Second quantization

Consider a complete set of single-particle wave functions ${\displaystyle \psi _{\alpha }(\mathbf {r} )}$ labeled by ${\displaystyle \alpha }$ (which may be a combined index of a number of quantum numbers). The following wave function

${\displaystyle \Psi [\mathbf {r} _{i}]=\prod _{i=1}^{N}\psi _{\alpha _{i}}(\mathbf {r} _{i})\equiv \psi _{\alpha _{1}}\otimes \psi _{\alpha _{2}}\otimes \cdots \otimes \psi _{\alpha _{N}}}$

represents an N-particle state with the ith particle occupying the single-particle state ${\displaystyle |{\alpha _{i}}\rangle }$. In the shorthanded notation, the position argument of the wave function may be omitted, and it is assumed that the ith single-particle wave function describes the state of the ith particle. The wave function ${\displaystyle \Psi }$ has not been symmetrized or anti-symmetrized, thus in general not qualified as a many-body wave function for identical particles. However, it can be brought to the symmetrized (anti-symmetrized) form by operators ${\displaystyle {\mathcal {S}}}$ for symmetrizer, and ${\displaystyle {\mathcal {A}}}$ for antisymmetrizer.

For bosons, the many-body wave function must be symmetrized,

${\displaystyle \Psi _{\rm {B}}[\mathbf {r} _{i}]={\mathcal {N}}{\mathcal {S}}\Psi [\mathbf {r} _{i}]={\mathcal {N}}\sum _{\pi \in S_{N}}\prod _{i=1}^{N}\psi _{\alpha _{\pi (i)}}(\mathbf {r} _{i})={\mathcal {N}}\sum _{\pi \in S_{N}}\psi _{\alpha _{\pi (1)}}\otimes \psi _{\alpha _{\pi (2)}}\otimes \cdots \otimes \psi _{\alpha _{\pi (N)}};}$

while for fermions, the many-body wave function must be anti-symmetrized,

${\displaystyle \Psi _{\rm {F}}[\mathbf {r} _{i}]={\mathcal {N}}{\mathcal {A}}\Psi [\mathbf {r} _{i}]={\mathcal {N}}\sum _{\pi \in S_{N}}(-1)^{\pi }\prod _{i=1}^{N}\psi _{\alpha _{\pi (i)}}(\mathbf {r} _{i})={\mathcal {N}}\sum _{\pi \in S_{N}}(-1)^{\pi }\psi _{\alpha _{\pi (1)}}\otimes \psi _{\alpha _{\pi (2)}}\otimes \cdots \otimes \psi _{\alpha _{\pi (N)}}.}$

Here ${\displaystyle \pi }$ is an element in the N-body permutation group (or symmetric group) ${\displaystyle S_{N}}$, which performs a permutation among the state labels ${\displaystyle \alpha _{i}}$, and ${\displaystyle (-1)^{\pi }}$ denotes the corresponding permutation sign. ${\displaystyle {\mathcal {N}}}$ is the normalization operator that normalizes the wave function. (It is the operator that applies a suitable numerical normalization factor to the symmetrized tensors of degree n; see the next section for its value.)

If one arranges the single-particle wave functions in a matrix ${\displaystyle U}$, such that the row-i column-j matrix element is ${\displaystyle U_{ij}=\psi _{\alpha _{j}}(\mathbf {r} _{i})\equiv \langle \mathbf {r} _{i}|\alpha _{j}\rangle }$, then the boson many-body wave function can be simply written as a permanent ${\displaystyle \Psi _{\rm {B}}={\mathcal {N}}\operatorname {perm} U}$, and the fermion many-body wave function as a determinant ${\displaystyle \Psi _{\rm {F}}={\mathcal {N}}\det U}$ (also known as the Slater determinant).[6]

First quantized wave functions involve complicated symmetrization procedures to describe physically realizable many-body states because the language of first quantization is redundant for indistinguishable particles. In the first quantization language, the many-body state is described by answering a series of questions like "Which particle is in which state?". However these are not physical questions, because the particles are identical, and it is impossible to tell which particle is which in the first place. The seemingly different states ${\displaystyle \psi _{1}\otimes \psi _{2}}$ and ${\displaystyle \psi _{2}\otimes \psi _{1}}$ are actually redundant names of the same quantum many-body state. So the symmetrization (or anti-symmetrization) must be introduced to eliminate this redundancy in the first quantization description.

In the second quantization language, instead of asking "each particle on which state", one asks "How many particles are there in each state?". Because this description does not refer to the labeling of particles, it contains no redundant information, and hence leads to a precise and simpler description of the quantum many-body state. In this approach, the many-body state is represented in the occupation number basis, and the basis state is labeled by the set of occupation numbers, denoted

${\displaystyle |[n_{\alpha }]\rangle \equiv |n_{1},n_{2},\cdots ,n_{\alpha },\cdots \rangle ,}$

meaning that there are ${\displaystyle n_{\alpha }}$ particles in the single-particle state ${\displaystyle |\alpha \rangle }$ (or as ${\displaystyle \psi _{\alpha }}$). The occupation numbers sum to the total number of particles, i.e. ${\textstyle \sum _{\alpha }n_{\alpha }=N}$. For fermions, the occupation number ${\displaystyle n_{\alpha }}$ can only be 0 or 1, due to the Pauli exclusion principle; while for bosons it can be any non-negative integer

${\displaystyle n_{\alpha }={\begin{cases}0,1&{\text{fermions,}}\\0,1,2,3,...&{\text{bosons.}}\end{cases}}}$

The occupation number states ${\displaystyle |[n_{\alpha }]\rangle }$ are also known as Fock states. All the Fock states form a complete basis of the many-body Hilbert space, or Fock space. Any generic quantum many-body state can be expressed as a linear combination of Fock states.

Note that besides providing a more efficient language, Fock space allows for a variable number of particles. As a Hilbert space, it is isomorphic to the sum of the n-particle bosonic or fermionic tensor spaces described in the previous section, including a one-dimensional zero-particle space C.

The Fock state with all occupation numbers equal to zero is called the vacuum state, denoted ${\displaystyle |0\rangle \equiv |\cdots ,0_{\alpha },\cdots \rangle }$. The Fock state with only one non-zero occupation number is a single-mode Fock state, denoted ${\displaystyle |n_{\alpha }\rangle \equiv |\cdots ,0,n_{\alpha },0,\cdots \rangle }$. In terms of the first quantized wave function, the vacuum state is the unit tensor product and can be denoted ${\displaystyle |0\rangle =1}$. The single-particle state is reduced to its wave function ${\displaystyle |1_{\alpha }\rangle =\psi _{\alpha }}$. Other single-mode many-body (boson) states are just the tensor product of the wave function of that mode, such as ${\displaystyle |2_{\alpha }\rangle =\psi _{\alpha }\otimes \psi _{\alpha }}$ and ${\displaystyle |n_{\alpha }\rangle =\psi _{\alpha }^{\otimes n}}$. For multi-mode Fock states (meaning more than one single-particle state ${\displaystyle |\alpha \rangle }$ is involved), the corresponding first-quantized wave function will require proper symmetrization according to the particle statistics, e.g. ${\displaystyle |1_{1},1_{2}\rangle =(\psi _{1}\psi _{2}+\psi _{2}\psi _{1})/{\sqrt {2}}}$ for a boson state, and ${\displaystyle |1_{1},1_{2}\rangle =(\psi _{1}\psi _{2}-\psi _{2}\psi _{1})/{\sqrt {2}}}$ for a fermion state (the symbol ${\displaystyle \otimes }$ between ${\displaystyle \psi _{1}}$ and ${\displaystyle \psi _{2}}$ is omitted for simplicity). In general, the normalization is found to be ${\textstyle {\sqrt {{1}/{N!\prod _{\alpha }n_{\alpha }!}}}}$, where N is the total number of particles. For fermion, this expression reduces to ${\displaystyle {\tfrac {1}{\sqrt {N!}}}}$ as ${\displaystyle n_{\alpha }}$ can only be either zero or one. So the first-quantized wave function corresponding to the Fock state reads

${\displaystyle |[n_{\alpha }]\rangle _{\rm {B}}=\left({\frac {1}{N!\prod _{\alpha }n_{\alpha }!}}\right)^{1/2}{\mathcal {S}}\bigotimes \limits _{\alpha }\psi _{\alpha }^{\otimes n_{\alpha }}}$

for bosons and

${\displaystyle |[n_{\alpha }]\rangle _{\rm {F}}={\frac {1}{\sqrt {N!}}}{\mathcal {A}}\bigotimes \limits _{\alpha }\psi _{\alpha }^{\otimes n_{\alpha }}}$

for fermions. Note that for fermions, ${\displaystyle n_{\alpha }=0,1}$ only, so the tensor product above is effectively just a product over all occupied single-particle states.

The creation and annihilation of a particle is implemented by the insertion and deletion of the single-particle state from the first quantized wave function in an either symmetric or anti-symmetric manner. Let ${\displaystyle \psi _{\alpha }}$ be a single-particle state, let 1 be the tensor identity (it is the generator of the zero-particle space C and satisfies ${\displaystyle \psi _{\alpha }\equiv 1\otimes \psi _{\alpha }\equiv \psi _{\alpha }\otimes 1}$ in the tensor algebra over the fundamental Hilbert space), and let ${\displaystyle \Psi =\psi _{\alpha _{1}}\otimes \psi _{\alpha _{2}}\otimes \cdots }$ be a generic tensor product state. The insertion ${\displaystyle \otimes _{\pm }}$ and the deletion ${\displaystyle \oslash _{\pm }}$ operators are linear operators defined by the following recursive equations

${\displaystyle \psi _{\alpha }\otimes _{\pm }1=\psi _{\alpha },\quad \psi _{\alpha }\otimes _{\pm }(\psi _{\beta }\otimes \Psi )=\psi _{\alpha }\otimes \psi _{\beta }\otimes \Psi \pm \psi _{\beta }\otimes (\psi _{\alpha }\otimes _{\pm }\Psi );}$

${\displaystyle \psi _{\alpha }\oslash _{\pm }1=0,\quad \psi _{\alpha }\oslash _{\pm }(\psi _{\beta }\otimes \Psi )=\delta _{\alpha \beta }\Psi \pm \psi _{\beta }\otimes (\psi _{\alpha }\oslash _{\pm }\Psi ).}$

Here ${\displaystyle \delta _{\alpha \beta }}$ is the Kronecker delta symbol, which gives 1 if ${\displaystyle \alpha =\beta }$, and 0 otherwise. The subscript ${\displaystyle \pm }$ of the insertion or deletion operators indicates whether symmetrization (for bosons) or anti-symmetrization (for fermions) is implemented.

The boson creation (resp. annihilation) operator is usually denoted as ${\displaystyle b_{\alpha }^{\dagger }}$ (resp. ${\displaystyle b_{\alpha }}$). The creation operator ${\displaystyle b_{\alpha }^{\dagger }}$ adds a boson to the single-particle state ${\displaystyle |\alpha \rangle }$, and the annihilation operator ${\displaystyle b_{\alpha }}$ removes a boson from the single-particle state ${\displaystyle |\alpha \rangle }$. The creation and annihilation operators are Hermitian conjugate to each other, but neither of them are Hermitian operators (${\displaystyle b_{\alpha }\neq b_{\alpha }^{\dagger }}$).

The boson creation (annihilation) operator is a linear operator, whose action on a N-particle first-quantized wave function ${\displaystyle \Psi }$ is defined as

${\displaystyle b_{\alpha }^{\dagger }\Psi ={\frac {1}{\sqrt {N+1}}}\psi _{\alpha }\otimes _{+}\Psi ,}$

${\displaystyle b_{\alpha }\Psi ={\frac {1}{\sqrt {N}}}\psi _{\alpha }\oslash _{+}\Psi ,}$

where ${\displaystyle \psi _{\alpha }\otimes _{+}}$ inserts the single-particle state ${\displaystyle \psi _{\alpha }}$ in ${\displaystyle N+1}$ possible insertion positions symmetrically, and ${\displaystyle \psi _{\alpha }\oslash _{+}}$ deletes the single-particle state ${\displaystyle \psi _{\alpha }}$ from ${\displaystyle N}$ possible deletion positions symmetrically.

Hereinafter the tensor symbol ${\displaystyle \otimes }$ between single-particle states is omitted for simplicity. Take the state ${\displaystyle |1_{1},1_{2}\rangle =(\psi _{1}\psi _{2}+\psi _{2}\psi _{1})/{\sqrt {2}}}$, create one more boson on the state ${\displaystyle \psi _{1}}$,

${\displaystyle {\begin{array}{rl}b_{1}^{\dagger }|1_{1},1_{2}\rangle =&{\frac {1}{\sqrt {2}}}(b_{1}^{\dagger }\psi _{1}\psi _{2}+b_{1}^{\dagger }\psi _{2}\psi _{1})\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {3}}}\psi _{1}\otimes _{+}\psi _{1}\psi _{2}+{\frac {1}{\sqrt {3}}}\psi _{1}\otimes _{+}\psi _{2}\psi _{1}\right)\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {3}}}(\psi _{1}\psi _{1}\psi _{2}+\psi _{1}\psi _{1}\psi _{2}+\psi _{1}\psi _{2}\psi _{1})+{\frac {1}{\sqrt {3}}}(\psi _{1}\psi _{2}\psi _{1}+\psi _{2}\psi _{1}\psi _{1}+\psi _{2}\psi _{1}\psi _{1})\right)\\=&{\frac {\sqrt {2}}{\sqrt {3}}}(\psi _{1}\psi _{1}\psi _{2}+\psi _{1}\psi _{2}\psi _{1}+\psi _{2}\psi _{1}\psi _{1})\\=&{\sqrt {2}}|2_{1},1_{2}\rangle .\end{array}}}$

Then annihilate one boson from the state ${\displaystyle \psi _{1}}$,

${\displaystyle {\begin{array}{rl}b_{1}|2_{1},1_{2}\rangle =&{\frac {1}{\sqrt {3}}}(b_{1}\psi _{1}\psi _{1}\psi _{2}+b_{1}\psi _{1}\psi _{2}\psi _{1}+b_{1}\psi _{2}\psi _{1}\psi _{1})\\=&{\frac {1}{\sqrt {3}}}\left({\frac {1}{\sqrt {3}}}\psi _{1}\oslash _{+}\psi _{1}\psi _{1}\psi _{2}+{\frac {1}{\sqrt {3}}}\psi _{1}\oslash _{+}\psi _{1}\psi _{2}\psi _{1}+{\frac {1}{\sqrt {3}}}\psi _{1}\oslash _{+}\psi _{2}\psi _{1}\psi _{1}\right)\\=&{\frac {1}{\sqrt {3}}}\left({\frac {1}{\sqrt {3}}}(\psi _{1}\psi _{2}+\psi _{1}\psi _{2}+0)+{\frac {1}{\sqrt {3}}}(\psi _{2}\psi _{1}+0+\psi _{1}\psi _{2})+{\frac {1}{\sqrt {3}}}(0+\psi _{2}\psi _{1}+\psi _{2}\psi _{1})\right)\\=&\psi _{1}\psi _{2}+\psi _{2}\psi _{1}\\=&{\sqrt {2}}|1_{1},1_{2}\rangle .\end{array}}}$

Starting from the single-mode vacuum state ${\displaystyle |0_{\alpha }\rangle =1}$, applying the creation operator ${\displaystyle b_{\alpha }^{\dagger }}$ repeatedly, one finds

${\displaystyle b_{\alpha }^{\dagger }|0_{\alpha }\rangle =\psi _{\alpha }\otimes _{+}1=\psi _{\alpha }=|1_{\alpha }\rangle ,}$

${\displaystyle b_{\alpha }^{\dagger }|n_{\alpha }\rangle ={\frac {1}{\sqrt {n_{\alpha }+1}}}\psi _{\alpha }\otimes _{+}\psi _{\alpha }^{\otimes n_{\alpha }}={\sqrt {n_{\alpha }+1}}\psi _{\alpha }^{\otimes (n_{\alpha }+1)}={\sqrt {n_{\alpha }+1}}|n_{\alpha }+1\rangle .}$

The creation operator raises the boson occupation number by 1. Therefore, all the occupation number states can be constructed by the boson creation operator from the vacuum state

${\displaystyle |n_{\alpha }\rangle ={\frac {1}{\sqrt {n_{\alpha }!}}}(b_{\alpha }^{\dagger })^{n_{\alpha }}|0_{\alpha }\rangle .}$

On the other hand, the annihilation operator ${\displaystyle b_{\alpha }}$ lowers the boson occupation number by 1

${\displaystyle b_{\alpha }|n_{\alpha }\rangle ={\frac {1}{\sqrt {n_{\alpha }}}}\psi _{\alpha }\oslash _{+}\psi _{\alpha }^{\otimes n_{\alpha }}={\sqrt {n_{\alpha }}}\psi _{\alpha }^{\otimes (n_{\alpha }-1)}={\sqrt {n_{\alpha }}}|n_{\alpha }-1\rangle .}$

It will also quench the vacuum state ${\displaystyle b_{\alpha }|0_{\alpha }\rangle =0}$ as there has been no boson left in the vacuum state to be annihilated. Using the above formulae, it can be shown that

${\displaystyle b_{\alpha }^{\dagger }b_{\alpha }|n_{\alpha }\rangle =n_{\alpha }|n_{\alpha }\rangle ,}$

meaning that ${\displaystyle {\hat {n}}_{\alpha }=b_{\alpha }^{\dagger }b_{\alpha }}$ defines the boson number operator.

The above result can be generalized to any Fock state of bosons.

${\displaystyle b_{\alpha }^{\dagger }|\cdots ,n_{\beta },n_{\alpha },n_{\gamma },\cdots \rangle ={\sqrt {n_{\alpha }+1}}|\cdots ,n_{\beta },n_{\alpha }+1,n_{\gamma },\cdots \rangle .}$

${\displaystyle b_{\alpha }|\cdots ,n_{\beta },n_{\alpha },n_{\gamma },\cdots \rangle ={\sqrt {n_{\alpha }}}|\cdots ,n_{\beta },n_{\alpha }-1,n_{\gamma },\cdots \rangle .}$

These two equations can be considered as the defining properties of boson creation and annihilation operators in the second-quantization formalism. The complicated symmetrization of the underlying first-quantized wave function is automatically taken care of by the creation and annihilation operators (when acting on the first-quantized wave function), so that the complexity is not revealed on the second-quantized level, and the second-quantization formulae are simple and neat.

The following operator identities follow from the action of the boson creation and annihilation operators on the Fock state,

${\displaystyle [b_{\alpha }^{\dagger },b_{\beta }^{\dagger }]=[b_{\alpha },b_{\beta }]=0,\quad [b_{\alpha },b_{\beta }^{\dagger }]=\delta _{\alpha \beta }.}$

These commutation relations can be considered as the algebraic definition of the boson creation and annihilation operators. The fact that the boson many-body wave function is symmetric under particle exchange is also manifested by the commutation of the boson operators.

The raising and lowering operators of the quantum harmonic oscillator also satisfy the same set of commutation relations, implying that the bosons can be interpreted as the energy quanta (phonons) of an oscillator. The position and momentum operators of a Harmonic oscillator (or a collection of Harmonic oscillating modes) are given by Hermitian combinations of phonon creation and annihilation operators,

${\displaystyle x_{\alpha }=(b_{\alpha }+b_{\alpha }^{\dagger })/{\sqrt {2}},\quad p_{\alpha }=(b_{\alpha }-b_{\alpha }^{\dagger })/({\sqrt {2}}\mathrm {i} ),}$

which reproduce the canonical commutation relation between position and momentum operators (with ${\displaystyle \hbar =1}$)

${\displaystyle [x_{\alpha },p_{\beta }]=\mathrm {i} \delta _{\alpha \beta },\quad [x_{\alpha },x_{\beta }]=[p_{\alpha },p_{\beta }]=0.}$

This idea is generalized in the quantum field theory, which considers each mode of the matter field as an oscillator subject to quantum fluctuations, and the bosons are treated as the excitations (or energy quanta) of the field.

The fermion creation (annihilation) operator is usually denoted as ${\displaystyle c_{\alpha }^{\dagger }}$ (${\displaystyle c_{\alpha }}$). The creation operator ${\displaystyle c_{\alpha }^{\dagger }}$ adds a fermion to the single-particle state ${\displaystyle |\alpha \rangle }$, and the annihilation operator ${\displaystyle c_{\alpha }}$ removes a fermion from the single-particle state ${\displaystyle |\alpha \rangle }$.

The fermion creation (annihilation) operator is a linear operator, whose action on a N-particle first-quantized wave function ${\displaystyle \Psi }$ is defined as

${\displaystyle c_{\alpha }^{\dagger }\Psi ={\frac {1}{\sqrt {N+1}}}\psi _{\alpha }\otimes _{-}\Psi ,}$

${\displaystyle c_{\alpha }\Psi ={\frac {1}{\sqrt {N}}}\psi _{\alpha }\oslash _{-}\Psi ,}$

where ${\displaystyle \psi _{\alpha }\otimes _{-}}$ inserts the single-particle state ${\displaystyle \psi _{\alpha }}$ in ${\displaystyle N+1}$ possible insertion positions anti-symmetrically, and ${\displaystyle \psi _{\alpha }\oslash _{-}}$ deletes the single-particle state ${\displaystyle \psi _{\alpha }}$ from ${\displaystyle N}$ possible deletion positions anti-symmetrically.

It is particularly instructive to view the results of creation and annihilation operators on states of two (or more) fermions, because they demonstrate the effects of exchange. A few illustrative operations are given in the example below. The complete algebra for creation and annihilation operators on a two-fermion state can be found in Quantum Photonics.[8]

Hereinafter the tensor symbol ${\displaystyle \otimes }$ between single-particle states is omitted for simplicity. Take the state ${\displaystyle |1_{1},1_{2}\rangle =(\psi _{1}\psi _{2}-\psi _{2}\psi _{1})/{\sqrt {2}}}$, attempt to create one more fermion on the occupied ${\displaystyle \psi _{1}}$ state will quench the whole many-body wave function,

${\displaystyle {\begin{array}{rl}c_{1}^{\dagger }|1_{1},1_{2}\rangle =&{\frac {1}{\sqrt {2}}}(c_{1}^{\dagger }\psi _{1}\psi _{2}-c_{1}^{\dagger }\psi _{2}\psi _{1})\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {3}}}\psi _{1}\otimes _{-}\psi _{1}\psi _{2}-{\frac {1}{\sqrt {3}}}\psi _{1}\otimes _{-}\psi _{2}\psi _{1}\right)\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {3}}}(\psi _{1}\psi _{1}\psi _{2}-\psi _{1}\psi _{1}\psi _{2}+\psi _{1}\psi _{2}\psi _{1})-{\frac {1}{\sqrt {3}}}(\psi _{1}\psi _{2}\psi _{1}-\psi _{2}\psi _{1}\psi _{1}+\psi _{2}\psi _{1}\psi _{1})\right)\\=&0.\end{array}}}$

Annihilate a fermion on the ${\displaystyle \psi _{2}}$ state, take the state ${\displaystyle |1_{1},1_{2}\rangle =(\psi _{1}\psi _{2}-\psi _{2}\psi _{1})/{\sqrt {2}}}$,

${\displaystyle {\begin{array}{rl}c_{2}|1_{1},1_{2}\rangle =&{\frac {1}{\sqrt {2}}}(c_{2}\psi _{1}\psi _{2}-c_{2}\psi _{2}\psi _{1})\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {2}}}\psi _{2}\oslash _{-}\psi _{1}\psi _{2}-{\frac {1}{\sqrt {2}}}\psi _{2}\oslash _{-}\psi _{2}\psi _{1}\right)\\=&{\frac {1}{\sqrt {2}}}\left({\frac {1}{\sqrt {2}}}(0-\psi _{1})-{\frac {1}{\sqrt {2}}}(\psi _{1}-0)\right)\\=&-\psi _{1}\\=&-|1_{1},0_{2}\rangle .\end{array}}}$

The minus sign (known as the fermion sign) appears due to the anti-symmetric property of the fermion wave function.

Starting from the single-mode vacuum state ${\displaystyle |0_{\alpha }\rangle =1}$, applying the fermion creation operator ${\displaystyle c_{\alpha }^{\dagger }}$,

${\displaystyle c_{\alpha }^{\dagger }|0_{\alpha }\rangle =\psi _{\alpha }\otimes _{-}1=\psi _{\alpha }=|1_{\alpha }\rangle ,}$

${\displaystyle c_{\alpha }^{\dagger }|1_{\alpha }\rangle ={\frac {1}{\sqrt {2}}}\psi _{\alpha }\otimes _{-}\psi _{\alpha }=0.}$

If the single-particle state ${\displaystyle |\alpha \rangle }$ is empty, the creation operator will fill the state with a fermion. However, if the state is already occupied by a fermion, further application of the creation operator will quench the state, demonstrating the Pauli exclusion principle that two identical fermions can not occupy the same state simultaneously. Nevertheless, the fermion can be removed from the occupied state by the fermion annihilation operator ${\displaystyle c_{\alpha }}$,

${\displaystyle c_{\alpha }|1_{\alpha }\rangle =\psi _{\alpha }\oslash _{-}\psi _{\alpha }=1=|0_{\alpha }\rangle ,}$

${\displaystyle c_{\alpha }|0_{\alpha }\rangle =0.}$

The vacuum state is quenched by the action of the annihilation operator.

Similar to the boson case, the fermion Fock state can be constructed from the vacuum state using the fermion creation operator

${\displaystyle |n_{\alpha }\rangle =(c_{\alpha }^{\dagger })^{n_{\alpha }}|0_{\alpha }\rangle .}$

It is easy to check (by enumeration) that

${\displaystyle c_{\alpha }^{\dagger }c_{\alpha }|n_{\alpha }\rangle =n_{\alpha }|n_{\alpha }\rangle ,}$

meaning that ${\displaystyle {\hat {n}}_{\alpha }=c_{\alpha }^{\dagger }c_{\alpha }}$ defines the fermion number operator.

The above result can be generalized to any Fock state of fermions.

${\displaystyle c_{\alpha }^{\dagger }|\cdots ,n_{\beta },n_{\alpha },n_{\gamma },\cdots \rangle =(-1)^{\sum _{\beta <\alpha }n_{\beta }}{\sqrt {1-n_{\alpha }}}|\cdots ,n_{\beta },1+n_{\alpha },n_{\gamma },\cdots \rangle .}$[9]

${\displaystyle c_{\alpha }|\cdots ,n_{\beta },n_{\alpha },n_{\gamma },\cdots \rangle =(-1)^{\sum _{\beta <\alpha }n_{\beta }}{\sqrt {n_{\alpha }}}|\cdots ,n_{\beta },1-n_{\alpha },n_{\gamma },\cdots \rangle .}$

Recall that the occupation number ${\displaystyle n_{\alpha }}$ can only take 0 or 1 for fermions. These two equations can be considered as the defining properties of fermion creation and annihilation operators in the second quantization formalism. Note that the fermion sign structure ${\displaystyle (-1)^{\sum _{\beta <\alpha }n_{\beta }}}$, also known as the Jordan-Wigner string, requires there to exist a predefined ordering of the single-particle states (the spin structure) and involves a counting of the fermion occupation numbers of all the preceding states; therefore the fermion creation and annihilation operators are considered non-local in some sense. This observation leads to the idea that fermions are emergent particles in the long-range entangled local qubit system.[10]

The following operator identities follow from the action of the fermion creation and annihilation operators on the Fock state,

${\displaystyle \{c_{\alpha }^{\dagger },c_{\beta }^{\dagger }\}=\{c_{\alpha },c_{\beta }\}=0,\quad \{c_{\alpha },c_{\beta }^{\dagger }\}=\delta _{\alpha \beta }.}$

These anti-commutation relations can be considered as the algebraic definition of the fermion creation and annihilation operators. The fact that the fermion many-body wave function is anti-symmetric under particle exchange is also manifested by the anti-commutation of the fermion operators.

The creation and annihilation operators are Hermitian conjugate to each other, but neither of them are Hermitian operators (${\displaystyle c_{\alpha }\neq c_{\alpha }^{\dagger }}$). The Hermitian combination of the fermion creation and annihilation operators

${\displaystyle \chi _{\alpha ,{\text{Re}}}=(c_{\alpha }+c_{\alpha }^{\dagger })/{\sqrt {2}},\quad \chi _{\alpha ,{\text{Im}}}=(c_{\alpha }-c_{\alpha }^{\dagger })/({\sqrt {2}}\mathrm {i} ),}$

are called Majorana fermion operators. They can be viewed as the fermionic analog of position and momentum operators of a "fermionic" Harmonic oscillator. They satisfy the anticommutation relation

${\displaystyle \{\chi _{i},\chi _{j}\}=\delta _{ij},}$

where ${\displaystyle i,j}$ labels any Majorana fermion operators on equal footing (regardless their origin from Re or Im combination of complex fermion operators ${\displaystyle c_{\alpha }}$). The anticommutation relation indicates that Majorana fermion operators generates a Clifford algebra, which can be systematically represented as Pauli operators in the many-body Hilbert space.