Probability of resonant ionization

[Malamala]

Hello! I am trying to make some predictions for an experiment in which we have a first $E_2$ transition in an atom driven by a laser, and then we have a second laser that is ionizing the molecule only if the first laser was resonant (i.e. if the atom was excited). For the purpose of the question we can assume that the ionization efficiency is 100% i.e. if the atoms gets excited, it will get ionized, too. I found this formula for the probability per second of the $E_2$ transition:

$$A_{E_2} = \frac{32\pi^5\alpha c a_0^4}{15 \lambda g'}S_{E_2} = \frac{1.11995 \times 10^{18}}{\lambda g'}S_{E_2}$$

where $\lambda$ is the transition wavelength in angstroms, $g'$ is the degeneracy of the excited state and $S_{E_2}$ is the E2 line strength in atomic units. $S_{E_2}$ is known theoretically to be 1.43 and in the end I get $A_{E_2} \sim 2\times 10^{-4} s^{-1}$. I also have the (pulsed) laser power which is 100 mW and beam area which is 1mm$^2$. Using the Plank's formula, I get that the number of photons per unit second per mm$^2$ is $\sim 10^{18}$. If I have a given atom, interacting with such a pulse, I would like to know the probability of being excited. However, it seems like I would need to plug in some sort of area of the atom. Basically the formula I would use is:

$$2 \times 10^{18} \times 10^{-4} \times A_{atom} \times mm^{-2}$$

But I am not sure what to use for that. I guess I can use ~$10^{-10}$m for the atomic radius, but that seems like a classical approximation. I was wondering if I am doing something wrong. Why do I get this extra area in the formula and how do I deal with it? Thank you!

 

[Twigg]

The formula you gave looks like an Einstein A coefficient. If that's true, that's the probability of spontaneous emission per second. I think you want the probability of stimulated absorption per second, which would be an Einstein B coefficient (or equivalently, the absorption cross section).

If you have the A coefficient, can you estimate the transition quadrupole moment? (I only suggest that because I think that you could do this for an E1 line: estimating the transition dipole moment from the Einstein A coefficient (the formulas for doing this are in Ch3 of Metcalf's laser cooling book)). If you have the transition moment, I think you should be able to calculate the Rabi rate, which (I think) is what you really want.

Also, for your pulsed laser power, is that 100mW average power or 100mW peak power? You'll need to know the pulse duration and the repetition rate.

It might not be obvious why you need the repetition rate, but notice that your Einstein A coefficient is very slow. It will take 5000s on average for spontaneous emission, so unless you are pulsing your laser once per hour you should assume that the atom-photon interaction will be coherent from one pulse to the next.

 

[Malamala]

The formula you gave looks like an Einstein A coefficient. If that's true, that's the probability of spontaneous emission per second. I think you want the probability of stimulated absorption per second, which would be an Einstein B coefficient (or equivalently, the absorption cross section).

If you have the A coefficient, can you estimate the transition quadrupole moment? (I only suggest that because I think that you could do this for an E1 line: estimating the transition dipole moment from the Einstein A coefficient (the formulas for doing this are in Ch3 of Metcalf's laser cooling book)). If you have the transition moment, I think you should be able to calculate the Rabi rate, which (I think) is what you really want.

Also, for your pulsed laser power, is that 100mW average power or 100mW peak power? You'll need to know the pulse duration and the repetition rate.

It might not be obvious why you need the repetition rate, but notice that your Einstein A coefficient is very slow. It will take 5000s on average for spontaneous emission, so unless you are pulsing your laser once per hour you should assume that the atom-photon interaction will be coherent from one pulse to the next.

@Twigg Thank you for you reply. I took that formula from here. I am attaching below a screenshot, too just in case. It looks like stimulated excitation probability, what do you think (the paper is not related to what I am doing, I was just looking for the formula).

For the laser you are absolutely right. In my case the laser has 10ns duration. So the energy would be $100 mJ/s \times 10^{-9}s = 10^{-10}J$ and using the plank formula i.e. $Nh\nu$ (in my case $\nu = 1695 nm$) I get $N \sim 10^{12}$ photons per area. Also in my case, we are doing collinear spectroscopy, so the atom beam stays in the interaction region for a few $\mu s$, so it is safe to assume that it interacts with only one laser pulse. But I am still not sure how to get the probability excitation per pulse from here.

 

[Twigg]

Thanks for the screenshot. The reason I suspect it's the spontaneous emission rate is because it's labeled as A (typically used for einstein A coefficient) and because the formula doesn't contain the intensity or frequency of the driving field. The stimulated rate should very much care about the exciting field intensity and frequency.

I don't have any idea where you could look up the formula for the Einstein B coefficient for a quadrupole transition. However, you could totally get the Rabi rate if you know the quadrupole moment tensor. Assuming the quantization axis is defined by the probe beam polarization, I believe based on this paper that the Rabi rate should be given by $$\hbar \Omega_{eg} = \frac{1}{2} \langle e | \sum_{i=x,y,z} Q_{iz} \nabla_i E_z | g \rangle$$ So, for the case of a traveling wave moving in the +x direction, $$\hbar \Omega_{eg} = \frac{1}{2}ikE_z \langle e | Q_{xz} | g \rangle$$ It's definitely worth double checking my work, because I'm no theorist. In any event, $E_z$ you can get from the average laser intensity, $k$ is just $\frac{2\pi}{\lambda}$ That should be enough for you to be able to solve the optical Bloch equations for your coherent pulse, for time ranging from 0 to 10ns (pulse duration). If you wanted to be more exact, you can substitute in the pulse intensity versus time (it's probably gaussian).

 

[Twigg]

I happened upon a very solid reference (see here) on how to get the Rabi rates for quadrupole transitions. Chapter 4 of this thesis is what you're looking for. To get the Rabi rate, just divide the off-diagonal Hamiltonian matrix element by $\hbar$.

 

[Malamala]

I happened upon a very solid reference (see here) on how to get the Rabi rates for quadrupole transitions. Chapter 4 of this thesis is what you're looking for. To get the Rabi rate, just divide the off-diagonal Hamiltonian matrix element by $\hbar$.

Thanks a lot for this! So is the formula I would need to use 4.62? I assume $n_k$ is the number of photons per pulse and I would choose a polarization, but I am not sure about that last term (the matrix element). Is that something that needs to be calculated by quantum chemistry methods? And once I get that term in 4.62 I would need to plug it in the Fermi golden's rule to get a rate? (or are you talking about a totally different equation?)

 

[Twigg]

Is that something that needs to be calculated by quantum chemistry methods?

Ordinarily, yes. But you already have some information from the A coefficient you found in the literature. The A coefficient should be proportional to the transition matrix element. The problem is I don't know the scaling factor off the top of my head. It should contain the density of photon states in the vacuum field and the fluctuations of the E-field gradient in the QED vacuum state. Sorry I don't have a reference for this. YI would recommend you practice doing this derivation (of the A coefficient) for an E1 transition to get some intuition.

 

[Malamala]

Ordinarily, yes. But you already have some information from the A coefficient you found in the literature. The A coefficient should be proportional to the transition matrix element. The problem is I don't know the scaling factor off the top of my head. It should contain the density of photon states in the vacuum field and the fluctuations of the E-field gradient in the QED vacuum state. Sorry I don't have a reference for this. YI would recommend you practice doing this derivation (of the A coefficient) for an E1 transition to get some intuition.

So I know there is this formula in Griffiths:

$$A = \frac{\omega^3\hbar}{\pi^2 c^3}B$$

Is this the one you are talking about?

 

[Twigg]

I had something in mind that was more involved. The relationship you quoted is actually a much quicker way to the answer you want.

Here's a good introduction on this subject. If it isn't clear, you cannot make the electric dipole approximation in your case. But that doesn't matter because the relationship $$A = \frac{\omega^3 \hbar}{\pi^2 c^3} B$$ is universal.

Once you have the B coefficient, all that's left is for you is to calculate the spectral energy density of the laser light, $u(\omega) = \frac{1}{2} \epsilon_0 E^2(\omega) = \frac{I(\omega)}{c}$ where $I(\omega)$ is the spectral intensity. (Edit: Note that for a laser beam, to get the spectral intensity, take your beam intensity (power / mode area) and then divide by the laser linewidth. This is an approximate recipe, but it's the most appropriate approximation when your laser is much less stable than your transition linewidth, as in this case.)

Then your transition rate (at frequency $\omega$, as per Fermi's golden rule) will be simply $B(\omega) u(\omega)$.

 

[Malamala]

I had something in mind that was more involved. The relationship you quoted is actually a much quicker way to the answer you want.

Here's a good introduction on this subject. If it isn't clear, you cannot make the electric dipole approximation in your case. But that doesn't matter because the relationship $$A = \frac{\omega^3 \hbar}{\pi^2 c^3} B$$ is universal.

Once you have the B coefficient, all that's left is for you is to calculate the spectral energy density of the laser light, $u(\omega) = \frac{1}{2} \epsilon_0 E^2(\omega) = \frac{I(\omega)}{c}$ where $I(\omega)$ is the spectral intensity.

Then your transition rate (at frequency $\omega$, as per Fermi's golden rule) will be simply $B(\omega) u(\omega)$.

Thank you, I got back to this! So I have A from the paper/screenshot above and I can easily get B using the formula in the last post, right? I am not sure how to get the spectral density, tho. I know the pulse power and duration. Can I do a FT assuming the central frequency is the correct one, see how spread the pulse is in frequency space (based on the 10 ns duration), and from there figure out the power at the center? But I am confused about the units. B is a probability/s and $u(\omega)$ should be $J/m^3$. But what I need in the end is the percentage of excited atoms per laser pulse (so I would need something unitless or maybe with units of $s^{-1}$). Is there an extra step I am missing?

 

[Malamala]

Ok so I read a bit more into this. I found this paper, and the term in the 2x2 Hamiltonian I would need is in Eq. (2). In my case, I can assume that the light goes in the x direction and it is z polarized. So (please let me know if this is wrong) I think that the profile of the electric field is:

$$E_z(x) = Ne^{-\frac{(x-ct)^2}{2\sigma^2}}$$
where I assumed that the profile of the laser pulse is a gaussian with a spread of $\sigma = t' \times c$ with $t' = 10$ns. From here I can easily calculate $\frac{\partial E_z}{\partial x}$ which means that the equation (2) reduces to:

$$W = -\frac{1}{6}Q_{13}\frac{\partial E_3}{\partial x_1}|_{x_1=0}$$
with $E_3=E_z$ and $x_1=x$. Then for the matrix element of the quadrupole I can use Equation (5). Based on equation (A12) that is non-zero only for $\Delta M = \pm 1$ and the reduced matrix element I already have from some previous quantum chemistry calculations. So by plugging all these in I would get a 2x2 Hamiltonian, with a time dependent term on the off-diagonal (coming from the first equation above) and I can in principle solve the SE for this numerically and get the population transfer for one pulse. Does this make sense? Would this give me the right answer?

 

[Twigg]

Sorry for the slow reply.

So I have A from the paper/screenshot above and I can easily get B using the formula in the last post, right?

Yep

I am not sure how to get the spectral density, tho. I know the pulse power and duration.

But I am confused about the units.

If your probe light is a gaussian beam, then you can use the power and beam diameter to calculate the intensity: $P = \frac{1}{2} \pi I w^2$ where $w$ is the beam waist (radius), $I$ is the average beam intensity, and $P$ is the beam power. (The factor of $\frac{1}{2}$ comes from integrating the gaussian beam profile.) Then you can use the formula I gave in post #9 to get spectral density: $u = \frac{I}{c}$.

You then have everything you need to calculate the transition rate: $\Gamma_{E2} = Bu$. (Here I'm leaving out the frequency dependence that comes from Fermi's golden rule, because everything is calculated at the resonant frequency.) You can then multiply by the pulse time to get the total transition probability: $P_{E2} = \Gamma_{E2} \times \Delta t$. Keep in mind that this is a perturbative approximation that only works in a small neighborhood of Hilbert space near the ground state. Real systems will have stimulated emission and Rabi flopping.

So by plugging all these in I would get a 2x2 Hamiltonian, with a time dependent term on the off-diagonal (coming from the first equation above) and I can in principle solve the SE for this numerically and get the population transfer for one pulse. Does this make sense? Would this give me the right answer?

I recommend this approach. It is non-perturbative and fully general, unlike the B-coefficient calculation.

Numerical simulation of the time-dependent SE will work in principle, but I would suggest a slightly different (but mathematically identical) approach. Transform your 2x2 Hamiltonian to the interaction picture (removing the time-independent diagonals). Then the off-diagonal of your 2x2 interaction picture Hamiltonian gives you the Rabi rate $\Omega = \frac{H_{12}}{\hbar}$ and you can use the much more intuitive optical Bloch equation: $$\frac{d}{dt} \vec{R} = \vec{W(t)} \times \vec{R}$$ where $\vec{R}$ is the usual Bloch vector and $\vec{W(t)} = \left[ \begin{matrix} \mathrm{Im}[\Omega(t)] & \mathrm{Re}[\Omega(t)] & \delta \end{matrix} \right]^T$ and $\delta$ is the difference between your laser frequency and the E2 transition frequency. I only recommend this because it is much easier to visualize, so you'll know for sure if your simulation is working (if the Bloch vector doesn't precess around $\vec{W}$, you know you messed up). Here's a good set of notes on the optical Bloch equations. I left out the spontaneous emission term because your pulse is much faster than the E2 lifetime, so its effect should be negligible.

 

[Malamala]

Sorry for the slow reply.


Yep



If your probe light is a gaussian beam, then you can use the power and beam diameter to calculate the intensity: $P = \frac{1}{2} \pi I w^2$ where $w$ is the beam waist (radius), $I$ is the average beam intensity, and $P$ is the beam power. (The factor of $\frac{1}{2}$ comes from integrating the gaussian beam profile.) Then you can use the formula I gave in post #9 to get spectral density: $u = \frac{I}{c}$.

You then have everything you need to calculate the transition rate: $\Gamma_{E2} = Bu$. (Here I'm leaving out the frequency dependence that comes from Fermi's golden rule, because everything is calculated at the resonant frequency.) You can then multiply by the pulse time to get the total transition probability: $P_{E2} = \Gamma_{E2} \times \Delta t$. Keep in mind that this is a perturbative approximation that only works in a small neighborhood of Hilbert space near the ground state. Real systems will have stimulated emission and Rabi flopping.


I recommend this approach. It is non-perturbative and fully general, unlike the B-coefficient calculation.

Numerical simulation of the time-dependent SE will work in principle, but I would suggest a slightly different (but mathematically identical) approach. Transform your 2x2 Hamiltonian to the interaction picture (removing the time-independent diagonals). Then the off-diagonal of your 2x2 interaction picture Hamiltonian gives you the Rabi rate $\Omega = \frac{H_{12}}{\hbar}$ and you can use the much more intuitive optical Bloch equation: $$\frac{d}{dt} \vec{R} = \vec{W(t)} \times \vec{R}$$ where $\vec{R}$ is the usual Bloch vector and $\vec{W(t)} = \left[ \begin{matrix} \mathrm{Im}[\Omega(t)] & \mathrm{Re}[\Omega(t)] & \delta \end{matrix} \right]^T$ and $\delta$ is the difference between your laser frequency and the E2 transition frequency. I only recommend this because it is much easier to visualize, so you'll know for sure if your simulation is working (if the Bloch vector doesn't precess around $\vec{W}$, you know you messed up). Here's a good set of notes on the optical Bloch equations. I left out the spontaneous emission term because your pulse is much faster than the E2 lifetime, so its effect should be negligible.

Thanks for the clarifications (and sorry for the late reply). So I tried to do the B version first. This is what I got. The lifetime of the state is 1 min, so $A = 1/60$. Also $\omega = 1695 nm$. Plugging everything in I get: $B \sim 10^{15}$. Then for the laser I have a pulse of 10 ns, beam size about 1 cm$^{2}$ and 100mW of power. So $I \sim 1000$ so $\Gamma \sim 10^{10}$ so the probability would be about 100, which is not physical. However, I imagined that given that this is a highly forbidden transition and that the laser power is not super high (even at 0.1 mW it seems like I would not be perturbative), this approach should work. Am I doing something wrong. Could it be due to the fact that I assume that all the power is on resonance, when in practice that is not the case. Basically, the linewidth of my laser is tens of MHz, while the transition is about 10 mHz. Would it be better to assume that the power is distributed as a gaussian around the resonant frequency, and the power actually available for the transition is the integral around a 10mHz region around that frequency?

 

[Malamala]

Just to explain better what I meant, for the pulse shape, assuming 10 MHz laser linewidth, and 10 mHz transition linewidth, I would need to calculate the integral:

$$\int_{-0.01}^{0.01}\frac{1}{10^{10}}\sqrt{2\pi}e^{-0.5\frac{x^2}{10^{20}}}dx$$
which is equal to $\sim 10^{-12}$. So now I get for the transition probability $10^{15}\times 10^4 \times 10^{-1}\times 10^{-8} \times 10^{-12} \times 10^{-8} = 10^{-10}$, where the powers above correspond to: B, $\omega$, $P$, $c$, the integral above and the pulse time. So this means that basically the number of excited atoms will be $N \times 10^{10}$, where N is the number if atoms per bunch. Is this right?

 

[Twigg]

First off, I made a mistake in the formulas I gave you for spectral energy density $u(\omega)$. What I gave you has units of $\mathrm{J/m^3}$, but it should have units of $\mathrm{J\cdot s/m^3}$. For a rough estimate, I think you can just divide by the laser linewidth. I don't know off the top of my head how best to incorporate the 10mHz lineshape. I don't think you need to incorporate the 10mHz linewidth, because the rate equations works like Fermi's golden rule and so it only evaluates the transition rate at the resonance frequency, ignoring off-resonant contributions. I'm not 100% confident though.

If you need to measure your laser linewidth and/or lineshape, a self-homodyne measurement over fiber is the standard technique (see here).

Plugging everything in I get: B∼1015. Then for the laser I have a pulse of 10 ns, beam size about 1 cm2 and 100mW of power. So I∼1000 so Γ∼1010 so the probability would be about 100, which is not physical.

Check your numbers? Even with the wrong formula I got different numbers. I got $B \approx 1.7\times10^{13}$. If it's helpful, I got $u \approx 3\times10^{-6} J/m^3$ (this is the part that's wrong, per my above explanation) and $\frac{\omega^3 \hbar}{\pi^2 c^3} \approx 10^{-15} \frac{J*s}{m^3}$

Using $10MHz$ as your laser linewidth, I get $u(\omega) \approx 0.5 \times 10^{-13}\frac{\mathrm{J\cdot s}}{\mathrm{m}^3}$

With the corrected formulae, and using your value for A, I get $\Gamma = 0.8\mathrm{s}^{-1}$.

So this means that basically the number of excited atoms will be N×1010, where N is the number if atoms per bunch. Is this right?

I don't follow. Is it a pulsed beam source for the atoms?

 

[Malamala]

First off, I made a mistake in the formulas I gave you for spectral energy density $u(\omega)$. What I gave you has units of $\mathrm{J/m^3}$, but it should have units of $\mathrm{J\cdot s/m^3}$. For a rough estimate, I think you can just divide by the laser linewidth. I don't know off the top of my head how best to incorporate the 10mHz lineshape. I don't think you need to incorporate the 10mHz linewidth, because the rate equations works like Fermi's golden rule and so it only evaluates the transition rate at the resonance frequency, ignoring off-resonant contributions. I'm not 100% confident though.

If you need to measure your laser linewidth and/or lineshape, a self-homodyne measurement over fiber is the standard technique (see here).


Check your numbers? Even with the wrong formula I got different numbers. I got $B \approx 5\times10^{-7}$. If it's helpful, I got $u \approx 3\times10^{-6} J/m^3$ (this is the part that's wrong, per my above explanation) and $\frac{\omega^3 \hbar}{\pi^2 c^3} \approx 10^{-15} \frac{J*s}{m^3}$

Using $10MHz$ as your laser linewidth, I get $u(\omega) \approx 0.5 \times 10^{-13}\frac{\mathrm{J\cdot s}}{\mathrm{m}^3}$

With the corrected formulae, and using your value for A, I get $\Gamma = 0.8\mathrm{s}^{-1}$.


I don't follow. Is it a pulsed beam source for the atoms?

I am a bit confused. Indeed $\frac{\omega^3 \hbar}{\pi^2 c^3} \approx 10^{-15} \frac{J*s}{m^3}$, and so we have $A = 10^{-15}B$, with $A = \frac{1}{60}$. Don't we get $B \sim 10^{14}$?

And yes, the atoms are bunched.

 

[Twigg]

Sorry, I typed the wrong number for $B$. I edited the post.

 

[Malamala]

Sorry, I typed the wrong number for $B$. I edited the post.

Oh, I see, I think I missed the $(2\pi)^3$ factor (and maybe some approximations along the way). What I have is consistent with you and I get the same $\Gamma$, when dividing my the laser linewidth. So the transition probability per bunch, assuming N atoms per bunch, would be: $N \times 0.8 \times 10^{-8}$ (for 10 ns pulse length)? I am not sure tho if using perturbation theory is valid in this case. The 0.8 is quite a big number, but the final $\sim 10^{-8}$ probability should be valid for PT (as it is very small). Which number should dictate the validity of the PT? Should I expect that doing a full Hamiltonian diagonalization give me the same result basically?

 

[Twigg]

It helps to take a step back and look at the foundations of perturbation theory. What's the main assumption in PT? That $\psi \approx \psi^0 + \epsilon \psi^1$ for $\epsilon \rightarrow 0$. Here, $\psi^0$ is the ground state. To translate this into more practical language, so long as you stay close to the south pole of the Bloch sphere, everything is peachy.

However, I still recommend doing the optical bloch equation calculation. The reason for this is because it's easy to make bad assumptions in the rate equations picture. It's always nice to have independent confirmation. The hardest part of the optical bloch calculation will be getting your rotating frame hamiltonian. Up to you!