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My (admittedly limited) understanding of the Schrodinger equation tells me that the vector differential operators are only meaningful over a differentiable phase space.

For example, if the dimensions of my phase space are the Euclidean coordinates of $N$ featureless point particles, then it makes sense that you can construct a $3N$ary field of some (algebraically) vector quantity, associate it with these coordinates, and apply differential operators. If the quantum wavefunction only governed these positions then the interpretation would be very direct.

But if the wavefunction allows particles to be created, then we might now have a future state in a $3(N+1)$ary field. In the finite case the interpretation breaks down.

Now, I know the wavefunction operates on an infinite-dimensional phase space, and if I strain my imagination I can just about conceive an association between this object and the particles in the universe (if I take it for granted that all particles are everywhere and the particle field is the sum of these, okay)...

...but I cannot conceive of a way this wavefunction can be continuous when particles are created or destroyed...

...and if it isn't continuous, how can we apply differential operators?

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First, you are not talking about a phase space, but configuration space.

Now, the space of wavefunctions of a single particle in 3D space is the space of Lebesgue square-integrable functions $L^2(\mathbb{R}^3)$ on the configuration space $\mathbb{R}^3$. So already for a single particle, there is no generic requirement for the wavefunction to be smooth, differentiable or even continuous.

However, in practice, you may deal with Hamiltonians that are differential operators, and hence one needs to require the existence of derivatives of the wavefunction, i.e. pass to the Sobolev space $W^{2,2}(\mathbb{R}^3)$ for which weak second derivatives exist. Still the wavefunction itself is an equivalence class of square-integrable functions, and the representants need not be differentiable in the strong sense.

Therefore, the wavefunction is already not necessarily smooth, differentiable or even continuous in the case of a single particle.

Introducing many particles and the possibility of their creation and annihilation, quantum mechanics deals with this by defining the one particle-space $H_1 := L^2(\mathbb{R}^3)$ and the $i$-th particle space as the $i$-th symmetrized tensor product $$ H_i = S^i (H_1)$$ for bosons or the antisymmetrized product for fermions. The total space of states then becomes the Fock space $\bigoplus_{i=0}^\infty H_i$. However, the proper framework for the annihilation of creation of arbitrary particles is quantum field theory, not quantum mechanics.

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Continuity of a wave-function is not a necessary condition, as already pointed out by @ACuriousMind.

For instance, consider a linearly dispersing particle in the presence of a $\delta$-function potential. The time-independent Schrödinger equation in this case is

$-i \frac{\partial}{\partial x} \psi(x) + V \delta(x) \psi(x) = E \psi(x)$

Typically, you have quadratically dispersing particles, in which case you can indeed find a continuous wave-function (See https://en.wikipedia.org/wiki/Delta_potential).

However, owing to the linear dispersion, for any $V\neq 0$, the wave-function must necessarily have a discontinuity at $x = 0$.

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How can quantum wavefunctions be smooth/continuous when particles are created/destroyed/changed?

Because particle creation etc merely involves a change of configuration. There is no magic. In gamma-gamma pair production you start with two photons, and you end up with an electron and a positron. Each has a wave nature, as evidenced by things like electron diffraction. When you annihilate an electron and a positron you get two gamma photons, and they have a wave nature too. Note that in atomic orbitals electrons "exist as standing waves". It's like you change the wave configuration from an open path to a closed path.

My (admittedly limited) understanding of the Schrodinger equation tells me that the vector differential operators are only meaningful over a differentiable phase space.

All you need to understand is that the Schrödinger equation equation is a wave equation.

For example, if the dimensions of my phase space are the Euclidean coordinates of N featureless point particles

There are no point particles. The E=hf photon has a wave nature, and so does the E=mc² electron. It's field is what it is. It isn't some speck that has a field, it is field. Think standing wave standing field, and there ain't no billiard ball in the middle. And if anybody objects to that, tell them it's quantum field theory, not quantum point-particle theory.

then it makes sense that you can construct a 3Nary field of some (algebraically) vector quantity

Yes, you can construct a standing field from a dynamical wave. That's what gamma-gamma pair production does. Then we talk about vector fields:

enter image description here

Public domain image by Fibonacci, see Wikipedia.

But if the wavefunction allows particles to be created, then we might now have a future state in a 3(N+1)ary field. In the finite case the interpretation breaks down.

Sorry, I don't know what you mean by this. But can I offer that the electron's electromagnetic field is dynamical, hence its magnetic moment. And its helical motion through a magnetic field. And the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics."

Now, I know the wavefunction operates on an infinite-dimensional phase space, and if I strain my imagination I can just about conceive an association between this object and the particles in the universe (if I take it for granted that all particles are everywhere and the particle field is the sum of these, okay)...

You should read up on weak measurement and stuff like this by Jeff Lundeen:

"With weak measurements, it’s possible to learn something about the wavefunction without completely destroying it. As the measurement becomes very weak, you learn very little about the wavefunction, but leave it largely unchanged. This is the technique that we’ve used in our experiment. We have developed a methodology for measuring the wavefunction directly, by repeating many weak measurements on a group of systems that have been prepared with identical wavefunctions. By repeating the measurements, the knowledge of the wavefunction accumulates to the point where high precision can be restored. So what does this mean? We hope that the scientific community can now improve upon the Copenhagen Interpretation, and redefine the wavefunction so that it is no longer just a mathematical tool, but rather something that can be directly measured in the laboratory."

but I cannot conceive of a way this wavefunction can be continuous when particles are created or destroyed.

Conceive of it changing direction. From straight to curved, or vice versa.

...and if it isn't continuous, how can we apply differential operators?

It's continuous.

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