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HumanistHippy t1_j9gs3bl wrote

It was the result of his equation that combined quantum mechanics with special relativity in order to describe the behavior of an electron moving at a relativistic speed.

The equation necessitated a "positron" mathematically. Unless the math was incorrect (which it wasn't), the "positron" had to be there even if we were unable to observe it at the time.

Source: CERN

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shimadon t1_j9in5h2 wrote

Good answer, a quick correction: even if the math is correct, it doesn't necessarily mean that everything the math predicts has to be real. That's indeed what happened with Dirac, but it doesn't have to be true for all mathematical models of the physical world.

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Holiday_Document4592 t1_j9jnxdn wrote

So how do we distinguish between the math that has predictive capability and the one that doesn't?

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mfb- t1_j9ka5v6 wrote

We run experiments to check.

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tpolakov1 t1_j9lef5p wrote

If a theory predicts something and we don't see it in an experiment, then it was a wrong prediction, or a wrong theory.

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[deleted] t1_j9ixno1 wrote

[deleted]

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Redingold t1_j9jmpip wrote

I want to expand on the bit about negative energy there.

The issue with having negative energy states allowed is that, generally, particles try to occupy the lowest energy states available to them. Electrons in an atomic orbital for example, will emit photons and move onto lower energy orbitals if those orbitals are empty. Having negative energy states, and, in particular, having negative energy states with no lower limit (you can see that as p gets larger and larger, -sqrt(p^(2)c^2 + m^(2)c^(4)) gets lower and lower without bound) means that electrons should all be shedding as much energy as they can as they cascade down towards negative infinity energy. This doesn't resemble the reality we live in, so something else must be happening.

What Dirac reasoned was that you could make it work if all the negative energy states were already filled, all the way down to the bottom. The positive energy states would be free and available for all the conventional electron physics we already know about, and the negative energy states would be blocked, so electrons won't have room to go barrelling down towards negative infinity.

This has an interesting implication, though. There's a gap between the positive and negative energy states: when p = 0, the energy jumps from E = mc^2 to E = -mc^(2) (so the gap is 2mc^2 wide, or wider if p is not 0), and there are no states at all in this region. If, however, you supplied that 2mc^2 worth of energy (or more), you could make an electron jump up from an occupied negative energy state, into a free positive energy state. This would leave a hole in the negative energy states, and this hole would act like it had a positive charge (because when an electric field is applied, the electron next to it moves into that space, leaving a new space where it was before, and then the next electron moves into that space, and so on, like how a bubble rises in water by the water falling down and pushing the bubble up. The hole would move in the opposite direction of the electrons, and so would have the opposite charge, i.e. positive). So by supplying at least 2mc^2 of energy, you could make an electron, and a hole that behaves like a positively charged electron. That's a positron.

Nowadays, modern physics doesn't really use this Dirac sea model, as it's called. Quantum field theory lets us treat the ground state as a vacuum containing no particles, not an infinite sea of negative energy particles, and it treats positrons as real particles, not holes where an electron isn't in the negative energy sea. The Dirac sea model also struggles to explain why we don't see the effects of having infinity negatively charged particles everywhere in the universe (although the energy of the vacuum isn't a solved problem in quantum field theory, either, see the cosmological constant problem). That said, the two are mathematically compatible to an extent.

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bazongaenthusiast OP t1_j9ln0xf wrote

Sorry, I forgot to mention it. I am a mechanical engineer who is very much interested in quantum physics. Thank you for your answer, it cleared up the doubt effortlessly.

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