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Perhaps I was mistaken, but I thought this was sort a necessary revelation when learning about QM. It's been years since I took Atomic Physics, so perhaps this

Perhaps I was mistaken, but I thought this was sort a necessary revelation when learning about QM. It's been years since I took Atomic Physics, so perhaps this is something much ... deeper? Was it a serious idea that upon measurement the particle literally lost it's wave properties? Like, there were actually two separate approaches to the math? I know we made some efforts to simplify in class, approximating as particles and such due to the drastically simplified math, but we all knew that was happening; the professor was reasonably explicit about it. EDIT: As in, I remember us going over the evolution of a waveform, and how boundaries affect the solutions, and how uncertainty causes the particles to have field distributions that just happen to be the wave solution of the particle. I'm completely serious about being confused here: QM is super easy to misunderstand, and I'd love to feel the eureka this article is trying to convey.

mjklin 12y ago
http://en.wikipedia.org/wiki/Collapse_of_the_wavefunction http://en.wikipedia.org/wiki/Collapse_of_the_wavefunction
- HCIdivision17 12y ago
  Yeah, the relevant snippit is here: However, we never observe collapse to a single eigenstate of a continuous-spectrum operator (e.g. position, momentum, or a scattering Hamiltonian), because such eigenfunctions are non-normalizable. In these cases, the wave function will partially collapse to a linear combination of "close" eigenstates (necessarily involving a spread in eigenvalues) that embodies the imprecision of the measurement apparatus. I suppose I just took away that the particle 'point' idea is likely just an asymptotic approximation of what the thing could be modeled as, but in reality - deep down at the atomic scales - it's really a 'linear combination of "close" eigenstates'.
  - tjradcliffe 12y ago
    Speculation: the person who wrote that has never heard of the Dirac delta function? Regardless of what happens in reality, the Dirac delta is precisely a normalized point-like eigenstate (infinite height, zero width, unit area.)
    - HCIdivision17 12y ago
      Reminds me now of the wonder I felt as after a few classes that had us doing integrals of various tedious geometries for EM fields, we come to QM - the really hard course - and we just pop a Dirac delta in to trivialize some integrals. It's like suddenly being able to say, "Enough! Just tell me the value right there!" Which turns out to reduce to your friendly neighborhood point particle model. (Pretty much by design, if I'm not forgetting my history correctly - it's sort of a made up tool that sort of abuses notation, like the Heaviside step function.) But deep down I never trusted that a particle was actually a basically zero dimensional-like point, though. It was just useful to think of it as one.
  - millstone 12y ago
    The key word there is continuous-spectrum. A free electron can have any energy, and therefore the energy eigenstates are not physically realizable, due to the normalization requirements. But a bound electron, like in a quantum harmonic oscillator, has quantized energies and therefore can assume those eigenstates exactly. Think about electron orbitals in molecules: very precise energies.
jeffwass 12y ago
I haven't yet read the actual papers linked to be article, but I'm confused what the big breakthrough is here, the article is very light on specifics. The generalized uncertainty principle is nothing but an application of the 'Schwartz Inequality' to two non-commuting operators. Specifically, if the commutator [A,B] = iC for operators A and B (C can be an operator or scalar, i is sqrt(-1) to preserve Hermitian-ness), then the product of uncertainties Delta A x Delta B >= <C>/2. Where Delta A = sqrt(<A^2> - <A>^2) and <A> is the expectation value of A for a given wave function, and [A,B] = AB - BA. The wave function itself is the fundamental driver behind the wave-particle duality. And the generalized uncertainty principle just follows downstream from mathematical application. I don't see why the article (not the papers though) is saying they're two unrelated things. FYI, this brings back annoying memories from one of my midterm exams in grad level quantum mechanics. Five questions to solve in one hour. And one question alone was to derive the generalized uncertainty relation from first principles.
- HCIdivision17 12y ago
  Ok, cool. That explains nicely the feeling I was getting. (And your note reminds me of some of the QM and modern physics tests I had as well - I didn't take grad work, but I still remember thinking, "Man, we spent hours deriving some of the stuff, and I've got... less than 10 minutes. Whelp...")
marcosdumay 12y ago
That equivalence was explicitly stated on the book I learned QM. The article reads too much like "uninformed journalist writing about physics", but the paper title suggests the journalist isn't misreporting it. Anyway, I get an error when I click on the paper. I still expect it to contain something else, besides this article's contents.
cynicalkane 12y ago
The mythical "wave or particle" interpretation of wave-particle duality is oddly persistent, and is not only pervasive in popular science but I've heard it repeated by serious physicists also, even though it's unsupported by the very math they work on.
- wellwell123 12y ago
  It's not clear to me what you think the "wave or particle" interpretation is, nor what you think is wrong with it, nor what you think you've heard "serious physicists" say.
- astrange 12y ago
  If someone believes an interpretation, can it really be mythical?
dnautics 12y ago
I remember something similar in my qchem class, that uncertainty basically falls out of the Fourier transform (functional equivalent of vectorial commutativity problem), and so the particle (center of probability of the waveform) automatically inherited uncertainty from the wave equation.
ChuckMcM 12y ago
Here is the paper: http://arxiv.org/pdf/1403.4687v2.pdf http://arxiv.org/pdf/1403.4687v2.pdf My quick read suggests that the authors are restating the relationship using their "Entropic View" notation which, as it turns out, has the same expression if you come at it from the uncertainty principle side or the wave equation side. That is nice from the point of view of validating their notation but it doesn't give us anything we didn't already know.
- drostie 12y ago
  Right. At least at Cornell (David Muller's quantum class) we were basically taught that wave-particle duality was the uncertainty principle from our earliest undergraduate experience. Basically, the very meaning of de Broglie's law (a particle with momentum p has a wavelength L = h / p) specifies a "particle wavelength", but that wavelength can only exist within that particle's position-uncertainty. Shrink that uncertainty and you necessarily have a small L to fit within there, which means a very high p, which means that little fluctuations e.g. in the direction of p translate to large uncertainties about p. About a quarter-way into the course it is revealed that de Broglie's law essentially identifies p-space with k-space (that is, the Fourier conjugate to x-space), and you then prove that all Fourier conjugate pairs have an uncertainty principle, quite independent of quantum mechanics. So the earliest statement of wave-particle duality already implies the uncertainty principle and this has been known for a long time.
obastani 12y ago
This seems to be a key reference (article titled "Complementarity and uncertainty"): http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0.pdf http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0... In particular, they have a description of why they find the traditional connection between uncertainty and wave-particle duality to be unsatisfactory. Unfortunately the technical details are beyond me, but hopefully someone else can sort them out? Edit: Okay, some more digging suggests the following story: (i) We have wave-particle duality => Heisenberg uncertainty, but not the other way around (ii) With entropic uncertainty [1] (stronger than Heisenberg uncertainty), the authors show that we can obtain wave-particle duality Again, can somebody more knowledgeable can clarify? :) [1] http://en.wikipedia.org/wiki/Entropic_uncertainty http://en.wikipedia.org/wiki/Entropic_uncertainty
- quarterwave 12y ago
  @obastani : very good comment. Aside: If |Psi> is a state with wavefunction <x|Psi>, then is there a linear operator A such that <x|A|Psi> = log(<x|Psi>)? Note: Logarithm of a complex function.
  - filmor 12y ago
    Well, the logarithm is all but linear. If there was such an A you'd have <x|A(|Psi> + |Chi>) = log(<x|(|Psi> + |Chi>)) = log(<x|Psi> + <x|Chi>) = log(<x|Psi>) log(<x|Chi>) /= <x|A|Psi> + <x|A|Chi>
    - throwaway183839 12y ago
      I don't think that log(a + b) = log(a) log(b) is a rule I'm familiar with!
      - filmor 12y ago
        You're right, I messed that one up. But as log(a + b) /= log(a) + log(b) in general, the result is still correct.

mjklin 12y ago

http://en.wikipedia.org/wiki/Collapse_of_the_wavefunction http://en.wikipedia.org/wiki/Collapse_of_the_wavefunction

HCIdivision17 12y ago
Yeah, the relevant snippit is here: However, we never observe collapse to a single eigenstate of a continuous-spectrum operator (e.g. position, momentum, or a scattering Hamiltonian), because such eigenfunctions are non-normalizable. In these cases, the wave function will partially collapse to a linear combination of "close" eigenstates (necessarily involving a spread in eigenvalues) that embodies the imprecision of the measurement apparatus. I suppose I just took away that the particle 'point' idea is likely just an asymptotic approximation of what the thing could be modeled as, but in reality - deep down at the atomic scales - it's really a 'linear combination of "close" eigenstates'.
- tjradcliffe 12y ago
  Speculation: the person who wrote that has never heard of the Dirac delta function? Regardless of what happens in reality, the Dirac delta is precisely a normalized point-like eigenstate (infinite height, zero width, unit area.)
  - HCIdivision17 12y ago
    Reminds me now of the wonder I felt as after a few classes that had us doing integrals of various tedious geometries for EM fields, we come to QM - the really hard course - and we just pop a Dirac delta in to trivialize some integrals. It's like suddenly being able to say, "Enough! Just tell me the value right there!" Which turns out to reduce to your friendly neighborhood point particle model. (Pretty much by design, if I'm not forgetting my history correctly - it's sort of a made up tool that sort of abuses notation, like the Heaviside step function.) But deep down I never trusted that a particle was actually a basically zero dimensional-like point, though. It was just useful to think of it as one.
- millstone 12y ago
  The key word there is continuous-spectrum. A free electron can have any energy, and therefore the energy eigenstates are not physically realizable, due to the normalization requirements. But a bound electron, like in a quantum harmonic oscillator, has quantized energies and therefore can assume those eigenstates exactly. Think about electron orbitals in molecules: very precise energies.

jeffwass 12y ago

I haven't yet read the actual papers linked to be article, but I'm confused what the big breakthrough is here, the article is very light on specifics. The generalized uncertainty principle is nothing but an application of the 'Schwartz Inequality' to two non-commuting operators. Specifically, if the commutator [A,B] = iC for operators A and B (C can be an operator or scalar, i is sqrt(-1) to preserve Hermitian-ness), then the product of uncertainties Delta A x Delta B >= <C>/2. Where Delta A = sqrt(<A^2> - <A>^2) and <A> is the expectation value of A for a given wave function, and [A,B] = AB - BA. The wave function itself is the fundamental driver behind the wave-particle duality. And the generalized uncertainty principle just follows downstream from mathematical application. I don't see why the article (not the papers though) is saying they're two unrelated things. FYI, this brings back annoying memories from one of my midterm exams in grad level quantum mechanics. Five questions to solve in one hour. And one question alone was to derive the generalized uncertainty relation from first principles.

HCIdivision17 12y ago
Ok, cool. That explains nicely the feeling I was getting. (And your note reminds me of some of the QM and modern physics tests I had as well - I didn't take grad work, but I still remember thinking, "Man, we spent hours deriving some of the stuff, and I've got... less than 10 minutes. Whelp...")

marcosdumay 12y ago

That equivalence was explicitly stated on the book I learned QM. The article reads too much like "uninformed journalist writing about physics", but the paper title suggests the journalist isn't misreporting it. Anyway, I get an error when I click on the paper. I still expect it to contain something else, besides this article's contents.

cynicalkane 12y ago

The mythical "wave or particle" interpretation of wave-particle duality is oddly persistent, and is not only pervasive in popular science but I've heard it repeated by serious physicists also, even though it's unsupported by the very math they work on.

wellwell123 12y ago
It's not clear to me what you think the "wave or particle" interpretation is, nor what you think is wrong with it, nor what you think you've heard "serious physicists" say.
astrange 12y ago
If someone believes an interpretation, can it really be mythical?

dnautics 12y ago

I remember something similar in my qchem class, that uncertainty basically falls out of the Fourier transform (functional equivalent of vectorial commutativity problem), and so the particle (center of probability of the waveform) automatically inherited uncertainty from the wave equation.

ChuckMcM 12y ago

Here is the paper: http://arxiv.org/pdf/1403.4687v2.pdf http://arxiv.org/pdf/1403.4687v2.pdf My quick read suggests that the authors are restating the relationship using their "Entropic View" notation which, as it turns out, has the same expression if you come at it from the uncertainty principle side or the wave equation side. That is nice from the point of view of validating their notation but it doesn't give us anything we didn't already know.

drostie 12y ago
Right. At least at Cornell (David Muller's quantum class) we were basically taught that wave-particle duality was the uncertainty principle from our earliest undergraduate experience. Basically, the very meaning of de Broglie's law (a particle with momentum p has a wavelength L = h / p) specifies a "particle wavelength", but that wavelength can only exist within that particle's position-uncertainty. Shrink that uncertainty and you necessarily have a small L to fit within there, which means a very high p, which means that little fluctuations e.g. in the direction of p translate to large uncertainties about p. About a quarter-way into the course it is revealed that de Broglie's law essentially identifies p-space with k-space (that is, the Fourier conjugate to x-space), and you then prove that all Fourier conjugate pairs have an uncertainty principle, quite independent of quantum mechanics. So the earliest statement of wave-particle duality already implies the uncertainty principle and this has been known for a long time.

obastani 12y ago

This seems to be a key reference (article titled "Complementarity and uncertainty"): http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0.pdf http://www.nature.com/nature/journal/v375/n6530/pdf/375367b0... In particular, they have a description of why they find the traditional connection between uncertainty and wave-particle duality to be unsatisfactory. Unfortunately the technical details are beyond me, but hopefully someone else can sort them out? Edit: Okay, some more digging suggests the following story: (i) We have wave-particle duality => Heisenberg uncertainty, but not the other way around (ii) With entropic uncertainty [1] (stronger than Heisenberg uncertainty), the authors show that we can obtain wave-particle duality Again, can somebody more knowledgeable can clarify? :) [1] http://en.wikipedia.org/wiki/Entropic_uncertainty http://en.wikipedia.org/wiki/Entropic_uncertainty

quarterwave 12y ago
@obastani : very good comment. Aside: If |Psi> is a state with wavefunction <x|Psi>, then is there a linear operator A such that <x|A|Psi> = log(<x|Psi>)? Note: Logarithm of a complex function.
- filmor 12y ago
  Well, the logarithm is all but linear. If there was such an A you'd have <x|A(|Psi> + |Chi>) = log(<x|(|Psi> + |Chi>)) = log(<x|Psi> + <x|Chi>) = log(<x|Psi>) log(<x|Chi>) /= <x|A|Psi> + <x|A|Chi>
  - throwaway183839 12y ago
    I don't think that log(a + b) = log(a) log(b) is a rule I'm familiar with!
    - filmor 12y ago
      You're right, I messed that one up. But as log(a + b) /= log(a) + log(b) in general, the result is still correct.