiation, and one simplifies the calculation by taking only the matrix elements of the dipole moment between the two electronic states. But this is an approximation. If the matrix element of the dipole moment vanishes (for example if the initial and final states have the same parity), then the line is called "forbidden", but this does not mean that a transition is impossible. It could proceed via quadrupole emission, with a much reduced intensity (by a factor $(a / \lambda)^2$).

When I think of it, it seems to me that the size of orbitals is much smaller than the wavelength of the emitted radiation if, and because, the speed of electron is much smaller than the speed of light.
In a H atom, an electron on 3d orbital can undergo a dipole allowed transition to 2p (Balmer α) followed by dipole allowed transition to 1s (Lyman α). Or it could undergo a quadrupole forbidden transition to 1s (Lyman β).
In a He⁺ ion, the electron on a 3d orbital will have the orbital 2 times smaller than in H, but the wavelength of the transition will be 4 times shorter. Therefore the size of the orbital will be a bigger fraction of wavelength, because the speed of the electron is bigger and a bigger fraction of speed of light. For a He⁺ ion, Lyman β (or whatever it is called in He⁺) would still be forbidden, but the branching ratio between allowed Balmer α and forbidden Lyman β would be more favourable to Lyman β. And both would be in hard ultraviolet.
For U⁹¹⁺ ion in 3d orbital, the speed of electron in 1s would be an appreciable fraction of speed of light. So the ratio of quadrupole transition (Kβ) to dipole (Lα) would be more favourable.
Likewise, for a 4f orbital, the allowed dipole transition is to 3d (Paschen α/Mα), but the forbidden transitions are quadrupole to 2p (Balmer β/Lβ) and octupole to 1s (Lyman γ/Kγ). I suspect that for H, the octupole line would be much weaker again than the already weak quadrupole one?
Correct?

 

[EigenState137]

When I think of it, it seems to me that the size of orbitals is much smaller than the wavelength of the emitted radiation if, and because, the speed of electron is much smaller than the speed of light.
In a H atom, an electron on 3d orbital can undergo a dipole allowed transition to 2p (Balmer α) followed by dipole allowed transition to 1s (Lyman α). Or it could undergo a quadrupole forbidden transition to 1s (Lyman β).
In a He⁺ ion, the electron on a 3d orbital will have the orbital 2 times smaller than in H, but the wavelength of the transition will be 4 times shorter. Therefore the size of the orbital will be a bigger fraction of wavelength, because the speed of the electron is bigger and a bigger fraction of speed of light. For a He⁺ ion, Lyman β (or whatever it is called in He⁺) would still be forbidden, but the branching ratio between allowed Balmer α and forbidden Lyman β would be more favourable to Lyman β. And both would be in hard ultraviolet.
For U⁹¹⁺ ion in 3d orbital, the speed of electron in 1s would be an appreciable fraction of speed of light. So the ratio of quadrupole transition (Kβ) to dipole (Lα) would be more favourable.
Likewise, for a 4f orbital, the allowed dipole transition is to 3d (Paschen α/Mα), but the forbidden transitions are quadrupole to 2p (Balmer β/Lβ) and octupole to 1s (Lyman γ/Kγ). I suspect that for H, the octupole line would be much weaker again than the already weak quadrupole one?
Correct?

Greetings,

Not correct. Atomic states cannot be described in terms of point charges in orbits about the nucleus. There is no physically meaningful concept of the "speed" of the electron within some semi-classical orbit.

The physical phenomenon of radiative decay of atomic excited states is not the same as that of synchroton radiation. Nor are the angular momentum selection rules associated with transitions between atomic states associated with synchroton radiation.

It also appears that you are formulating atomic transitions within some kind of quasi-Borh framework in which the only property of significance is the radial size of the orbital. That would effectively be the principal quantum number, $n$. Transitions between the electronic states of atoms and molecules are governed by changes in angular momentum, not by principal quantum number. There is no selection rule for $\Delta n$.

You assert that

In a H atom, an electron on 3d orbital can undergo a dipole allowed transition to 2p (Balmer α) followed by dipole allowed transition to 1s (Lyman α). Or it could undergo a quadrupole forbidden transition to 1s (Lyman β).

What quantitative prediction does your model provide for the relative probabilities of those electric dipole and electric quadrupole transitions? Compare that prediction to the Einstein $A$ coefficients given in the NIST Atomic Spectra Database.ES

 

[snorkack]

The physical phenomenon of radiative decay of atomic excited states is not the same as that of synchroton radiation. Nor are the angular momentum selection rules associated with transitions between atomic states associated with synchroton radiation.

It also appears that you are formulating atomic transitions within some kind of quasi-Borh framework in which the only property of significance is the radial size of the orbital. That would effectively be the principal quantum number, $n$. Transitions between the electronic states of atoms and molecules are governed by changes in angular momentum, not by principal quantum number. There is no selection rule for $\Delta n$.

Oh, there is. The selection rule for transitions is Δn<=-1.
I specifically mentioned that the pure Bohrian orbit (l=n) is unphysical, but that I concentrated on the orbitals where l=n-1 - and look towards the limit where both n and l are big.
Look at the starting case of 3d orbital - n=3, l=2.
Because of the selection rule Δn<=-1, n'<=2. Note that n<=0 is also forbidden because unphysical, so n' can only be 2 or 1.
Because of the selection rule for dipole emission, Δl=+-1, l' might be 1 or 3. But because n'<=2, and l<=n-1, the orbital n'=2, l'=3 is unphysical, leaving only l'=1. And n'=1, l'=1 is also unphysical. Which means that for 3d, the only allowed dipole transition is to 2p (n'=2, l'=1)
Agreed so far?

 

[EigenState137]

Oh, there is. The selection rule for transitions is Δn<=-1.
I specifically mentioned that the pure Bohrian orbit (l=n) is unphysical, but that I concentrated on the orbitals where l=n-1 - and look towards the limit where both n and l are big.
Look at the starting case of 3d orbital - n=3, l=2.
Because of the selection rule Δn<=-1, n'<=2. Note that n<=0 is also forbidden because unphysical, so n' can only be 2 or 1.
Because of the selection rule for dipole emission, Δl=+-1, l' might be 1 or 3. But because n'<=2, and l<=n-1, the orbital n'=2, l'=3 is unphysical, leaving only l'=1. And n'=1, l'=1 is also unphysical. Which means that for 3d, the only allowed dipole transition is to 2p (n'=2, l'=1)
Agreed so far?

Greetings,

No, not agreed at all. To start with, selection rules are the same for both photon absorption and photon emission. That by itself negates your $\Delta n$ claim. It is also worth mentioning that the rigorous selection rule is for $\Delta J$ not $\Delta l$.

If you want to consider high Rydberg states, that is high $n$, then you have $\Delta n$ as a propensity rule, not a selection rule. This is clearly consistent with observed radiative recombination emission lines in the radio frequency domain from nebular environments with values of $n\approx 100$ in the case of hydrogen.

Ultimately the fundamental points are:
1. The Borh model is a zero-order approximation of atomic structure.
2. Your assertions are not consistent with the extensive, available body of objective observational and experimental data.ES

 

[EigenState137]

In a He+ ion, the electron on a 3d orbital will have the orbital 2 times smaller than in H, but the wavelength of the transition will be 4 times shorter. Therefore the size of the orbital will be a bigger fraction of wavelength, because the speed of the electron is bigger and a bigger fraction of speed of light. For a He+ ion, Lyman β (or whatever it is called in He+) would still be forbidden, but the branching ratio between allowed Balmer α and forbidden Lyman β would be more favourable to Lyman β. And both would be in hard ultraviolet.

Greetings,

This is the most obvious prediction, qualitative though it is, of your proposed treatment of atomic structure. It is subject to quantitative scrutiny which can be done taking advantage of the NIST Atomic Spectra Database.

We begin with clarification of the transitions under discussion. We want to compare the Einstein coefficients $A_{ki}$ in units of reciprocal seconds where $i$ specifies the lower state and $k$ specifies the upper state. Thus we wish first to determine the values of $A_{ki}$ for the following sets of transitions in HI:

$3d \; {}^{2}\textrm{D}_{5/2,3/2}\rightarrow 1s \; {}^{2}\textrm{S}_{1/2}\; \; \; (E2)\; \; \; [1]$

$3d \; {}^{2}\textrm{D}_{5/2,3/2}\rightarrow 2p \; {}^{2}\textrm{P}_{3/2,1/2}\; \; \; (E1)\; \; \; [2]$

I have used standard spectroscopic notation to describe the energy states and have specified the transition as either electric-dipole allowed (E1) or electric quadrupole allowed (E2). Looking up the the values of $A_{ki}$ for transitions [1] and [2] above we find their ratio to be $\approx 10^{-5}$ exact values depending on the specific values of $J$.

Doing the same for HeII we expect:

$3d \; {}^{2}\textrm{D}_{5/2,3/2}\rightarrow 1s \; {}^{2}\textrm{S}_{1/2}\; \; \; (E2)\; \; \; [3]$

$3d \; {}^{2}\textrm{D}_{5/2,3/2}\rightarrow 2p \; {}^{2}\textrm{P}_{3/2,1/2}\; \; \; (E1)\; \; \; [4]$

Looking up the the values of $A_{ki}$ for transitions [3] and [4] above we find for transition [4] $A_{ki}\approx 10^{9}$ but we find no data for transition [3]. That is to say that there is no available data to indicate that transition [3] has ever been observed.

For a He+ ion, Lyman β (or whatever it is called in He+) would still be forbidden, but the branching ratio between allowed Balmer α and forbidden Lyman β would be more favourable to Lyman β.

Your prediction is not consistent with the available quantitative data.ES

 

[EigenState137]

Greetings,

Oh, there is. The selection rule for transitions is Δn<=-1.

If that is a selection rule you will have no difficulty providing us with an definitive reference--Herzberg, Townes and Schawlow, Corney, ...

Please do so.ES

 

[hutchphd]

Oh, there is. The selection rule for transitions is Δn<=-1.

This is much ado about nothing.
Every scattering event requires a localized electron. If it is nearly free we put it in a big box and typically require cyclic boundary conditions: the spectrum is not really continuous but since the box is very large it doesn't matter and k is usually taken as continuous.
For the cyclotron, again the path is huge on an atomic scale and so any granularity caused by wavelength again does not matter for any real life condition creating synchrotron radiation. The continuum limit is just fine and the number of available states scales with volume.
So the issue is of no consequence.

 

[snorkack]

Greetings,

No, not agreed at all. To start with, selection rules are the same for both photon absorption and photon emission. That by itself negates your $\Delta n$ claim.

If you choose to group them together.
It is a true and vital insight that selection rules between photon absorption and emission must be symmetric. However, when the absorption and emission are viewed as separate if related processes, then they can have distinct selection rules, and do.
For a hydrogen-like atom, where all states with the same n are degenerate, transitions where Δn=0 are forbidden because zero energy.
Also because all states where n'>n have higher energy than the initial state (because of degeneracy), transitions where Δn>0 are absorptions, not emissions.
Only emissions where Δn<=-1 are allowed.
Furthermore, since states where n'<1 are unphysical, there is a selection rule
n-1>=n'>=1

It is also worth mentioning that the rigorous selection rule is for $\Delta J$ not $\Delta l$.

That much is true, however...
Look at the state 3d
It is true that the state

3d ²D_3/2 might possibly have dipole transitions to any state where J =1/2... and that includes the state 1s ²S_1/2
However, the transitions with spin flip do seem to be appreciably suppressed as well.
In any case, for state
3d ²D_5/2
there is simply no 1s state with total J of 3/2. So dipole transition to n=1, for this spin state, is still forbidden.

If you want to consider high Rydberg states, that is high $n$, then you have $\Delta n$ as a propensity rule, not a selection rule. This is clearly consistent with observed radiative recombination emission lines in the radio frequency domain from nebular environments with values of $n\approx 100$ in the case of hydrogen.

But recombination produces high n states in a variety of l and J states. Which of course allows for multiple transitions.
My point was to have a look specifically at the states which have high n, and the highest possible l and J for the given n. That is, l=n-1, J=n-1/2.
And I argue that dipole selection rules in that case (ΔJ=+-1, Δn<=-1, J'<=n'-1/2, J=n-1/2) allow only one dipole transition, which is Δn=-1, ΔJ=-1.

 

[EigenState137]

Greetings,

However, when the absorption and emission are viewed as separate if related processes, then they can have distinct selection rules, and do.

Viewing them as separate processes is totally artificial, and no the selection rules are not distinct. Atomic selection rules are about angular momentum and parity. And once again my question posed above: for atomic bound-bound transitions, please provide a citation to the scholarly literature supporting your assertion that a $\Delta n$ selection rule exists.

For a hydrogen-like atom, where all states with the same n are degenerate, transitions where Δn=0 are forbidden because zero energy

This never happens so your point is artificial and meaningless. That also is inconsistent with your desire to selectively choose the single highest $l$ state from some high $n$ state. The substates are either degenerate or they are not.

3d 2D3/2 might possibly have dipole transitions to any state where J =1/2... and that includes the state 1s 2S1/2
However, the transitions with spin flip do seem to be appreciably suppressed as well.
In any case, for state
3d 2D5/2
there is simply no 1s state with total J of 3/2. So dipole transition to n=1, for this spin state, is still forbidden.

I neither stated nor implied that $\Delta J$ is the only selection rule. $\Delta S$ selection rules depend on the angular momentum coupling in the states of interest.

But recombination produces high n states in a variety of l and J states. Which of course allows for multiple transitions.
My point was to have a look specifically at the states which have high n, and the highest possible l and J for the given n. That is, l=n-1, J=n-1/2.
And I argue that dipole selection rules in that case (ΔJ=+-1, Δn<=-1, J'<=n'-1/2, J=n-1/2) allow only one dipole transition, which is Δn=-1, ΔJ=-1.

Your point is contrived. At high $n$ the substates are quasi-degenerate and very likely to mix.

Enough about selection rules until you provide the requested literature citation in support of your assertion.

There is however a more fundamental problem. From post 11 above with emphasis added:

When I think of it, it seems to me that the size of orbitals is much smaller than the wavelength of the emitted radiation if, and because, the speed of electron is much smaller than the speed of light.
In a H atom, an electron on 3d orbital can undergo a dipole allowed transition to 2p (Balmer α) followed by dipole allowed transition to 1s (Lyman α). Or it could undergo a quadrupole forbidden transition to 1s (Lyman β).
In a He+ ion, the electron on a 3d orbital will have the orbital 2 times smaller than in H, but the wavelength of the transition will be 4 times shorter. Therefore the size of the orbital will be a bigger fraction of wavelength, because the speed of the electron is bigger and a bigger fraction of speed of light

Atomic states cannot be described in terms of point charges in orbits about the nucleus. There is no physically meaningful concept of the "speed" of the electron within some semi-classical orbit.

The crux of my concern is that you are breaking both atomic spectroscopy and quantum mechanics. If this discussion is to prove instructive you need to support your statements with something more substantive than mere repetitions.ES