If I understand correctly, there is a small probability the same electron to be found anywhere in the universe.

Suppose that an anti-electron collides with an electron, annihilating it and producing two photons. Assuming the speed of light limit is correct, right after the collision, the probability that the electron is found at a distance of $d$ from the collision must still be non-zero for a time of at least $d/c$.

But wouldn't this mean that it's still possible for the annihilated electron to still be present somewhere else and collide with another anti-electron, meaning that the same electron was annihilated twice?

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    $\begingroup$ What would that mean for energy conservation? Also, why stop at twice? $\endgroup$ – G. Smith Aug 14 '19 at 0:00
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    $\begingroup$ Stefan, non-relativistic quantum mechanics (QM) is (unsurprisingly) incompatible with Special Relativity (SR). Reconciling QM with SR eventually led to quantum field theory (QFT). What is the context of your question, QM or QFT? $\endgroup$ – Alfred Centauri Aug 14 '19 at 0:11
  • $\begingroup$ @AlfredCentauri if I understand your comment correctly, I could say that given A, B and C QM predicts X but SR proves that X is impossible, and QFT predicts Y which neither QM or SR can prove is impossible, right? If so, can I say that in the context of A, B and C QM is wrong and QFT can't be proven wrong so far, so QM is an invalid context for a question about A, B and C and a context that we so far consider valid is QFT? $\endgroup$ – Blueriver Aug 14 '19 at 18:33
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    $\begingroup$ Nice, a double-free vulnerability in the universe $\endgroup$ – Addison Crump Aug 15 '19 at 3:40
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    $\begingroup$ Have you heard of Wheeler’s one-electron universe postulate? Of course the same electron gets annihilated twice: there’s only one electron anyway! $\endgroup$ – Gilbert Aug 16 '19 at 13:45

No, you have to apply the superposition principle consistently. Schematically, the initial state is $$|\text{electron here} \rangle + |\text{electron there} \rangle$$ where I've dropped normalization constants, and a $+$ denotes quantum superposition. Now suppose a lot of positrons come through, so electron states get annihilated, $$|\text{electron here} \rangle \mapsto |\text{some gamma rays here} \rangle,$$ $$|\text{electron there} \rangle \mapsto |\text{some gamma rays there} \rangle.$$ What you're essentially claiming is that the final state is $$|\text{some gamma rays here } \textbf{and} \text{ some gamma rays there}\rangle$$ but if you just apply linearity, the final state is actually $$|\text{some gamma rays here} \rangle + |\text{some gamma rays there} \rangle.$$ This reasoning implies you can't get double the gamma rays, no matter how severe the speed of light delay or any other delays are. Instead the superposition of electron positions can at best turn into a superposition of gamma ray positions. The mistake you made is essentially forgetting that the electromagnetic field behaves quantum mechanically too. (It's a forgivable mistake, which was made by many in the early days of quantum mechanics, leading to precisely the same kinds of paradoxes.)

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    $\begingroup$ +1, this is one of the easiest-to-understand quantum mechanics answer I've seen on this site. $\endgroup$ – jpa Aug 14 '19 at 9:05
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    $\begingroup$ Please tell me if I'm wrong: Once the positron hit the electron at a location L, the probability the electron is anywhere else not L drops to zero. - It's basic probability. The probability an observed, previous event happened the way it happened is 1. - Also, Morpheus lied to Neo in the Mero's elevator. $\endgroup$ – Mindwin Aug 14 '19 at 12:12
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    $\begingroup$ @Mindwin That's not an ideal way to put it because it treats the positron as being some magical object not subject to quantum mechanics (just as I complained the OP was treating gamma rays). The rules for everything are the same. In the absence of measurements, a superposition just turns into another superposition. $\endgroup$ – knzhou Aug 15 '19 at 1:10
  • $\begingroup$ @knzhou I thought that photons did not have a position operator. So is |some gamma rays here> a valid state in QFT? $\endgroup$ – Dale Aug 15 '19 at 11:30
  • $\begingroup$ @Dale There are technical restrictions on how well one can localize photons, but that doesn't make it meaningless to speak of the location of anything having to do with the electromagnetic field. For example, when I turn on the lights I'm pretty sure the photons produced are in my house. $\endgroup$ – knzhou Aug 16 '19 at 0:53

The nice answer by knzhou considers a situation in which the electron is annihilated with probability $1$. The following answer considers a situation in which the final state is a superposition of already-annihilated and not-yet-annihilated.

Choose two points in space, $x_1$ and $x_2$, that are arbitrarily far away from each other. Consider an electron in the state $$ |\psi\rangle = |1\rangle + |2\rangle, $$ where

  • $|1\rangle$ is a state in which the electron is tightly localized near $x_1$, and an antielectron is approaching $x_1$.

  • $|2\rangle$ is a state in which the electron is tightly localized near $x_2$, and an antielectron is approaching $x_1$. (This is not a typo! The antielectron is approaching $x_1$ in both $|1\rangle$ and $|2\rangle$.)

When the antielectron reaches the point $x_1$,

  • $|1\rangle$ becomes $|1'\rangle$, a state with two photons propagating outward from $x_1$.

  • $|2\rangle$ becomes $|2'\rangle$, a state with one electron at $x_2$ and one antielectron receding from $x_1$.

Time-evolution is linear, so the final state overall is $$ |\psi'\rangle = |1'\rangle + |2'\rangle. $$ This is a superposition of "electron already annihilated" and "electron still present at $x_2$," so there is still a non-zero probability $$ \frac{\langle 2'|2'\rangle}{\langle\psi'|\psi'\rangle} $$ that an antielectron can annihilate the electron at $x_2$.

Be careful, though: this does not mean that the same electron can be annihilated twice! The electron is only annihilated once, but it can be in a state that is a superposition of already-annihilated and not-yet-annihilated. If we measure some observable to tell us how many times the electron was annihilated (say, by counting pairs of outgoing photons), the answer will be either $0$ or $1$, with probabilities calculated as shown above.

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What you need to realize, is that quantum mechanics throws the concept of local realism out of the window. When you measure one entangled particle, you instantly change the state of the other.

With your electron, annihilation with a localized positron is a measurement of locality. And the result of that measurement instantly changes the entire wave function of the electron (I'm using the Copenhagen Interpretation here). The other possible annihilation is another measurement of locality on the same wave function, and if either annihilation succeeds, the other won't.

Note that this does not mean that you can communicate any information faster than light: You cannot tell from the results of the measurements which measurement happened first on an entangled state, you can only tell that the two measurements influence each other.

If you are uneasy about this "spooky action at a distance", congrats, you are in good company. Einstein had similar concerns. He and others came up with the EPR paradox to show that quantum mechanics could not be complete because it predicted such "spooky action at a distance", which simply could not be. Unfortunately for these three brilliant physicists, later experiments have shown that the predictions of quantum mechanics are indeed correct, and that no theory with local realism can adequately describe reality.

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  • $\begingroup$ I appreciate your plain-language response. It's the only one I'm competent to read. No doubt contained within the other answers and firmly implied in your is this: the death of an electron here does not "prevent" its annihilation elsewhere. That would be a causal relationship. Like statistics, we cannot state what can happen in any particular case -- all we can say is that it cannot die in two places. $\endgroup$ – user220778 Aug 15 '19 at 8:24
  • $\begingroup$ When you measure one annihilation that inherently makes another such event probabilistically impossible: in that sense aren’t you communicating with the future event that “hey because I happened , you can’t !” $\endgroup$ – Jay D Aug 16 '19 at 22:54
  • $\begingroup$ @JayD The point is, that you cannot decide which event happened first. Neither one needs to be in the future of the other. If I send two entangled photons in opposite directions so that all experiments done on them are further apart than the notion of future/past allows, and I measure their polarizations, I will find that the measurements influence each other. However, the results are so symmetric that I cannot tell from the results which of the two photons was allowed to travel further before measuring it. Something non-local happens, but it doesn't have a direction of cause and effect to it. $\endgroup$ – cmaster - reinstate monica Aug 17 '19 at 7:34

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