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Correct me if I'm wrong, but when two or more particles become bonded together, the particle they make together can have its own wavefunction that behaves just like any other. This happens when you go from quarks to baryons, baryons to nuclei, nuclei to atoms, and atoms to molecules. We've gotten things as big and complex as buckyballs to act like waves. When the bigger structure starts acting like a wave, what happens to the wavefunctions of the component particles? Do they just merge? Do they temporarily cease to exist?

Please explain your answer qualitatively. Equations won't help me.

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Consider a non-relativistic, spin-less, elementary particle. The wavefunction of such a particle is an assignment of a complex number $\psi(\mathbf x)$ to each point in space. The interpretation of this quantity is that $|\psi(\mathbf x)|^2$ is the probability density for finding the particle near the point $\mathbf x$.

A composite system of two such particles also has a wavefunction; however, the composite wavefunction is not a function of one position but rather of two. That is, it is a function $\Psi(\mathbf x,\mathbf y)$ which is interpreted such that $|\Psi(\mathbf x,\mathbf y)|^2$ is the probability density for finding the first particle near $\mathbf x$ and the second particle near $\mathbf y$.

It is generally not the case that the composite wavefunction can be separated into the product of two single particle wavefunctions. That is, we typically cannot find $\psi_1$ and $\psi_2$ such that $\Psi(\mathbf x,\mathbf y) = \psi_1(\mathbf x)\psi_2(\mathbf y)$. This means that for a general state of a composite system, it doesn't make sense to talk about the state of the first or second particle individually because they are inextricably tied together.

That being said, we can do the following. If we let $\mathbf R = (\mathbf x + \mathbf y)/2$ (more generally, we could let $\mathbf R$ be the center of mass) and $\Delta = (\mathbf x - \mathbf y)/2$, we can rewrite the full wavefunction as $\Phi(\mathbf R,\Delta) = \Psi(\mathbf R+\Delta,\mathbf R-\Delta)$. It may then be possible to approximate the total wavefunction as $\Phi(\mathbf R,\Delta) \simeq \phi_1(\mathbf R)\phi_2(\Delta)$. The interpretation of this is the following: $\phi_1(\mathbf R)$ is the "extrinsic" wavefunction of the composite particle - for a bound state, $\phi_1(\mathbf R)$ would be the probability amplitude for the state's average position to be at $\mathbf R$. The other part, $\phi_2(\Delta)$, would encode the internal structure of the state. As long as your experiments don't probe this internal structure, then it may be sufficient to treat the composite particle as a single particle with wavefunction $\phi_1(\mathbf R)$.

If you are familiar with the hydrogen atom problem, this is precisely what we do; we find that $\phi_1(\mathbf R)$ is the wavefunction of a free particle. The hydrogen atom wavefunctions that we usually discuss (consisting of the electronic orbitals) are in fact the $\phi_2(\mathbf \Delta)$ part of the composite wavefunction, because we're typically not interested in the former part.


When we model two particles suddenly interacting with one another, we typically do the following. First, we assume that the initial states of the two particles can be disentangled from one another - i.e. $\Psi(\mathbf x,\mathbf y,t=0)=\psi_1(\mathbf x)\psi_2(\mathbf y)$. At this moment in time, it makes sense to say that the two particles have individual wavefunctions. However, as time progresses forward the wavefunction loses this property and becomes an inseparable, joint function of $\mathbf x$ and $\mathbf y$. Such a state is called entangled, and it no longer makes sense to ask about the individual wavefunctions of the constituent parts.

I know you aren't looking for equations, but as a cartoon example of a composite wavefunction (where I ignore questions of normalization for simplicity), consider the following example for two particles in 1D:

$$\Psi(x,y,t) = e^{-(x-x_0)^2} e^{-(y-y_0)^2} e^{-t} + e^{-(x+y)^2}(1-e^{-t})$$

At $t=0$, this becomes $\Psi(x,y,0)=e^{-(x-x_0)^2} e^{-(y-y_0)^2}$. Because it is clearly a product state, we can say that the first particle has wavefunction $e^{-(x-x_0)^2}$ and the second has wavefunction $e^{-(y-y_0)^2}$. After a long time, however, the wavefunction becomes $\Psi(x,y,t\rightarrow \infty) = e^{-(x+y)^2}$, which cannot be written as the product of two single-particle wavefunctions. As such, there's no way to talk about the wavefunction of the first or second particle individually.

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  • $\begingroup$ Is this the reason it's said that everything in the universe is described by one, ludicrously complex wave function? $\endgroup$
    – zucculent
    Aug 8, 2021 at 21:53
  • $\begingroup$ @zucculent I've added a section to my answer. To answer your follow-up, yes, that's the idea. Whether that assumption yields a useful and accurate model is not entirely clear. $\endgroup$
    – J. Murray
    Aug 8, 2021 at 21:59
  • $\begingroup$ Looking at the edited section, what I'm getting is that the reason composite particles can be treated as having their own wavefunctions is that the binding force makes the component particles likely to be found next to each other when measured, so when you zoom out, their position values will be so close that it looks like a wavefunction with one position variable instead of two. $\endgroup$
    – zucculent
    Aug 8, 2021 at 22:15
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    $\begingroup$ @zucculent That’s more or less right. States in which the components are far away from one another exist, of course - scattering states are an example, as are states in which the constituents are so far apart as to be effectively free - but the stable bound states tend to be characterized by a high degree of spatial localization. $\endgroup$
    – J. Murray
    Aug 8, 2021 at 23:09
  • $\begingroup$ Do things change when we're talking about momentum? I'm pretty sure the extra mass protons and neutrons have is from the quarks zipping around inside them, so it doesn't seem like the momentums of bound particles are necessarily as closely tied. Although being so closely bound together in space mandates that the constituent particles similar if not identical average momentums across time. $\endgroup$
    – zucculent
    Aug 8, 2021 at 23:15
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We can take this question all the way to the limit: People conjecture that there exists a wavefunction describing the entire universe. Such a wavefunction must be extremely complicated, because it has to contain all of the possible degrees of freedom in the entire universe, but that does not detract from it being a wavefunction.

Qualitatively, we expect the correct wavefunctions that describe bigger and bigger things to act exactly as we observe those bigger things. So it is indeed correct that the wavefunction for a bowling ball won't seem to be acting so much like a "wave" in the usual sense, but people still believe there to be a wavefunction capable of describing all of the bowling ball's properties.

Relatedly: physicists don't believe there to be an upper limit to the size of an object that may diffract through a double slit - it's just that the properties of the required slit are hard to come by when you get much larger than, say (if I recall correctly), viruses.

So what happens to the wavefunctions of the individual particles? In the prevailing theory, there is no individual wavefunction for each particle. Instead, there is a composite wavefunction for the entire universe. If you want to describe an individual particle, you have to do some mathematical operation that is physically equivalent to ignoring the rest of the universe. When you do this process with a sufficiently isolated particle, it seems as though we have a wavefunction that describes the particle as existing all on its own, with nothing else in the universe to affect it. This will only ever be an approximation (a good approximation in clever experiments and a poor approximation in most other scenarios), while to be perfectly correct we need to look at how the wavefunction of the entire universe evolves instead of how an individual particle's wavefunction evolves (at any point in time, we can repeat the "ignoring" operation to see what the individual particle seems to be doing).

Now, let's assume the question already presupposed all of the above and wanted to know what happens to sufficiently isolated individual particles when they are brought together. For example, let's say there are two individual particles infinitely far away from each other in an otherwise-empty universe, which are gradually brought closer together. If the two particles do not have any means of interacting, the overall system will simply be described by the two individual wavefunctions - it is as if each particle is invisible to the other. If the two particles do have a means of interacting, they will evolve jointly in such a way that one will not be able to describe either each particle on its own, which will be a manifestation of quantum entanglement. Stronger/longer interations in general lead to more entanglement and thus less ability to describe each particle on its own. This evolution happens continuously, through a "unitary" evolution that preserves the total probability of all possible events that can occur. With some additional mathematical machinery and a description of all of the possible particles and their interactions, one can build up an entire description of the universe (at least, one should be able to reproduce all quantum field theories).

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  • $\begingroup$ This doesn't answer my question. I'm asking what happens to the wavefunctions of the particles that make up the bigger structure, not how big and complex a wavefunction can get. $\endgroup$
    – zucculent
    Aug 8, 2021 at 21:13
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    $\begingroup$ Agree with zucculent: answer doesn't really address the question. $\endgroup$
    – Gert
    Aug 8, 2021 at 22:09
  • $\begingroup$ @zucculent ah but there's the crux: everything is described by the evolution of the big wavefunction. I've added more detail along the lines of assuming we're approximating isolated systems $\endgroup$ Aug 9, 2021 at 2:23
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I want to point out a simple thing. If the particles do not interact, the original wavefunctions just coexist. The total wavefunction is just the tensor product of the original two.

If they do interact, their respective Hamiltonians change due to the presence of the other particle. Each Hamiltonian will now have new solutions. Now IF you were to track this in time, each particle can be assigned a new wavefunction at each point in time, until they come to rest.

IF on the other hand, you look at two equilibrium situations, one with the particles "far away" from another, and one were they are bound, the connection is absolutely unclear. This is typically the approach, since the time evolution is either uninteresting or impossible to solve.

But in principle, if such a hypothetical tracking were possible, you could always seperate the wavefunctions each particle is in. Just that these new wavefunctions certainly are not the original wavefunctions.

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