What does the Pauli Exclusion Principle mean if time and space are continuous?

Assuming time and space are continuous, identical quantum states seem impossible even without the principle. I guess saying something like: the closer the states are the less likely they are to exist, would make sense, but the principle is not usually worded that way, it's usually something along the lines of: two identical fermions cannot occupy the same quantum state

  • 5
    Your assertion in the second sentence needs to be backed up with physics, not gut feeling. – Jon Custer Oct 25 '16 at 13:02
  • 22
    bound systems come with discrete states even in continous spacetimes – Christoph Oct 25 '16 at 13:04
  • 6
    @JonCuster no it doesn't, because that's the point of the question. – Nathaniel Oct 26 '16 at 10:12
  • 1
    My understanding is that is is more like "two identical fermions within the same local system (same atom, same molecule ...) cannot occupy the same quantum state". In other words, the Pauli Exclusion Principle only applies to small scale things, and certainly not to things of cosmic scale. – Kevin Fegan Oct 26 '16 at 10:50
  • 1
    @KevinFegan but what is a "local system"? In white dwarves for example, probably one of the systems in which Pauli's principle has the strongest effect, the electrons are effectively in a (degenerate) gas state, so not bound by any atom, molecule, or anything else. Unless you want in that case to consider the whole star as the "local system", but that's a bit of a stretch I think (can't you also consider the whole universe as a "local system"?) – glS Oct 26 '16 at 19:22
up vote 31 down vote accepted

The other answer shows nicely how one may interpret the Pauli exclusion principle for actual wavefunctions. However, I want to address the underlying confusion here, encapsulated in the statement

If time and space are continuous then identical quantum states are impossible to begin with. in the question.

This assertion is just plainly false. A quantum state is not given by a location in time and space. The often used kets $\lvert x\rangle$ that are "position eigenstates" are not actually admissible quantum states since they are not normalized - they do not belong to the Hilbert space of states. Essentially by assumption, the space of states is separable, i.e. spanned by a countably infinite orthonormal basis.

The states the Pauli exclusion principle is usually used for are not position states, but typically bound states like the states in a hydrogen-like atom, which are states $\lvert n,\ell,m_\ell,s\rangle$ labeled by four discrete quantum numbers. The exclusion principle says now that only one fermion may occupy e.g. the state $\lvert 1,0,0,+1/2\rangle$, and only one may occupy $\lvert 1,0,0,-1/2\rangle$. And then all states at $n=1$ are exhausted, and a third fermion must occupy a state of $n > 1$, i.e. it must occupy a state of higher energy. This is the point of Pauli's principle, which has nothing to do with the discreteness or non-discreteness of space. (In fact, since the solution to the Schrödinger equation is derived as the solution to a differential equation in continuous space, we see that non-discrete space does not forbid "discrete" states.)

  • 4
    This is a really important point: physicists do this clever trick where you quietly assume that you can use an uncountable basis, and you can't actually do that because the underlying maths falls apart horribly. – tfb Oct 25 '16 at 13:48
  • 2
    One lingering point of uncertainty here is that the Pauli exclusion principle is invoked for more than just the levels of hydrogen. For example, for why you don't fall through the floor (example site only) - there's a hand-wavy "electrons can't occupy the same quantum state, so they can't occupy the same 'spot', so therefore they can't occupy the same space, hence the electron clouds of your feet don't fall through those of the floor." There's an implicit assumption that location equals quantum state there. – R.M. Oct 25 '16 at 16:03
  • 5
    @JgL Note the "uncountable" in tfb's comment. The $\lvert x\rangle$ really don't form a basis of the Hilbert space in the standard mathematical sense - any Hilbert/Schauder basis should be countable, and Hamel bases are rather useless. – ACuriousMind Oct 25 '16 at 16:36
  • 1
    What about time though? If time is continuous wouldn't simultaneity be impossible? – Yogi DMT Oct 25 '16 at 17:53
  • 1
    @JgL I didn't say infinite-dimensional, I said uncountable, which is a very different thing: infinite but countable bases are one thing, infinite but uncountable ones are very different. Sliding between these two things without being clear about it the 'clever physicist's trick' I referred to, and is a mathematical horror. – tfb Oct 25 '16 at 19:01

Real particles are never completely localised in space (well except in the limit case of a completely undefined momentum), due to the uncertainty principle. Rather, they are necessarily in a superposition of a continuum of position and momentum eigenstates (a wave packet).

Pauli's Exclusion Principle asserts that they cannot be in the same exact quantum state, but a direct consequence of this is that they tend to also not be in similar states. This amounts to an effective repulsive effect between particles.

You can see this by remembering that to get a physical two-fermion wavefunction you have to antisymmetrize it. This means that if the two single wavefunctions are similar in a region, the total two-fermion wavefunction will have nearly zero probability amplitude in that region, thus resulting in an effective repulsive effect.

To more clearly see this consider the simple 1-dimensional case, and two fermionic particles with partially overlapping wavefunctions. Let's call the wavefunction of the first and second particle $\psi_A(x)$ and $\psi_B(x)$, respectively:

The properly antisymmetrized wavefunction of the two fermions will be given by: $$ \Psi(x_1,x_2) = \frac{1}{\sqrt2}\left[ \psi_A(x_1) \psi_B(x_2)- \psi_A(x_2) \psi_B(x_1) \right]. $$ For any pair of values $x_1$ and $x_2$, $\lvert\Psi(x_1,x_2)\rvert^2$ gives the probability of finding one particle in the position $x_1$ and the other particle in the position $x_2$. Plotting $\lvert\Psi(x_1,x_2)\rvert^2$ we get the following:

As you can clearly see for this picture, for $x_1=x_2$ the probability vanishes, as an immediate consequence of Pauli's exclusion principle: you cannot find the two identical fermions in the same position state. But you also see that the more $x_1$ is close to $x_2$ the smaller is the probability, as it must be due to the wavefunction being continuous.

Addendum: Can the effect of Pauli's exclusion principle be thought of as a force in the conventional $F=ma$ sense?

The QM version of what is meant by force in the classical setting is an interaction mediated by some potential, like the electromagnetic interaction between electrons. This is in practice an additional term in the Hamiltonian of the system, which says that certain states (say, same charges very close together) correspond to high-energy states and are therefore harder to reach, and vice versa for low-energy states.

Pauli's exclusion principle is conceptually entirely different: it is not due to an increase of energy associated with identical fermions being close together, and there is no term in the Hamiltonian that mediates such "interaction" (important caveat here: this "exchange forces" can be approximated to a certain degree as "regular" forces).

Rather, it comes from the inherently different statistics of many-fermion states: it is not that identical fermions cannot be in the same state/position because there is a repulsive force preventing it, but that there is no physical (many-body) state associated with them being in the same state/position. There simply isn't: it's not something compatible with the physical reality described by quantum mechanics. We naively think of such states because we are used to think classically and cannot really wrap our heads around what the concept of "identical particles" really means.

Ok, but what about things like degeneracy pressure then? In some circumstances, like in dying stars, Pauli's exclusion principle really seems to behave like a force in the conventional sense, contrasting the gravitational force and preventing white dwarves from collapsing into a point. How do we reconcile the above described "statistical effect" with this?

What I think is a good way of thinking about this is the following: you are trying to squish a lot of fermions into the same place. However, Pauli's principle dictates a vanishing probability of any pair of them occupying the same position.

The only way to reconcile these two things is that the position distribution of any fermion (say, the $i$-th fermion) must be extremely localised at a point (call it $x_i$), different from all the other points occupied by the other fermions. It is important to note that I just cheated for the sake of clarity here: you cannot talk of any fermion as having an individual identity: any fermion will be very strictly confined in all the $x_i$ positions, provided that all the other fermions are not. The net effect of all this is that the properly antisymmetrized wavefunction of the whole system will be a superposition of lots of very sharp peaks in the high dimensional position space. And it is at this point that Heisenberg's uncertainty comes into play: very peaked distribution in position means very broad distribution in the momentum, which means very high energy, which means that the more you want to squish the fermions together, the more energy you need to provide (that is, classical speaking, the harder you have to "push" them together).

To summarize: due to Pauli's principle the fermions try so hard to not occupy the same positions, that the resulting many-fermion wavefunction describing the joint probabities becomes very peaked, highly increasing the kinetic energy of the state, thus making such states "harder" to reach.

Here (and links therein) is another question discussing this point.

  • Extending this a bit, the phrase the closer the states are the less likely they are to exist is expressed mathematically for multi-particle systems by the pair correlation function. So while the wording isn't exactly as the OP phrased it, the content is expressed. – garyp Oct 25 '16 at 13:22
  • 1
    Thanks, this is a good answer as well. I only chose the other one because it addressed my particular misunderstanding a little bit more. – Yogi DMT Oct 25 '16 at 14:37
  • One thing I've often wondered about this is: does the effective repulsiveness constitute a force in the normal F=ma sense of the word? And why don't we include this force in the list of fundamental forces? – spraff Oct 26 '16 at 11:39
  • @spraff I think that is a very interesting question. Edited post to try to answer it (but see also the many other questions on that topic) – glS Oct 26 '16 at 13:57
  • 3
    As a non-physicist (math/cs) this is probably the best and most helpful explanation I have ever read for explaining the PEP and particularly it's implied apparent continuous/discrete paradox. The 3D graph was especially helpful, thnx. – RBarryYoung Oct 26 '16 at 18:18

protected by Qmechanic Oct 25 '16 at 20:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.