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Thread: Particles as wave packets - why they don't dissipate?

  1. #1 Particles as wave packets - why they don't dissipate? 
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    Particles in quantum mechanics are often seen as wave packets - linear superpositions of plane waves summing to a localized excitation.
    But wave packets dissipate - for example passing such single photon through a prism, its different plane waves should choose different angles - such single photon would dissipate: its energy would be spread on a growing area ... while we know that in reality its energy remains localized: will be finally adsorbed as a whole by e.g. a single atom.
    Analogously for different particles like electron - any dependence on momentum while scattering would make such wave packet dissipating (e.g. indivisible elementary charge).

    How is this problem of dissipating particles solved in quantum mechanics?
    Aren't there some additional (nonlinear?) mechanisms needed to hold particles together, make these wave packets maintaining their shapes - being so called solitons?


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    Hi, they do in fact "dissipate". Width of gaussian wavepacket does increase lineary with time. However these are probabilistic wavepackets. Therefore when you perform measurement of position of particle you will get localized delta function in space. With time this delta function becomes gaussian with increasing width. That does not mean that particle itself is dissipating though. If you perform measurement again, the gaussian wavepacket will breakdown into another delta function which designates position of the particle. In time this becomes gaussian again and so on.


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    So what is objectively happening here: a single excited atom produces single optical photon, which comes through a prism and finally is absorbed by another single atom?
    Both initially and finally this energy is localized inside an atom.
    It is also localized just after photon production - when it is imagined as a compact wave packet.

    Does this energy dissipate when the packet dissipates? How it is finally gathered together into a single atom?
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    Ok this is tricky. The excited atom produces optical photon with lorentzian distribution in frequency and space (this is because selfenergy implicates exponential decay). Now the uncertaintity in frequency of photon is given by unceraintity of atomic transition (which is againt lorentzian emision line). This is very important, it says that properties of photon are given by it`s origin! Now until we perform measurement of energy of photon we do not know what was the exact energy of atomic transition, we know only probabilistic distributions.
    Now when this wavepacket comes through prism you are completely right that it will spread the wavepacket in different dirrections according to frequency components of wavepacket. Now when you put atom behind and look wheter it will be excited or not you are basicaly performing measurement of the photon. For simplicity lets assume that the second atom has perfectly sharp atomic transition.
    If atom will not be excited after measurement that means that original atomic transition had different energy therefore photon has different energy and cannot be absorbed by second atom.
    If second atom will be excited the original atomic transition had exact energy as atomic transition of second atom and photon had correct energy to excite it. There is a very fine line between saying that the "wavepacket" is photon and wavepacket is probabilistic description of photon!
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    Quote Originally Posted by Gere View Post
    If second atom will be excited the original atomic transition had exact energy as atomic transition of second atom and photon had correct energy to excite it.
    Assuming the same energy differences of both atoms - if they are close to each other, this energy is still localized in the wavepacket and so can be easily absorbed.

    But what if this wave packet has already dissipated - energy of this photon is no longer in a tiny place, but spread over a large area - how the final atom can gather all this energy from a large area?
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    Wavepacket is probabilistic description. It does not mean that energy is spreading in space.
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    So what objectively is particle - just superposition of plane waves, or there is also some additional mechanism maintaining their shapes (making them solitons)?
    Should we search for soliton particle models, like Skyrme's?
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    Quote Originally Posted by Gere View Post
    Wavepacket is probabilistic description.
    That depends on the interpretation. It's always probabilistic from the perspective of the observer, but the many-worlds interpretation provides a deterministic objective view of the wavefunction. I've been discussing this on the Physics Forum in the thread "Quantum Physics > Double split experiment question", so I won't repeat it here.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  10. #9  
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    Quote Originally Posted by Gere View Post
    Ok this is tricky. The excited atom produces optical photon with lorentzian distribution in frequency and space (this is because selfenergy implicates exponential decay). Now the uncertaintity in frequency of photon is given by unceraintity of atomic transition (which is againt lorentzian emision line). This is very important, it says that properties of photon are given by it`s origin! Now until we perform measurement of energy of photon we do not know what was the exact energy of atomic transition, we know only probabilistic distributions.
    Now when this wavepacket comes through prism you are completely right that it will spread the wavepacket in different dirrections according to frequency components of wavepacket. Now when you put atom behind and look wheter it will be excited or not you are basicaly performing measurement of the photon. For simplicity lets assume that the second atom has perfectly sharp atomic transition.
    If atom will not be excited after measurement that means that original atomic transition had different energy therefore photon has different energy and cannot be absorbed by second atom.
    If second atom will be excited the original atomic transition had exact energy as atomic transition of second atom and photon had correct energy to excite it. There is a very fine line between saying that the "wavepacket" is photon and wavepacket is probabilistic description of photon!
    This is an interesting thought experiment. Would what you are saying mean that the dispersion caused by a prism would have the effect of making the probability of absorption by the second atom lower than it would have been for a photon travelling directly, with no prism in between?
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    Quote Originally Posted by Gere View Post
    For simplicity lets assume that the second atom has perfectly sharp atomic transition.
    If the emission has a Lorentzian distribution of a given width, would the corresponding absorption also be a Lorentzian distribution of the same width? Microscopic reversibility suggests that it would.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  12. #11  
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    Quote Originally Posted by KJW View Post
    Quote Originally Posted by Gere View Post
    For simplicity lets assume that the second atom has perfectly sharp atomic transition.
    If the emission has a Lorentzian distribution of a given width, would the corresponding absorption also be a Lorentzian distribution of the same width? Microscopic reversibility suggests that it would.
    Might this not depend on the radiation environment of the emitting and receiving atoms? I seem to recall the uncertainty in the energy gap is related to the lifetime of the excited state - which I would have thought would be influenced by both the spontaneous and the stimulated emission rates, with the latter being influenced by the radiation flux to which the atom is exposed. But it was a long time ago and my memory may be faulty.
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    Quote Originally Posted by KJW View Post
    If the emission has a Lorentzian distribution of a given width, would the corresponding absorption also be a Lorentzian distribution of the same width? Microscopic reversibility suggests that it would.
    Yes you are of course right. I was just trying to make a point that putting second atom in the way is basically momentum measurement which does projection onto eigenstates. In real case the absorption line would serve as "characteristic function of detector" or whatever is correct english expression. The result of experiment would be given by convolution of characteristic function of detector with signal if that makes sense. (In simplified case characteristic function is just delta function.) Sorry for my sloppiness.


    Quote Originally Posted by exchemist View Post
    Might this not depend on the radiation environment of the emitting and receiving atoms? I seem to recall the uncertainty in the energy gap is related to the lifetime of the excited state - which I would have thought would be influenced by both the spontaneous and the stimulated emission rates, with the latter being influenced by the radiation flux to which the atom is exposed. But it was a long time ago and my memory may be faulty.
    Yes you are exactly right. Quanitative description would be like this. Let be coherent Hamltonian and - imaginary part of selfenergy. Full Hamiltonian is then . Let be wavefunction of excited state (I will omit increase in population of ground state). Then time evolution in Schrodinger picture is like this.



    therefore

    .

    Fourier transform of this gives Lorentzian.



    From population of exited state one can say that time of life of excited state is .



    Quote Originally Posted by exchemist View Post
    This is an interesting thought experiment. Would what you are saying mean that the dispersion caused by a prism would have the effect of making the probability of absorption by the second atom lower than it would have been for a photon travelling directly, with no prism in between?
    Yes of course because you are basicaly saying that you will change direction of every photon (here meant planewave) with wrong frequency. While atom would in principle be able to absorb it thanks to width of absorption line the photon simply won`t be there. Since real state of light is superposition of planewave photons you are sending probabilistic wave components in wrong direction.

    There is one more freakish aspect though. If you will spread the wave packet by prism you can then rejoin different components of that wavepacket with some mirrors. What is freakish is that these probabilistic waves will interfere even if you send one photon (here meant real state of light) at a time.
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  14. #13  
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    Quote Originally Posted by Gere View Post
    Quote Originally Posted by KJW View Post
    If the emission has a Lorentzian distribution of a given width, would the corresponding absorption also be a Lorentzian distribution of the same width? Microscopic reversibility suggests that it would.
    Yes you are of course right. I was just trying to make a point that putting second atom in the way is basically momentum measurement which does projection onto eigenstates. In real case the absorption line would serve as "characteristic function of detector" or whatever is correct english expression. The result of experiment would be given by convolution of characteristic function of detector with signal if that makes sense. (In simplified case characteristic function is just delta function.) Sorry for my sloppiness.


    Quote Originally Posted by exchemist View Post
    Might this not depend on the radiation environment of the emitting and receiving atoms? I seem to recall the uncertainty in the energy gap is related to the lifetime of the excited state - which I would have thought would be influenced by both the spontaneous and the stimulated emission rates, with the latter being influenced by the radiation flux to which the atom is exposed. But it was a long time ago and my memory may be faulty.
    Yes you are exactly right. Quanitative description would be like this. Let be coherent Hamltonian and - imaginary part of selfenergy. Full Hamiltonian is then . Let be wavefunction of excited state (I will omit increase in population of ground state). Then time evolution in Schrodinger picture is like this.



    therefore

    .

    Fourier transform of this gives Lorentzian.



    From population of exited state one can say that time of life of excited state is .



    Quote Originally Posted by exchemist View Post
    This is an interesting thought experiment. Would what you are saying mean that the dispersion caused by a prism would have the effect of making the probability of absorption by the second atom lower than it would have been for a photon travelling directly, with no prism in between?
    Yes of course because you are basicaly saying that you will change direction of every photon (here meant planewave) with wrong frequency. While atom would in principle be able to absorb it thanks to width of absorption line the photon simply won`t be there. Since real state of light is superposition of planewave photons you are sending probabilistic wave components in wrong direction.

    There is one more freakish aspect though. If you will spread the wave packet by prism you can then rejoin different components of that wavepacket with some mirrors. What is freakish is that these probabilistic waves will interfere even if you send one photon (here meant real state of light) at a time.
    Excellent, many thanks, how interesting. I have to confess that 40 years on the physical picture has stayed with me rather better than the maths.

    Regarding your last point about the single photon, it has seemed to me for while that the only way to rationalise this sort of thing physically is to consider matter to be made fundamentally of the waves, and that it is the detection of matter that is observed to be only in whole quanta, i.e. particles. But I've no idea whether such a view would be considered sound by a physicist.
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