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The difference between bosons and fermions lies in the way swapping two of them works. That sounds like what Wikipedia says is the rigorous version of the Paul

The difference between bosons and fermions lies in the way swapping two of them works. That sounds like what Wikipedia says is the rigorous version of the Pauli exclusion principle[1]. I was trying to get close to the not-rigorous version (1st paragraph of the link) in terms that are easily understandable without having taken a university QM course. I guess can/can't be in the same place at the same time would be a better approximation of it? [1] https://en.wikipedia.org/wiki/Pauli_exclusion_principle https://en.wikipedia.org/wiki/Pauli_exclusion_principle

danbruc 10y ago
Loosely peaking the wave function of a quantum mechanical system specifies for each possible state of the system the probability of finding the system in that state. Actually it is not the probability but the probability amplitude, a complex number from which you can derive the probability by squaring it. Assume you have two identical fermions, say two electrons, the first one in state x and the second one in state y. State means everything required to fully describe the particle, for example position and spin. Therefore x stands for the first electron being in a specific position and having a specific spin, similarly for y and the second electron. Let A(x, y) be the probability amplitude for finding the first electron in state x and the second electron in state y, i.e. the first argument of A is the state of the first electron, the second argument is the state of the second electron. Now swap the two electrons, take the first and put it where the second one is, take the second one and put it where the first one was. Also change the spins as necessary. The probability amplitude is now A(y, x), the first electron is now in state y, the second electron is now in state x. The important thing is now that the two electrons are identical, you can not tell the difference between the situations before and after swapping the two electrons. Had you painted one electron blue and one red, then you could easily tell the difference, but without that you can not. That was the entire point of swapping the electrons, bringing them into exactly the state of the other one. But if you can not distinguish the two situations, then it better be the case, that they have the same probability, i.e. A(x, y) = A(y, x). But that is not quite right, A is the probability amplitude, not the probability. It turns out that there are actually two valid possibilities, A(x, y) = A(y, x) and A(x, y) = -A(y, x). As mentioned at the beginning, you get the probability by squaring the probability amplitude, so that the minus sign in the second case vanishes. The first possibility is how bosons (particles with integer spin, for example photons and gluons) behave, the second one is how fermions (particles with half integer spin, for example quarks and electrons but also helium-3) behave. Now we finally arrive at the important point, what happens if both electrons are in the same state, i.e. if the first electron is in state x and the second electron is also in state x. Then the probability amplitude is A(x, x) and we have to satisfy A(x, x) = -A(x, x) because electrons are fermions. But there is only one complex number identical to its negative and that is of course zero. Therefore the probability amplitude and in consequence the probability obtained by squaring the probability amplitude are both zero, which means that the probability of finding the system in the state where the first electron is in state x and the second electron is also in state x, is zero. The two electrons or more generally two fermions can never be in the exact same state.

danbruc 10y ago

Loosely peaking the wave function of a quantum mechanical system specifies for each possible state of the system the probability of finding the system in that state. Actually it is not the probability but the probability amplitude, a complex number from which you can derive the probability by squaring it. Assume you have two identical fermions, say two electrons, the first one in state x and the second one in state y. State means everything required to fully describe the particle, for example position and spin. Therefore x stands for the first electron being in a specific position and having a specific spin, similarly for y and the second electron. Let A(x, y) be the probability amplitude for finding the first electron in state x and the second electron in state y, i.e. the first argument of A is the state of the first electron, the second argument is the state of the second electron. Now swap the two electrons, take the first and put it where the second one is, take the second one and put it where the first one was. Also change the spins as necessary. The probability amplitude is now A(y, x), the first electron is now in state y, the second electron is now in state x. The important thing is now that the two electrons are identical, you can not tell the difference between the situations before and after swapping the two electrons. Had you painted one electron blue and one red, then you could easily tell the difference, but without that you can not. That was the entire point of swapping the electrons, bringing them into exactly the state of the other one. But if you can not distinguish the two situations, then it better be the case, that they have the same probability, i.e. A(x, y) = A(y, x). But that is not quite right, A is the probability amplitude, not the probability. It turns out that there are actually two valid possibilities, A(x, y) = A(y, x) and A(x, y) = -A(y, x). As mentioned at the beginning, you get the probability by squaring the probability amplitude, so that the minus sign in the second case vanishes. The first possibility is how bosons (particles with integer spin, for example photons and gluons) behave, the second one is how fermions (particles with half integer spin, for example quarks and electrons but also helium-3) behave. Now we finally arrive at the important point, what happens if both electrons are in the same state, i.e. if the first electron is in state x and the second electron is also in state x. Then the probability amplitude is A(x, x) and we have to satisfy A(x, x) = -A(x, x) because electrons are fermions. But there is only one complex number identical to its negative and that is of course zero. Therefore the probability amplitude and in consequence the probability obtained by squaring the probability amplitude are both zero, which means that the probability of finding the system in the state where the first electron is in state x and the second electron is also in state x, is zero. The two electrons or more generally two fermions can never be in the exact same state.