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This page:

https://blog.afach.de/?p=62

Discusses the error Neil deGrasse Tyson made when talking about electronic transitions (video included there). Tyson clearly said in his Cosmos series that electrons disappear from one level and appear in the other one.

The guy in the page discusses that this is wrong, because there is no proof of that, and because it breaks simplest rules of relativity, and because it contradicts the simplest quantum model.

Is his argument correct? And why would Tyson do such a horrible mistake?

For more information please visit the link.

Brandon Enright
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The Quantum Physicist
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    And why would Neil Tyson do such a horrible mistake? Oversimplification. Btw, I've skimmed article and there is a mistake: the wavefunction is not zero only at infinity. – jinawee Aug 03 '14 at 15:11
  • Perhaps this page http://en.wikipedia.org/wiki/Atomic_electron_transition will help clarify "reality" ? – Carl Witthoft Aug 03 '14 at 15:12
  • @jinawee Where else? – The Quantum Physicist Aug 03 '14 at 15:20
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    For the hydrogen atom in the ground state, the wavefunction is zero at the center of mass. In general, you would have to use numerical methods, see this chart: http://teacher.pas.rochester.edu/PHY237/Exams/Exam2/Exam2_files/image008.png Anyway, I don't think "to teleport between energy levels" has sense in modern QM, since energy levels aren't identified with any position. – jinawee Aug 03 '14 at 15:24
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    @jinawee: the graphs you show are the probability of finding the electron between $r$ and $r + dr$, which is $P(r) = \psi\psi^4\pi r^2$. The wavefunction* is not zero at $r = 0$. For example the $1s$ is $\psi(r) = Ae^{-r/a_0}$ and this is actually a maximum at $r = 0$. – John Rennie Aug 03 '14 at 16:07
  • As JohnRennie said, the correct graph would be: http://ojensen.files.wordpress.com/2010/08/hydrogen_functions.png – jinawee Aug 03 '14 at 16:19
  • Not my field of specialty, and I don't have the book the blogger references, but ... I'm fairly sure that those calculation are for an atom in the presence of an externally imposed electromagnetic field which is treated as a perturbation, meaning that we still label the states as the (pure) atomic they most resemble. But the Hamiltonian is changed so that there is a time-evolution operator between them; which is emphatically not the case for a pure, pristine, ground-state, hydrogen-like atom in an empty universe. So, there is a little bit of apples and oranges going on there. – dmckee --- ex-moderator kitten Aug 04 '14 at 21:01

2 Answers2

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Tyson claims that an electron disappears from one orbital and appears in another and claims that this is like going from the second floor of a building to the fourth floor without existing in between. This doesn't actually happen. What happens instead is that each possible state of the system has a continuous amplitude associated with it. In a transition from state 1 to state 2, the amplitude of state 1 continuously decreases over time while that of state 2 continuously increases. These states overlap in space and so there is no mysterious magical mystery about an electron hopping from one place to another or anything like that. There is just continuous evolution of continuous amplitudes of the discrete set of states in which the electron can be found upon measurement.

The confusion about this kind of issue is a result of the mess that is usually made of discussing how quantum mechanics should be understood when it comes to measurement. The standard story is that the state of a quantum system jumps or collapses into one of the possible outcomes on measurement, but this is false. A more accurate description goes like this. A measurement has to create information that can be freely copied so that where is was previously present only in one system it latterly becomes present in many. It is not necessarily true that the system that was originally measured has its state copied. The state can be wiped instead, it can be reset to some default state. But the measurement result itself has to be able to be copied, otherwise you can't discuss it because to know about it you have to copy it into your brain. And information that can be copied in this way is discrete and the copying process prevents interference, see

http://arxiv.org/abs/1212.3245.

Even this paper is not as clear as it should be. The author should state that in fact the state doesn't jump or collapse, it just evolves according to the appropriate equations of motion. The observer is often present in multiple states after the measurement but he can't experience any state other than the one he is in because the copying process prevents interference between the different versions of the measurement result.

See also

http://arxiv.org/abs/quant-ph/0104033.

alanf
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  • I can't follow your argument following "A more accurate...". But the argument is probably too long for a short paragraph. And I haven't read the papers. But: is it not true that the time evolution you talk about refers to an ensemble of identically prepared systems? The argument does not apply to a single atom. My understanding, which could be just wrong, is that no time evolution can be described for a single atom. (?) Saying that it's mysteriously instantaneous is the best we can do. (?) – garyp Sep 09 '14 at 12:54
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    No. The time evolution describes the evolution of a single system. The square amplitudes can in general only be measured by measurements over multiple systems. However, there is no explanation of a single particle interference experiment in which each particle only goes through one slit at a time, see "The Fabric of Reality" by David Deutsch, Chapter 2. – alanf Sep 09 '14 at 14:55
  • +1 Good answer: I gave a quantitative version of your answer for a very like question to this one; see http://physics.stackexchange.com/a/86486/26076 – Selene Routley Sep 15 '14 at 12:13
  • Thank you for putting forward the time evolution theory for the transition between quantum states. I believe it is the correct theory but I am routinely and massively downvoted when I try to make that case. One of the consequences of the time evolution picture is that there is an oscillating charge distribution during the transition. I claim that all the ordinary thermal interactions between matter and radiation (including the black-body spectrum) are explained by applying Maxwell's equation to that oscillating charge. Do you agree? – Marty Green Oct 04 '16 at 18:54
  • @MartyGreen No. Maxwell's equations don't include quantisation. Also, the black body spectrum is explained by quantisation. Read a basic book on statistical mechanics. – alanf Oct 05 '16 at 08:31
  • @alanf So you believe that there is a time evolution and you believe there is oscillating charge...but you don't think you'd get the correct value for the black-body spectrum if you applied Maxwell's equation to the oscillating charge that you get from purely mechanical considerations? – Marty Green Oct 05 '16 at 17:59
  • @martygreen The observable for charge does not represent the motion of a charge in the same way the Maxwell equations do. For each possible state it predicts the probability of finding the charge in that state. The charge does not move back and forth between those states. On a short enough scale there is no single fact of the matter about where the charge is. This is one reason why Maxwell's equations do not apply. To understand some of the basics of quantum mechanics, see "The Fabric of Reality" and "The Beginning of Infinity" by David Deutsch and https://www.youtube.com/watch?v=uTVLkk3bQIs. – alanf Oct 07 '16 at 07:47
  • To understand something about quantum field theory which raises other issues see "Quantum field theory for the gifted amateur" by Lancaster and Blundell. – alanf Oct 07 '16 at 07:47
  • @alanf what about the square of the wave function? You believe there is a superposition that evolves in time. You take the square of the wave function and treat it as though it represents a charge denstiy EVEN THOUGH YOU THINK IT REPRESENTS SOMETHING ELSE. Treat it as charge density. Then why can't you apply Maxwell's Equations to that "charge density"? Of course you can. – Marty Green Oct 07 '16 at 12:06
  • @MartyGreen The square amplitude of the wave function does not represent charge density. In a particular relative state, any particular electron will be detected as being in some particular state, not as being in all of the places where the square amplitude was non-zero. As for why you can't use Maxwell's equations, they represent charge in terms of continuous distributions that do not exist in the real world and so the equations are false. There are quantum replacements for those equations of motion that are different, see the QFT book in my previous comment. – alanf Oct 07 '16 at 12:27
  • I know you think the square of the wave amplitude doesn't represent charge density, but if you PRETEND that it does, then you can apply Maxwell's equations to that "pretend" charge density. And then you will get a "pretend" radiation. Now: what makes you so sure that the "pretend" radiation answer you get is false? Have you ever calculated it or seen anyone else calculate it? – Marty Green Oct 07 '16 at 12:33
  • @MartyGreen Maxwell's equations contradict quantum electrodynamics. Maxwell's equations predict that stable atoms are impossible and so are known to be false. Why would I use a theory that is known to be false to calculate stuff that can be calculated with a theory that has not been refuted? – alanf Oct 07 '16 at 13:18
  • So you haven't done the calculation. So you don't know if it gives the right or wrong answer. – Marty Green Oct 07 '16 at 13:29
  • And by the way, if you apply Maxwell's Equations to the ground state of the hydrogen atom IN THE WAY I SUGGESTED ( by "pretending" the square of the wave function gives you the charge density) you get the right answer...the atom is stable. So you are incorrect when you claim Maxwell's equations (applied in the way I suggest) predict that stable atoms are impossible. – Marty Green Oct 07 '16 at 13:30
  • Im familiar with 3D software, and would like to model a physically accurate model of atoms, how should the transition look like? Teleport? – eromod Dec 31 '17 at 15:53
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Your questions reminds me of the movie "a serious men" by the Coen brothers (you should watch it regardless of the main character being a college physics professor). As some of the comments said, it is an oversimplification or a much more complex phenomenon. The only way to get some idea of what is happening is to understand the equations of quantum mechanics. But even if you do, there still is the problem of interpreting what they really mean. And not all physicists agree on that. That is a subject of study in foundations or philosophy of physics. There is still no agreement in the physics community on how to interpret the meaning of the equations of quantum mechanics. Much less in how to explain them to the general public!

  • Balogni! “If you can't explain it to a six year old, you don't understand it yourself.”

    ― Albert Einstein

     – eromod Dec 31 '17 at 15:54