Atomic dipole force

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Hello! I read that if we have a 2 level system (say an atom) and we make it interact with a laser whose frequency is detuned from the resonant transition frequency of the system, the atom experiences an effective potential/force. In the classical description it makes sense, but quantum mechanically I am not sure what is going on. How does the atom interacts with the laser, given that it is far detuned? I initially thought that this is due Raman stimulated emission, such that the light can excite a virtual level and make it decay by stimulated emission back to the ground state. But it seems like that violates the conservation of energy. For example if we have a photon traveling in ##+z## direction doing the excitation and the one emitted by stimulated emission goes in ##-z## (if the emitted one is in the ##+z##, the net force would be zero and we would have no effective force), for the whole atom-laser system, we are just as in the beginning (atom in the ground state and 2 quanta coherent in the field), but it seems that now the atom gained a momentum, equal to twice the momentum of the photon, and hence a kinetic energy and I am not sure where this is coming from. In Raman transition, usually we have 2 lasers either of different frequency or different power. But in all the examples I see for dipole force they use a build-up cavity with the atoms trapped at the center. But in that case we have the same power and frequency detuning, as it is basically the same laser, reflected from some mirrors and I am not sure how we get energy conservation. Can someone help me understand this? Thank you!
 

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  • #2
Twigg
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Think about the mixed photon-atom states (the "dressed states"). The dressed states energy is shifted by the dipole interaction, and this shift scales with the intensity. Another way to say this is that the dressed states energy shift scales with photon density (intensity ##= h\nu c \times## photon density). So, if there's more photons smacking into an atom one on side than on the other side (gradient in photon density), there's a net force.
 
  • #3
hutchphd
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If I understand the original question the answer is simply momentum conservation. In the Center of Momentum frame the free atom's initial state will not be a stationary nucleus, and the correct Quantum description needs to include this and the energy.
This momentum conservation is the same as classical (but of course it is in a QM framework)
 
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If I understand the original question the answer is simply momentum conservation. In the Center of Momentum frame the free atom's initial state will not be a stationary nucleus, and the correct Quantum description needs to include this and the energy.
This momentum conservation is the same as classical (but of course it is in a QM framework)
I am not sure I understand this. Initially, for an atom with momentum p (say in the lab frame) and 2 photons, the total energy would be ##2E_\gamma + p^2/2m## assuming non-relativistic motion. After the interaction, we get the 2 photons back, the atom is back in the ground state, but now the atom received a momentum recoil equal to twice the photon momentum. So after the interaction the energy is ##2E_\gamma + (p+2p_\gamma)^2/2m##. What am I missing?
 
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hutchphd
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I'm confused. Two photons? I may not be understanding
 
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I'm confused. Two photons? I may not be understanding
Oh sorry, in the example I gave (which I am not sure if that is what is actually happening, this is what I am trying to figure out actually) you have one photon that gets absorbed by the atom and a second one that produces the stimulated emission of the first photon (you need 2 photons for a Raman transition).
 
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Twigg
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I am not sure I understand this. Initially, for an atom with momentum p (say in the lab frame) and 2 photons, the total energy would be ##2E_\gamma + p^2/2m## assuming non-relativistic motion. After the interaction, we get the 2 photons back, the atom is back in the ground state, but now the atom received a momentum recoil equal to twice the photon momentum. So after the interaction the energy is ##2E_\gamma + (p+2p_\gamma)^2/2m##. What am I missing?

You had it right in your first post: the net momentum gained is zero after one absorption and one emission. An electric field with constant intensity will not result a dipole force, it will only change the energy of the atom by a constant amount ("light shift"). When the intensity varies in space, then you will see a force (gradient of intensity -> gradient of energy = a force). You need a gradient to make a dipole force.
 
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You had it right in your first post: the net momentum gained is zero after one absorption and one emission. An electric field with constant intensity will not result a dipole force, it will only change the energy of the atom by a constant amount ("light shift"). When the intensity varies in space, then you will see a force (gradient of intensity -> gradient of energy = a force). You need a gradient to make a dipole force.
Actually now I am confused about this picture of photons absorption and emission. I would expect that, if I have an intensity gradient, the force would be from the region with high intensity, towards the one with low intensity. However, the direction depends on whether the light is blue or red detuned (and there is no force on resonance, even if there is a gradient in the laser intensity). How exactly should I think about this force in this case. How can the atom be pushed towards a region with more photons?
 
  • #9
Twigg
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I think the confusion about how atoms can be accelerated from regions of low photon intensity to regions of high photon intensity stems from comparing the atom-photons system to kinetic theory of gases. As you alluded to, the effect of a photon on the atom's energy levels is given by a "light shift" (aka AC Stark shift). The energy of the ground state (in the rotating frame, for convenience) is given by $$E_g = -\frac{\hbar \delta}{2} - \frac{\hbar \sqrt{\delta^2 + \Omega^2}}{2}$$ where ##\Omega## is the Rabi rate on resonance and ##\delta = \omega - \omega_0## is the detuning. Note that for resonant light, we have ##E_g = - \frac{\hbar\Omega}{2}##. So yes, there can still be a dipole force on resonance, but keep in mind that when you're on resonance the atoms won't be mostly in the ground state anymore, so you have to calculate the force on the excited state separately. If you're deeply saturated (excited and ground state equally populated), then yes there will be no force because the atoms will spend equal time sampling equal and opposite forces. In the far-detuned case, there is an approximate expression, which I believe is what you had in mind when you pointed out the change in sign of the force with detuning: $$\Delta E_g \approx - \frac{\hbar \Omega^2}{4\delta}$$ where ##\Delta E_g## refers to only the change in energy.

The more photons there are in a given space, the higher ##\Omega## is. If the effect of those photons is to reduce the energy of the ground state (as it is for red detuned photons), then more photons means lower energy. The atom will naturally go to the lowest energy. For the dipole force, I tend not to think about absorption and emission, but instead just changes in potential energy. Less pitfalls that way.

Edit: There are very clever manipulations of the dipole force where you DO have to think about absorption and emission, but these are special cases. Their characteristic feature is that they involve non-monochromatic light or short pulses (or both, as in frequency chirps). Examples of these include the bichromatic force and the rapid adiabatic passage force (ARP force).
 
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I think the confusion about how atoms can be accelerated from regions of low photon intensity to regions of high photon intensity stems from comparing the atom-photons system to kinetic theory of gases. As you alluded to, the effect of a photon on the atom's energy levels is given by a "light shift" (aka AC Stark shift). The energy of the ground state (in the rotating frame, for convenience) is given by $$E_g = -\frac{\hbar \delta}{2} - \frac{\hbar \sqrt{\delta^2 + \Omega^2}}{2}$$ where ##\Omega## is the Rabi rate on resonance and ##\delta = \omega - \omega_0## is the detuning. Note that for resonant light, we have ##E_g = - \frac{\hbar\Omega}{2}##. So yes, there can still be a dipole force on resonance, but keep in mind that when you're on resonance the atoms won't be mostly in the ground state anymore, so you have to calculate the force on the excited state separately. If you're deeply saturated (excited and ground state equally populated), then yes there will be no force because the atoms will spend equal time sampling equal and opposite forces. In the far-detuned case, there is an approximate expression, which I believe is what you had in mind when you pointed out the change in sign of the force with detuning: $$\Delta E_g \approx - \frac{\hbar \Omega^2}{4\delta}$$ where ##\Delta E_g## refers to only the change in energy.

The more photons there are in a given space, the higher ##\Omega## is. If the effect of those photons is to reduce the energy of the ground state (as it is for red detuned photons), then more photons means lower energy. The atom will naturally go to the lowest energy. For the dipole force, I tend not to think about absorption and emission, but instead just changes in potential energy. Less pitfalls that way.

Edit: There are very clever manipulations of the dipole force where you DO have to think about absorption and emission, but these are special cases. Their characteristic feature is that they involve non-monochromatic light or short pulses (or both, as in frequency chirps). Examples of these include the bichromatic force and the rapid adiabatic passage force (ARP force).
Thank you for this explanation! I understand the math of it, and if I think in terms of potential energy it makes perfect sense. However, I am not sure how the interaction between atom and laser happens (in normal cases, not the special ones you mentioned). That potential energy (or force) which lowers (or rises) dressed states, in a QFT formalism, must come from an exchange of particles (as any physical interaction). I am not sure how the interaction happens here at a particle level. Do the photons actually get absorbed (they are far detuned but maybe the flux is big enough)? Do you have stimulated Raman transition? How does the atom know about the existence of the EM field i.e. how does the atom experience the potential energy created by the EM field? As you said, in my mind there is something along the lines of kinetic theory of gases, which obviously doesn't work for blue detuning. But I am just not sure I understand what exactly happens at a particle level and how should I think about it.
 
  • #11
Twigg
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Sorry for the slow reply. I slowed down a bit, reviewed your question and I believe found a satisfying answer to your question. Sorry it took me 4 replies before I actually read things carefully and got it right o0) Hope I didn't cause you too much headache

For example if we have a photon traveling in +z direction doing the excitation and the one emitted by stimulated emission goes in -z (if the emitted one is in the +z, the net force would be zero and we would have no effective force), for the whole atom-laser system, we are just as in the beginning (atom in the ground state and 2 quanta coherent in the field), but it seems that now the atom gained a momentum, equal to twice the momentum of the photon, and hence a kinetic energy and I am not sure where this is coming from.

Your example requires that you have two counterpropagating laser optical modes for the atom to interact with, since you have photons going in both +z and -z directions. The simplest case is a standing wave in z: $$\vec{E}(\vec{x},t) = \vec{E_0} (\cos(kz + \omega t) + \cos(-kz + \omega t)$$

So where does the momentum go? The +k photon mode loses one photon worth of momentum (that's -1 photons times ##+\hbar k##), the -k photon mode gains one photon worth of momentum (that's +1 times ##-\hbar k##), and the atom gains 2 photons worth of momentum in the +z direction, as you said. Overall, that's ##(-1)(+\hbar k) + (+1) (-\hbar k) + 2\hbar k = -2\hbar k + 2\hbar k = 0##.

The key here is that the atom can interact with multiple optical modes while staying in the ground state at the end of the day. If you want to think of it as a Raman transition, then it'd be like a "lambda" (two low energy states, one high energy state) system where the atomic excited state is the go-between. One photon ends up switching sides from the ##+z## mode to the ##-z## mode.

I am not sure how we get energy conservation
I couldn't find an answer to this in Cohen-Tannoudji. Here's what I can say:

The total Hamiltonian of the atom-laser system is given by $$H = H_L + \frac{p^2}{2m}+ H_a + V$$ where ##H_L## is the Hamiltonian of the photons before interaction with the atom, ##\frac{p^2}{2m}## is the Hamiltonian associated with the kinetic energy of the atom, ##H_a## is the atomic structure Hamiltonian before interaction, and ##V## is the interaction term. Per the previous example, one photon goes from the +z mode to the -z mode with the same wavenumber k, so there is no change in energy associated with ##H_L## since the number of photons and their frequency hasn't changed. Also, the atom gained a momentum of ##+2\hbar k##, so if we assume the atom started at rest then ##\frac{p^2}{2m}## increases by ##\frac{4k^2}{2m}##. The energy associated with ##H_a## doesn't change because the atom stays in the ground state the whole time. By process of elimination, that leaves a change in the interaction term ##V##.

When the atom gains momentum ##2\hbar k##, there is an associated Doppler shift such that the detuning changes as $$\delta \rightarrow \delta_{\pm} = \delta \pm kv$$ where ##\delta_+## refers to the detuning of the beam that propagates in +z, and vice versa for ##\delta_-##. In this case, ##v = p/m## and ##k = p/\hbar##, so $$\delta_{\pm} = \delta \pm \frac{p^2}{m \hbar}$$. Throw in a factor of ##\hbar## and you see an energy scale of ##\hbar \delta_{\pm} = \hbar \delta \pm \frac{p^2}{2m}##. You can use plug this into the expression for the E-field of the standing wave, evaluate ##\langle V \rangle = \langle -d \cdot E\rangle##, and that energy shift will compensate for the change in kinetic energy. I'd do the derivation here, but unfortunately I don't have time to finish that right now. If you're still interested, let me know in a reply and I'll get around to it when I can.
 

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