This year we give thanks for the spin-statistics theorem. (Previously we gave thanks for the Lagrangian of the Standard Model of particle physics, and for Hubble’s Law.)

You will sometimes hear physicists explain that elementary particles come in two types: bosons, which have a spin of 0, 1, 2, or some other integer, and fermions, which have a spin of 1/2, 3/2, 5/2, or some other half-integer. That’s true, but it’s hiding what’s important and emphasizing what’s auxiliary.

When it comes to classifying elementary particles, it’s not really the spin that’s important, it’s the statistics. And really, the word “statistics” in this context makes something deep and wonderful sound dry and technical. A boson is a particle that obeys Bose statistics: when you take two identical bosons and switch them with each other, the state you end up with is indistinguishable from the state you started with. Which only makes sense, really; if you exchange two identical particles, what else could you get? The answer is, Fermi statistics: when you take two identical fermions and switch them with each other, you get minus the state you started with. Remember that the real world is based on quantum mechanics, in which the state of a system is described by a wave function that tells you what the probability of obtaining various results for certain observations would be; when we say “minus the state you started with,” we mean that the wave function is multiplied by -1.

This difference in “statistics” seems a bit esoteric and removed from one’s everyday life, but in fact it is arguably the most important thing in the universe. This simple difference in what happens to the state of two particles when you interchange them underlies the most blatant features of how particles behave in the macroscopic world. Think of two identical particles that are in the same quantum state: sitting in the same place, doing the same thing, right on top of each other. If those two particles are bosons, that’s cool; we can switch them and get the same state, which just makes sense. But if they’re fermions, we have a problem; the two particles are purportedly in the same state, but if we switch them (which doesn’t really do anything, as they are in the same place) the state becomes minus what it used to be — seemingly a contradiction.

This seeming puzzle has a simple solution: in the real world, two identical fermions can never occupy the same quantum state! That’s the Pauli exclusion principle, and it has a simple translation into everyday English: fermions take up space. Electrons, which are fermions, can’t just be piled on top of each other as densely as we like; some of them would have to be in the same state, and that can’t happen. That’s why atoms take up a certain amount of space, which in turn is why ordinary material objects don’t simply collapse into themselves. Fermions — electrons, quarks, neutrinos, etc. — are matter particles, constituting the “stuff” of which the objects of our world are comprised.

Bosons, on the other hand, have no problem being in the same quantum state. So they will happily pile on top of each other. This is also important to our everyday lives. Bosons — photons, gravitons, gluons, etc. — are force particles, which pile on top of each other to form the classical force fields that hold fermions together. When you see light — a classical electromagnetic wave made of photons — or are held to the ground by gravity — a classical field made of gravitons — it’s only possible because of Bose statistics.

So the important distinction between bosons and fermions is not the “integer spin”/”half-integer spin” distinction, it’s the “pile on top of each other”/”take up space” distinction. The fact that these sets of features come hand-in-hand is the content of the spin-statistics theorem: particles that pile on have integer spins, particles that take up space have half-integer spins. Which is a deep and beautiful result that relies on the fact that nature is fundamentally quantum rather than classical, and on the topology of the group of rotations in three (or more) spatial dimensions, and on the features of relativistic field theory. None of which I’m going to explain right here, but John Baez has a fun “proof” of the theorem using ribbons which is worth checking out.

Rather, I will just reiterate that if the fermions comprising a turkey didn’t take up space, it would hardly constitute a filling meal; and if the gravitons from the Earth didn’t pile up to form a classical field, the traditional football game really wouldn’t work at all. So for the spin-statistics theorem, we should all be thankful.

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28 Responses to Thanksgiving

  1. Sean says:

    And meanwhile, our thoughts go out to those affected by the horrific attacks in India.

  2. ioresult says:

    What about a Bose-Einstein condensate? I hear the particles are made of Fermions, but they act as Bosons?

  3. MedallionOfFerret says:

    Yep, much more horrific to be attacked in a hotel or train station in India, rather than while attending weddings in Afghanistan or northern Pakistan. Bose statistics vs. Fermi statistics, right? Dulce et decorum est pro patria mori!

    …but I really did like the post.

  4. Neil B says:

    Heh, there’s something in the Navy called a “Bosun’s Mate” (really.) Maybe it’s a misspelling of “boson”, and the Navy likes them because they can simultaneously share a bunk, to save space?

    OK really: is there a way to prove the Exclusion Principle from more fundamental considerations, so it isn’t “ad hoc”?

    Also, electrons are fermions and so they can’t be squished together indefinitely. Sure they electrostatically repel as well, but that would be a continuum and more force could incrementally overcome it to any degree (in principle.) So it seems to me that the “degeneracy” force keeping a white dwarf from collapsing even smaller should count as a separate “fundamental” force? Sure, the electrons are charged but like I said, this is an additional effect. (What if electrons were not charged but otherwise the same, they still couldn’t be compressed indefinitely.) And, by definition one trait of a “force: it can oppose other forces (like gravity). The EM isn’t enough to oppose the gravity entirely by itself. Since degeneracy pressure isn’t counted as a “fundamental force”, I wonder why.

  5. harshpotatoes says:

    There are ways to prove the exclusion principle from more fundamental principles. its actually kinda fun to go through. consider that fermions must have antisymmetric wavefunctions (bosons have symmetric, although the proof for this is basically noticing in the beginning you could have two types of wavefunctions, and one was called a fermion the other a boson), so that under the parity operator the sign of the wave function is reversed. so: f(x,y) is the wavefunction, P is the parity operator. P(f(x,y)) = -f(y,x) which implies that f(x,x) = 0, meaning that the probability two fermions are at the same position is zero.

    My reasoning for thinking that the degeneracy pressure isn’t counted as a “fundamental force” because there are no force carriers mediatting the force. (ie no bosons like photons gluons or gravitons etc). so the force is purely quantum mechanical. thats my thought, but i still have much to learn.

  6. Robert says:

    Funny you bring this up. I have recently thought about the spin statistics theoreom because I am supposed to teach it in our Mathematical Quantum Mechanics class (I realise this is QFT and not QM). However, I have not been able to find an elementary way to present it. I can show without problems what a 360 degrees rotation is represented by (+1 for integer spins, -1 for half integer spins, any phase in 2+1 dimensions). But how do I connect this to the statistics of the multi-particle wave function? I find John Baez’ explanation not satisfactory (it it vague exactly at that point), the wikipedia article is completely mysterious (what is R?), many QFT textbooks show that the wrong commutation relations do not lead to a local theory but they do this only for free fields as they use the explicit form of the two point function.

    And finally, there is the explanation of having two identical particles and rotating them into each other. But to do that I have to rotate two particles by 180 degrees and there is no a priori connection to rotating one particle by 360 degrees (for more than one dimensional representations).

    Anybody having a better explanation? Please contact me as !!!

  7. DP says:

    Posts like this are why I read this blog. I’ve never seen the principles explained in quite that way before, and it increased my understanding. “Thanks!” for Cosmic Variance.

  8. Vibi says:

    So in the undiscovered supersymmetric world fermionic partners should be the force carries while bosonic partners should make up matter!!

  9. Neil B says:

    Harshpotatoes, thanks and also your explanation of the essentially “semantic” reason for not calling degeneracy pressure “fundamental” is likely appropriate. However, the trouble remains that it is still “insinuated” in the framing of the issue that all forces between particles derive from the typical fundamental four: EM, grav, strong, weak. OK, call the exclusion force “quantum mechanical”, not mediated etc, but it is still ipso facto “a force.” See, I care about the proper use of language.
    And BTW if exclusion is a matter of principle (?), what determines the magnitude of the force/s derived from it?

    Also, food for strange thoughts: nuclei can have integer or half-integer spins (from the combination of effects despite each nucleon being a fermion.) This is basically the idea behind Einstein-Bose condensate (although put forth as the spin of “atoms”) but then if so fundamental, why the need to cool the stuff so much? (I mean, to get them to intermingle per se more so than to achieve a single quantum state – after all photons pack together without having to be in the same state – but pardon any middle-brow confusion.) Messiness caused by the associated electrons?

    In any case, I don’t see much talk about distribution of fermionic v. bosonic nuclei in various stars at high density; pre-neutronium stage. Yet it seems it ought to affect the shrinkage and collapse of the star.

  10. miller says:

    I think degeneracy pressure is basically caused by the fact that if we add more and more electrons into the same amount of volume, they must occupy higher and higher energy states because the lower states have already been filled. Therefore, in absence of greater forces, it requires less energy for the electrons to remain spread out. The magnitude of this “force” is equal to the gradient in the energy required for those higher energy states.

  11. Pieter Kok says:

    The exclusion principle is not a force because it is a consequence of the structure of quantum mechanics, rather than a (set of) terms in the Lagrangian. Also, I do not like the “derivation” of the exclusion principle by harshpotatoes (no offense): you have shifted the problem to showing that wave functions are either symmetric or anti-symmetric. There is no intrinsic reason why this must be so (at least, not in quantum mechanics). For example, there is currently a lot of research going on about anyons, particles that do not have (half-) integer spin. Their wavefunctions have no particular symmetry.

  12. Callum M. says:

    @Neil B:

    The “Bosun” is a corruption of the term “Boatswain”, who is the chief maintenance officer aboard a ship. He’s also the one who “pipes” the captain and other ranking officers aboard. 🙂

    Just a random spot of information there 😀

  13. Lawrence Crowell says:

    The exclusion principle is a topological aspect of QFT. Given the Fermionic field Y then Y^2 = 0, which is topologically the same as d^2 = 0. Topology further comes in with BRST, for SUSY operators Q or Q^* we have Q^2 = 0 and states |Y> are determined by ker(Q)/im(Q) so that a physical state is NOT determined by |Y> = Q|X>.

    The exclusion principle is of course not a force, for a force is something which is determined locally or by some local gauge/phase change. It just means that for two fermions, say electrons in an atom with eigenvalues E,l,s, etc cannot occupy the same state (Y^2 = 0), which is a topological obstruction that prevents any “force” from changing an energy state &E = F*&x so two electrons occupy the same quantum state.

    We can be thankful for this, for it makes chemistry possible. All that tryptophan would not be possible otherwise. 🙂

    Lawrence B. Crowell

  14. Chris W. says:

    (Off-topic…) Speaking of fermions and forces, I look forward to a bit of coverage on CV of the article that appeared in Science last week (21 Nov 2008, Vol. 322. no. 5905, pp. 1224 – 1227):

    Ab Initio Determination of Light Hadron Masses

    We present a full ab initio calculation of the masses of protons, neutrons, and other light hadrons, using lattice quantum chromodynamics. ……. …. …. Our results completely agree with experimental observations and represent a quantitative confirmation of this aspect of the Standard Model with fully controlled uncertainties.

  15. MZ says:

    This was great. I wish you would write a whole series like this. “An Intuitive Explanation of the Universe.”

  16. harshpotatoes says:

    Ah, so going back to my derivation of the pauli exclusion. More reasoning behind again comes back to the parity operator. If P is the parity operator, f(x,y) is the wavefunction, then we note that P^2(f(x,y)) = f(x,y). Apply the parity operator twice, which switches the particles twice, leaving them exactly where the were in the beginning. Assuming that the parity operator is hermitian, meaning it is observable, means that its eigenvalues can only be real, meaning there are only two possibilities: 1 or -1. (squareroot of 1 is either 1 or -1). which leads to the conclusion that in this simplified case of two particles in one dimension there are two types of wavefunctions, one we will call bosons and the other fermions.

    So i guess coming back to this possibility of anyons, either the parity operator is not an observable for them, or they can’t exist in one dimension.

    however, even this simple proof, while not a generalized proof for any dimension with any number of particles, it does show that there is a type of wavefunction which has zero probability when particles are at the same location.

  17. Amiya Sarkar says:

    Great! It reminded me of my passion in particle physics. The passion is still in my heart but it got in a Boson state piling up with other disciplines of science. I think, may be wistfully, that all these fermions and bosons are mere wave functions which may somehow be explained in ‘string theory’ terms.
    The bloody mayhem in India, Bose’s homeland, is very unfortunate. I prefer to call them ‘stupid cowards’ and not ‘terrorists’.
    Thanks again for the mindboggling post, in no ‘uncertain’ terms!

  18. Jimbo says:

    For Robert,
    Look up the 2002 paper by Arthur Broyles on the arxiv. He recites some of
    the history of the SST, from Feynman’s bemoaning that, other than Pauli’s 1940 QFT derivation, no simple, intuitive derivation existed. Broyles provided such a simple derivation from the properties of wave functions alone, sans QFT. It was promptly attacked by Duck & Sudarshan, but to my knowledge, is the only such `simple’ SST derivation to date. Why it has not made it to textbook status, I do not know.

  19. Fermi-Walker Public Transport says:

    The Arthur Broyles paper came out in 1999,
    here is a link:

  20. Chris W. says:

    As a matter of fact, Broyles did post a preprint in 2002 (last revised in Feb 2003). Its subject is also quite interesting:

    One Loop Vacuum Polarization without Infinities (hep-th/0207069)

    A technique for avoiding infinite integrals in the calculation of the one-loop diagram contribution to the vacuum polarization component of an atomic energy level is presented. This makes renormalization unnecessary. Infinite integrals do not occur because, as it is shown, no delta functions are required for the Green’s functions. Thus there are none to overlap. This procedure is shown to produce the same formula as the one obtained by dimensional renormalization.

    (But I digress…)

  21. Pingback: Ars Mathematica » Blog Archive » Thanks for Spin-Statistics Theorem

  22. Carl Brannen says:

    “The answer is, Fermi statistics: when you take two identical fermions and switch them with each other, you get minus the state you started with.”

    This is true only for the state vector formalism of quantum mechanics. In density matrix formalism, the arbitrary complex phases are eliminated and exchanging two particles leaves one with the same density matrix state.

    Since density matrices and state vectors give identical results in quantum mechanics, it is not possible to say which one is fundamental. Consequently, it cannot be said that swapping fermions changes a state.

    From the density matrix point of view, a state vector is a sort of square root of a density matrix. It is not surprising that one must supply a phase (in analogy with the sign that arises with a square root). But if the fundamental nature is the density matrix (which can represent states that state vectors cannot and therefore are more general), then the minus sign that shows up when fermions are swapped is just an accident of mathematics, not a fundamental part of the physics. Instead, the difference between fermions and bosons is that the occupation numbers for fermions can only be 0 or 1 while bosons can count higher.

    Google “nine formulations of quantum mechanics” for more info.

  23. Pat Dennis says:

    Sean, this post was crystal-clear, as was the one on Hubble’s Law…. but I’d sure like to have seen something a little more accessible on the Lagrangian of the Standard Model of particle physics. Four semesters of college physics, and calculus through elementary differential equations, didn’t – in my case, at least – even come close to preparing me for that one!!

  24. Neil B says:

    Carl Brannen, isn’t it so that the “density matrix” is a sort of kludge representing uncertain knowledge of a wave function, it isn’t considered to be a “true” representation of a given actual WF. For example, I may have a source that emits linear polarized light, and I don’t know the details of how they are made (maybe, I don’t have a distinct linear filter available, maybe my assistant keeps turning it around, etc.) But must I think that “real waves” can literally not have a definite polarization (unless entangled)? So I wonder what physical significance there would have to be to a DM formulation, considering it is “a way of looking at things” more than an actual situation of thing.

  25. Carl Brannen says:


    You can use density matrices in something like 5 very much different ways. One of them is as you’ve described. But you can also take the pure states that are used in state vectors and convert them into “pure density matrices”. These are exactly equivalent to the usual state vectors in terms of information content, except that state vectors have arbitrary complex phases that have no physical interpretation. (The physical uses of complex phase are shared with between density matrices, otherwise they wouldn’t work.)

    “So I wonder what physical significance there would have to be to a DM formulation, considering it is “a way of looking at things” more than an actual situation of thing.”

    If you’re interested in the actual thing (or ontology), rather than mathematical ways of describing activity, then you’re probably interested in Bohmian mechanics. I’ve linked to a paper that discusses density matrices with respect to Bohmian mechanics which was the first reference I saw on arXiv in 30 seconds, but there are a lot of more complete articles. It’s an active field.

    From the density matrix point of view, state vectors are a kluge that is done so that a quantum state can be converted from its natural bilinear condition, to an unnatural, but mathematically convenient linear one. The mathematical convenience is “linear superposition”, which is a principle that allows you to describe a large number of quantum states from taking linear combinations of other states, for example, the usual basis states in spin-1/2 of spin up and spin down.

    Linear superposition (as opposed to interference) does not translate into a literal experiment that can be performed in the lab. That would be like taking, for example, combining a spin up electron with a spin down electron and expecting to get an electron oriented with spin in the x direction. You will not get this as a result. Instead, you will get two electrons and to model them you will require a more complicated state vector, i.e. ++, +-, -+, — as basis. Of course interference works fine in density matrix calculations just like with state vectors.

    The bilinearity of density matrices can be thought of as describing a quantum state in terms of how it acts as an operator. The pure density matrices are projection operators. This makes density matrices, rather than state vectors, the natural way to define the “Consistent Histories” interpretation of quantum mechanics (which generalizes QM to quantum cosmology). Read the article on consistent histories at Wikipedia, or do a search for arXiv articles.

    So density matrices have an interpretation of quantum mechanics that is devoted to their use and avoids the state vector formalism. At the very least this makes density matrices the equal of state vectors but for those who prefer quantum mechanics that naturally fits into quantum cosmology, density matrices are superior. And their sign does not change when you swap particles in them.

    Mathematically, this comes about because a density matrix is made from copies of the same state vector, a bra and a ket. When you swap two particles, you get a minus sign for the bra and a minus sign for the ket. Since (-1)(-1) = +1, the density matrix is unchanged.