I'm not a quantum scientist, but I don't think this is the correct view amplitude. If a configuration has amplitude -1 for electron located in some place, it doesn't mean that there's a positron there.

I am a quantum physicist and I can confirm that’s not how probability amplitudes work . The amplitudes are just complex numbers. Remember that complex numbers describe oscillations. A number r * exp(i phi) is an oscillation with amplitude r and current phase phi. Quantum states are wave functions, hence the complex numbers describing the oscillations. Amplitude -1 is just the bottom of the oscillation. Then on top of that you have that probabilities are the squares of the amplitude (for which I’m not sure if have a concise explanation).

I'm not a quantum physicist either. The amplitude would have to be j or -j to get a probability of -1 (probability = amplitude squared?). Does +/-j correspond to a positron?

No. The probability of an event is never, ever negative, even in quantum. We use the complex scalar product, c x c^*, to get probabilities, and those are necessarily positive even for complex numbers. The same is true for the generalization to functions required in quantum mechanics.

If your wave function at point (0,0,0) were i x delta(0) (i.e. a point-like particle perfectly located at the three-dimensional origin), the probability of finding the particle there isn't i^2, but i x i^* = i x (-i) = 1.

Thanks!

Surely that probability should be i x delta(0) x (-i) x delta(0) so, you know, your everyday squared Dirac delta function.

I secretly did the integration over space to go from the probability density to the probability, so the deltas are integrated out :P

Doing calculations over text-based comments is so painful that I gloss over a lot, in the expectation that this isn't the right place to talk about anything but vague ideas anyway.

No, any electron amplitude with absolute value (squared) equal to one corresponds to a probability density of 1 to find an electron. You can have a positron wavefunction if you have a positron, but take care that you don’t have both, or they will (even if with a vanishingly small probability) meet and, by participating in a process with energy comparable to mass, boot you right out of nonrelativistic quantum mechanics into quantum field theory.

I hate to tell people to learn A before they can try B, but here I really, really don’t know of a way to start thinking about QFT with its positrons and annihilation and so on without getting hopelessly confused unless you’re already comfortable with normal wavefunctions-and-Schrödinger’s-equation QM. You’ll still be confused even if you are comfortable with it, mind you, it’s just that then you’ll at least have a ghost of a chance of getting your confusion down to manageable levels.

> ghost of a chance ... down to manageable levels

Argh. Manageable as in non-divergent, so take a small step back to regularize your confusion, and get haunted by a <https://en.wikipedia.org/wiki/Ghost_(physics)>.

I think you're on your way to coming up with the idea of an expected value. The additional step of reasoning to get there is to think about what number we'd use to describe an outcome that had a 50% probability of going "forwards" and a 50% probability of going "backwards." If 0 makes sense as an answer to that, you're doing a probability-weighted average of outcomes. Quantum mechanics is full of expected values, and you can interpret wavefunctions just fine with them.