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Can two electrons be in the same state, when they belong to two different atoms, which are "far enough" (whatever that means) apart from each other? With "same state" I mean that (as far as specifiable) the states are really identical, except for the position of the electrons.

More specifically:
I am still not clear of how the separation of particles is taken into account for the Pauli exclusion principle. E.g. in a crystal the electrons seem close enough for the exclusion principle to become meaningful, for particles "a universe apart" from each other, it seems that it is kind of redundant (is there specific maths to that?), but what about everything in between?

(This is closely related to the talk Brian Cox gave, see this question: physics.stackexchange.com/q/18527/16689 )

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Most of this question has already been answered there physics.stackexchange.com/q/18527/16689 please tell us if you need more details –  FraSchelle Jun 26 '13 at 12:48
    
I knew the things that were stated in the answer, but I am still not clear of how the separation of particles is taken into account for the Pauli exclusion principle. E.g. in a crystal the electrons seem close enough for the exclusion principle to become meaningful, for particles "a universe apart" from each other, it seems that it is kind of redundant (is there specific maths to that?), but what about everything in between? –  Jack Jun 26 '13 at 12:56
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That sounds as if it is up to me if the electrons can be in the same state or not. If I describe the electron on Mars and on Earth with the same wave-function, then automatically the Pauli exclusion principle needs to hold. If I don't feel like writing it as one wave function, then it may not? (Yeah, sure I am also looking for the physical meaning behind it, I just want the math to support it) –  Jack Jun 26 '13 at 13:05
    
Sure, I know, but the question is weather I should describe the two particles on Mars and Earth with one antisymmetric wave function in the first place –  Jack Jun 26 '13 at 13:53
    
I put more detailed discussion in my answer. Tell me there if you want more details. I erased my previous comments since they were redundant. –  FraSchelle Jun 26 '13 at 14:32
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3 Answers 3

Here's a long answer leading up to a mathematical measure for the relevance of the Pauli principle. The quantum state of an electron is not just determined by its energy, angular momenta (quantities that lead to good or approximate quantum numbers). Position is also part of the equation. Now, position in QM is not just a simple vector like it is in classical mechanics, it's ruled by a probability distribution. Such a probability distribution might have an exotic form, but it must satisfy a few elementary rules for [physically relevant] probability distributions:

  1. The probability of the particle being inside some given volume is then given by the integral of the probability distribution over that volume. Note that the probability of the particle being at one specific position is zero (some might say it is infinitesimally small).

    You can see this is true by considering that the amount $N$ of positions the particle could be in is infinite (if space is dense, which we assume it is) and you're wondering if is in $1$ particular position. The chance of that being true is $1/N = 1/\infty = 0$.

  2. The probability of the particles being anywhere at all must be $1$. We can say with a hundred percent certainty that the particle is somewhere. This is expressed mathematically by requiring the integral of the probability over the entire volume to be $1$.
  3. The probability distribution must be well-behaved. This usually means it must be continuous (can't have any jumps) and must vanish for infinitely large arguments (this is a requirement that comes along with the previous one, since a non-vanishing distribution would not be integrable to $1$ over all space).

Now, in QM the probability distribution is given by the squared modulus of the wavefunction, $|\psi(\vec{r},t)|^2$. If $\psi(\vec{r},t)$ is an exact wavefunction, it contains all the information about the particle(s) it describes. Let's assume it is exact and it describes the state of electron numero uno. Furthermore, let's assume electron numero duo is described by the exact wavefunction $\phi(\vec{r},t)$.

So we surely have two probability distributions linked to this situation. One being $|\psi(\vec{r},t)|^2$, the other $|\phi(\vec{r},t)|^2$. However, there's another probability we might be concerned with: what's the probability that electron numero duo is within a certain volume, given the probability distribution of electron numero uno? This is a conditional probability, which we can calculate as the integral over the product of both wavefunctions where the first one is complex conjugated:

$$\langle\phi(\vec{r},t)|\psi(\vec{r},t)\rangle = \int_{V}{\phi^*(\vec{r},t)\psi(\vec{r},t) d\vec{r}}$$

I've used the conventional notation for this probability in the left hand side of the equation. The right hand side is sometimes called the overlap integral because the integrand is a measure for how strongly both wavefunctions overlap. Note that it becomes the normal probability for a single particle if we take the overlap integral of one wavefunction with itself.

This helps our intuition about the Pauli principle. If the overlap integral is $1$, the electrons must differ in at least one quantum number, e.g. their spin. If it is close to $1$, the electrons should be highly unlikely to be found in the same state. The smaller the overlap integral becomes, the less strictly the Pauli exclusion principle applies. This sounds strange because in its well-known form [no two identical fermions may occupy the same quantum state simultaneously], the Pauli principle looks very strict and law-like. But that's the problem with the well-known form.

Now, we've been assuming the wavefunctions to be exact. This would have been fine if the particles had no way of interacting with each other. However they do. So we need to model that interaction. Electron numero uno influences numero duo and vice versa (little actual English in that sentence). So we'll now treat both electrons as one single system, described by an abstract wavefunction of two spatial arguments $\Psi(\vec{r}_1,\vec{r}_2,t)$.

Let's see what happens if we interchange both electrons. This is sometimes pictured as if we "make them switch places", which makes it seem like we're switching two little balls. Of course this is a bad picture because of the delicate nature of positions in QM we talked about earlier. A better way to think about it is to imagine we changed all the quantum numbers (actually just the entire state except for the probability distributions) of both electrons. So let's do that. We'll denote this switch as a switch of the spatial arguments which suggests the first picture, but this is simply for notational ease.

The operation of interchanging the states is a permutation, we'll call the operator carrying out this task $\hat{P}$. The effect of this operator is clear. We'll drop the time dependence since it is inconsequential here:

$$\hat{P}\Psi(\vec{r}_1,\vec{r}_2) = \Psi(\vec{r}_2,\vec{r}_1).$$

The probability distribution $|\Psi(\vec{r}_2,\vec{r}_1)| = |\Psi(\vec{r}_1,\vec{r}_2)|$ must be unchanged by this operation, so this must mean that

$$\Psi(\vec{r}_2,\vec{r}_1) = \text{e}^{i\delta}\Psi(\vec{r}_1,\vec{r}_2).$$

Furthermore, switching the electrons again should return the wavefunction to its original form, meaning that

$$\text{e}^{2i\delta} = 1$$

or

$$\text{e}^{i\delta} = \pm 1.$$

So this means the act of "switching" the electrons either leaves the wavefunction unchanged, or it yields a sign change. Note that nothing in the above discussion relies on the fact that these are electrons we're describing. So this holds for two general particles. Now, the first option (no sign change upon switching) is associated with bosons, the second one with fermions. So for our electrons a sign change occurs. That's the antisymmetry Oaoa already stressed in his answer. We get a form for the wavefunction out of this property:

$$\Psi(\vec{r}_1,\vec{r}_2) = A\left[\psi(\vec{r}_1)\phi(\vec{r}_2)-\psi(\vec{r}_2)\phi(\vec{r}_1)\right]$$

Note that this wavefunction vanishes if $\psi = \phi$, i.e. if the states of both electrons are the same.

Now using this antisymmetric wavefunction we can find out what the relation is between Pauli's principle (or the antisymmetry of the two-electron wavefunction) and the distance between the electrons. To do this, let's calculate the average squared distance between them. This is given by (let's assume 1D for simplicity of notation)

$$\langle\Psi(x_1,x_2)|(x_1-x_2)^2|\Psi(x_1,x_2)\rangle \equiv \langle(x_1-x_2)^2\rangle = \langle x_1^2\rangle + \langle x_2^2\rangle - 2 \langle x_1x_2\rangle$$

The "unmixed" expectation values can be calculated quite easily and are of course equal. Together they yield

$$\langle x_1^2\rangle + \langle x_2^2\rangle = \langle\psi|x^2|\psi\rangle + \langle\phi|x^2|\phi\rangle.$$

The "mixed" expectation value yields

$$\langle x_1x_2\rangle = \langle\psi|x|\psi\rangle\langle\phi|x|\phi\rangle - |\langle\psi|x|\phi\rangle|^2.$$

Note that the second term in this corresponds to the fermionic permutation $|\psi\rangle \leftrightarrow|\phi\rangle$. These results yield the following expression for the expectation value of the squared distance between the electrons:

$$\langle(x_1-x_2)^2\rangle = \langle\psi|x^2|\psi\rangle + \langle\phi|x^2|\phi\rangle - 2\langle\psi|x|\psi\rangle\langle\phi|x|\phi\rangle + 2|\langle\psi|x|\phi\rangle|^2.$$

What can be learnt from this? Well, for distinguishable particles (say an electron and a muon), which cannot be in the same state, we get the same expectation value except for the last term. This term is called the exchange term. It seems the electrons are on average farther away from each other than, say, an electon and a muon. How much farther is determined by the exchange term. Note that the exchange term is zero if there is no overlap between the wavefunctions (the states are entirely different).

This confirms our earlier intuition, only now we have a mathematical prescription to back it up. We could say that if the average squared distance between the electrons is far greater than the exchange term, the Pauli principle does not contribute much. If the exchange term is comparable to the other terms, the Pauli principle is a lot stricter.

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Nice answer ! (It also clearly shows to me that I was missing the main point of the question... whatever :-) You may insist a bit more on your next-to-last equation about the exchange $\psi\rightarrow\phi$ for the second term. Also, note there is a factor $2$ lacking on your last formula before the exchange terms. Actually, I call the difference of the two last terms the exchange terms, but I'm not sure if I'm using normal convention... Thanks again for this beautiful answer. –  FraSchelle Jun 26 '13 at 17:14
    
@Oaoa Thanks! :-) Good catch with the 2's, I caught a mistake as well while changing that (I had $x^2$ in the exchange term in my second-to-last equation). I'm not entirely sure I know what you would have me insist on in in that equation though? –  Wouter Jun 26 '13 at 17:28
    
@Oaoa I can understand that :) They're reasonable suggestions so I've tweaked those things in my answer ;-) –  Wouter Jun 26 '13 at 18:03
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So, the Pauli principle reads

  • the wave function of a system of identical integer-spin particles has the same value when the positions of any two particles are swapped. Particles with wavefunctions symmetric under exchange are called bosons;

  • the wave function of a system of identical half-integer spin particles changes sign when two particles are swapped. Particles with wavefunctions anti-symmetric under exchange are called fermions.

from the Wikipedia page about the spin-statistic theorem. The above statement implies the Pauli exclusion principle, we will show how.

So, now let us imagine that our universe is not larger than the system Earth-Moon (for understanding commodity only). Let us describe one electron -- one fermion, one spin $1/2$ particle, whatever you prefer -- on the Earth with a position-time dependent wave function $\phi_{E}\left(x_{1}\right)$. For commodity I will note $x\equiv\left(x,y,z,t\right)$ for all space-time parameter. The same for the electron on the Moon, described by $\phi_{M}\left(x_{2}\right)$. Indices $1$ and $2$ are for latter use, I can choose the frame of reference I want by the way. Note that they are highly redundant with the indices $E$ and $M$. The Pauli exclusion principle then states that the full wave function for the two electrons is

$$\Psi\left(x_{1},x_{2}\right) = \phi_{E}\left(x_{1}\right)\otimes\phi_{M}\left(x_{2}\right)-\phi_{M}\left(x_{1}\right)\otimes\phi_{E}\left(x_{2}\right)$$

where the indices $E$ and $M$ are now meaningless. Since the tensor product $\otimes$ is commutative, one has in particular

$$\Psi\left(x_{1}=x_{2}\right) = 0$$

that's the Pauli exclusion principle at work !

If you try to put words on the mathematical expression for $\Psi\left(x_{1},x_{2}\right)$, it could be:

taking the wave-function $\phi_{E}$ of the one electron we previously identified as being on the Earth and the wave-function $\phi_{M}$ of the other electrons we previously identified as being on the Moon, and making them interchanging their places results in an opposite sign for the tensor product $\phi_{E}\otimes\phi_{M}$ describing them all together

What do we mean by the indices $E$ and $M$ are meaningless ? Well, if you know that one of the electron is on the Moon whereas the second one is actually on the Earth, there is no way to find $\Psi\left(x_{1}=x_{2}\right)$, since the two electrons will never ever have a chance to meet each other. (I hope that the distance Earth-Moon was sufficient to convince yourself, otherwise do that in between any stars you prefer until being convinced ...)

The key point behind is that we deal with indiscernible particles, when the indices $E$ and $M$ are not making any physical sense. These indices are just there to gives sense to $\Psi\left(x_{1},x_{2}\right)$, and properly define the antisymmetry of the total wave function.

A really more clever way to think would be: if you really insist in describing the two electrons (one on the Earth, one on the Moon) by a unique wave-function, then this wave-function $\Psi\left(x_{1},x_{2}\right)$ has the property

$$\Psi\left(x_{1},x_{2}\right)=-\Psi\left(x_{2},x_{1}\right)$$

and in particular $\Psi\left(x_{1}=x_{2}\right) = 0$ and that's it. There is nothing mystical behind. It's a mathematical help because the notion of indistinguishability is really hard to understand by hands.

So, what's the reason for antisymmetry ? Well, from the beginning, I assume that one can locate one electron. This makes no sense in quantum mechanics, which is just about the density probability to find something somewhere sometimes, the famous $\Psi\left(x\right)$. So $\Psi\left(x_{1},x_{2}\right)$ is just about the probability to find one electron at space-time position $x_{1}$ and a second electron at space-time position $x_{2}$.

Please tell us if you have a better idea to model this thing :-)

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Please also see the answer @Wouter gave somewhere else on this page physics.stackexchange.com/a/69267/16689 about the decrease of of the exchange interaction with length, and a nice introduction to the exchange interaction concept. –  FraSchelle Jun 26 '13 at 17:24
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If you can express the two electrons with a single wavefunction then the Pauli exclusion principle will apply. However apart from the very simplest systems the electrons will interact with their environment and become entangled with it. Now you no longer have a wavefunction of two electrons, but a wavefunction of two electrons plus the gazillion other particles the electrons have interacted with. In such a complex system there will no longer be a simple correlation between the spins of the electrons.

For an object as small as an atom or molecule the interaction between the electrons is far greater than the interaction with the rest of the universe. However if you attempted to construct a two electron system of macroscopic dimensions the interaction with the environment would be dominant and you'd lose correlation between the electron spins.

Brian Cox's comments make for good TV, but would apply only in unrealistically simplified systems.

Response to comment:

I think you might be mixing up ideas about entanglement and the Pauli exclusion principle. If two electrons, one on Earth and one on Mars, are entangled and I make a measurement on the electron on earth then the electron on mars is instantly affected. However this doesn't cause us any problems because no information is transmitted faster than light. The superluminal behavior has actually been measured experimentally.

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Hm, are you saying it's rather interactions with a lot of other stuff that may allow two electrons to be in the same state as their spatial separation? But anyhow, is there a quantitative way of telling under which circumstances the Pauli principle becomes redundant? What if we look at a case that is somewhere nicely in between an atom and some situation, where electrons really do not seem to be relatable any more? –  Jack Jun 26 '13 at 15:17
    
Btw, if I remember correctly, Brian Cox did specifically say "everything is connected" and changing the energy of one electron instantly(!) causes electrons across the universe to change their energy, because of the Pauli exclusion principle. This would really be information traveling faster than the speed of light. So (if again I am not mistaken) he wasn't just not being accurate, he was talking complete nonsense. Where is the Feynman of our age? –  Jack Jun 26 '13 at 15:25
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@Jack Don't be so quick to judge that statement as complete nonsense. If I have two points, A and B, separated by exactly 1 million light years, and I move A 1 meter closer to B, then the separation between A and B instantly becomes 1 million ly minus 1 m, whether or not B knows this, and any measurable quantity that depends on that separation will instantly change in the sense that when it is eventually measured it will reflect that change. –  Chris White Jun 26 '13 at 15:37
    
@Jack: I've edited my answer to respond to your comment. –  John Rennie Jun 26 '13 at 15:47
    
@Chris White I don't see any correlation of what you just said to what Brian Cox said. You can just have your own electron, measure it whenever you like, and according to Brian Cox you could always measure a different energy as soon as somewhere (could be across the universe) another electron's energy has changed. Actually this isn't true, I have read somewhere that he (Brian Cox) wrote, that this energy difference would always arise, but will be "too small to be measured", which just reminds me of Pauli (hehe) "Not even wrong" (it cannot even be falsified...) –  Jack Jun 26 '13 at 16:00
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