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How does an Electron 'jump' from one orbit to the other?

  1. Oct 9, 2009 #1
    Well, the question is that what happen when an electron changes orbits? if it cannot have states in between those orbits, how can it 'move through' those restricted areas? Do the go through those states? or just disappear and reappear without any time lapse, at two different places??

    please help? can it be explained without QFT?
  2. jcsd
  3. Oct 9, 2009 #2


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    Staff: Mentor

    An atomic electron does not have discrete spatial orbits like the orbits of planets around the sun. Its states have discrete energies, and probability distributions for position. See for example


    for graphs of the probability distributions. You can see that the probability distributions for the various states overlap a lot. So don't think of a change in energy as corresponding to a spatial "jump."
  4. Oct 13, 2009 #3
    But isn't it the case that the allowable atomic energy levels are solutions to the steady state wave equation?

    What about the time dependent case of a transition from one energy level to another? Does the electron probability distribution morph smoothly from one distribution to another during a change in energy levels?
  5. Oct 13, 2009 #4
    It's as you say. An electron can move slowly from one energy state to another, or relatively fast. Once raised to a metastable state (a relatively stable raised energy state), the probability that it will be found in the lower energy state increases slowly over time. This is a slow transition.

    The difference manifests as bandwidth of the emitted radiation. The metastable transition has the higher uncertainty in the time of emission and produces the more monochromatic radiation.
  6. Oct 14, 2009 #5
    In that case I'm wondering what the probability density for the electron looks like during transitions.

    We've all seen the 3d spatial probability densities for electron orbitals, but what about a 3d visualisation of the probability density during a transition? Has anyone seen such a thing?
  7. Oct 14, 2009 #6
    I think perhaps your question is better phrased, "How fast does a quantum transition occur?". In which case, I don't know that there is a definitive answer... other than infinitely fast.

    Remember, when you talk about changing states in an atomic system, you have to talk about changes quantum mechanically. Therefore the changes between states do not occur continuously, but in discrete quantum intervals. If you add energy to a system, it has to come in the form of a packet of energy (say you are adding energy with light in which case you can talk about the interaction in terms of photons of discrete energy [tex]\hbar\omega[/tex]). Therefore it doesn't make much sense to talk about the time 'between' transitions, since there is no resolvable energy state between the point at which you are in state [tex]|a>[/tex] and after the transition when you are in state [tex]|b>[/tex], since the exchange in energy had to happen in a discrete quantum of energy.
    Last edited: Oct 14, 2009
  8. Oct 14, 2009 #7
    Instead, consider the state (3/5)|a> - (4/5)i|b> .
  9. Oct 14, 2009 #8
    Perhaps you could elaborate a bit?
  10. Oct 14, 2009 #9
    The probability of finding the electron in the raised energy state decays (to first approximation) exponentially. This means there is a characteristic time tau.

    Probability(E2) = e-t/tau

    Have you wondered, "how wide is the spectral line?" of some particular orbital transition. Spectral line width is dependent on a few things. There is broadening of the line due to Doppler. The atoms are moving around thermally. Some are approaching and some are receding from the sensor. Some photons are red shifted and some blue.

    But the line width we are interested in, is due to the transition time, tau. The longer the transition time the more monochromatic the radiation.

    I could guess at the superposition a|E1> + b|E2> as a function of time, but I'm not the one to say anything much further on this subject without getting into trouble if I haven't already,

    so maybe someone who knows this stuff will show up. :smile:
    Last edited: Oct 14, 2009
  11. Oct 14, 2009 #10
    I think you are confusing the concept of lifetime with transition time. The probability does indeed change through time, but that doesn't mean the state is changing, only the probability of finding the system in a specific state.

    If an atom is excited via a transition to an unstable state, it will decay over time, but if you measure the state of the atom during that lifetime, there is no point at which you will find the atom in a partial transition state (that is to say having energy between that of state a and b). It will either be excited or it won't, the probability of finding it in such a state is the time dependent part.

    It may be that I have misinterpreted the question here. Is what is being asked what does the time evolution of the transition probability look like? I was under the impression the question was "What does the probability look like during photon absorption/emission"

    Also I still don't understand your previous comment about the (3/5)|a> - (4/5)i|b> state, could you please explain what the point of this is?
  12. Oct 14, 2009 #11


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    Staff: Mentor

    During the transition you have a state which is a linear combination of the initial and final states:

    [tex]\Psi(x,t) = a(t)\Psi_{initial}(x,t) + b(t)\Psi_{final}(x,t)[/tex]

    where a and b vary with time such that at the beginning of the transition a = 1 and b = 0, and at the end of the transition a = 0 and b = 1. The probability distribution oscillates or "sloshes" with frequency [itex]f = \Delta E / h[/itex].

    Doing the actual calculation for, say, a = b = 0.5 is kind of messy for actual atomic orbitals, but you can get the basic idea by doing the calculation for two states of the one-dimensional infinite square well that everybody does as their first example or exercise in solving the Schrödinger equation. Find the probability distribution [itex]\Psi^*(x,t) \Psi(x,t)[/itex] for, say, [itex]\Psi(x,t) = a \Psi_1(x,t) + b \Psi_2(x,t)[/itex] and see what you get!
    Last edited: Oct 15, 2009
  13. Oct 14, 2009 #12
    Thanks for bailing me out jtbell, but isn't this the case of a hybrid orbital, rather than an orbital transition? It's perfectly consistant with the question, so this begs the question, what exactly was the question the questioner had in mind?
  14. Oct 15, 2009 #13
    Apparently someone has seen such a thing. This link talks about attosecond spectroscopy.

    http://www.attoworld.de/attoworld/slowmotion.html [Broken]

    I've been trying to find answers to why their youtube video of the transit looks like concentric rings and not the orbital cloud that everyone has been talking about.

    Here is the animation:

    That company and scientists claim that the transit is within the attosecond domain of time. Any imaging device that is slower than that would produce "blur," is their implication.
    Last edited by a moderator: May 4, 2017
  15. Oct 15, 2009 #14


    Staff: Mentor

    This doesn't make sense to me. The state determines everything that we can measure about the system including the probability, so if the probability has changed then the state must have changed.
  16. Oct 15, 2009 #15
    I assume you mean the total state vector which would be a (normalized) linear combination of the eigenstates of the system. In which case you are indeed correct.

    My use of the term state was more in the sense of, at the time of measurement either the atom is excited or it isn't. While the wave function tells you the probability of finding the atom in one of its respective states (excited or ground lets say), when you actually perform the measurement on the system it can't remain in more than one state, but must be in one of the possible states (i.e. those with non-zero probability).

    So when I said the state isn't changing I meant that just becuase the probability is changing through time doesn't imply that the actual photon absorption/emission process has occured yet, it simply means that you are more likely to find the atom in an excited/unexicted state the longer you wait (barring any additional external interactions).

    For Example:
    Consider a two level atom in the presense of an electromagnetic field. Then the state vector (or wave function) of the system is given by [tex]\Psi(\vec{r},t)=C_{1}(t)\psi_{1}(\vec{r},t)+C_{2}(t)\psi_{2}(\vec{r},t)[/tex] where [tex]C_1,C_2[/tex] are the (normalized) probabilities (well [tex]|C|^2[/tex] really) of finding the system in state 1 or 2 respectively, and are in general time dependent. [tex]\psi_1,\psi_2[/tex] are the eigenstates of hamiltonian for a two level system. Now if we start with an atom in the ground state then [tex]C_1[/tex] (we will call state 1 the ground state) will be equal to 1 and [tex]C_2=0[/tex]. Once we switch on the radiation field the probability of finding the system in state 2 now increases with time (lets ignore the effects of the atom reradiating for the moment). However just becuase the probability of finding the system in state 2 increases with time does NOT imply that a measurement on that system always will yield an atom in state 2 (hence the atom hasn't changed it state in such a case, its still in the ground state). The only time at which it becomes certain that the system is in state 2 and there has infact been a transition is when [tex]C_2=1[/tex] and [tex]C_1=0[/tex]. At this point we can stay for sure that the atom has changed its state from state 1 to state 2. Of course in practice this doesn't happen since the atom will generally reradiate before the probability reaches 1.
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