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Second Quantization:
The Source of the Difference in Particle
Statistics for Bosons and Fermions

Bosons and Fermions

Bosons are particles for which there is no limit on the number which may occupy a particular quantum state. But at most only one fermion may occupy a particle quantum state. There is a cute image that helps one remember which behavior prevails for each. The image is of a tavern in which there is a communal table in the middle and booths around the walls. The communal table is occupied by bosons (allusion to gregarious bozo carousers) to any number. The booths have limited occupation and are occupied by fermions (allusion to firm uptight, upright customers). Given that fermions have two spin states the image of two to a booth fits.

Field Quantization

This material is to establish the quantization of a field and the particles it contains. Let ψ(r, t) be the wavefunction for a field where r stands for the set of coordinates of a point. The dynamics of the field are given by the time-dependent Schrödinger equation

−(ħ/i)(∂ψ/∂t) = − (ħ²/(2m))∇²ψ + V(r)ψ

where ħ is Planck's constant divided by 2π.

Consider ψ(r) expressed as a linear combination of a set of function {φn(r); n=0, 1, …, M}; i.e.,

ψ(r, t) = Σnbn(t)φn(r)

The set of functions to be used are the solutions to the time-independent Schrödinger equation

− (ħ²/(2m))∇²φn(r) + V(r)φn(r) = Enφn(r)

The solutions are orthogonal and can therefore be made into an orthonormal set; i.e.,

∫φn(r) φm(r)dr = δnm

where δnm=1 if n=m and is zero otherwise.

When the linear combination is put into the time-dependent equation the result is

−(ħ/i)Σn(dbn/dt)φn(r) = ΣnEnφn(r)

This equation can be multiplied by φm(r) and integrated over space. Because of the orthonorality of the set of φn(r) functions the result is

−(ħ/i)(dbm/dt) = Embm(t)

For particles the expression −(ħ²/(2m))∇²ψ + V(r)ψ is the Hamiltonian function; for a field it is the density of the Hamitonian function. The Hamiltonian for a field is given by its expected value; i.e.,

H = ∫ψ*( − (ħ²/(2m))∇² + V(r))ψdr

where ψ* is the complex conjugate of ψ.

The quantity E(r)ψ can replace −(ħ²/(2m))∇² + V(r))ψ(r) in the integral to give

H = ∫ψ*(r)E(r)ψ(r)dr

If ψ is replaced by the linear combination of the φn functions and the integration carried out the result is

H = Σnb*nEnbn

This Hamiltonian is in the nature of a sum of the Hamiltonians of a set of harmonic oscillators. Consider the harmonic oscillator for a particular value of n. It is an oscillator with a frequency of En/ħ.


Let Bn− denote the operator corresponding to bn and Bn+ the operator corresponding to bn*. Bn+ is the operator adjoint to Bn−. The reason for this notation is that Bn+ represents the creation of an additional particle in the field and Bn− the annilhilation of a particle in the field.

The Hamiltonian operator is then

H = ΣnBn+EnBn−

Let the bracket symbol [P, Q] denoted the commutator of two operations, P and Q; i.e.,

[P, Q] = PQ − QP

The set of operators Bn− and Bn+ are required to satisfy the following canonical quantification conditions

[Bn−, Bm−] = 0^
[Bn+, Bm+] = 0^
[Bn−, Bm+] = δnm

where 0^ denotes the zero operation; i.e., the operation that generates the zero function.

From these it follows that

[Bn−, H^] = EnBn−
[Bn+, H^] = EnBn+

These conditions along with the previously derived

−(ħ/i)(dbn/dt) = Enbn(t)

make the results consistent with the Heisenberg equation for the dynamics of the operators

−(ħ/i)(dBn−/dt) = [Bn−, H^]
−(ħ/i)(dBn+/dt) = [Bn+, H^]

Eigenfunctions and Eigenvalues

Consider the operator Bn+Bn−. It is so significant for the analysis that it is given a special name and symbol. It is called the number operator and is denoted as Nn

Let L be an eigenfunction of Nn and λ its eigenvalue; i.e.,

NnL = (Bn+Bn−)L = λL

In the webpage Operator Relations it is shown that (Bn−)L is also an eigenfunctions of Nn but with an eigenvalue which is 1 less than that of L. From this it follows that λ must be an integer and hence the eigenvalues of Nn are the nonnegative integers 0, 1, 2, 3, …. If n is the eigenvalue of Nn then the eigenvalue of (Bn−)L is (n−1). This is why Bn− is the annilhilator operator; it diminishes the number of particles in the field by one.

Similarly it can be shown that (Bn+)L is an eigenfunction of Nn but with an eigenvalue that is one unit more than that of L. Thus Bn+ is a creation operator that creates an additional particle for a field. This is the case of bosons.


For the case of fermions it is the anticommutator of two operations that must be considered. The anticommutator of two operations, P and Q, is defined as

{P, Q} = PQ + QP

The operator corresponding to bn will be denoted as βn− and that of bn* as βn+. The canonical quantification conditions to be satisfied by βn− and βn are

n−, βm−} = 0^
n+, βm+} = 0^
n−, βm+} = δnm

Note that

n−, βn+ = 1^
implies that
βn−βn+ = 1^ − βn+βn−

where 1^ is the identity operator.

Now consider

n+βn−)(βn+βn−) = βn+n−βn+n−
= βn+(1^ − βn+βn−n−
= (βn+βn−) − βn+n+βn−n−
= (βn+βn−) − (βn+βn+)(βn−βn−)

But (βn+βn+) and (βn−βn−) are both zero. Thus

n+βn−)(βn+βn−) = (βn+βn−)

Let L be an eigenfunction of (βn+βn−) and λ its eigenvalue. Then

n+βn−)L = λL
n+βn−)(βn+βn−)L = (βn+βn−)λL = λ²L

n+βn−)(βn+βn−)L =
n+βn−)L = λL

This means that

λ² = λ
and hence
λ must equal 0 or 1

Thus a fermion state can have at most one particle. This is the Pauli exclusion principle.

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