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There appears to be a very basic error in this writeup. I say there appears to be because it seems to be so easy to misunderstand language used in casual descri

by gpsx 8y ago

There appears to be a very basic error in this writeup. I say there appears to be because it seems to be so easy to misunderstand language used in casual descriptions of quantum mechanics. Neither I nor the writer is a lawyer. My apologies if I misunderstood what was written. My experience here in general is when something seems very clearly wrong, it is my misunderstanding. None the less... Quantum mechanics does not say a particle can be in two places at the same time. A wave function does not say a particle is at point A AND point b. The wave function says the particele is at point A OR point B. To take this a step further, if the particle has an electric field, then the wave function would look like the following: (particle at A AND electric field around point A) OR (particle at B AND electric field around point B) The wave function would NOT be: (particle at A OR particle at B) AND (electric field around A OR electric field around B) And also not: (particle at A AND particle at B) AND (electric field around A and B) Presumably the same applies to a gravitational field.

comicjk 8y ago
Quantum mechanics only explicitly says what will happen when you do a measurement, not how to visualize what happens between measurements. Many different interpretations can be useful for visualizing the wavefunction. I actually like the AND interpretation (the final one you described) because it helps in my field, electrons in molecules (aka chemistry). For example, if you add an extra electron to a benzene ring, the new charge acts symmetrical even though measuring it at high precision would show it to be localized off center at some unpredictable position.
candiodari 8y ago
Isn't this the mistake that the double slit experiment tests ? Particles very much are at point A AND point B, and those electric fields, for instance, interfere with each other. Also with the outside world. That would not happen in the OR case, only in the AND case.
- krastanov 8y ago
  Neither the OR nor the AND descriptions are great as they give you just a partial (wrong) intuition of what the mathematical model says. Moreover, the mathematical model can always be wrong (although improbable in this case). But to address your comment: the OR description is a bit better than the AND one because the possiblities described by the wave function are excluding one another (when a good basis is chosen). However neither the OR nor the AND description actually mention the fact that the "probabilities" interfere (which is what the double slit experiment tests). That experiment disproves both of these simplified descriptions. Another reason I like the OR intuition better if I have to chose between these two wrong simplifications: if you do the double slit experiment with one photon at a time and repeat it multiple time then you do see only one photon at a time hit the screen (the OR intuition). But the image you get if you wait and average over many photons is indeed the interference pattern. TLDR: OR acts more like probabilistic but not interfering particles. AND acts more like waves. Both are wrong, but in different contexts one or the other might be a convenient simplification.
  - candiodari 8y ago
    Thanks for the explanation.
  - whatshisface 8y ago
    I an duty-bound by the code of the secret society of MWI enthusiasts to point out that actually, no phenomena inconsistent with waves ("AND") has ever been observed, unless you count the philosophical discomfort of the idea that the superposition could also extend to the experimenter.
drdeca 8y ago
I like this description of it : https://www.smbc-comics.com/comic/the-talk-3 https://www.smbc-comics.com/comic/the-talk-3 key quote: 'Superposition doesn't mean "and", but it also doesn't mean "or".' 'It means a complex linear combination of a 0 state and a 1 state. You should think of it as a new ontological category: a way of combining things that doesn't really map onto any classical concept.'
- SuoDuanDao 8y ago
  I like that quote myself, but one thing I don't understand is - are the amplitudes always of a magnitude 1? Because if qbits are really always the same magnitude, why represent them as two-dimensional? Wouldn't the math be easier if we converted it to polar form and just had one measurement per superposition? Sorry if this is a dumb question, I'm not formally trained in any of this stuff...
  - drdeca 8y ago
    There are 2 complex numbers, one of them the coefficient of |0> and |1>. The sum of the square of the magnitudes of these numbers is 1. Alternatively/equivalently , the magnitude of the vector taken as a whole, is 1. So if you have (a+bi)|0> + (c+di)|1>, the magnitude of this vector is a^2 + b^2 + c^2 + d^2 = 1 Edit: however, the vector only has to have magnitude 1 if it is describing the entire system, I think?
  - millstone 8y ago
    It's a good question. The cartoon says that qubits are "unit vectors in 2D Hilbert space." However because the length is 1, you can think of it as a vector in 2D projective Hilbert space. Projective space is what you get when you throw away a vector's length and only talk about its angles. Some intuition: if you have a particle and you look everywhere for it, your chance of finding it must be 100%. It can't be 50% (where could it be?) and it can't be 200% (duh). QM in one dimension is formulated as "a 1d complex unit vector", i.e. a complex number of magnitude 1, which is routinely represented in polar form (a complex exponential). QM in two dimensions is a "2d complex unit vector" which is really 3d: you get three linearly independent components, and then the fourth one is decided for you. You can think of this as spherical coordinates, but complex numbers are much easier to manipulate, so that's what we do.
ummonk 8y ago
Quantum mechanics says there is a wavefunction. How you choose to interpret that wavefunction in terms of particles (“and” vs “or”) is really about your point of view. Note though that quantum mechanics is itself just a simplified way of modeling what happens due to field theory. Under field theory, fields are central, and particles are just phenomena that happen under fields. It is of course possible that in a more fundamental theory, particles are central and their interaction just results in field-like behavior.
pdonis 8y ago
> A wave function does not say a particle is at point A AND point b. The wave function says the particle is at point A OR point B. No, it doesn't. It doesn't say either of those things. It says the particle is in a superposition of being at point A and being at point B. There is no simple logical connective in ordinary language that captures that. However, for the purposes of trying to figure out the gravitational field of the particle, being in a superposition of being at point A and being at point B is functionally equivalent to being in both positions at the same time, in the sense that we don't know how to treat either one quantum mechanically. See below. > if the particle has an electric field, then the wave function would look like the following... No, it wouldn't. It wouldn't look like any of the possibilities you wrote down. It would look like a superposition of "particle at A and electric field around point A" and "particle at B and electric field around point B". But since we know how to quantize the electromagnetic field, we know how to treat such superpositions quantum mechanically. We don't have a corresponding way to treat superpositions of gravitational fields quantum mechanically.
- gpsx 8y ago
  > There is no simple logical connective in ordinary language that captures that Yes, I'd agree with this. I was trying to say something aproximately true to convey a point. In your post you did agree with my salient point - the electric field moves with the electron, it doesn't do something funny when the elestron has some spread in its wavefuntion for position. This seemed to be the what the article was talking about, that the gravitational field behaved differently for a superposition of states. I will try a more concrete example. Suppose you have an electron gun and shoot a single elctron at a metal plate. Suppose the wavefunction is such that it is uniform over the surface of the plate (So the electron has an equal probability of hitting anywhere on the plate.) What does the damage to the plate look like? Is it spread uniformly over the plate? No. It is a single small mark at some specific place on the plate. The electron is not smeared over the plate, the probability of finding the electron is smeared over the plate. The electron is still a point particle (to the extent of our current laws of physics) and it will only interact with the plate at one point. (Some people will not like my language and tell me I am wrong. But it is true the electron will hit the plate at a single point - not smeared out like the wavefunction is.) In actuality the electron didn't cause the damage to the plate. The electric field did. The electric field moves with the electron almost like you would expect in classical physics. It's not like the electric field is smeared around because the electron is in a superposition of states. I'd expect gravity to work the same way. But seeing how we don't have a theory it is possible it could violate the above idea. To me that would be a pretty weird violation too. (This short writeup doesn't really do that justice.) I'd bet against it.

by gpsx 8y ago

comicjk 8y ago

Quantum mechanics only explicitly says what will happen when you do a measurement, not how to visualize what happens between measurements. Many different interpretations can be useful for visualizing the wavefunction. I actually like the AND interpretation (the final one you described) because it helps in my field, electrons in molecules (aka chemistry). For example, if you add an extra electron to a benzene ring, the new charge acts symmetrical even though measuring it at high precision would show it to be localized off center at some unpredictable position.

candiodari 8y ago

Isn't this the mistake that the double slit experiment tests ? Particles very much are at point A AND point B, and those electric fields, for instance, interfere with each other. Also with the outside world. That would not happen in the OR case, only in the AND case.

krastanov 8y ago
Neither the OR nor the AND descriptions are great as they give you just a partial (wrong) intuition of what the mathematical model says. Moreover, the mathematical model can always be wrong (although improbable in this case). But to address your comment: the OR description is a bit better than the AND one because the possiblities described by the wave function are excluding one another (when a good basis is chosen). However neither the OR nor the AND description actually mention the fact that the "probabilities" interfere (which is what the double slit experiment tests). That experiment disproves both of these simplified descriptions. Another reason I like the OR intuition better if I have to chose between these two wrong simplifications: if you do the double slit experiment with one photon at a time and repeat it multiple time then you do see only one photon at a time hit the screen (the OR intuition). But the image you get if you wait and average over many photons is indeed the interference pattern. TLDR: OR acts more like probabilistic but not interfering particles. AND acts more like waves. Both are wrong, but in different contexts one or the other might be a convenient simplification.
- candiodari 8y ago
  Thanks for the explanation.
- whatshisface 8y ago
  I an duty-bound by the code of the secret society of MWI enthusiasts to point out that actually, no phenomena inconsistent with waves ("AND") has ever been observed, unless you count the philosophical discomfort of the idea that the superposition could also extend to the experimenter.

drdeca 8y ago

I like this description of it : https://www.smbc-comics.com/comic/the-talk-3 https://www.smbc-comics.com/comic/the-talk-3 key quote: 'Superposition doesn't mean "and", but it also doesn't mean "or".' 'It means a complex linear combination of a 0 state and a 1 state. You should think of it as a new ontological category: a way of combining things that doesn't really map onto any classical concept.'

SuoDuanDao 8y ago
I like that quote myself, but one thing I don't understand is - are the amplitudes always of a magnitude 1? Because if qbits are really always the same magnitude, why represent them as two-dimensional? Wouldn't the math be easier if we converted it to polar form and just had one measurement per superposition? Sorry if this is a dumb question, I'm not formally trained in any of this stuff...
- drdeca 8y ago
  There are 2 complex numbers, one of them the coefficient of |0> and |1>. The sum of the square of the magnitudes of these numbers is 1. Alternatively/equivalently , the magnitude of the vector taken as a whole, is 1. So if you have (a+bi)|0> + (c+di)|1>, the magnitude of this vector is a^2 + b^2 + c^2 + d^2 = 1 Edit: however, the vector only has to have magnitude 1 if it is describing the entire system, I think?
- millstone 8y ago
  It's a good question. The cartoon says that qubits are "unit vectors in 2D Hilbert space." However because the length is 1, you can think of it as a vector in 2D projective Hilbert space. Projective space is what you get when you throw away a vector's length and only talk about its angles. Some intuition: if you have a particle and you look everywhere for it, your chance of finding it must be 100%. It can't be 50% (where could it be?) and it can't be 200% (duh). QM in one dimension is formulated as "a 1d complex unit vector", i.e. a complex number of magnitude 1, which is routinely represented in polar form (a complex exponential). QM in two dimensions is a "2d complex unit vector" which is really 3d: you get three linearly independent components, and then the fourth one is decided for you. You can think of this as spherical coordinates, but complex numbers are much easier to manipulate, so that's what we do.

ummonk 8y ago

Quantum mechanics says there is a wavefunction. How you choose to interpret that wavefunction in terms of particles (“and” vs “or”) is really about your point of view. Note though that quantum mechanics is itself just a simplified way of modeling what happens due to field theory. Under field theory, fields are central, and particles are just phenomena that happen under fields. It is of course possible that in a more fundamental theory, particles are central and their interaction just results in field-like behavior.

pdonis 8y ago

> A wave function does not say a particle is at point A AND point b. The wave function says the particle is at point A OR point B. No, it doesn't. It doesn't say either of those things. It says the particle is in a superposition of being at point A and being at point B. There is no simple logical connective in ordinary language that captures that. However, for the purposes of trying to figure out the gravitational field of the particle, being in a superposition of being at point A and being at point B is functionally equivalent to being in both positions at the same time, in the sense that we don't know how to treat either one quantum mechanically. See below. > if the particle has an electric field, then the wave function would look like the following... No, it wouldn't. It wouldn't look like any of the possibilities you wrote down. It would look like a superposition of "particle at A and electric field around point A" and "particle at B and electric field around point B". But since we know how to quantize the electromagnetic field, we know how to treat such superpositions quantum mechanically. We don't have a corresponding way to treat superpositions of gravitational fields quantum mechanically.

gpsx 8y ago
> There is no simple logical connective in ordinary language that captures that Yes, I'd agree with this. I was trying to say something aproximately true to convey a point. In your post you did agree with my salient point - the electric field moves with the electron, it doesn't do something funny when the elestron has some spread in its wavefuntion for position. This seemed to be the what the article was talking about, that the gravitational field behaved differently for a superposition of states. I will try a more concrete example. Suppose you have an electron gun and shoot a single elctron at a metal plate. Suppose the wavefunction is such that it is uniform over the surface of the plate (So the electron has an equal probability of hitting anywhere on the plate.) What does the damage to the plate look like? Is it spread uniformly over the plate? No. It is a single small mark at some specific place on the plate. The electron is not smeared over the plate, the probability of finding the electron is smeared over the plate. The electron is still a point particle (to the extent of our current laws of physics) and it will only interact with the plate at one point. (Some people will not like my language and tell me I am wrong. But it is true the electron will hit the plate at a single point - not smeared out like the wavefunction is.) In actuality the electron didn't cause the damage to the plate. The electric field did. The electric field moves with the electron almost like you would expect in classical physics. It's not like the electric field is smeared around because the electron is in a superposition of states. I'd expect gravity to work the same way. But seeing how we don't have a theory it is possible it could violate the above idea. To me that would be a pretty weird violation too. (This short writeup doesn't really do that justice.) I'd bet against it.