130_notes.dvi

The total Hamiltonian we are aiming at, is the integral of the Hamiltonian density.

H=

∫

d^3 xH

When we integrate over the volume only products likeeikρxρe−ikρxρwill give a nonzero result. So
when we multiply one sum overkby another, only the terms with the samekwill contribute to the
integral, basically because the waves with different wave number areorthogonal.

1

V

∫

d^3 x eikρxρe−ik

′ρxρ

=δkk′

H =

∫

d^3 xH

H = −

∂Aμ

∂x 4

∂Aμ

∂x 4

∂Aμ

∂x 4

= −

1

√

V

∑

k

∑^2

α=1

ǫ(μα)

(

ck,α(0)

ω

c

eikρxρ−c∗k,α(0)

ω

c

e−ikρxρ

)

H = −

∫

d^3 x

∂Aμ

∂x 4

∂Aμ

∂x 4

H = −

∫

d^3 x

1

V

∑

k

∑^2

α=1

(

ck,α(0)

ω

c

eikρxρ−c∗k,α(0)

ω

c

e−ikρxρ

) 2

H = −

∑

k

∑^2

α=1

(ω

c

) (^2) [
−ck,α(t)c∗k,α(t)−c∗k,α(t)ck,α(t)

]

H =

∑

k

∑^2

α=1

(ω

c

) 2 [

ck,α(t)c∗k,α(t) +c∗k,α(t)ck,α(t)

]

H =

∑

k,α

(ω

c

) (^2) [
ck,α(t)c∗k,α(t) +c∗k,α(t)ck,α(t)

]

This is theresult we will use to quantize the field. We have been carefulnot to commutec
andc∗here in anticipation of the fact that they do not commute.

It should not be a surprise that the terms that made up the Lagrangian gave a zero contribution
becauseL=^12 (E^2 −B^2 ) and we know that E and B have the same magnitude in a radiation field.

(There is one wrinkle we have glossed over; terms with~k′=−~k.)

33.4 Canonical Coordinates and Momenta

We now have theHamiltonian for the radiation field.

H=

∑

k,α

(ω

c

) (^2) [
ck,α(t)c∗k,α(t) +c∗k,α(t)ck,α(t)

]