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When dealing with a particle as represented by a probability field, how is mass distributed across the field? Would the mass be averaged across the field?

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marked as duplicate by WillO, Jon Custer, Aaron Stevens, Kyle Kanos, ZeroTheHero Nov 2 '18 at 11:50

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    $\begingroup$ I'm not quite sure what you mean by "how is mass distributed across the field". Would you edit your question and add additional detail for what you have in mind? Are you thinking of mass as an observable with an associated basis? Or is it something else? $\endgroup$ – Alfred Centauri Nov 1 '18 at 1:39
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Probability fields are not physical. They do not represent a "smearing out" of the particle in question$^*$. Therefore, you don't need to wonder how the mass is distributed throughout space. Also, your probabilities don't even have to pertain to the measurement of a particle at a certain location in space. You could talk about probabilities of measuring other observables, like certain values for energy for example. In other words, these "probability fields" do not exist in physical space. (See below for a more in depth explanation.)

Furthermore, it doesn't make a lot of sense to say that a particle is "in a superposition", since that superposition depends on what basis you choose to work in. It would make more sense if you specified the basis you are referring to.


In general, your system can be described by an abstract state vector $|\psi\rangle$. If we want to know the probability of measuring the particle to be at some position$^\dagger$ $x$, we choose to express $|\psi\rangle$ in the position basis $$|\psi\rangle=\int_{-\infty}^\infty|x\rangle\langle x|\psi\rangle\ dx=\int_{-\infty}^\infty\psi(x)|x\rangle\ dx$$

This gives us the wavefunction, $\psi(x)$. The probability of measuring the particle to be at a position between $x$ and $x+dx$ is given by $|\psi(x)|^2$, and this is most likely what you are referring to. It does cover all space, but it is not a physical field. We started with an abstract vector $|\psi\rangle$ and then decided to express it in terms of position eigenvectors (in the position basis spanned by the $|x\rangle$ vectors).

However, we don't need to work in the position basis. We could also work in the momentum basis spanned by the momentum eigenvectors $|p\rangle$: $$|\psi\rangle=\int_{-\infty}^\infty|p\rangle\langle p|\psi\rangle\ dp=\int_{-\infty}^\infty\psi(p)|p\rangle\ dp$$ This gives us a function $\psi(p)$, whose squared magnitude $|\psi(p)|^2$ tells us the probability of measuring the particle to have a momentum between $p$ and $p+dp$. In this example it is easier to see we aren't working with physical entities, since (as far as I know of) "momentum space" is not somewhere we can physically be.

Furthermore, both of the above examples expresses the state of the particle as a superposition of basis vectors. In general a superposition is just a sum, and you can argue that the state vector is always in a superposition (since $|\psi\rangle = |\psi\rangle+0$). Therefore, it is always beneficial to express what "type of superposition" you are expressing your state vector as.


$^*$ I see a lot of layman explanations of QM that say something along the lines of "Particles can exist at multiple positions at once." I find this to be inaccurate. A better way to say it is "Particles have no definite position until we measure them to be there." (Of course there are other interpretations of QM, but this is the "traditional" interpretation).

$^\dagger$Technically finding the particle at an exact position is not physical, but I will not get into it here.

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  • $\begingroup$ "In other words, these "probability fields" do not exist in physical space." I see your asterisk but I thought that position (maybe not exact position) would be one property we could say is in physical space ..... but I believe you about the exact position part. $\endgroup$ – PhysicsDave Nov 1 '18 at 1:40
  • $\begingroup$ @PhysicsDave Of course measurements of physical locations of particles exist in physical space. But the state vector of the particle is not physical. Even if we determine $\psi(x)$ it is not a physical thing (although I am not trying to make a philosophical point). You could argue it is related to physical space, just so far as it tells you the probability of measuring the particle to be at some position in space. But, as you can see from the momentum (or any other observable) example, these state vectors / wave functions are not physical themselves. $\endgroup$ – Aaron Stevens Nov 1 '18 at 1:45
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    $\begingroup$ @PhysicsDave, recall that the wavefunction for $N$ particles lives in $3N$ dimensional configuration space and not physical space. $\endgroup$ – Alfred Centauri Nov 1 '18 at 1:56
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    $\begingroup$ Aaron, I even see non-layman explanations using the "particle in multiple positions at once" which has always seemed wrong to me. I'm glad I'm not the only one that finds it so. $\endgroup$ – Alfred Centauri Nov 1 '18 at 2:02
  • $\begingroup$ Ok thanks Aaron, but it's getting over my head, I'm an amateur at QM, just one course in 3rd year. I can sort of see the what your saying that each particle gets 3 dimensions to account all interactions (?). I appreciate your answers. $\endgroup$ – PhysicsDave Nov 1 '18 at 2:04

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