From Classical Mechanics to Quantum Field Theory

34 From Classical Mechanics to Quantum Field Theory. A Tutorial

(ii)a†acts as multiplication byz∗:

a†:f(z∗)=〈z|f〉→〈z|a†|f〉=z∗〈z|f〉=z∗f(z∗). (1.157)

We leave to the reader the proof that the operators∂z∂∗,z∗are one the adjoint of
the other with respect to the measure (1.154) and that they satisfy:

[ ∂

∂z∗,z

∗]=I.

We can now represent inHBFany operatorAby means of an integral repre-
sentation:

(Aψ)(z∗)=

∫

dμ(z′)A(z∗,z′)ψ(z′∗), (1.158)

with a kernelA(z∗,z′) given by:

A(z∗,z′)=

∑∞

m,k=0

Amk

(z′)k

√

k!

(z∗)m

√

m!

=〈z|A|z′〉,Amn≡〈m|A|n〉. (1.159)

To prove these relations, we notice that, for any|ψ〉=

∑∞

n=0cn|n〉∈H:

ψ(z∗)=〈z|ψ〉=

∑∞

n=0

cn〈z|n〉=

∑∞

n=0

cn

(√z∗)n

n!

and

(Af)(z∗)=〈z|A|f〉=

∑∞

n=0

cn〈z|A|n〉=

∑∞

n,m=0

cn〈z|m〉〈m|A|n〉

=

∑∞

n,m=0

cnAmn

(z∗)m

√

m!

.

We can now use the identity

δkn=

∫

dμ(z′)

(z′)k

√

k!

(z′∗)n

√

n!

,

to write:

(Af)(z∗)=

∑∞

n,m,k=0

cnAmkδkn

(z∗)m

√

m!

=

∫

dμ(z′)

∑∞

n,m,k=0

cnAmk

(z′)k

√

k!

(z′∗)n

√

n!

(z∗)m

√

m!

=

∫

dμ(z′)

⎡

⎣

∑∞

m,k=0

Amk

(z′)k

√

k!

(z∗)m

√

m!

⎤

⎦

(∞

∑

n=0

cn

(z′∗)n

√

n!

)

Using such integral representation, it is very simple to calculate the trace of
an operator, since we can write:

TrH[A]=

∑∞

n=0

〈n|A|n〉=

∫

dμ(z)

∑∞

n=0

〈n|A|z〉〈z|n〉=

∫

dμ(z)

∑∞

n=0

〈z|n〉〈n|A|z〉

=

∫

dμ(z)e−|z|

2

〈z|

(∞

∑

n=0

|n〉〈n|

)

A|z〉=

∫

dμ(z)〈z|A|z〉.