If particles are also waves cant we invert them by making it reflect off a fixed boundary? When a pulse on a rope gets reflected off a fixed boundary, it phase shifts 180° and inverts itselfs.

If particles are also waves, will doing an equivalent thing create antiparticles?
Is it possible for the quantum field to have non conductive properties in certain conditions and thus allow for the invertion of particles?
 A: The  elementary particles are described by quantum mechanical equations , wave equations but the solutions are not waves in the  (x,y,z, t) of the particle but of the probability to find the particle at (x,y,z,t). The wave nature is in the probability of interaction distributions, which means a single event interaction of a particle has the footprint we expect of a classical particle.
A way to understand the particle /wave duality can be  the single electron double slit experiment: a single electron scattering  on a double slit, see my answer here.


So it is not a classical particle behavior because even though the energy is carried by the single electron, its (x,y) is controlled by a probability distribution; and it is not the classical wave, i.e. a single electron that is "waving" its mass all over the screen interference pattern. Each electron is a quantum mechanical entity.

A: Here is a simulation of the Schrödinger equation I made a while ago:

In the gif you see the Schrödinger equation for a single, one-dimensional particle. The particle moves on the left-to-right axis and the other two axes show the real and imaginary parts of the wavefunction; the value of the wavefunction consists of complex numbers. If you have never heard of them I invite you to learn more about them because they are used extensively in quantum physics and also other areas of math and physics. The distance$^\dagger$ from the red line to the axis shows the probability to find a particle at that certain location: a large distance means a high probability.
The boundary conditions is a fixed boundary just like in your example. What you see after the reflection is just the same particle but travelling in the other direction. No particles are created/destroyed. Regular quantum mechanics mostly works with a fixed number of particles; if you want to model particle creation/annihilation you have to resort to Quantum Field Theory, which is even more complicated than quantum mechanics.
Also the creation of antiparticles is limited by strict rules. To name an example a photon can be used to create an electron/anti-electron pair. An anti-electron is also called a positron. They come in a pair because charge and something called "lepton number" has to be conserved. Also the photon has to have enough energy to ensure energy conservation. It also has to happen near a nucleus because otherwise momentum cannot be conserved: we can position ourselves such that the electron/positron together have zero momentum but a lone photon can never have zero momentum.
$\dagger$ actually the distance squared but this is just a detail.
A: I see your intuition that cancelling waves are analogous to particle-antiparticle annihilation, but quantum state of an antiparticle is not just a $180^\circ$ phase shift of the quantum state of a particle.
A: In quantum mechanics , particles are described by wave functions, which represent the probability that a particle is found in vicinity of point (x,y,z). Now, suppose you have a barrier whose task is to just give a shift of 180 degrees to the particle or make wavefunction to its additive inverse, the schrodinger equation will remain unchanged because it is same for +f or -f where f is the wavefunction.
Although, if you have a barrier, which shifts the time coordinate, although then it would mean that the potential barrier particle is facing is momentum dependent and the wave function of particle would no longer be described by schrodinger equation. At this point, such a potential barrier allows creation of antiparticles whose probability of finding is in opposite direction of time.
A: There is a very nice answer by @annav, I would like to add something interesting. A very nice example of what you are asking for is when a photon (or classically light) gets reflected off a surface (for example reflecting off a mirror or the wall).
Now there are mainly two theories explaining reflection:

*

*elastic scattering (specular reflection from a mirror), in this case the photon just changes angle, but the photon does not cease to exist, it is the same photon reflecting back from the mirror, only the angle (momentum vector) has changed. In this case, your question, whether the photon is turned into an anti-particle by reflection is, no.

https://en.wikipedia.org/wiki/Specular_reflection

https://www.youtube.com/watch?v=ahsmaodguTo
In the image (please look at the video) above they did something truly phenomenal, recording light bounce off a mirror at 100 billion frames per second.


*absorption and re-emission (diffuse reflection off the wall), in this case the theory is that the photon gets absorbed by the atoms/molecules in the surface of the material, and consequently gets re-emitted. The original photon ceases to exist. The newly emitted photon could have the same quantum characteristics, as the old one (energy, phase etc), only its momentum vector is different (or the newly emitted photon could have completely different energy level etc). In this case, you could imply that the newly emitted and original photon are each other's antiparticle, and you could interpret this as being right about turning photons into their antiparticle by reflection. Theoretically we do say, that photons are their own anti-particles.

https://en.wikipedia.org/wiki/Diffuse_reflection
So the answer to your question is that in certain cases of reflection, when a photon gets absorbed (ceases to exist), and re-emitted (a new photon), you could interpret this as a procedure that turned a photon into its own antiparticle (but only because the widely accepted theory is that photons are their own antiparticles).
