Quantum Mechanics for Mathematicians

that satisfy the so-called canonical commutation relations (CCR)

[aj,a†k] =δjk 1 , [aj,ak] = [a†j,a†k] = 0

The simple change in the harmonic oscillator problem that takes one from
bosons to fermions is the replacement of the bosonic annihilation and creation
operators (which we’ll now denoteaBandaB†) by fermionic annihilation and
creation operators calledaFandaF†, and replacement of the commutator

[A,B]≡AB−BA

of operators by the anticommutator

[A,B]+≡AB+BA

The commutation relations are now (ford= 1, a single degree of freedom)

[aF,a†F]+= 1 , [aF,aF]+= 0, [a†F,a†F]+= 0

with the last two relations implying thata^2 F= 0 and (a†F)^2 = 0
The fermionic number operator

NF=a†FaF

now satisfies

NF^2 =aF†aFa†FaF=a†F( 1 −a†FaF)aF=NF−a†F

2

a^2 F=NF

(using the fact thata^2 F=a†F

2

= 0). So one has

NF^2 −NF=NF(NF− 1 ) = 0

which implies that the eigenvalues ofNFare just 0 and 1. We’ll denote eigen-
vectors with such eigenvalues by| 0 〉and| 1 〉. The simplest representation of the
operatorsaFanda†Fon a complex vector spaceHFwill be onC^2 , and choosing
the basis

| 0 〉=

(

0

1

)

, | 1 〉=

(

1

0

)

the operators are represented as

aF=

(

0 0

1 0

)

, a†F=

(

0 1

0 0

)

, NF=

(

1 0

0 0

)

Since

H=

1

2

(a†FaF+aFa†F)

is just^12 the identity operator, to get a non-trivial quantum system, instead we
make a sign change and set

H=

1

2

(a†FaF−aFa†F) =NF−

1

2

1 =

( 1

2 0

0 −^12

)