P = Pr + iPi

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(5.31) The reason for the v'2 factor will become evident soon. For each of the two pairs ofreal variables (qr,Pr) and (qi,Pi) in (5.31), we have a pair of driven harmonic oscillator equations qr= - , and where (5.34) The equations of motion (5.32) and (5.33) can be obtained from a Hamiltonian such as (5.24). As they are all independent variables, the four equations of motion can be

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(5.32)

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(5.33)

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obtained from a Hamiltonian given by the sum of these two separate Hamiltonians; that is, 11. - 2m

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. 81i Pr = - 8qr'

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(5.35)

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with the equations of motion given by 811. . qr = -8 ' Pr and (5.36)

. 81i Pi = - -8 . qi In terms of the complex variables Q and p, (5.35) is given simply by

. qi

= -8' Pi

(5.37)

11. =

P*P w2 2m +m Q*Q+ Q*F+c.c.,

(5.38)

where c.c. stands for complex conjugate, showing that 11. is real as a Hamiltonian should be. Now, from (5:31), we find that 8 Pr and 1 8 1 8

= V2 8P + V2 8P* '

8 8Pi 8 8qi

V2 8P

V2 8P*

(5.39)

81818 qr = V2 8Q + V2 8Q* '

V2 8Q - V2 8Q* .

181i 1 811. -28Q - 28Q*

1 811. 1 811.

(5.40)

Substituting (5.39) and (5.40) in (5.36) and (5.37), we obtain

Pr V2 =

(5.41)

. qi

V2 =

28P + 28P* '

1 811.

1 811.

V2 = 28Q

28Q* .

(5.42)

Differentiating (5.29) and (5.30) with respect to time and substituting in (5.41) and (5.42), we find the complex version of Hamilton equations of motion: 811. . Q = 8P*'

= - 8Q*'

(5.43)

From the Poisson brackets of the real canonical variables, we can work out those of the complex ones:

[Q, P*]PB =

~[qr + i%Pr + ipi]PB

2~ ([qr,Pr}PB -i [qr.Pi}PB +i [%Pr}PB + [qi,Pi}PB)

---- ------- ------- ------

(5.44)

A SINGLE POINT DIPOLE

Analogously,

[Q,PjPB =0,

[Q, Q*jpB = 0,

[P, P*]PB = O.

(5.45)

Now comparing (5.27) and (5.28) with (5.22) and (5.23), we find that we can make the following: correspondences

ck, m= 1,

(5.46) (5.47) (5.48) (5.49) (5.50)

Qj{k) = - ck Vj,

Pj{k)

Fj{k) = 27r 2 ckp Ej{k),

where.N is an arbitrary constant, and

=.N Bj , .N

(5.51) as these are continuous rather than discrete variables. So this interesting analogy tells us that each pair of field variables Vj{k) and Bj{k) corresponds to a Hamiltonian of the sort (5.38) by way of (5.46)-(5.50). Now, provided that we restrict k to a half space, pairs corresponding to different values of j and k are independent. Thus, the Hamiltonian for our single electric dipole interacting with the radiation field is just the sum (integral) of each of these individual Hamiltonians for each pair Vj{k) and Bj{k):

= .N2 2

2 d3 k " " ' { Bj{k) ~

2 + Vj{k) -

1 7r2P Ej{k) Vj{k)

+ c.c. } + H atm ,

(5.52) where we have added the Hamiltonian H atm describing the free evolution of the atom (dipole) and the prime on the integral sign is to remind us that this integral is over half of the space rather than over the entire space, as usual. Choosing the arbitrary constant so that.N2 = 27r2 , we make the Hamiltonian coincide with the total energy; that is, the Hamiltonian in real space is given by (5.53) Up to now everything is classical. We have shown that -7rv'2vj / ck and 7rv'2Bj are canonical variables and we have found their Hamiltonian. Then to quantize the radiation field, we can just apply the ordinary canonical quantization procedure as in 4. We take the commutator of the quantized variables as their classical Poisson bracket times in. Then, defining (5.54)

LET MATTER BEl

so that (5.55) and

d k ck

~ {S}{k) . Sj{k) + V}{k) . Vj{k) + H.c.}

d k ck

~ {a}{k)aj{k) + aj{k)a}{k) }

(5.56)

as for a continuum of harmonic oscillators, we find that

D{r) = -i

J~~~ ~ j{k) {aj{k) J

a}{ -k)} e

ik r .

(5.57)

B{r) = -i

J J~~~ ~ ~

1\ j{k) {aj{k)

+ a}{ -k)} e ik .r .

(5.58)

We have quantized the electromagnetic field, but what about matter Here matter is represented by a single atom that is approximated by a point electric dipole p. How can we quantize this dipole What is its free evolution3 Hamiltonian iIatm To address these questions, we must first understand how the dipole approximation arises. Suppose for simplicity that the atom is hydrogen.4 Assume that it is at rest. As the nucleus is much heavier than the electron, we can think of the nucleus as fixed at the origin. The electron, on the other hand, will move about and after quantization its position will be described by the position operator f. Then the charge density associated with this atom is given by

Patm{r)

e oCr) - e oCr -

(5.59)

where -e is the charge of the electron. Now if r = Irl is much largeii than the atomic size, we can approximate oCr - f) by

oCr -

f) =

(211')3

eirk.([r/rJ-[r/rJ)

3That is, uncoupled to the field. 4This is not a very restrictive assumption. Rydberg atoms used in many cavity QED experiments can be very well approximated by a scaled model of hydrogen. In a Rydberg atom one of the electrons is in a very high energy level. As the lower-energy electrons are much closer to the nucleus, they partially shield the nucleus charge so that the outer electron experiences an effective Coulomb potential similar to that in a hydrogen atom. sThis is the case when we look at the atom's charge distribution from a macroscopic distance, then we cannot see all the details of its charge distribution (i.e., the several multipole terms in its multipole expansion). But more important here, this is also the case when the wavelength .>. of the electromagnetic radiation interacting with the atom is much larger than the size of the atom (e.g., the Bohr radius for a hydrogen atom). For we can easily see, then, using a Fourier argument, that only r "" ).. will contribute significantly to the interaction Hamiltonian with the radiation field.