Time scale of the atom


by Niles
Tags: atom, scale, time
Niles
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#1
Feb27-12, 11:56 AM
P: 1,863
Hi

I am reading about the force of a coherent EM-beam acting upon an atom, and I have a question in this regard. It is regarding the explanation on page 150 of this book, starting from "The geometric approximation of atom optics is valid when": http://books.google.dk/books?id=SUBH...20atom&f=false. It is only the first part of that page.

As far as I understand, what they try to tell us is that in order to treat the atom as a classical particle, the time it takes for the internal state to change (1/Gamma) has to be very short compared to the time it takes for the external dynamics to change. That is at least what the inequality says.

Physically I don't see why this condition must be satisfied. Does it simply mean that the atom has to be in equilibrium at all times?

Best regards,
Niles.
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M Quack
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#2
Mar1-12, 07:57 AM
P: 640
I can't see the page of the book, but your understanding is correct.

Another way to look at this is to consider the energy scale of things.
[itex] E = \hbar \omega = \frac{h}{T}[/itex]

If the time scale of the atom is much faster than the external perturbation, then the perturbation will be on a much smaller energy scale than what's going on in the atom. Hence the atom will not be affected significantly.

This is of course not an exact relation, it just gives you a rule of thumb of what is relevant and what can be neglected.
Niles
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#3
Mar1-12, 11:32 AM
P: 1,863
Thanks for replying. That relation makes good sense. Maybe this link works better (it is on page 150 of the book): http://books.google.dk/books?id=SUBH...page&q&f=false

The authors mention that this corresponds to the internal atomic dynamics following the center-of-mass motion of the atom adiabatically. When I hear "adiabatic", I would believe that it means that the atom is in the same internal state during its center-of-mass motion. The above inequality states that the internal dynamics are very fast, so isn't it wrong to say that there is adiabatic following? It is the opposite of adiabatic following, since the internal state changes very fast during the external motion (?).

Thanks in advance.

Best,
Niles.

M Quack
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#4
Mar1-12, 11:57 AM
P: 640

Time scale of the atom


OK, that link works.

adiabatic in this context means without transfer of energy.

Note the formula they give,
[itex]\omega_{\mathrm{rec}} = \frac{\hbar k^2}{2m} \ll \Gamma[/itex]

If you multiply both sides (all 3 sides :-) ) by [itex]\hbar[/itex], the formula compares 3 energies.

[itex]\frac{\hbar^2 k^2}{2m}[/itex] is a kinetic energy

[itex]\hbar \omega[/itex] is the energy of an oscillator or wave (written as such to derive a characteristic time scale from the energy)

[itex]\hbar \Gamma[/itex] is the width of an emission line (for example), which is finite because of the finite life time of the initial state.
Niles
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#5
Mar1-12, 12:51 PM
P: 1,863
Quote Quote by M Quack View Post
Note the formula they give,
[itex]\omega_{\mathrm{rec}} = \frac{\hbar k^2}{2m} \ll \Gamma[/itex]

If you multiply both sides (all 3 sides :-) ) by [itex]\hbar[/itex], the formula compares 3 energies.

[itex]\frac{\hbar^2 k^2}{2m}[/itex] is a kinetic energy

[itex]\hbar \omega[/itex] is the energy of an oscillator or wave (written as such to derive a characteristic time scale from the energy)

[itex]\hbar \Gamma[/itex] is the width of an emission line (for example), which is finite because of the finite life time of the initial state.
Ah, thanks for making that clear.



Quote Quote by M Quack View Post
adiabatic in this context means without transfer of energy.
I have to admit I still don't fully understand "adiabatic" in this context. So the internal dynamics follows the external COM-motion without any energy transfer(?). I'm not even sure I know what that means.

Does it refer to the fact that the spontaneously emitted photons (characterized by [itex]\Gamma^{-1}[/itex]) are emitted *much* faster than the atom moves, so their effect is zero on average since they are emitted so often?


Thanks in advance.

Best regards,
Niles.
M Quack
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#6
Mar2-12, 02:42 AM
P: 640
Quote Quote by Niles View Post
I have to admit I still don't fully understand "adiabatic" in this context. So the internal dynamics follows the external COM-motion without any energy transfer(?). I'm not even sure I know what that means.
Yes. Below, I try to give a more intuitive example.

Does it refer to the fact that the spontaneously emitted photons (characterized by [itex]\Gamma^{-1}[/itex]) are emitted *much* faster than the atom moves, so their effect is zero on average since they are emitted so often?
It's more like the external perturbations are so weak that they do not excite any transitions in the atom. Whenever there is an emission line with a characteristic frequency or energy, there is a corresponding absorption or excitation process. What goes up must come down, so if the atom gets excited by some external stimulus it will eventually emit a photon and drop back into its ground state. That is very non-classical behavior, so the classical approximation is only good if you avoid that happening.

The concept of adiabatic changes is not limited to the quantum world.

Think of the characteristic frequency of the atom as a resonance. A weight on a spring for example, has a natural frequency. Hold the spring in your hand and let the weight bounce up and down. If you move your hand rapidly, the oscillation amplitude will increase. If you move your hand very slowly (on a time scale much slower than the oscillation period), the amplitude will not change noticably.
Physics Monkey
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#7
Mar3-12, 05:11 PM
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To use M Quack's example, imagine that you have a spring in your hand with a weight attached. The spring has some damping and the you can move your hand around in a spatially varying gravitational field.

You know that the equilibrium position of the spring depends on the local gravitational field. You also know that because of the damping the spring will reach this equilibrium position in a time that is roughly 1/(decay rate) provided you hold your hand still.

But suppose your hand does move slowly. A question you could ask is, how slowly should your hand move so that the spring is always in local equilibrium. It's reasonable to suppose that you would want a large decay rate compared to the timescale of hand motion so that you are effectively sitting in one place for much longer than it takes to reach equilbrium.

However, I'm also not sure this is precisely what this book is talking about. Looking at page 150 just above section 6.3, the book states that [itex] \omega_{rec} \ll \Gamma [/itex] which implies that [itex] 1/\Gamma \ll 1/\omega_{rec} [/itex]. However, just below that they state the internal timescale [itex] 1/\Gamma [/itex] should be MUCH SLOWER than the external timescale [itex] 1/\omega_{rec} [/itex]. I would interpret much slower to mean the internal time scale is longer than the external timescale, which gives the opposite inequality. Perhaps I misread or misunderstood or perhaps they meant much shorter?
Niles
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#8
Mar3-12, 06:01 PM
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Quote Quote by M Quack View Post
It's more like the external perturbations are so weak that they do not excite any transitions in the atom. Whenever there is an emission line with a characteristic frequency or energy, there is a corresponding absorption or excitation process. What goes up must come down, so if the atom gets excited by some external stimulus it will eventually emit a photon and drop back into its ground state. That is very non-classical behavior, so the classical approximation is only good if you avoid that happening.
I'm sorry, but I have to admit that I don't agree with the bolded part. In order to exert a force on the atom, the photons have to induce transitions in the atom.



Quote Quote by Physics Monkey View Post
However, I'm also not sure this is precisely what this book is talking about. Looking at page 150 just above section 6.3, the book states that [itex] \omega_{rec} \ll \Gamma [/itex] which implies that [itex] 1/\Gamma \ll 1/\omega_{rec} [/itex]. However, just below that they state the internal timescale [itex] 1/\Gamma [/itex] should be MUCH SLOWER than the external timescale [itex] 1/\omega_{rec} [/itex]. I would interpret much slower to mean the internal time scale is longer than the external timescale, which gives the opposite inequality. Perhaps I misread or misunderstood or perhaps they meant much shorter?
Thanks for that. I also paused when I read "much slower" for the first time. It must be a mistake by them, because the inequality is correct. I believe they also make another error when saying that (page 150, top): "... ensures that the internal quasi-stationary state of the atom follows the center-of-mass dynamics adiabatically". If they by "quasi-stationary" mean "equilbrium", then I agree.

Thanks for both your explanations of the mass-spring-hand system. It is a good analogy of the system

Best wishes,
Niles.
M Quack
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#9
Mar4-12, 02:21 AM
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Quote Quote by Niles View Post
I'm sorry, but I have to admit that I don't agree with the bolded part. In order to exert a force on the atom, the photons have to induce transitions in the atom.
No, you can perfectly accelerate a Na+ ion in a weak electric field without exciting any transtions. Mass spectrometers and residual gas analyzers do that all over the place every day.


Thanks for that. I also paused when I read "much slower" for the first time. It must be a mistake by them, because the inequality is correct. I believe they also make another error when saying that (page 150, top): "... ensures that the internal quasi-stationary state of the atom follows the center-of-mass dynamics adiabatically". If they by "quasi-stationary" mean "equilbrium", then I agree.
I also thinki it is a mistake in the book. 1/Gamma is not a time scale, it is an energy scale.

As for the mass-spring example, just try it out. Btw, it also works with a pendulum. If you wait much longer than the damping time, the everything will be in the ground state, no matter where you started. But if you move slowly compared to the oscillation period, the state of the oscillator will not change significantly, even if it is already in motion.
M Quack
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#10
Mar4-12, 04:22 AM
P: 640
Just found this here:

http://en.wikipedia.org/wiki/Adiabatic_theorem

Obviously, in a classical system there is no energy gap.
Niles
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#11
Mar4-12, 07:40 AM
P: 1,863
Quote Quote by M Quack View Post
No, you can perfectly accelerate a Na+ ion in a weak electric field without exciting any transtions. Mass spectrometers and residual gas analyzers do that all over the place every day.
I see, I was specifically talking about slowing atoms with light.

Quote Quote by M Quack View Post
I also thinki it is a mistake in the book. 1/Gamma is not a time scale, it is an energy scale.
I would say 1/Gamma is a time scale, if Gamma is the inverse of the lifetime of the excited state. I believe that is how Gamma is usually defined.


I think it is correct to say that in our case, the adiabatic approximation ultimately means that the external perturbation does not perturb the resonance condition of the atom.

Best,
Niles.
M Quack
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#12
Mar4-12, 08:41 AM
P: 640
Quote Quote by Niles View Post
I would say 1/Gamma is a time scale, if Gamma is the inverse of the lifetime of the excited state. I believe that is how Gamma is usually defined.
Oops. you are right. Too much back and forth between Gamma and 1/Gamma.
Niles
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#13
Mar5-12, 05:07 AM
P: 1,863
Thanks to both of you, I learned a lot from this.

Best,
Niles.
M Quack
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#14
Mar5-12, 05:43 AM
P: 640
My pleasure. It is always nice to see a qualified question that is not just a homework problem :-)


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