Canonical Quantization and the Particle Statistics for Bosons and Fermions
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Canonical Quantization and
the 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

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

(iħ)(∂ψ/∂t) = H^ψ;

where ħ is Planck's constant divided by 2π and H^ is the Hamiltonian operator for the system. The equations of the anaysis are greatly simplified by using the natural system of units in which ħ is equal to unity.

The Existence of So-called Creator,
Annihilator and Counting Operators

Here a field is a function over space and an operator is a function that take a field function as an input and returns a field function as an output. The commutator [P, Q] of two operators P and Q is defined as

[P, Q] = PQ − QP

Let B be an operator and B* as its adjoint (conjugate) operator. Suppose the operators B and B* are such that their commutator [B, B*] is equal to the identity operator; i.e..

[B, B*] = BB* − B*B = I

where I is the identity operator. This is called canonical quantification.

This means that

BB* = B*B + I

Let Ψ be an eigenfunction of B*B; i.e.,

B*BΨ = λΨ

Now consider BΨ. From the above

B(B*BΨ) = λBΨ
but this means
(BB*)BΨ = λBΨ
so BΨ is an
eigenfunction of BB*.

However, from canonical quantification BB* = B*B + I so

(BB*)BΨ = (B*B + I)BΨ = (B*B)BΨ + BΨ = λBΨ
and hence
(B*B)BΨ = (λ−1)BΨ

So BΨ is also an eigenfunction of B*B but with a eigenvalue of (λ−1).

Likewise this means that BnΨ is an eigenfunction of B*B with an eigenvalue of (λ−n).

Note that B*B is Hermitian since (B*B)*=B*(B*)*=B*B. The eigenvalues of the eigenfunctions of B*B must therefore be nonnegative so there must be some integer m such that (λ−m) is equal to zero. Thus the eigenvalues of all of the eigenfunctions of B*B must be nonnegative integers.

Now consider B*Ψ. From the previous.

B*(B*B)Ψ = λB*Ψ
but from
canonical quantification
B*(BB*−I)Ψ = λB*Ψ
B*(BB*)Ψ − B*Ψ = λB*Ψ
and hence
(B*B)BΨ = λB*Ψ + B*Ψ = (λ+1)B*Ψ

Thus B*Ψ is an eigenfunction of B*B but with an eigenvalue of (λ+1).

Thus the cononical quantification of B*B results in B* being a creation operator and B being an annihilator operator. Since all of the eigenvalues of B*B are nonnegative integers B*B can be called the number operator, a sort-of counting operator.

So for a state Ψ there can be any number of particles occupying it. The nature of such particles is open to question. This can only be answered by linking B*B to the Hamiltonian operator. When the physical system is a harmonic oscillator there are operators B and B* such that H^=B*B+½I. It follows that the eigenvalues of H^, which are the energies of the system, are of the form (n+½), where n is an integer. Thus the so-called particles are merely the number of energy quanta of the system.

This is the case that applies to bosons. So boson statistics are linked to the canonical quantification condition that the commutator of a field operator with its adjoint (complex conjugate) is equal to the identity operator. This was the case for 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

Let F be an operator and F* be its adjoint. The canonical quantification conditions to be satisfied by F and F* are

{F, F*} = FF* + F*F = I
{F, F} = 0^

Note that {F, F*)=I implies that

FF* = I − F*F

The condition that {F, F}=0^ and hence {F*, F*}=0^ is new for the case of the anti-commutator, but [P, P]=0^ is automatically satisfied for the commutator. The conditions that {F, F}=0^ and {F*, F*}=0^ imply that FF=0^ and F*F*=0^.

Consider now

(F*F)(F*F) = F*(FF*)F = F*(I − F*F)F = F*F − (F*F*)(FF)
but both
F*F* and FF
are equal to 0^
(F*F)(F*F) = F*F

Let Φ be an eigenfunction of F*F.

F*FΦ = λΦ
(F*F)(F*FΦ) = λ(F*F)Φ = λ²Φ
(F*F)(F*FΦ) = FF*Φ = λΦ

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.


What we find is that the assumption of canonical quantification for the commutator and anti-commutator is sufficient to assure the existence of the creator, annihilator and number operator without any reference to the physics of the particles.

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