path: root/content/know/concept/second-quantization/index.pdc

---
title: "Second quantization"
firstLetter: "S"
publishDate: 2021-02-26
categories:
- Quantum mechanics
- Physics

date: 2021-02-26T10:04:16+01:00
draft: false
markup: pandoc
---

# Second quantization

The **second quantization** is a technique to deal with quantum systems
containing a large and/or variable number of identical particles.
Its exact formulation depends on
whether it is fermions or bosons that are being considered
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)).

Regardless of whether the system is fermionic or bosonic,
the idea is to change basis to a set of certain many-particle wave functions,
known as the **Fock states**, which are specific members of a **Fock space**,
a special kind of [Hilbert space](/know/concept/hilbert-space/),
with a well-defined number of particles.

For a set of $N$ single-particle energy eigenstates
$\psi_n(x)$ and $N$ identical particles $x_n$, the Fock states are
all the wave functions which contain $n$ particles, for $n$ going from $0$ to $N$.

So for $n = 0$, there is one basis vector with $0$ particles,
for $n = 1$, there are $N$ basis vectors with $1$ particle each,
for $n = 2$, there are $N (N \!-\! 1)$ basis vectors with $2$ particles,
etc.

In this basis, we define the **particle creation operators**
and **particle annihilation operators**,
which respectively add/remove a particle to/from a given state.
In other words, these operators relate the Fock basis vectors
to one another, and are very useful.

The point is to express the system's state in such a way that the
fermionic/bosonic constraints are automatically satisfied, and the
formulae look the same regardless of the number of particles.


## Fermions

Fermions need to obey the Pauli exclusion principle, so each state can only
contain one particle. In this case, the Fock states are given by:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            n &= 0:
            \qquad \ket{0, 0, 0, ...}
            \\
            n &= 1:
            \qquad \ket{1, 0, 0, ...} \quad \ket{0, 1, 0, ...} \quad \ket{0, 0, 1, ...} \quad \cdots
            \\
            n &= 2:
            \qquad \ket{1, 1, 0, ...} \quad \ket{1, 0, 1, ...} \quad \ket{0, 1, 1, ...} \quad \cdots
        \end{aligned}
    }
\end{aligned}$$

The notation $\ket{N_\alpha, N_\beta, ...}$ is shorthand for
the appropriate [Slater determinants](/know/concept/slater-determinant/).
As an example, take $\ket{0, 1, 0, 1, 1}$,
which contains three particles $a$, $b$ and $c$
in states 2, 4 and 5:

$$\begin{aligned}
    \ket{0, 1, 0, 1, 1}
    = \Psi(x_a, x_b, x_c)
    = \frac{1}{\sqrt{3!}} \det\!
    \begin{bmatrix}
        \psi_2(x_a) & \psi_4(x_a) & \psi_5(x_a) \\
        \psi_2(x_b) & \psi_4(x_b) & \psi_5(x_b) \\
        \psi_2(x_c) & \psi_4(x_c) & \psi_5(x_c)
    \end{bmatrix}
\end{aligned}$$

The creation operator $\hat{c}_\alpha^\dagger$ and annihilation
operator $\hat{c}_\alpha$ are defined to live up to their name:
they create or destroy a particle in the state $\psi_\alpha$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \hat{c}_\alpha^\dagger \ket{... (N_\alpha\!=\!0) ...}
            &= J_\alpha \ket{... (N_\alpha\!=\!1) ...}
            \\
            \hat{c}_\alpha \ket{... (N_\alpha\!=\!1) ...}
            &= J_\alpha \ket{... (N_\alpha\!=\!0) ...}
        \end{aligned}
    }
\end{aligned}$$

The factor $J_\alpha$ is sometimes known as the **Jordan-Wigner string**,
and is necessary here to enforce the fermionic antisymmetry,
when creating or destroying a particle in the $\alpha$th state:

$$\begin{aligned}
    J_\alpha = (-1)^{\sum_{j < \alpha} N_j}
\end{aligned}$$

So, for example, when creating a particle in state 4
of $\ket{0, 1, 1, 0, 1}$, we get the following:

$$\begin{aligned}
    \hat{c}_4^\dagger \ket{0, 1, 1, 0, 1}
    = (-1)^{0 + 1 + 1} \ket{0, 1, 1, 1, 1}
\end{aligned}$$

The point of the Jordan-Wigner string
is that the order matters when applying the creation and annihilation operators:

$$\begin{aligned}
    \hat{c}_1^\dagger \hat{c}_2 \ket{0, 1}
    &= \hat{c}_1^\dagger \ket{0, 0}
    = \ket{1, 0}
    \\
    \hat{c}_2 \hat{c}_1^\dagger \ket{0, 1}
    &= \hat{c}_2 \ket{1, 1}
    = - \ket{1, 0}
\end{aligned}$$

In other words, $\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$,
meaning that the anticommutator $\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$.
You can verify for youself that
the general anticommutators of these operators are given by:

$$\begin{aligned}
    \boxed{
        \{\hat{c}_\alpha, \hat{c}_\beta\} = \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\} = 0
        \qquad \quad
        \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\} = \delta_{\alpha\beta}
    }
\end{aligned}$$

Each single-particle state can only contain 0 or 1 fermions,
so these operators **quench** states that would violate this rule.
Note that these are *scalar* zeros:

$$\begin{aligned}
    \boxed{
        \hat{c}_\alpha^\dagger \ket{... (N_\alpha\!=\!1) ...} = 0
        \qquad \quad
        \hat{c}_\alpha \ket{... (N_\alpha\!=\!0) ...} = 0
    }
\end{aligned}$$

Finally, as has already been suggested by the notation, they are each other's adjoint:

$$\begin{aligned}
    \matrixel{... (N_\alpha\!=\!1) ...}{\hat{c}_\alpha^\dagger}{... (N_\alpha\!=\!0) ...}
    = \matrixel{...(N_\alpha\!=\!0) ...}{\hat{c}_\alpha}{... (N_\alpha\!=\!1) ...}
\end{aligned}$$

Let us now use these operators to define the **number operator** $\hat{N}_\alpha$ as follows:

$$\begin{aligned}
    \boxed{
        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
    }
\end{aligned}$$

Its eigenvalue is the number of particles residing in state $\psi_\alpha$
(look at the hats):

$$\begin{aligned}
    \hat{N}_\alpha \ket{... N_\alpha ...}
    = N_\alpha \ket{... N_\alpha ...}
\end{aligned}$$


## Bosons

Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state.
The Fock states are therefore as follows:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            n &= 0:
            \qquad \ket{0, 0, 0, ...}
            \\
            n &= 1:
            \qquad \ket{1, 0, 0, ...} \quad \ket{0, 1, 0, ...} \quad \ket{0, 0, 1, ...} \quad \cdots
            \\
            n &= 2:
            \qquad \ket{1, 1, 0, ...} \quad \ket{1, 0, 1, ...} \quad \ket{0, 1, 1, ...} \quad \cdots
            \\
            &\qquad\:\:\:
            \qquad \ket{2, 0, 0, ...} \quad \ket{0, 2, 0, ...} \quad \ket{0, 0, 2, ...} \quad \cdots
        \end{aligned}
    }
\end{aligned}$$

They must be symmetric under the exchange of two bosons.
To achieve this, the Fock states are represented by Slater *permanents*
rather than determinants.

The boson creation and annihilation operators $\hat{c}_\alpha^\dagger$ and
$\hat{c}_\alpha$ are straightforward:

$$\begin{gathered}
    \boxed{
        \begin{aligned}
            \hat{c}_\alpha^\dagger \ket{... N_\alpha ...}
            &= \sqrt{N_\alpha + 1} \: \ket{... (N_\alpha \!+\! 1) ...}
            \\
            \hat{c}_\alpha \ket{... N_\alpha ...}
            &= \sqrt{N_\alpha} \: \ket{... (N_\alpha \!-\! 1) ...}
        \end{aligned}
}\end{gathered}$$

Applying the annihilation operator $\hat{c}_\alpha$ when there are zero
particles in $\alpha$ will quench the state:

$$\begin{aligned}
    \boxed{
        \hat{c}_\alpha \ket{... (N_\alpha\!=\!0) ...} = 0
    }
\end{aligned}$$

There is no Jordan-Wigner string, and therefore no sign change when commuting.
Consequently, these operators therefore satisfy the following:

$$\begin{aligned}
    \boxed{
        [\hat{c}_\alpha, \hat{c}_\beta] = [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] = 0
        \qquad
        [\hat{c}_\alpha, \hat{c}_\beta^\dagger] = \delta_{\alpha\beta}
    }
\end{aligned}$$

The constant factors applied by $\hat{c}_\alpha^\dagger$ and $\hat{c}_\alpha$
ensure that $\hat{N}_\alpha$ keeps the same nice form:

$$\begin{aligned}
    \boxed{
        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
    }
\end{aligned}$$


## Operators

Traditionally, an operator $\hat{V}$ simultaneously acting on $N$ indentical particles
is the sum of the individual single-particle operators $\hat{V}_1$ acting on the $n$th particle:

$$\begin{aligned}
    \hat{V}
    = \sum_{n = 1}^N \hat{V}_1
\end{aligned}$$

This can be rewritten using the second quantization operators as follows:

$$\begin{aligned}
    \boxed{
        \hat{V}
        = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}_1}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
    }
\end{aligned}$$

Where the matrix element $\matrixel{\alpha}{\hat{V}_1}{\beta}$ is to be
evaluated in the normal way:

$$\begin{aligned}
    \matrixel{\alpha}{\hat{V}_1}{\beta}
    = \int \psi_\alpha^*(\vec{r}) \: \hat{V}_1(\vec{r}) \: \psi_\beta(\vec{r}) \dd{\vec{r}}
\end{aligned}$$

Similarly, given some two-particle operator $\hat{V}$ in first-quantized form:

$$\begin{aligned}
    \hat{V}
    = \sum_{n \neq m} v(\vec{r}_n, \vec{r}_m)
\end{aligned}$$

We can rewrite this in second-quantized form as follows.
Note the ordering of the subscripts:

$$\begin{aligned}
    \boxed{
        \hat{V}
        = \sum_{\alpha, \beta, \gamma, \delta}
        v_{\alpha \beta \gamma \delta} \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
    }
\end{aligned}$$

Where the constant $v_{\alpha \beta \gamma \delta}$ is defined from the
single-particle wave functions:

$$\begin{aligned}
    v_{\alpha \beta \gamma \delta}
    = \iint \psi_\alpha^*(\vec{r}_1) \: \psi_\beta^*(\vec{r}_2)
    \: v(\vec{r}_1, \vec{r}_2) \: \psi_\gamma(\vec{r}_1)
    \: \psi_\delta(\vec{r}_2) \dd{\vec{r}_1} \dd{\vec{r}_2}
\end{aligned}$$

Finally, in the second quantization, changing basis is done in the usual way:

$$\begin{aligned}
    \hat{c}_b^\dagger \ket{0}
    = \ket{b}
    = \sum_{\alpha} \ket{\alpha} \braket{\alpha}{b}
    = \sum_{\alpha} \braket{\alpha}{b} \hat{c}_\alpha^\dagger \ket{0}
\end{aligned}$$

Where $\alpha$ and $b$ need not be in the same basis.
With this, we can define the **field operators**,
which create or destroy a particle at a given position $\vec{r}$:

$$\begin{aligned}
    \boxed{
        \hat{\Psi}^\dagger(\vec{r})
        = \sum_{\alpha} \braket{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger
        \qquad \quad
        \hat{\Psi}(\vec{r})
        = \sum_{\alpha} \braket{\vec{r}}{\alpha} \hat{c}_\alpha
    }
\end{aligned}$$


## References
1.  L.E. Ballentine,
    *Quantum mechanics: a modern development*, 2nd edition,
    World Scientific.