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?
 A: 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.


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*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.

*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.
A: 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.
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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).
A: 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.
A: 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.  
A: 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.
