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In another post, we discussed the oscillating charge in a hydrogen atom and the weight of opinion seemed to be that there is indeed an oscillating charge when you consider the superposition of the 1s and 2p states. One of the correspondents (freecharly) went a little farther and said that Schroedinger believed this oscillating charge to be the source of radiation. I wonder if the actual calculation bears this out? Specifically, in the case of the hydrogen atom in this particular superposition, do you get the correct decay times for the superposition of states if you apply Maxwell's equations to the oscillating charge and assume that as the system loses energy by radiation, the "probability" flows from the 2p to the 1s state in accordance with the amount of energy remaining in the system?

EDIT: Some people are objecting in different ways to the basic premise of the question, so let me make it a little more specific: I am not asking if hydrogen atoms ACTUALLY EXIST in a particular superposition of these states. (I may ask that in another question.) What I am asking here is IF you take (just to be specific) a 50-50 superposition of the 1s and 2p states, and apply Maxwell's equations to the oscillating charge, AND you assume that as the atom radiates the probability drains from the excited state to the ground state in such a way as to maintain conservation of energy...IF you do all those things, do you get a result that is consistent with standard QM?

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Marty Green
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    You've got another problem - if you just initialize on the $2p$ state, which should give you the strongest emission, there's no charge oscillation and therefore no radiation. (Similarly, if you initialize in, say, $\sqrt{0.01}|1s⟩+\sqrt{0.99}|2p⟩$, the oscillation amplitude will be very small, and it will take a long while to get up to speed.) That said, this question would improve a fair bit if you specify the mechanism you have in mind for "as the system loses energy the probability flows from the 2p to the 2s state". Are you just assuming Schrödinger dynamics? Or something else? – Emilio Pisanty Nov 19 '16 at 17:56
  • The moot point Marty is that the hydrogen atom isn't really in a superposition of 1s and 2p states, it's in neither state. It's akin to a 1s state moving back and forth with 10.2ev worth of energy, but it isn't actually a 1s state. In similar vein it isn't actually a 2p state either. – John Duffield Nov 19 '16 at 18:28
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    You are posing a question ("I wonder if...") without any effort to research an answer. For someone of such high rep, that is a bad example to newcomers. Moreover, this is not a discussion site. – sammy gerbil Nov 19 '16 at 18:30
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    I am not concerned about the example I set for newcomers. – Marty Green Nov 19 '16 at 20:26
  • In the time-dependent perturbation theory to describe the transition probability between the energy eigenstates the time dependence of the square of the superposed energy eigenfunctions (corresponding to the oscillating dipole moment) is used to calculate the transition probability. See QM textbooks. – freecharly Nov 20 '16 at 04:33
  • So does the semi-classical calculation give the same answer or not? Post an answer if you can. – Marty Green Nov 20 '16 at 13:07
  • @Marty Green - The semiclassical calculation of transition probabilities under an external perturbation doesn't explicitly use the energy loss or gain by radiation. It gives the transition probability as a function of time. Also, there is no semiclassical theory for the spontaneous emission, except for the probability derived from the Einstein coefficients. – freecharly Nov 20 '16 at 15:20
  • When I said "the semi-classical calculation", I didn't mean the one in the textbooks. I meant the one I outlined in the question. Spontaneous emission has nothing to do with it because I specified the particular superposition. And the calculation I outlined specifically called for the use of the radiation loss in adjusting the s vs p coefficients in the course of the transition. – Marty Green Nov 20 '16 at 15:30
  • @Marty Green - I think that your question is interesting, but I don't know any "semi-classical" calculation of the decay time related to the energy loss due to the electromagnetic wave emission. In the usual quantum mechanical picture, there is also the problem what energy to assign to an arbitrary superposition of two energy eigenstates. The superposition with complex coefficients $c_1$ and $c_2$ only yields the probabilities $|c_{1}|^2$ and $|c_{2}|^2$ for measuring the energies $E_1$ and $E_2$, respectively, in this superposition state. – freecharly Nov 20 '16 at 17:53
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    @freecharly I am glad you find my question interesting. I had hoped my intention was clear that the calculation ought to be carried out by assuming the ACTUAL energy of the superposition should be taken as |c1|2E1 + |c2|2 E2, with c1 growing at the expense of c2 as the radiation carries energy away from the system. I think it's a fairly straightforward calculation and it's odd to me that, assuming you are knowledgeable in this territory, you aren't aware of anyone having published a calculation of this type. – Marty Green Nov 20 '16 at 18:00
  • @Marty Green - I understand your assumption. It is probably at odds with orthodox QM. I am not a professional expert neither in the history nor in conventional quantum theory or its alternative interpretations. But I think that there might be alternative interpretations to the so-called "Copenhagen interpretation" that might be closer to reality. So I keep an open mind in this matter. – freecharly Nov 20 '16 at 18:12
  • @MartyGreen Why do you think it's odd that it hasn't been done before? This is not a natural model (because it predicts no decay for a pure 2s state, and it posits a completely ad hoc back-action mechanism). That said, it is indeed an interesting question, and there are areas that use similar models (without the ad hoc back action), which can be justified under some suitable conditions. If I have time I'll write it up. – Emilio Pisanty Nov 22 '16 at 09:20
  • The superposition of a 2s with a 1s state is a spinning magnet if the two states have opposite spins. So it radiates classically. – Marty Green Nov 22 '16 at 12:45

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I suspect the results would be correct (at least, approximately), as Barut developed his "self-field electrodynamics" (see, e.g. http://phys.lsu.edu/~jdowling/publications/Barut89b.pdf) and claimed results very close to those of QED. In self-field electrodynamics, radiation is created by charge density related to the wave function in a standard way (for the Dirac field).

akhmeteli
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I am disappointed that no one in this discussion group has been able to post a definitive answer as to whether the semi-classical calculation, applying Maxwell's equations to the quantum-mechanically oscillating charge, gives the correct result for the emission of radiation from an excited hydrogen atom. I appreciate akhmeteli's reference to a related publication but it does not directly address this question. So I am going to have to answer this question myself to the best of my ability by demonstrating a "back-of-the-envelope" type calculation.

I said I wanted to consider the 50-50 superposition of the 1s and 2p states. So first we need to know the maximum dipole moment of the superposition. I found the result on this University of Texas website by Prof. Richard Fitpatrick. I think I am interpreting it correctly when I say that the maximum charge displacement is 0.4 angstroms (about 75% of the standard radius of the ground state).

Then we need the frequency of the oscillation. Of course, this is the difference frequency corresponding to the 10.5 eV energy difference of the states, or 1.6 x 10^16 rad/sec.

Now we can calculate the acceleration. The easiest way to do this is to pretend it is uniform circular motion and use w^2*r. I get an acceleration of 10^22/m-sec^2. (Since it is actually harmonic motion and not circular, this will give us an error factor of 2 in the final result.)

Now I just plug this acceleration into the Larmour formula. You can find the Larmour formula anywhere on the internet, but I have simply converted all the physical constants into numerical values, and it comes to

   **Radiated Power  =   6 x 10^-54 a^2**

You can see that when I plug my value for acceleration into this formula I get a total radiate power of 6 x 10^-10 watts. This we divide by 2 to account for the harmonic motion vs circular.

Is this the correct power? We have to convert to "transition time" to find out. The total energy of the excited state is 10.5 (call it 10) eV which comes to 1.6 x 10^-18 Joules. Dividing the energy by the power, we get the lifetime of the excited state as just about 5 nanoseconds. Or maybe I'm wrongt about the energy and I should be taking it as half (because of the superposition) which would then give me a lifetime of 2.5 nanoseconds. Something like that.

This may not be exact but I think it's pretty much in the ballpark.

Marty Green
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  • Hard to believe how quickly the downvotes come in on this answer. Is there something wrong with my physics? – Marty Green Nov 21 '16 at 14:36
  • In the meantime I have found that in chapter 9.3.2 Lifetime of an Excited State in the the textbook of D. Griffith "Introduction to Quantum Mechanics" there is, for the case of a harmonic oscillator, a calculation of the power of emitted radiation according to QM and according to classical electrodynamics. – freecharly Nov 21 '16 at 18:19
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    You should not forget that the "lifetime", which is also given in the linked website to be 1.6ns for this transition, is actually the time until a large number $N_0$ of excited H atoms has decayed to $N_0/e$. This lifetime, which assumes a sudden quantum jump for the transition, is not the same as the one you are calculating with the emitted energy consideration. – freecharly Nov 21 '16 at 18:38
  • Whether it's exactly the same lifetime or not, it's pretty close. And the "sudden quantum jump" is of course, an assumption just like my continuous radiation. Both assumptions are models which guide us as how to do the calculation, and neither seems to be experimentally provable versus the other one. – Marty Green Nov 21 '16 at 18:59
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    Actually, it should be possible to experimentally differentiate between these models. If the classical radiation model were correct, then all atoms should decay with the same time. Quantum mechanically some decay much earlier and some much later. You also see this in radioactive decay. While the half-life gives you the decay of a large number of atoms to a half, there are many atoms that decay only after an orders of magnitude longer time. – freecharly Nov 21 '16 at 19:08
  • You gotta think real careful about just how you'd set up an experiment to demonstrate this supposed difference. I don't think you can do it. – Marty Green Nov 21 '16 at 19:18
  • I shortly googled and found this article on lifetime measurements of excited states: http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-14754.pdf – freecharly Nov 21 '16 at 20:12
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    This answer shows that you get answers in the ballpark order of magnitude, but that is not as unexpected as you think - both are (relatively) simple theories with few dimensional constants, being used in relatively similar ways, so there isn't all that much room for them to disagree. In any case, you're missing the important part: for the model to be of any use at all, it needs to directly predict an exponential decay, and it's not clear that it can (because the oscillation amplitude of the dipole decreases as the superposition changes). – Emilio Pisanty Nov 22 '16 at 21:24
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    I also don't know where your disappointment comes from. The question as posed is interesting but it asks for a fairly substantial calculation, which is barely explored in the literature (and for good reasons), so answering this takes time. – Emilio Pisanty Nov 22 '16 at 21:26