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 NN single-particle energy eigenstates ψn(x)\psi_n(x) and NN identical particles xnx_n, the Fock states are all the wave functions which contain nn particles, for nn going from 00 to NN.

So for n=0n = 0, there is one basis vector with 00 particles, for n=1n = 1, there are NN basis vectors with 11 particle each, for n=2n = 2, there are N(N ⁣ ⁣1)N (N \!-\! 1) basis vectors with 22 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 need to obey the Pauli exclusion principle, so each state can only contain one particle. In this case, the Fock states are given by:

n=0:0,0,0,...n=1:1,0,0,...0,1,0,...0,0,1,...n=2:1,1,0,...1,0,1,...0,1,1,...\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 Nα,Nβ,...\Ket{N_\alpha, N_\beta, ...} is shorthand for the appropriate Slater determinants. As an example, take 0,1,0,1,1\Ket{0, 1, 0, 1, 1}, which contains three particles aa, bb and cc in states 2, 4 and 5:

0,1,0,1,1=Ψ(xa,xb,xc)=13!det ⁣[ψ2(xa)ψ4(xa)ψ5(xa)ψ2(xb)ψ4(xb)ψ5(xb)ψ2(xc)ψ4(xc)ψ5(xc)]\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 c^α\hat{c}_\alpha^\dagger and annihilation operator c^α\hat{c}_\alpha are defined to live up to their name: they create or destroy a particle in the state ψα\psi_\alpha:

c^α...(Nα ⁣= ⁣0)...=Jα...(Nα ⁣= ⁣1)...c^α...(Nα ⁣= ⁣1)...=Jα...(Nα ⁣= ⁣0)...\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 JαJ_\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 α\alphath state:

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

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

c^40,1,1,0,1=(1)0+1+10,1,1,1,1\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:

c^1c^20,1=c^10,0=1,0c^2c^10,1=c^21,1=1,0\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, c^1c^2=c^2c^1\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger, meaning that the anticommutator {c^2,c^1}=0\{\hat{c}_2, \hat{c}_1^\dagger\} = 0. You can verify for youself that the general anticommutators of these operators are given by:

{c^α,c^β}={c^α,c^β}=0{c^α,c^β}=δαβ\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:

c^α...(Nα ⁣= ⁣1)...=0c^α...(Nα ⁣= ⁣0)...=0\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:

...(Nα ⁣= ⁣1)...c^α...(Nα ⁣= ⁣0)...=...(Nα ⁣= ⁣0)...c^α...(Nα ⁣= ⁣1)...\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 N^α\hat{N}_\alpha as follows:

N^α=c^αc^α\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 ψα\psi_\alpha (look at the hats):

N^α...Nα...=Nα...Nα...\begin{aligned} \hat{N}_\alpha \Ket{... N_\alpha ...} = N_\alpha \Ket{... N_\alpha ...} \end{aligned}


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

n=0:0,0,0,...n=1:1,0,0,...0,1,0,...0,0,1,...n=2:1,1,0,...1,0,1,...0,1,1,...2,0,0,...0,2,0,...0,0,2,...\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 c^α\hat{c}_\alpha^\dagger and c^α\hat{c}_\alpha are straightforward:

c^α...Nα...=Nα+1...(Nα ⁣+ ⁣1)...c^α...Nα...=Nα...(Nα ⁣ ⁣1)...\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 c^α\hat{c}_\alpha when there are zero particles in α\alpha will quench the state:

c^α...(Nα ⁣= ⁣0)...=0\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:

[c^α,c^β]=[c^α,c^β]=0[c^α,c^β]=δαβ\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 c^α\hat{c}_\alpha^\dagger and c^α\hat{c}_\alpha ensure that N^α\hat{N}_\alpha keeps the same nice form:

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


Traditionally, an operator V^\hat{V} simultaneously acting on NN indentical particles is the sum of the individual single-particle operators V^1\hat{V}_1 acting on the nnth particle:

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

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

V^=α,βαV^1βc^αc^β\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 αV^1β\matrixel{\alpha}{\hat{V}_1}{\beta} is to be evaluated in the normal way:

αV^1β=ψα(r)V^1(r)ψβ(r)dr\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 V^\hat{V} in first-quantized form:

V^=nmv(rn,rm)\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:

V^=α,β,γ,δvαβγδc^αc^βc^δc^γ\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 vαβγδv_{\alpha \beta \gamma \delta} is defined from the single-particle wave functions:

vαβγδ=ψα(r1)ψβ(r2)v(r1,r2)ψγ(r1)ψδ(r2)dr1dr2\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:

c^b0=b=ααα|b=αα|bc^α0\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 α\alpha and bb 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 r\vec{r}:

Ψ^(r)=αα|rc^αΨ^(r)=αr|αc^α\begin{aligned} \boxed{ \hat{\Psi}^\dagger(\vec{r}) = \sum_{\alpha} \Inprod{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger \qquad \quad \hat{\Psi}(\vec{r}) = \sum_{\alpha} \Inprod{\vec{r}}{\alpha} \hat{c}_\alpha } \end{aligned}


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