There is no intermediate state and if you want ot resort to the idea of a (virtual) intermediate state where does it say that this has a lifetime of a femtosecond.

Matthew

Please react; we also take our time to go over your question. Reminds me of my minot master thesis 40 years ago: the feasibility of tfa in aromatics

If you are considering real intermediate states (not clear from your statement) then you are not dealing with tow photon absorption but with two one photon absorptions which dynamics has been studied at length in the field of laser physics.

It seems to me you are considering virtual transitions of electrons between real states, aren't you? Virtual states are a misuse of language. I agree with Harry.

Two-photon absorption (2PA), not to be confused with cascaded one-photon absorption (or excited state absorption), may be "resonantly enhanced" if one of the two photons involved in the absorption process has an energy close to the energy difference between the ground state and some other state of the system (what I referred to as the "intermediate state") See, for example: http://prl.aps.org/abstract/PRL/v33/i3/p128_1. My question is whether the population of these "intermediate states" will change the two-photon absorption rate?

Wei-Chun Chu, can we not simply apply perturbation theory with different initial populations of the system? In this case, the 2PA rate will surely be different simply because the population distribution has changed. But will the 2PA cross-section be effected at all?

This is (quite) vague: what is "close" to the energy of an excisted state? Mind you with some Boltzman activation it is a real transition

This appears to be the case of 2 colour two photon excitation. Yes in this case the population of the intermediate state would influence the TPE rate. IN a single colour 2 photon case it does not.

Harry ten Brink: "Close" meaning as small an energy difference as possible without resulting in actual one-photon absorption into the intermediate state, so no less than several linewidths.

Prem Bisht: How can the two photons being the same color or not (degenerate vs. nondegenerate) make a difference?

Matthew: It was from similar experiments on REMPI. IN two colour experiments, we generally come across a state that can be prepared, generally a laser that is fired to the matching transition, hence the transition has a lifetime even longer than fs.The other photon now takes it to the desired level. In the present question, I see this sort of querry.

Matthew

what compound are you talking about? Line width only has a meaning in atoms and very small molecules and what frequency; I was thinking of electronic transitions

As I mentioned at lab temperature molecules are in excited vib-rot states according to the Boltzmann distribution and thus can have a transition at a frequency quite smaller that the pure electronic transition

Prem

what you describe is NOT TFA, as dr Hirlemann pointed out above already and the questioneer agreed to; please read comments before you answer

Wei-Chun Chu

Exactly. I was hoping someone knew the answer, but I suppose I could calculate it.

Prem Bisht

Though I am not very familiar with it, I believe REMPI is akin to cascaded one-photon absorption where the final state is in the continuum. In this case the electron may stay in the first excited state for that state's lifetime. However, two-photon absorption is an "instantaneous" process.

Harry ten Brink

An atom, for a simple case.

What do YOU want to do then?

Matt,

As seen in the PRL article you cited, the *presence* of intermediate states in the sum-over-states can greatly affect the 2PA rate; however, it isn't clear what you mean by the "populations" of these intermediate states. Please ask your question anew, to clear the air for both you and us.

Although Eq(1) in the PRL paper seems to be the result of a time-independent perturbative approach, it stems in fact from a time-dependent fromalism. And, as sugested in this series of posts, it is instructive to have such a "pulsed" (rather than cw-laser) situation in mind. Mathematically, the coefficients corresponding to the intermediate state(s) will be non-zero during the pulse, and will vary in time; the squares of these coefficients are typically (and losely) called populations. Presumably, the atom starts in the ground state; i.e., typical initial conditions with pop(gr-state)=1 and all other pop=0. I would, in fact, say something than runs a bit counter to what you were asking; namely, the 2-photon process affects these populations, rather than the other way around.

Hope this didn't widen the confusion;

waiting for Matt's restating of the original question.

Cheers.

Ion

It is now 40 years that I studied TFA; what I still recall is that in the perturbation approach you can always include the sum of alll states in between the brackets; indeed in this way a state that is close to a frequency/energy will enhance the transition strength. So these atates are not virtual bu are virtual in the sense that there is NO one photon transition to that state. So indeed what has the questioneer in mind with him asking on the population of that state?

For what kind of TPA are you asking? The use of the term TPA in English is often confusing, since the same term is used in case of resonant and non-resonant TPA. The appropriate terminology should be TPA (or two quantum absorption, TQA) for a “pure” two-photon absorption through a VIRTUAL intermediate state i.e. a third order nonlinear optical phenomenon in which a molecule absorbs two photons AT THE SAME TIME. In presence of a REAL state the appropriate term is resonant TPA or two-step (step-wise) absorption. Please define your case. If it is a “pure” TPA the population of the intermediate state is zero (or the sum of the normalised population of the initial and the final state is 1. If you have a resonant absorption you need the parameters of all states involved (absorption crossection, lifetime etc.) to calculate the time-dépendent populations or you can do it experimentaly in some simple cases

Dear Matthew,

There is another important point i would like to add. A two-photon transition is always in a three level system. The first photon couples the ground state and the intermediate state (whether there is population transfer or not) and the second photon couples the intermediate state with the excited state. The frequency detuning of the first photon from the intermediate state determines the so-called "enhancement factor", therefore at zero detuning the enhancement is maximal (stepwise excitation).

There is a very informative derivation of the excitation probability which transforms the problem into frequency domain (PRA 60, 1287 (1999)) This is more or less the foundation of quantum coherent control.

I have computer code that simulates TPA for broadband pulses, including resonant and non-resonant terms. If you explain a bit more about your experiment i might be able to adjust the code for you.

Itan

as discussed earlier, two-photon absorption does not imply an intermediate state

Harry

In two-photon transitions the ground state and excited state are not coupled via a dipole transition (electric dipole forbidden due to selection rules). That's why an intermediate state is absolutely necessary. Whether this intermediate state is resonant with your laser is a different issue.

This point is frequently overlooked which is why i stress it again: a two-photon transition is a three level problem. When you do the math you see immediately that without an intermediate state the two-photon transition rate is zero.

In true TPA there is NO resonant intermediate state. Which intermediate state do you propose then?

The process is described by second-order perturbation theory, with thus no dipole selection rules

From the nonlinear optics point of view it is part of a nonparametric chi(3) process. This means it involves 3 fields to give rise to (4th) another field.

Itan

In the mean time I was searching for my master thesis on TPA but cannot find it (anymore)

So went via Google to find an MIT lecture and found, quote, what I mentioned earlier on the intermediate state [f>:

"In reality, [f> represents one of a complete set of eigenstates which have non- vanishing dipole matrix elements with [a>.

The full description is on page 1 below equation 9.2

And what is wrong with this publication of the godfather of "opto-electronics"?

Harry

The MIT lecture notes state exactly what i was trying to say. In Eq. 9.11 you see that the excitation probability depends on the detuning from the intermediate state. If there was no intermediate state the the equation would not make sense. Therefore, there must be an intermediate state. In Yariv's paper you can see also that there must be an intermediate state for exactly the same reason.

I would like to emphasize again, i don't mean that there is population transfer to the intermediate state. The only thing i am saying is that an intermediate state is needed in order to couple the ground and excited states.

Your remark is absolutely correct that in reality there are many intermediate states, however, in practice the closest intermediate state causes the largest enhancement and the others can be neglected.

I have have a recent paper about TPA in a new and surprising configuration. It seems that there are specialists in this field here. If anyone is interested, i would love to get some feedback (whether positive or negative) on this work.

http://www.nature.com/nphoton/journal/v7/n1/full/nphoton.2012.299.html

http://arxiv.org/abs/1207.5934

Itan

I just do not see the necessity: without any explan ation the unity operator is put in between the two dipole terms and that workd out well. However, none of the two references tell us WHY it is done

Harry

When you write "operator is put in between the two dipole terms", which dipole terms do you mean? The answer is: one dipole term that couples the ground state with the intermediate state and the other dipole term couples the intermediate state with the excited state. Hence an intermediate state is absolutely necessary.

A full derivation of this equation can be found in F. H. M. Faisal Theory of multi photon processes Ch. 2. There you can see that the derivation assumes the existence of an intermediate state.

No:]

what is done is bracketing the sum of the eigenstate operator in between whihc equals the unit operator that si a trick that is not explained but works out because it indeed couples the ground state with all excited eigenstates and all excited state with the one reached with TPA. However, none explains this choice and why a second-order perturbation as such does not work

Harry,

I don't really know what else i can say, i thought that my arguments were convincing but apparently you are not convinced easily.

If you are in Amsterdam soon maybe we can have a cup of coffee and i'll show you the derivation i know from beginning to end. Maybe then my argument will be clearer.

Itan

Can you then please attach your arguments in mathematical form. I have not seen these yet. This would be somewhat sooner than making an appointment, though one my very last meetings were at the VU in 2010!!

It just happens that the subject has a very high value of youth sentiment to me, as I mentioned before

Harry

No problem, this is the main topic of my PhD so I also like it very much. Give me a couple of days and i'll get my argument in a mathematical form.

Itan

OK I really look forward to it

Sorry to take so long to respond.

Indeed, I am referring to actual two-photon absorption, where both photons are absorbed simultaneously, and neither photon is resonant with the ground-to-intermediate state transition.

Perhaps I can rephrase my question a different way. In the perturbation theory calculation of the two-photon transition rate, one sums over all intermediate states which have a non-zero transition dipole moment with the ground state. Must one consider whether or not a particular intermediate state is occupied to include it in this summation? For example, in a lithium atom, must the summation include the two innermost electronic states (n=1), when calculation two-photon absorption of the single valence electron (n=2)? Even though those states are occupied? (Certainly their contribution will be small considering the large detuning) Or does the fact that those states are occupied mean that those intermediates states are blocked, and cannot contribute to the two-photon transition?

Matthew

My response would be that all the introduction of the intermediate states is a "mathematical trick" and the full sum all of the orthogonal eigenstate/operators of the time-independent (say atomic) Hamiltonian is nedded so these add up to the unity operator, to be placed between the two dipole operators, see the publication of Yariv in IEEE. However, in practice it means that the transition is mostly or completely dominant by one single level because it is the least "detuned" from the laser frequesncy and has a signifant value with iinitial anf final state. It does not matter whether this state is occupied or not.

oops gone before I rerally finished it: here is the correct response

Matthew

My response would be that all the introduction of the intermediate states is a "mathematical trick" and the full sum all of the orthogonal eigenstate/operators of the time-independent (say atomic) Hamiltonian is nedded so these add up to the unity operator, to be placed between the two dipole operators, see the publication of Yariv in IEEE. However, in practice it means that the transition is mostly or completely dominant by one single level because it is the least "detuned" from the laser frequency and has a significant value with initial and final state. It is irrelevant whether this state is occupied or not, because the formulation is not related to actual eigen states.

Harry,

I believe the introduction of the intermediate states is more than simply a "mathematical trick". As I've said before, when one of the photon's has energy close to the ground-to-intermediate state transition energy (but not exactly resonant with it) the two-photon absorption rate will be greatly enhanced. The intermediate states have a physical effect.

Certainly, strictly speaking, one must include them all of the states in a full summation in the perturbation approach. However, certain terms in this summation, corresponding to individual intermediate states, may become zero. This will happen, for example, if the transition dipole moment between the ground and a particular intermediate state is zero. Therefore, that particular intermediate state will not contribute to the two-photon transition.

Matthew

What you mean I think is whether a one-photon excitation of the most important intermadiate state would affect the TPA: my opnion is that this has no effect becasue the two-photn transition does NOT include an actual state

Harry

I would have to agree with Matthew on this one. I will find a good reference that shows this easily or just write the derivation myself.

Matthew

If an intermediate level is occupied then the transition to it is not allowed and will also not participate in the two-photon excitation. The atom is actually excited to the intermediate state for a very short time (due to the uncertainty principle) before going to the upper state. Therefore, if the transition to the intermediate state is not possible (for any reason) then the excitation to the upper state is zero.

Matthew

I thought a bit more about your question and i think that the answer is a bit more complicated than what i wrote before. For the example you gave on Li i am sure that you don't need to take into account the n=1 state because a one photon transition to that level is impossible. However, if the 2p state is occupied then the laser can drive the transition back to the ground state. Therefore the excitation dynamics is a bit more involved.

So to answer your question again, i don't have a definite answer. In practical terms this is a very unique situation that doesn't happen often, but i guess you already knew that :)

Itan,

Thank you very much for your answers.

My example of Li, as you pointed out, was a poor choice, but it seems you understood my intention. In general, do occupied energy states need to be included in the summation (assuming the transition dipole moment is non-zero) for calculating the two-photon transition rate?

This question has relevance to my own research involving two-photon absorption in semiconductors. Here, since within the band gap there are no intermediate states, the intermediate states which dominate the two-photon absorption rate are within the valence and conduction bands themselves. (See Wherrett 1984, http://www.opticsinfobase.org/josab/abstract.cfm?uri=josab-1-1-67). There are then two possible pathways: valence-valence-conduction, and valence-conduction-conduction. However, the valence band is filled. So if what you're saying is correct, that occupied intermediate states do not contribute to the two-photon transition rate, then only the valence-conduction-conduction pathway is possible. I have not seen mention of this in the literature on the subject, though it is possible I've overlooked it.

Hi,

I don't know if this will be helpful for this discussion, but I would like to support the viewpoint that intermediate states are important for the two-photon absorption. We have observed it in ab-initio calculations of the hydrogen molecular ion (http://pra.aps.org/abstract/PRA/v83/i2/e023412). We considered mid-infrared lasers. Two-photon absorption brings the population from one vibrational state to another vibrational state within n=1. However, the intermediate state can only be allowed in n=2, n=4, etc. As you said, as the population of the intermediate state is so fast, we used it in this case to avoid dissociation (n=2, the one that contributes more in the sum of intermediate states, is dissociative in H_2^+).

This is very interesting, I hope you get your answer.

Thanks Antonio, i think that we have a fairly broad support for the importance of the intermediate state.

Matthew

In your case my intuition tells me that if the valence band is filled then it will not contribute to the two-photon transition, but my expertise is limited to atomic and molecular systems.

Why are you considering two-photon transitions? Can't you excite these transitions with one photon? The excitation would then be orders of magnitude higher. The system you are describing is interesting, maybe you can tell us more about it (i need a broad range of references for a paper we want to publish).

Mattew,

A convenient measure of the two-photon absorption is provided by the stationary population of the final (second excited) state. You can solve the system non-perturbatively by allowing the field to be classical and looking for the stationary solution of the appropriate expectation values of the Heisenberg operators, including populations.

If there are losses (and, if fact, they are always present), all initial populations and coherences will die out, so that the stationary populations will not depend on the initial ones. However, the stationary population of the intermediate (first excited) state need not be exactly zero, but will depend on two damping constants as well as possible coherent or incoherent pumping. The same is true about the population of the (second) excited state. In this sense these two populations (and hence

the intermediate-state population and two-photon absorption) are connected.

If you want, I can do some calculations and figures for you in the following week.

But there is a *huge* amount of works about similar systems in quantum optical literature, look for papers of, e.g., Joseph H. Eberly and Girish S. Agarwal of eighties and nineties.

Regards,

Maciej

Itan

I like to remind you of your promise, some time ago, to put your ideas on TPA on paper for me/us

Harry,

Unfortunately i didn't have time yet to do the derivation myself. I did however find two references that will hopefully convince you.

In the first paper (PRA 60, 1287 (1999)) a general equation for multiphoton transitions is derived. This is specifically for short pulse excitation. If you look at the second page you will see that they sum over all intermediate states. The physical interpretation of this process is that for a very short time the atom is actually excited to the intermediate state even for a large detuning. This is possible due to the uncertainty. If there were no intermediate states then there was also no excitation. This concept is also seen in the equation itself where you can see that the transition amplitude depends on the detuning from resonance. If you believe this equation then you would have to agree that if there was no intermediate state then the excitation probability would go to zero. This makes perfect sense to me because the ground and excited states cannot be excited by a dipole field so there must be a mediating intermediate transition.

I know that i've repeated most of the arguments i've given before but to me this makes perfect sense. If you are still not convinced the i suggest you read chapter 2 of "theory of multiphoton processes" by F. Faisal. He derives the TPA equation from first principles. To be honest this chapter is pretty difficult and i am not sure i could reproduce this derivation. Maybe it is more appealing for you.

I hope this help. I need to submit my PhD thesis in the coming 9 months so by then i hope to have the full derivation. I'll be happy to send you a copy when it's done.

dear Itan

Thank you for your time to answer me with such an extensive letter, given that you have little time left before your "promotie".

Unfortunately I have no access to physics abstrac (anymore, after retiring). As for the extremely -hort pulse explnation, yes the Heissenberg relation provides the uncertainty in time versus energy and one can explain the virtual transition this way too. I will await your theis and would welcome a copy; wish you success at the defence

Harry,

No problem.

The PRA i was mentioning can be found at http://www.weizmann.ac.il/home/feyaron/PDFfiles/PRAmultiphoton.pdf

"In two-step process the 2cd quatum should be absorbed during the lifetime of the excited molecule at the intermediate level. For two-photon process the simultaneous action of two light quanta is essential: the irradiated molecule gets excited only after the simultaneous absorption of two photons. Even more intensity laser radiation is needed (109-1011 W/cm2) " D.N. Nikogosyan, Meas. Sci.Technol.18 (2007)R1-R29.

For a fixed fluence and variable duration. We can compare the two processes thanks to the probabilities per illuminate surface. The intrinsic process (two photon) is dominant regarding the defect induced-process (two-step). G. PETITE, Lasers et Technologies Femtosecondes (2005) 319.

Karima

Please read the total thread before you participate. The question was NOY about a real 2-step excitation but whether the use of intermediate states is essential in explaining the simultaneous absorption of two photons in real TPA, that is a situation where the two photons do not give rise to a real excitation of an intermediate state that can be reached by that photon

I've just gave some references and underlined the difference between two processes. In G. PETITE, Lasers et Technologies Femtosecondes (2005) 319. We can find some formula and more explanation and for sure the answer. (The probability per illuminate surface unit for two photons process depends on the population in the valence band, the fluence, the pulse duration… but not on the population in the intermediate state).

Karima

You do not give any details in your answer as asked in the original question and especially the long paragraph added to the question

 Itan and Harry

Will TPA occur if there is no intermediate state between the initial state and final state? It means that the intermediate state is below the intial state or above the final state. In the case, energy detuning is larger than two-photon energy.

Yes: two-photon absorption can occur when the light is intense enough; it is a transition that is strictly forbidden for a single photon.

 Harry， Thanks a lot. For the case of big energy detuning, two photon Rabi oscillation is very weak, although two photon absorption can occur. Right?