# Second quantization

Second quantization is a powerful procedure used in quantum field theory for describing the many-particle systems by quantizing the fields using a basis that describes the number of particles occupying each state in a complete set of single-particle states. This differs from the first quantization, which uses the single-particle states as basis.

## Introduction

The starting point of this formalism is the notion of indistinguishability of particles that bring us to use determinants of single-particle states as a basis of the Hilbert space of N-particles states[clarification needed]. Quantum theory can be formulated in terms of occupation numbers (number of particles occupying one determined energy state) of these single-particle states. The formalism was introduced in 1927 by Dirac.[1]

### The occupation number representation

Consider an ordered and complete single-particle basis $\left\{| \nu_1 \rang, | \nu_2 \rang, | \nu_3 \rang, ...\right\}$, where $| \nu_i \rang$ is the set of all states $\nu$ available for the single particle. In an N-particle system, only the occupied single-particle states play a role. So it is simpler to formulate a representation where one just counts how many particles there are in each orbital $| \nu \rang$. This simplification is achieved with the occupation number representation. The basis states for an N-particle system in this representation are obtained simply by listing the occupation numbers of each basis state, $|n_{\nu_1}, n_{\nu_2}, n_{\nu_3},\dots \rang$, where $\sum_j n_{\nu_j} = N$ The notation means that there are $n_{\nu_j}$ particles in the state $\nu_j$. It is therefore natural to define the occupation number operator $\hat{n}_{\nu_j}$ which obeys

$\hat{n}_{\nu_j}|n_{\nu_j} \rang=n_{\nu_j}|n_{\nu_j} \rang$

For fermions $n_{\nu_j}$ can be 0 or 1, while for bosons it can be any non negative number

$n_{\nu_j}= \begin{cases} \ 0, 1. &\text{fermions}\\ 0,1,2,... &\text{bosons} \end{cases}$

The space spanned by the occupation number basis is denoted the Fock space.

## Creation and annihilation operators

The creation and annihilation operators are the way to connect the first and second quantizations. It is fundamental for the many-body theory that every operator can be expressed in terms of annihilation and creation operators. Originally constructed in the context of the quantum harmonic oscillator, these operators are the most general form to describe quantum fields.[2] Depending on the nature of the fields we can use two different approaches:

### Bosons

Given the occupation number, we introduce the annihilation $b_{\nu_j}$ and creation ${b^{\dagger}}_{\nu_j}$ operators that lowers(raises) the occupation number in the state $| \nu_j \rang$ by 1,

$b_{\nu_j}|\dots,n_{\nu_{j-1}}, n_{\nu_j}, n_{\nu_{j+1}},\dots \rang=\sqrt{n_{\nu_j}}|\dots,n_{\nu_{j-1}}, n_{\nu_j}-1, n_{\nu_{j+1}},\dots \rang$
${b^{\dagger}}_{\nu_j}|\dots,n_{\nu_{j-1}}, n_{\nu_j}, n_{\nu_{j+1}},\dots \rang=\sqrt{n_{\nu_j}+1}|\dots,n_{\nu_{j-1}}, n_{\nu_j}+1, n_{\nu_{j+1}},\dots \rang$

Since bosons are symmetric in the single-particle state index $\nu_j$ we demand that $b_{\nu_j}$ and ${b^{\dagger}}_{\nu_j}$ commute, So, we can obtain the mean properties of these operators:

$\begin{matrix} [{b^{\dagger}}_{\nu_j},{b^{\dagger}}_{\nu_k}] = 0 & [b_{\nu_j},b_{\nu_k}]=0 & [b_{\nu_j},{b^{\dagger}}_{\nu_k}]=\delta_{\nu_j\nu_k}\\ {b^{\dagger}}_{\nu_j}|n_{\nu_j}\rang=\sqrt{n_{\nu_j}+1}|n_{\nu_j}+1 \rang & b_{\nu_j}|n_{\nu_j}\rang=\sqrt{n_{\nu_j}}|n_{\nu_j} -1\rang & b_{\nu_j}|0\rang=0\\ {b^{\dagger}}_{\nu_j}b_{\nu_j}|n_{\nu_j} \rang=n_{\nu_j}|n_{\nu_j} \rang &\left({b^{\dagger}}_{\nu_j}\right)^{n_{\nu_j}}|0 \rang=\sqrt{(n_{\nu_j})!}|n_{\nu_j} \rang & n_{\nu_j}=0,1,2,\dots\\ \end{matrix}$

and therefore identify the first and second quantized states,

$\hat{S}_+|\psi_{n_{\nu_1}}(\bold{r}_1)\rang|\psi_{n_{\nu_2}}(\bold{r}_2)\rang\dots |\psi_{n_{\nu_N}}(\bold{r}_N)\rang= {b^{\dagger}}_{n_{\nu_1}}{b^{\dagger}}_{n_{\nu_2}}\dots{b^{\dagger}}_{n_{\nu_N}}|0\rang$

with $\hat{S}$ the symmetrization operator. Here, both contain N-particle state-kets completely symmetric in the single-particle state index $\psi_{\nu_j}$. Because the creation and annihilation operators of the quantum harmonic oscillator obey these properties, one can classify the field associated to it as bosonic.

### Fermions

Fermions have similar annihilation $c_{\nu_j}$ and creation ${c^{\dagger}}_{\nu_j}$ operators:

$\begin{matrix} c_{\nu_j}|\dots,1,\dots \rang= |\dots,0,\dots \rang & c_{\nu_j}|\dots,0,\dots \rang=0 \\ {c^{\dagger}}_{\nu_j}|\dots,0,\dots \rang=|\dots,1,\dots\rang& {c^{\dagger}}_{\nu_j}|\dots,1,\dots \rang=0 \end{matrix}$

(Note that the only permitted number of occupation are 0 or 1.) To maintain the fundamental fermionic antisymmetry upon exchange of orbitals, the operators must anticommute, rather than commute:

$\begin{matrix} \{ {c^{\dagger}}_{\nu_j},{c^{\dagger}}_{\nu_k}\} = 0 & \{c_{\nu_j},c_{\nu_k}\}=0 & \{c_{\nu_j}, {c^{\dagger}}_{\nu_k}\}=\delta_{\nu_j\nu_k}\\ \left({c^{\dagger}}_{\nu_j}\right)^2=\left(c_{\nu_j}\right)^2=0 & {c^{\dagger}}_{\nu_j}c_{\nu_j}|n_{\nu_j} \rang=n_{\nu_j}|n_{\nu_j} \rang & n_{\nu_j}=0,1\\ \end{matrix}$

Where we have used the anticommutator $\{,\}$ Therefore one can identify the first quantized states in terms of the second quantized:

$\hat{S}_-|\psi_{n_{\nu_1}}(\bold{r}_1)\rang|\psi_{n_{\nu_2}}(\bold{r}_2)\rang\dots |\psi_{n_{\nu_N}}(\bold{r}_N)\rang= {c^{\dagger}}_{n_{\nu_1}}{c^{\dagger}}_{n_{\nu_2}}\dots{c^{\dagger}}_{n_{\nu_N}}|0\rang$

with $\hat{S}_-$ the Antisymmetrizer operator. Here, both contain N-particle state-kets completely anti-symmetric in the single-particle state index $\psi_{\nu_j}$ in accordance with Pauli exclusion principle.

## Quantum field operators

Defining ${a^{(\dagger)}}_{\nu}$ as a general annihilation(creation) operator that could be either fermionic $({c^{(\dagger)}}_{\nu})$ or bosonic $({b^{(\dagger)}}_{\nu})$, the real space representation of the operators defines the quantum field operators $\Psi(\bold{r})$ and $\Psi^{\dagger}(\bold{r})$ by

$\Psi(\bold{r})=\sum_{\nu} \psi_{\nu} \left( \bold{r} \right) a_{\nu}$
$\Psi^{\dagger}(\bold{r})=\sum_{\nu} {\psi^*}_{\nu} \left( \bold{r} \right) {a^{\dagger}}_{\nu}$

Second quantization operators, while the coefficients $\psi_{\nu} \left( \bold{r} \right)$ and ${\psi^*}_{\nu} \left( \bold{r} \right)$ are the ordinary first quantization wavefunctions. Loosely speaking, $\Psi^{\dagger}(\bold{r})$ is the sum of all possible ways to add a particle to the system at position r through any of the basis states $\psi_{\nu}\left(\bold{r}\right)$. Since $\Psi(\bold{r})$ and $\Psi^{\dagger}(\bold{r})$ are second quantization operators defined in every point in space they are called quantum field operators. They obey the following fundamental commutator and anti-commutator,

$\left[\Psi(\bold{r}_1),\Psi^\dagger(\bold{r}_2)\right]=\delta (\bold{r}_1-\bold{r}_2)$ boson fields,
$\{\Psi(\bold{r}_1),\Psi^\dagger(\bold{r}_2)\}=\delta (\bold{r}_1-\bold{r}_2)$ fermion fields.

In homogeneous systems it is often desirable to transform between real space and the momentum representations, hence, the quantum fields operators in Fourier basis yields:

$\Psi(\bold{r})={1\over \sqrt {V}} \sum_{\bold{k}} e^{i\bold{k\cdot r}}a_{\bold{k}}$
$\Psi^{\dagger}(\bold{r})={ 1\over \sqrt{V}} \sum_{\bold{k}} e^{-i\bold{k\cdot r}}{a^{\dagger}}_{\bold{k}}$