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?
Answer
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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