Pauli Principle

Prerequisites: Spin, Canonical Quantum Mechanics. Full path (19 pages)

Why is it interesting?
Intuitive
Concrete
Abstract

Why is it interesting?

It is the fact that the electrons cannot all get on top of each other that makes tables and everything else solid. Richard Feynman

Intuitive

According to Pauli, not only do electrons each insists on taking up a finite amount of room, they have to have their own rooms. Thus, the quantum Hall problem may be described as a sort of housing crisis, or as the problem of assigning office space at the Institute for Theoretical Physics to visitors who do not want to share offices. http://xxx.lanl.gov/pdf/cond-mat/9501022v2

Concrete

Identical particles in quantum mechanics cannot be labelled, so exchanging two of them must give the same physical state. For two particles with coordinates $x_i = (\vec r_i, \sigma_i)$ (position and spin component) this means $\psi(x_2,x_1) = e^{i\alpha}\psi(x_1,x_2)$ , and exchanging twice gives back the original function, so $e^{i\alpha} = \pm 1$ . (This last step holds in three space dimensions; in two dimensions it fails, see the Abstract section.) Particles with the $+$ sign are bosons, those with the $-$ sign are fermions. Electrons, protons, neutrons and all other particles with half-integer spin are fermions:

$\psi(x_1, x_2) = -\psi(x_2, x_1) .$

Setting $x_1 = x_2$ gives $\psi(x,x) = 0$ . Two fermions are never found at the same position with the same spin component.

For non-interacting particles we build states from single-particle states $\phi_a$ . The antisymmetric two-particle state is

$\psi(x_1,x_2) = \frac{1}{\sqrt2}\left[\phi_a(x_1)\phi_b(x_2) - \phi_b(x_1)\phi_a(x_2)\right] ,$

and for $N$ particles the Slater determinant

$\psi(x_1,\ldots,x_N) = \frac{1}{\sqrt{N!}} \det\begin{pmatrix} \phi_{a_1}(x_1) & \cdots & \phi_{a_1}(x_N) \\ \vdots & & \vdots \\ \phi_{a_N}(x_1) & \cdots & \phi_{a_N}(x_N)\end{pmatrix} .$

If two of the $a_i$ are equal, two rows are equal and the determinant vanishes. This is Pauli's exclusion principle: no two identical fermions occupy the same single-particle state.

Atoms. In the hydrogen-like approximation an electron state is labelled by $(n, l, m_l, m_s)$ with $m_s = \pm\frac12$ . Each orbital $(n,l,m_l)$ therefore holds at most two electrons, and a shell with principal quantum number $n$ holds $2\sum_{l=0}^{n-1}(2l+1) = 2n^2$ electrons. Filling the states from the bottom gives the structure of the periodic table. In the helium ground state both electrons are in the 1s orbital; the spatial wave function is symmetric, so the spin state must be the antisymmetric singlet $(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)/\sqrt2$ .

Fermi gas. Put $N$ free electrons in a box of volume $V$ . Each momentum state holds two electrons (two spin states), and there are $V\,d^3k/(2\pi)^3$ states in $d^3k$ . In the ground state all states up to the Fermi wavenumber $k_F$ are filled:

$N = 2\,\frac{V}{(2\pi)^3}\,\frac{4\pi}{3}k_F^3 \quad\Rightarrow\quad k_F = (3\pi^2 n)^{1/3}, \quad n = N/V .$

Even at zero temperature the electrons have kinetic energies up to $E_F = \hbar^2k_F^2/2m$ . The resulting degeneracy pressure is what keeps white dwarfs from collapsing and contributes to the stiffness of solids.

Second quantization. Fermionic creation and annihilation operators obey anticommutation relations

$\{a_i, a_j^\dagger\} = \delta_{ij} , \qquad \{a_i,a_j\} = \{a_i^\dagger,a_j^\dagger\} = 0 .$

In particular $(a_i^\dagger)^2 = 0$ : creating a second fermion in the same state gives zero, and the occupation numbers $a_i^\dagger a_i$ have eigenvalues $0$ and $1$ only.

Abstract

If $\mathcal H$ is the one-particle Hilbert space, the $N$ -fermion space is the antisymmetric subspace $\Lambda^N\mathcal H \subset \mathcal H^{\otimes N}$ , the image of the projector $\frac{1}{N!}\sum_{\pi\in S_N}\operatorname{sgn}(\pi)\,U_\pi$ . The fermionic Fock space is the exterior algebra $\mathcal F(\mathcal H) = \bigoplus_{N\geq 0}\Lambda^N\mathcal H$ . For $f\in\mathcal H$ the creation operator $a^\dagger(f)$ is $f\wedge\cdot$ (suitably normalized), and the operators satisfy the canonical anticommutation relations

$\{a(f), a^\dagger(g)\} = \langle f, g\rangle , \qquad \{a(f),a(g)\} = 0 .$

Since $a^\dagger(f)^2 = 0$ and $\|a^\dagger(f)\| = \|f\|$ , the creation and annihilation operators are bounded, unlike in the bosonic case. The exclusion principle in its general form, valid also for interacting and entangled states, is a statement about the one-particle density matrix $\gamma = \operatorname{tr}_{N-1}|\Psi\rangle\langle\Psi|$ , normalized to $\operatorname{tr}\gamma = N$ : its eigenvalues (occupation numbers of natural orbitals) lie in $[0,1]$ .

Why only symmetric and antisymmetric states occur depends on the dimension. For $N$ identical particles in $\mathbb R^d$ the configuration space is $\left((\mathbb R^d)^N\setminus\Delta\right)/S_N$ , with $\Delta$ the coincidence set. For $d\geq3$ its fundamental group is $S_N$ , whose one-dimensional unitary representations are the trivial one (bosons) and the sign representation (fermions). For $d = 2$ the fundamental group is the braid group, which allows anyons with arbitrary exchange phase.

Which particles are fermions is fixed by the spin-statistics theorem: in a relativistic quantum field theory in $3+1$ dimensions with positive energy and local (anti)commutativity of fields at spacelike separation, fields of half-integer spin must be quantized with anticommutators and fields of integer spin with commutators. In non-relativistic quantum mechanics this connection is an input. In the path integral fermionic fields are Grassmann valued, and $\theta^2 = 0$ for a Grassmann variable is the counterpart of $(a^\dagger)^2 = 0$ .

The exclusion principle is also essential for the stability of bulk matter. Dyson and Lenard proved that the ground state energy of a system of electrons and nuclei with Coulomb interactions is bounded below by a constant times the total particle number, and this bound uses the fact that the electrons are fermions. Dyson showed that a comparable system of charged bosons would have a ground state energy that falls faster than linearly in $N$ .

Pauli Principle

Why is it interesting?

Intuitive

Concrete

Abstract

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