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what's the type of movement when the electron jump from level to another when gaining or losing a quantum of energy is it actually jump or move in a specific movement?

mar yam
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    What do you mean by "type of movement"? Are you aware that quantum mechanics does not really mesh with our classical notion of motion? – ACuriousMind Nov 16 '17 at 01:50

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Not only don't they "move" (as per BillyKalfus's answer), they don't even "jump". And jump's the far more common misconception (also as per Billy's answer:). Here's what more precisely happens...

Consider two states, (a) an initial state $\left|\alpha\right>$ where the electron, if measured, will definitely (with probability $1.0$) be found in a lower-energy state, and (b) a final state $\left|\beta\right>$ where the electron, if measured, will definitely (with probability $1.0$) be found in a higher-energy state.

Now, by "jump" you mean to suggest that there's some time $t_0$ such that the electron's state is $\left|\psi\right>=\left\{{\left|\alpha\right>,\; t<t_0\atop \left|\beta\right>,\; t>t_0}\right.$. That is, the electron's state discontinuously jumps from $\left|\alpha\right>$ to $\left|\beta\right>$ at time $t_0$.

But that's not what happens. Instead, $\left|\psi\right>$ is some smooth function of time that continuously evolves from $\left|\alpha\right>$ to $\left|\beta\right>$. That is, $\left|\psi(t)\right>=f(t)\left|\alpha\right>+(1-f(t))\left|\beta\right>$, for some $f(t)=\left\{{1.0,\; t<<t_0\atop 0.0,\; t>>t_0}\right.$ that smoothly goes from $1.0$ to $0.0$. So a measurement at an early time will more likely find the electron in state $\left|\alpha\right>$, and at a later time will more likely find it in state $\left|\beta\right>$.

Note that the the electron is never measured in some intermediate-energy state. It's always measured either low-energy or high-energy, nothing in-between. But the probability of measuring low-or-high slowly and continuously varies from one to the other. So you can't say there's some particular time at which a "jump" occurs. There is no "jump".

  • Do you think the OP (along with the vast majority of readers) would keep reading your answer past the unfamiliar mathematical symbols of the Dirac's ket notation with likely no idea of what it means $\left|\alpha\right>$ ? – safesphere Nov 16 '17 at 05:20
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    @safesphere Since you recognize Dirac notation, I assume you understand my answer. So, if you think it's too abstract for the likely audience, why don't >>you<< post an answer, explaining the same ideas in easier-to-understand language? To paraphrase your own comment: if all you're going to do is complain, why should anybody bother reading your complaint? –  Nov 16 '17 at 07:40
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    It was a question, not a complaint. You answer is actually very good (and I assume correct). Too bad most people may stop reading on the third line after seeing math symbols meaningless to most. Perhaps a one-line explanation could help, something like, "This is how the state (set of parameters) of a particle is noted in quantum mechanics." But it's totally up to you. – safesphere Nov 16 '17 at 07:59
  • Do we ever have an electron whose energy state has probability 1.0? In other words, no other energy level is possible? I thought (as a non-expert) that maybe all electrons exist with multiple probabilities, even if some of these are vanishingly small. It’s only when we observe it that we can say for sure what its energy level was at that moment. Or have I misunderstood this? – Chappo Hasn't Forgotten Nov 16 '17 at 12:30
  • @Chappo We can have an electron >>state<< that has probability 1.0 of being observed with a particular energy when measured. And there are other states with probability 1.0 for other different energies. And there are still other states in which there are non-zero probabilities for several different energies, i.e., we can't be sure what the energy measurement outcome will be. In the above answer, as per the OP's question, we're considering two different certain-energy (probability 1.0) states. –  Nov 16 '17 at 21:37
  • @safesphere It's entirely beyond my ability to devise, in your words, "a one-line explanation, something like, `This is how the state (set of parameters) of a particle is noted in quantum mechanics.'" Please illustrate a short explanation like that. As you say, I'm sure it would be helpful to lots of people. So, since I'm not capable of doing it, maybe you could please illustrate it for us. Thank you. –  Nov 16 '17 at 22:40
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    @JohnForkish Yes, I already have. The sentence you quoted. Just a line to explain what the brackets mean. – safesphere Nov 17 '17 at 03:22
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Good question. The atom emits a line spectrum, light in narrow ranges of frequencies. So the source must be an oscillator that is going through millions of oscillations.

One can regard the atom in transition as being in a superposition of the initial and the final stationary states (for an expression, see the answer by John Forkosh). The charge density of such a state is not stationary, see for example this animated gif of the density of a particle in a box: https://commons.wikimedia.org/wiki/File:Particle_in_a_box_(time_evolution).gif

So it will radiate energy with a frequency given by the energy difference between the two states. The contribution of the excited state to the wave function will diminish exponentially: a damped harmonic oscillator.

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The electrons don't "move" in the sense that you are probably thinking. They are instantaneously in one state or another. The transition is called a quantum leap and occurs instantly.

Billy Kalfus
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