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It seems you can make a beam of particles in identical state and you can perform the double slit experiment and get the fringes corresponding to the de Broglie waves of the entire system. Double slit experiment has been successfully demonstrated using large systems like buckyballs and the current record is doing it with 810 atom molecules. And that's more than 5000 protons, 5000 neutrons and 5000 electrons collectively behaving as a single particle.

What blows my mind is what makes these large systems behave a single particle instead of a collection of atoms?

How does nature decides how these systems should be grouped? Eg. a single entity for the 810 atom system, instead of 9 90 atom entities or any other kind of groupings.

That 810 atom molecule is probably large enough to be seen well in an electron microscope. But it can still make it a slit it can't even fit into. Why can it do that instead of just crashing into the slit and breaking up or clogging the slit?

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Obviously, the slits must be larger than the molecules passing through them, otherwise you would be right and the molecules would just break up and nothing could be seen.

Apart from that, it should not be such a huge surprise that a many-particle system effectively behaves like one single particle. This happens all the time in physics. If you throw a ball, the $\sim 10^{23}$ particles in it effectively move like one solid body, and it is sufficient to describe the motion by the motion of the center of mass. The same is true in quantum mechanics for bound systems like molecules. Nature decides what stays together by the binding forces, that's the point.

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  • $\begingroup$ But when we think in terms of wavefunctions and de Broglie wavelengths and frequencies, it's hard to see how these things add up. How does the tensor product of the wavefunctions of these many particles make the whole thing behave like a single particle? That's my question why asked 'How?' $\endgroup$
    – Calmarius
    Commented Aug 17, 2017 at 14:26
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    $\begingroup$ The wave function of this system will not be a tensor product, because the atoms in the molecule are strongly entangled (bound together). But you should expect that, really like in a classical solid body, the motion of center of mass decouples from the rest. Something like $\psi(x_1,...,x_N) = \psi_{cm}(X) \otimes \psi_{internal}(...)$ where $X$ is the center of mass coordinate. $\endgroup$
    – Luke
    Commented Aug 17, 2017 at 14:55
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It is very often the case that the lowest energy mode of a many-body system is collective precisely in the sense that the system functions as a single particle.

A nice illustration of this - valid in classical and quantum mechanics - is a long chain of identical masses connected to nearest neighbours by a spring. The potential energy for such a system is $$ U=\frac{k}{2}\sum_{j}\left(q_j-q_{j+1}\right)^2\, . $$ For $n$ of these, with cyclic boundary condition so that $q_{n+1}=q_1$, the normal coordinates are $$ Q_k=\sqrt{\frac{m}{n}}\sum_j e^{2\pi i k j/n} q_j \tag{1} $$ with normal frequencies $\omega_k=2\sqrt{\frac{\alpha}{m}}\sin(\vert k\vert \pi/n)$. The lowest energy mode thus corresponds to $k=0$ (it's a pure translation mode) and the normal coordinate is $$ Q_0\sim \sum_j q_j\, . $$ In this translational mode, all particles "collectively move as one" to the right (or the left). The same conclusion applies if the first and last masses are attached to fixed wall, although of course in this case the lowest energy mode does not have $\omega=0$, but nevertheless all masses oscillate together.

On the purely quantum side of the business the ground state of many systems is well described by a coherent state of this system. For instance, spin-S coherent states have the property that $\langle \vec S\rangle = S\langle \vec \sigma\rangle$, i.e. the average values of the total spin of this state is $S$ times the average value of the spin in any one state.

It seems to be somehow built into Nature that the lowest energy state are very symmetrical, leading naturally to (1) or to coherent states.

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