130_notes.dvi

33.3 The Hamiltonian for the Radiation Field

We now wish tocompute the Hamiltonian in terms of the coefficientsck,α(t). This is an
important calculation because we willuse the Hamiltonian formalism to do the quantization
of the field. We will do the calculation using the covariant notation (while Sakuraioutlines an
alternate calculation using 3-vectors). We have already calculatedthe Hamiltonian density for
a classical EM field.

H=Fμ 4

∂Aμ

∂x 4

+

1

4

FμνFμν

H =

(

∂A 4

∂xμ

−

∂Aμ

∂x 4

)

∂Aμ

∂x 4

+

1

4

(

∂Aν

∂xμ

−

∂Aμ

∂xν

)(

∂Aν

∂xμ

−

∂Aμ

∂xν

)

H = −

∂Aμ

∂x 4

∂Aμ

∂x 4

+

1

2

(

∂Aν

∂xμ

∂Aν

∂xμ

−

∂Aν

∂xμ

∂Aμ

∂xν

)

Now letscompute the basic element of the above formulafor our decomposed radiation field.

Aμ =

1

√

V

∑

k

∑^2

α=1

ǫ(μα)

(

ck,α(0)eikρxρ+c∗k,α(0)e−ikρxρ

)

∂Aμ

∂xν

=

1

√

V

∑

k

∑^2

α=1

ǫ(μα)

(

ck,α(0)(ikν)eikρxρ+c∗k,α(0)(−ikν)e−ikρxρ

)

∂Aμ

∂xν

= i

1

√

V

∑

k

∑^2

α=1

ǫ(μα)kν

(

ck,α(0)eikρxρ−c∗k,α(0)e−ikρxρ

)

∂Aμ

∂x 4

= −

1

√

V

∑

k

∑^2

α=1

ǫ(μα)

ω

c

(

ck,α(0)eikρxρ−c∗k,α(0)e−ikρxρ

)

We have all the elements to finish the calculation of the Hamiltonian. Before pulling this all together
in a brute force way, its good to realize thatalmost all the terms will give zero. We see that

the derivative ofAμis proportional to a 4-vector, saykνand to a polarization vector, sayǫ(μα). The
dot products of the 4-vectors, eitherkwith itself orkwithǫare zero. Going back to our expression
for the Hamiltonian density, we can eliminate some terms.

H = −

∂Aμ

∂x 4

∂Aμ

∂x 4

+

1

2

(

∂Aν

∂xμ

∂Aν

∂xμ

−

∂Aν

∂xμ

∂Aμ

∂xν

)

H = −

∂Aμ

∂x 4

∂Aμ

∂x 4

+

1

2

(0−0)

H = −

∂Aμ

∂x 4

∂Aμ

∂x 4

The remaining term has a dot product between polarization vectorswhich will be nonzero if the
polarization vectors are the same. (Note that this simplification is possible because we have assumed
no sources in the region.)