# Why do the laws of quantum mechanics not allow to put a pile of fermions all at a same place?

While i was reading the book "THE PARTICLE AT THE END OF UNIVERSE", the author said that we can not place a of pile fermions in a same place because laws of quantum mechanics do not allow that. Therefore, I got a doubt: why does not quantum mechanics allow that?

The reason fermions and bosons behave differently is the spin-statistic theorem, a rather subtle result from relativistic quantum field theory that does not hold in all of quantum mechanics and crucially requires special relativity.

1. The issue is actually not about fermions being in the same place (quantum objects do not have a well-defined place), but about being in the same quantum mechanical state. Any system of $N$ particles in quantum mechanics is described by a state that can be decomposed into sum of products of states of the individual particles, and in the simplest case, the non-entangled one, it just looks like $$\lvert \psi_\text{total}\rangle = \lvert \psi_1\rangle_1 \otimes \lvert \psi_2\rangle_2\otimes\cdots\otimes \lvert \psi_N\rangle_N.$$ On these states, it makes sense to define an "exchange operator" $S_{ij}$ that switches the state of the $i$-th and $j$-th particle, e.g. $$S_{12} (\lvert \psi_1\rangle_1\otimes \vert \psi_2\rangle_2) = \lvert \psi_2\rangle_1\otimes \lvert \psi_1\rangle_2.$$ In terms of wavefunctions, this amounts to having a wavefunction with $N$ arguments and swapping the $i$-th and $j$-th argument. If the particles are "indistinguishable", then that means that this exchanged operator should do nothing to the state, i.e. applying the exchange operator to a state of indistinguishable particles should change nothing about the physical state. Therefore, since physical states are actually rays in Hilbert space, we have that $$S_{ij} \lvert \psi_\text{indist}\rangle = c\lvert \psi_\text{indist}\rangle$$ for some complex number $c\in\mathbb{C}$, and since intuitively exchanging particles twice should return the original state, we have that $S_{ij}^2 = 1$ and therefore $c = \pm 1$. $c=1$ means the state is symmetric under exchange and the particles are called bosons, $c=-1$ means the state is anti-symmetric under exchange and the particles are called fermions. Fermions cannot be put into the same state because $$S_{12} (\lvert \psi\rangle_1 \otimes \lvert \psi\rangle_2) = -\lvert \psi \rangle_1 \otimes \lvert \psi\rangle_2$$ implies that $\lvert \psi\rangle_1\otimes\lvert\psi\rangle_2 = 0$, i.e. there are no non-trivial antisymmetric states with two particles in the same state.

Interestingly, this is not the full story, since in two dimensions there can be indistinguishable particles which are neither, called anyons. For an explanation see this answer.

2. We could simply observe that electrons etc. are fermions since they obey Pauli exclusion and leave it at that. In this ad hoc approach, the question "Why can we not put fermions into the same state?" has no deeper answer.

However, in relativistic quantum field theory, the question does have a deeper answer (or, more concretely, we can answer the question of why certain particles are fermions and others are not). The spin-statistics theorem tells us that particles with integer spin are necessarily bosons and particles with half-integer spin are necessarily fermions, and it relies crucially on relativity - in non-relativistic quantum field theory, fermions with integer spin and bosons with half-integer spin are consistently possible. The proof of this theorem is subtle and I don't think there is consensus about which version of it is canonical, but Streater/Wightman's "PCT, Spin, Statistics and all that" is a good place to start.

• Down-vote. Not for being incorrect, but for being so short as to be not helpful (other than supplying a link). Attributing the reason to the spin-statistics thm is pretty much the same as attributing to exchange symmetry. I'd like to see this answer improved by reference to exchange symmetry and the role of special relativity. And maybe the usual disclaimer about describing rather than explaining the universe. – garyp Aug 18 '17 at 14:24
• As you mention elsewhere, 'fermion' is synonymous with 'can't be pulled in the same spot'; the spin-statistics theorem just tells you about the rotational properties of those particles, but that's pretty tangential to the question as posed. – Emilio Pisanty Aug 18 '17 at 14:44
• @garyp I added an explanation of exchange symmetry, but I'm afraid a discussion of the actual role of relativity would require carefully writing out the proof of the theorem itself. – ACuriousMind Aug 18 '17 at 15:01
• @ACuriuouMind we don't just ask you, we demand of you that very careful writing!!!!! Yes, please. – hyportnex Aug 18 '17 at 15:03
• Given the way the OP has asked his question I for one do not believe that he will understand a word of what you ACuriousMind wrote. So here is the challenge to you and also to @garyp and Emilio Pisanty, would it be possible to describe what you wrote in plain English so that an intelligent high-school student could also understand it. It is so fundamental to the way our world is built that it deserves a very careful and simple description. – hyportnex Aug 18 '17 at 15:17

Laws, postulates and principles are a physics terminology referring to the extra axioms necessary so that a mathematical model, in this case quantum mechanics, fits observations and data and predicts successfully new set ups.

So that is why it is called the Pauli exclusion principle, because it is an observational fact, starting with electrons protons and neutrons when they were first detected experimentally, that two of the same species cannot occupy the same quantum mechanical state. " So the answer to your questions is "because it is axiomatically assumed so that mathematical models fit the data.

We have observed the atoms, and that electrons fill up different energy levels obeying to the Pauli exclusion principle. Thus our model of hydrogen, for example, in order to fit the data has to obey this law. Please note that if this law was not into effect the world as we know it would not exist, there would be no chemistry, which depends on the energy levels of atoms and molecules and the way they fit together. All electrons would lie at the lowest binding energy level. There would be no nuclear table of elements, whose regularities are explained by the shell model and the Pauli exclusion principle ( your fermion law).

It is not that they don't do it because quantum mechanics "does not allow". The rules of QM (as all the physics laws) were developed according to observed behavior of natural world. The laws are not the cause of anything. They are just rules used to describe the reality in a simple and consistent way. The behavior of fermions was observed to be as it is and rules were developed to describe this behavior.

• While technically true, all laws of physics are made to fit observations. When people ask (and answer) a "Why?" question in physics they are generally asking whether there exists a theoretical model in which the statement that is to be explained is derived rather than assumed. – ACuriousMind Aug 18 '17 at 13:55
• While in general true about people asking a why question, here the OP question was not why the fermions don't "pile up" but why quantum mechanics does not allow this pile up. A theoretical model can indeed explain the behavior of the fermions but the fermions do not behave this way due to the model. I think this is a point which should be clarified whenever it comes up. – nasu Aug 22 '17 at 0:56

Identical Fermions are described by anti-symmetric wave functions (further explanations of this go too far here), so if you have $N$ fermions, the wave function must obey for any permutation $\sigma$ that

$$\psi(t, x_1,...,x_N) = \mathrm{sign}(\sigma) \psi(t, x_{\sigma(1)},...,x_{\sigma(N)}),$$ or to say it briefly, interchanging the labels of two particles gives you a minus sign. This directly implies that if you insert the same position more than once, you get zero, because

$$\psi(t, x,x, x_3, x_4, ...,x_N) = - \psi(t, x,x,x_3,x_4,...,x_N) = 0.$$

So you see that fermions cannot be in the same state and as a special case of this, not at the exact same place.

• It should be pointed out that without quantum field theory, the statement "identical fermions are described by anti-symmetric wave functions" is an ad hoc assumption whose sole purpose is precisely to explain Pauli exclusion, i.e. "fermions' wavefunctions are antisymmetric" is precisely equivalent to "fermions can't occupy the same state", so it is doubtful whether this actually constitutes an explanation. – ACuriousMind Aug 18 '17 at 14:00
• Well, you can also view it like this: Already the physics of the Schrödinger equation tells you that wave functions of identical particles have to be symmetric or antisymmetric, so the alternative between bosons and fermions arises. And we see that electrons and other particles matter is made of can only behave like fermions. – Luke Aug 18 '17 at 14:03

Fermions have a location and mass so they fill space. Bosons however carry force. Massless bosons can pile up in one spot, fermions cannot unless you make cooper pairs in which fermions behave like bosons. This is seen in superconductivity.

• What? A Cooper pair consists of a pair of electrons with anti-symmetrically correlated momenta (i.e., their momenta sum to zero). The 2 electrons aren't located particularly close to one another, and Cooper pair formation is rather dynamic, so a given pair doesn't have a long life-time. And although multiple Cooper pairs can have the same quantum state, that doesn't somehow permit the individual electrons composing the pairs to bypass Pauli exclusion. I assume you know these things, but others who don't know them could be misled by your answer. – PM 2Ring Aug 19 '17 at 11:10
• FWIW, there's a nice post about Cooper pairs here. – PM 2Ring Aug 19 '17 at 11:10

## protected by Qmechanic♦Aug 18 '17 at 14:52

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