Categories: Physics, Quantum mechanics.

# 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).

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,
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 systemâ€™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. 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:

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:

Finally, as has already been suggested by the notation, they are each otherâ€™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:

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:

$\begin{aligned} \hat{c}_b^\dagger \Ket{0} = \Ket{b} = \sum_{\alpha} \Ket{\alpha} \Inprod{\alpha}{b} = \sum_{\alpha} \Inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0} \end{aligned}$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 $\vec{r}$:

## References

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