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d3 rψ — (r) f ’∇2 • (r) = d3 r f ’∇2 ψ — (r) • (r) , (3.21)

without any intermediate steps involving Fourier transforms.
The equations (3.8), (3.11)“(3.13), (3.15), and (3.16) were all derived by using the
expansions of the ¬eld operators in cavity modes, but once again the ¬nal forms are
independent of the size and shape of the cavity. Consequently, these results are valid
in free space.

B Reciprocal space
The rather strange looking result (3.16) becomes more understandable if we note that
the decomposition (3.8) into positive- and negative-frequency parts applies equally
well in reciprocal space, so that A (k) = A(+) (k) + A(’) (k). The Fourier transforms
of eqns (3.12) and (3.16) with respect to r and r yield respectively

(±) (±)
Ai (k) , Aj (k ) = 0 (3.22)

∆⊥ (k)
(+) (’) ij
(2π)3 δ k ’ k .
Ai (k) , Aj (’k ) = (3.23)
2 0c k
This reciprocal-space commutation relation does not involve any strange operators, like
’∇2 , but it is still rather complicated. Simpli¬cation can be achieved by noting
Field quantization

that the circular polarization unit vectors eks ”see Appendix B.3.2”are eigenvectors
of ∆⊥ (k) with eigenvalue unity:

∆⊥ (k) (eks )j = (eks )i . (3.24)

By forming the inner product of both sides of eqns (3.22) and (3.23) with e— and
ek s and remembering F† (k) = F (’k), one ¬nds

[as (k) , as (k )] = a† (k) , a† (k ) = 0 (3.25)
s s

as (k) , a† (k ) = δss (2π) δ k ’ k ,

2 0 ωk
e— · A(+) (k)
as (k) = (3.27)

and ωk = ck. The operators as (k), combined with the Fourier transform relation (3.5),
provide a replacement for the cavity-mode expansions (3.9) and (3.10):

d3 k
as (k) eks eik·r ,
A (r) = (3.28)
3 2 0 ωk
(2π) s

d3 k ωk
E(+) (r) = i as (k) eks eik·r . (3.29)
3 20
(2π) s

The number operator

d3 k
a† (k) as (k)
N= (3.30)
(2π) s

N, a† (k) = a† (k) , [N, as (k)] = ’as (k) , (3.31)
s s

and the vacuum state is de¬ned by as (k) |0 = 0, so it seems that a† (k) and as (k) can
be regarded as creation and annihilation operators that replace the cavity operators
aκ and a† . However, the singular commutation relation (3.26) exacts a price. For
example, the one-photon state |1ks = a† (k) |0 is an improper state vector satisfying
the continuum normalization conditions
1k s |1ks = δss (2π) δ k ’ k . (3.32)

Thus a properly normalized one-photon state is a wave packet state

d3 k
¦s (k) a† (k) |0 ,
|¦ = (3.33)
(2π) s

where the c-number function ¦s (k) is normalized by
Field quantization in the vacuum

d3 k 2
|¦s (k)| = 1 . (3.34)
(2π) s

The Fock space HF consists of all linear combinations of number states,

d3 k1 d3 kn
¦s1 ···sn (k1 , . . . , kn ) a†1 (k1 ) · · · a†n (kn ) |0 ,
|¦ = ··· ··· s s
3 3
(2π) (2π) s1 sn
d3 k1 d3 kn
|¦s1 ···sn (k1 , . . . , kn )|2 < ∞
··· ··· (3.36)
3 3
(2π) (2π) s1 sn

and n = 0, 1, . . ..

3.1.3 Energy, momentum, and angular momentum
A The Hamiltonian
The expression (2.105) for the ¬eld energy in a cavity can be converted to a form
suitable for generalization to free space by ¬rst inverting eqn (3.10) to get

aκ = ’i d3 rE κ (r) · E(+) (r) . (3.37)
ωκ V

The next step is to substitute this expression for aκ into eqn (2.105) and carry out the
sum over κ by means of the completeness relation (2.38); this calculation leads to

ω κ a† aκ
Hem = κ

(r) ∆⊥ (r ’ r ) Ej
(’) (+)
d3 r d3 r Ei
=2 (r ) . (3.38)
0 ij

Since the free-¬eld operator E(+) (r ) is transverse, the in¬nite volume limit is

d3 rE(’) (r) · E(+) (r) .
Hem = 2 (3.39)

This can also be expressed as

d3 rA(’) (r) · ’∇2 A(+) (r) ,
Hem = 2 0 c2 (3.40)

by using eqn (3.15). A more intuitively appealing form is obtained by using the plane-
wave expansion (3.29) for E(±) to get

d3 k
a† (k) as (k) .
Hem = ωk (3.41)
(2π) s
Field quantization

B The linear momentum
The cavity model does not provide any expressions for the linear momentum and
the angular momentum, so we need independent arguments for them. The reason for
the absence of these operators is the presence of the cavity walls. From a mechanical
point of view, the linear momentum and the angular momentum of the ¬eld are not
conserved because of the immovable cavity. Alternatively, we note that one of the
fundamental features of quantum theory is the identi¬cation of the linear momentum
and the angular momentum operators with the generators for spatial translations and
rotations respectively (Bransden and Joachain, 1989, Secs 5.9 and 6.2). This means
that the mechanical conservation laws for linear and angular momentum are equivalent
to invariance under spatial translations and rotations respectively. The location and
orientation of the cavity in space spoils both invariances.
Since the cavity model fails to provide any guidance, we once again call on the
correspondence principle by quoting the classical expression for the linear momentum
(Jackson, 1999, Sec. 6.7):

P= 0E ⊥ —B
d3 r

0E — (∇ — A) .
d3 r
= (3.42)

The vector identity F— (∇ — G) = Fj ∇Gj ’ (F · ∇) G combined with an integration
by parts and the transverse nature of E (r) provides the more useful expression

P= d3 rEj (r) ∇Aj (r) . (3.43)

The initial step in constructing the corresponding Schr¨dinger-picture operator is to
replace the classical ¬elds according to

A (r) ’ A (r) = A(+) (r) + A(’) (r) , (3.44)
E (r) ’ E (r) = E(+) (r) + E(’) (r) . (3.45)

The momentum operator P is then the sum of four terms, P = P(+,+) + P(’,’) +
P(’,+) + P(+,’) , where

(σ) („ )
d3 rEj ∇Aj for σ, „ = ± .
P(σ,„ ) = (3.46)

Each of these terms is evaluated by using the plane-wave expansions (3.28) and (3.29),
together with the orthogonality relation, e— ·eks = δss , and the re¬‚ection property”
see eqn (B.73)”e’k,s = e— , (s = ±) for the circular polarization basis. The ¬rst result
is P(+,+) = P(’,’)† = 0; consequently, only the cross terms survive to give

d3 k k
a† (k) as (k) + as (k) a† (k) .
P= (3.47)
s s
3 2
(2π) s
Field quantization in the vacuum

This is analogous to the symmetrical ordering (2.106) for the Hamiltonian in the cavity
problem, so our previous experience suggests replacing the symmetrical ordering by
normal ordering, i.e.
d3 k
a† (k) as (k) .
P= k (3.48)
(2π) s

From this expression and eqn (3.41), it is easy to see that [P, Hem ] = 0 and [Pi , Pj ] = 0.
Any observable commuting with the Hamiltonian is called a constant of the motion,
so the total momentum is a constant of the motion and the individual components Pi
are simultaneously measurable.
By using the inverse Fourier transform,

2 0 ωk
e— · d3 re’ik·r A(+) (r) ,
as (k) = (3.49)

which is the free-space replacement for eqn (3.37), or proceeding directly from eqn
(3.46), one ¬nds the equivalent position-space representation

(’) (+)
(r) ∇Aj
d3 rEj
P=2 (r) . (3.50)

C The angular momentum
Finally we turn to the classical expression for the angular momentum (Jackson, 1999,
Sec. 12.10):

J= d3 r r — [ 0 E (r, t) — B (r, t)] . (3.51)

Combining B = ∇ — A with the identity F — (∇ — G) = Fj ∇Gj ’ (F · ∇) G allows
this to be written in the form J = L + S, where

L= d3 r Ej (r — ∇) Aj (3.52)

S= d3 r E — A . (3.53)

Once again, the initial guess for the corresponding quantum operators is given by
applying the rules (3.44) and (3.45), so the total angular momentum operator is

J = L + S, (3.54)

where the operators L and S are de¬ned by quantizing the classical expressions L and
S respectively.
Field quantization

The application of the method used for the linear momentum to eqn (3.52) is com-
plicated by the explicit r-term, but after some e¬ort one ¬nds the rather cumbersome
expression (Simmons and Guttmann, 1970)

d3 k
i ‚
Mi† (k) k —
L=’ Mi (k) ’ HC
2 ‚k
d3 k ‚

= ’i (k) k —
3 Mi Mi (k) , (3.55)
M (k) = as (k) eks . (3.56)

In this case a substantial simpli¬cation results from translating the reciprocal-space
representation back into position space to get
2i 0 (’) (+)
(r) r — ∇ Aj
d3 rEj
L= (r) . (3.57)
A straightforward calculation using eqn (3.39) shows that L is also a constant of the
motion, i.e. [L, Hem ] = 0. However, the components of L are not mutually commuta-
tive, so they cannot be measured simultaneously.
The quantization of eqn (3.53) goes much more smoothly, and leads to the normal-
ordered expression
d3 k
sa† (k) as (k)
S= 3k s
(2π) s
a† (k) a+ (k) ’ a† (k) a’ (k) ,
= 3k (3.58)


where k = k/k is the unit vector along k. Another use of eqn (3.49) yields the equiv-
alent position-space form

d3 rE(’) — A(+) .
S=2 (3.59)

The expression (3.54) for the total angular momentum operator looks like the
decomposition into orbital and spin parts familiar from quantum mechanics, but this
resemblance is misleading. For the electromagnetic ¬eld, the interpretation of eqn
(3.54) poses a subtle problem which we will take up in Section 3.4.

D The helicity operator
It is easy to show that S commutes with P and with Hem , and further that

[Si , Sj ] = 0 . (3.60)

Thus S, P, and Hem are simultaneously measurable, and there are simultaneous eigen-
vectors for them. In the simplest case of the improper one-photon state |1ks =
Field quantization in the vacuum

a† (k) |0 , one ¬nds: Hem |1ks = ωk |1ks , P |1ks = k |1ks , k — S |1ks = 0, and
k · S |1ks = s |1ks . Thus |1ks is an eigenvector of the longitudinal component k · S
with eigenvalue s and an eigenvector of the transverse components k — S with eigen-
value 0. For the circular polarization basis, the index s represents the helicity, so S is
called the helicity operator.

E Evidence for helicity and orbital angular momentum
Despite the conceptual di¬culties mentioned in Section 3.1.3-C, it is possible to devise
experiments in which certain components of the helicity S and the orbital angular mo-
mentum L are separately observed. The ¬rst measurement of this kind (Beth, 1936)
was carried out using an experimental arrangement consisting of a horizontal wave
plate suspended at its center by a torsion ¬ber, so that the plate is free to undergo
twisting motions around the vertical axis. In a simpli¬ed version of this experiment,
a vertically-directed, linearly-polarized beam of light is allowed to pass through a
quarter-wave plate, which transforms it into a circularly-polarized beam of light (Born
and Wolf, 1980, Sec. 14.4.2). Since the experimental setup is symmetrical under ro-
tations around the vertical axis (the z-axis), the z-component of the total angular
momentum will be conserved.
We will use a one-photon state

ξs a† (k) |0 ,
|ψ = ξs |1ks = (3.61)
s s

with k = ku3 directed along the z-axis, as a simple model of an incident light beam
of arbitrary polarization. A straightforward calculation using eqn (3.55) for Lz shows
that Lz |1ks = 0; consequently, Lz |ψ = 0 for any choice of the coe¬cients ξs . In
other words, states of this kind have no z-component of orbital angular momentum.
The particular choice
1 1
= √ [|1k+ + |1k’ ] = √ a† (k) + a† (k) |0
|ψ (3.62)

2 2

de¬nes a linearly-polarized state which possesses zero helicity, i.e. Sz |ψ lin = 0. Due to
the action of the quarter-wave plate, the incident linearly-polarized light is converted
into circularly-polarized light. Thus the input state |ψ lin changes into the output state
|ψ cir = |1k,s=+ . The output state |ψ cir has helicity Sz = + , but it still satis¬es
Lz |ψ cir = 0. Since the transmitted photon carries away one unit (+ ) of angular
momentum, conservation of angular momentum requires the plate to acquire one unit

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