Download Optical Faraday Rotation Abstract

Optical Faraday Rotation
P. R. Berman
Michigan Center for Theoretical Physics and Physics Department,
University of Michigan, Ann Arbor, Michigan
48109-1040
(Date textdate; Received textdate; Revised textdate; Accepted textdate; Published textdate)
Abstract
Three calculations of optical Faraday rotation are presented. In all cases, a linearly polarized
…eld is incident on a medium of harmonic oscillators in the presence of a longitudinal magnetic …eld.
The rotation of the plane of polarization of the …eld is evaluated using (1) classical oscillators and
the Lorentz force equation, (2) quantum oscillators and the Heisenberg equations of motion, and
(3) quantum oscillators and a Schrödinger equation approach. It is shown that a simple argument,
based on the assumption that a circularly polarized …eld drives either
m = 1 or
m =
1
transitions on absorption (m is the magnetic quantum number), leads to an incorrect result for the
Verdet constant.
1
I.
INTRODUCTION
Faraday optical rotation refers to the rotation of the of the plane of polarization of an
optical …eld as it propagates in a dielectric medium in which a static magnetic …eld is
applied along the direction of propagation of the …eld.1 In his Optics, Sommerfeld, gives a
simple derivation of the Faraday rotation, modeling the atoms in the medium as classical
oscillators.2
Sommerfeld explains the e¤ect as arising from a di¤erence in the index of refraction
for the left (n+ ) and right (n ) circularly polarized components of the …eld.2 He uses the
Lorentz force equation to calculate the response of the oscillators to the applied …elds and
then obtains the polarization of the medium. From the polarization, he extracts the indices
of refraction associated with the left and right circularly components of the optical …eld and
…nds that the di¤erence in indices is given approximately by
n
n+ =
B0 !
N e3
;
2
nmo 0 (! 20 ! 2 )2
where N is the oscillator density, ! 0 is the natural frequency,
(1)
e the charge, and mo the
mass of an oscillator, B0 is the magnitude of the external magnetic …eld, ! is the frequency
of the optical …eld,
0
is the vacuum permittivity, and
N e2 =mo 0
1+ 2
!0 !2
n=
1=2
(2)
is the index of refraction of the medium of oscillators in the absence of the magnetic …eld.
For a sample having length L, the di¤erence in indices leads to a rotation angle
=
!
c
n
n+
2
L=
N e3 B0 L
!2
2nm2o c 0 (! 20 ! 2 )2
(3b)
= V LB0 = 0 ;
where
0
(3a)
is the vacuum permeability and
! n
n+
2c
B0 = 0
e 0 dn
=
!
2mo c d!
V =
=
N e3 0
!2
2nm2o c 0 (! 20 ! 2 )2
(4a)
(4b)
is known as the Verdet constant. It is seen that the Verdet constant is proportional to the
dispersion of the medium.
2
FIG. 1: Energy level scheme of an oscillator. The ground state has n = 0 and L = 0; while the
…rst excited state has n = 1 and L = 1. A longitudinal magnetic …eld splits the magnetic sublevels
of the L = 1 manifold.
Although the Faraday e¤ect is mentioned in many optics textbooks and studied experimentally in advanced undergraduate laboratories,3 discussions of the Faraday e¤ect in textbooks and laboratory manuals tend to be qualitative in nature. Often, there is no detailed
derivation of the Faraday rotation angle. An exception is the book of Rossi,4 which contains
a derivation that is essentially the same as Sommerfeld’s. As far as I can tell, quantummechanical derivations of Faraday rotation are not readily available in standard optics texts.
Of course there is a vast literature on the Faraday e¤ect, but many of these articles are rather
specialized, including applications involving speci…c molecules,5 nonlinear e¤ects,6 the role
of optical pumping,7 etc.. It might be useful, therefore, to present a quantum calculation for
a medium composed of harmonic oscillators, when a weak, linearly polarized …eld propagates
in the medium in the presence of a longitudinal magnetic …eld. The results should mirror
that of the classical calculation; however, paradoxically, the predictions based on a simple
quantum model seem to be at odds with the classical results.
The argument for the quantum calculation can be formulated for the level scheme of Fig.
1. Shown are the two lowest lying electronic state manifolds of a three-dimensional harmonic
oscillator having natural frequency ! 0 . The ground state has electronic quantum number
n = 0 and angular momentum L = 0, while the …rst excited state has n = 1 and L = 1:
If the incident …eld is x-polarized, one can view the …eld as an equal superposition of left
and right circularly polarized …elds or, equivalently, as an equal superposition or
3
+
and
radiation, where the
radiation induces
1 transitions on absorption and m is the
m=
magnetic quantum number. In this limit,
1=2
N e2 =mo 0
1+ 2
!2
!
n =
(5)
;
where
! = !0
are the frequencies of the
m=
eB0
2mo
(6)
1 transitions from the ground to excited state, including
the Zeeman shift of the levels. From Eqs. (5) and (6), it follows that
n
n+
! 0 B0
N e3
;
nm2o 0 (! 20 ! 2 )2
which di¤ers signi…cantly from Eq. (3a) if ! 0
(7)
!. Clearly something is wrong with this
approach.
It is the purpose of this article to explain the breakdown in the reasoning that led to Eq.
(7) and to provide a classical and two (relatively) simple quantum calculations of Faraday
rotation. I begin by reviewing the classical calculation based on the Lorentz force equation
(essentially equivalent to that given by Sommerfeld) and then present the analogous quantum
calculation using a Heisenberg operator approach. Finally, I use a Schrödinger picture
approach that helps to isolate the role of the
m =
1 transitions. Such an approach
allows one to uncover the fallacy in the reasoning that led to Eq. (7). Finally, I point out
an interesting feature related to the choice of the atom-…eld interaction Hamiltonian.
II.
CLASSICAL CALCULATION
The medium is assumed to be composed of a uniform density of harmonic oscillators,
each having charge
e. The force on an oscillator is given by
F=
e (E + v
B)
mo ! 20 r;
(8)
where E is the electric …eld vector in the medium,
E (Z; t) =
1
[Ex (Z)^
x + Ey (Z)^
y] exp [i (nkZ
2
!t)] + c.c.,
(9)
B is the magnetic induction,
B = B0^
z;
4
(10)
! = kc, n is the index of refraction of the medium given by Eq. (2), r is the displacement
of an oscillator from its equilibrium position produced by the applied …elds, and v = r_ . The
…eld incident on the medium is polarized in the x
^ direction, but acquires a y component as
it propagates through the medium. The displacement r(t) can be written as
r(t) = x+ (t)^
x + y+ (t)^
y + c.c.,
(11)
where x+ (t) and y+ (t) are positive frequency components of the displacement, varying as
exp ( i!t). The …eld amplitudes Ex (Z) and Ey (Z) are assumed to vary slowly in an optical
wavelength , and it is assumed that the dipole approximation, r
; allowing one to
evaluate both the amplitude and phase of the optical …eld at the equilibrium position of the
oscillator.
For these …elds, the classical equations of motion for x+ and y+ of an oscillator that is
located at position Z and subjected to the Lorentz force (8) are
x•+ + 2 x_ + + ! 20 x+ =
y•+ + 2 y_ + + ! 20 y+ =
where
eEx (Z) exp [i (nkZ
2mo
eEy (Z) exp [i (nkZ
2mo
eB0 y_ +
mo
!t)] eB0 x_ +
+
;
mo
!t)]
(12a)
(12b)
is a radiative decay rate. These equations are identical to those obtained by
Sommerfeld,2 except that I include radiative decay, which allows for a steady-state solution.
It is assumed that the magnitude of the electric …eld changes negligibly as a
function of Z in the medium, owing to the fact that the …eld is detuned by an amount
= !0
!
; eB0 =mo . Although the …eld magnitude remains constant, the plane of
polarization rotates as the …eld propagates in the medium.
Sommerfeld solves Eqs. (12) by introducing the circular components x+
iy+ ; however
to obtain the rate of change of the Faraday rotation angle, d =dZ, it is su¢ cient to calculate
the rotation angle d in a distance dZ for which the rotation angle is much less than unity.8
In this limit, one can take Ex (Z)
Ex (0)
E0 in the interval dZ; moreover, one can neglect
the in‡uence of the y component of the …eld on the dynamics of the x coordinate and take
as the equations of motion
x•+ + 2 x_ + + ! 20 x+ =
y•+ + 2 y_ + + ! 20 y+ =
eE0 ei(nkZ !t)
2mo
eEy (Z)ei(nkZ
2mo
5
(13a)
!t)
+
eB0 x_ +
:
mo
(13b)
The steady state solution of these equations (that is, the solution after all transients have
died away) is
x+ (Z; t) =
y+ (Z; t) =
where the limit
1
eE0 ei(nkZ !t)
;
2
2mo
!0 !2
1
e2 B0 E0 ei(nkZ
eEy (Z)ei(nkZ !t)
+
2mo
! 20 ! 2
2m2o
(14a)
!t)
(i!)
(! 20
! 0 has been taken, under the assumption that ; ! 0 ; !
2;
!2)
(14b)
.
To calculate the Faraday rotation, one must relate the …elds to the polarization in the
medium. The positive frequency component of the polarization is given by
P+ (Z; t) =
N er+ (t) =
N e [x+ (Z; t)^
x + y+ (Z; t)^
y]
(15)
and the wave equation in the medium can be written as
@ 2 E+
@Z 2
where I have set B =
0H
1 @ 2 E+
1 @ 2 P+
=
;
2
c2 @t2
@t2
0c
and D =
0E
(16)
+ P to arrive at this equation. Using Eqs. (16),
(15) and (14), one …nds that the wave equation for the x component of the …eld (9) leads
to an equation,
N e2 E0
1
;
(17)
2
!2
0 mo ! 0
which reproduces Eq. (2) for the index of refraction in the absence of the applied magnetic
n2
1 E0 =
…eld. Using a similar approach for the y component of the …eld and canceling terms using
Eq. (17), one …nds
iN e3 k 2 B0 E0
!
(18)
2;
2
2
mo 0
(! 0 ! 2 )
where the assumption that Ey (Z) varies slowly in an optical wavelength was used [i.e. the
2ink
@Ey (Z)
@Z
@ 2 Ey (Z)=@Z 2 term was neglected]. With boundary condition Ey (0) = 0; the solution of this
equation for Ey (dZ) is
N e3 kB0 E0
!
dZ;
(19)
2nm2o 0 (! 20 ! 2 )2
which implies a di¤erential change in the rotation angle for the …eld polarization given by
Ey (dZ)
d
!2
Ey (dZ)
N e3 B0
=
=
:
dZ
E0 dZ
2nm2o c 0 (! 20 ! 2 )2
(20)
For a medium having length L; one …nds a rotation angle
=
N e3 B0 L
!2
;
2nm2o c 0 (! 20 ! 2 )2
in agreement with Eq. (3a).
6
(21)
III.
HEISENBERG EQUATIONS OF MOTION
The Hamiltonian for an oscillator having charge
is
H=
e in the presence of the applied …elds
[p + eA (r;Z; t)]2 1
+ mo ! 20 r2 ;
2mo
2
(22)
where the vector potential,
(23a)
A (Z; t) = AE (Z; t) + AB (r) ;
1
[Ex (Z)^
x + Ey (Z)^
y] exp [i (nkZ
2i!
B0
[y^
x x^
y] ,
2
AE (Z; t) =
AB (r) =
!t)] + c.c. ;
(23b)
(23c)
is written in Coulomb gauge. Interactions with the vacuum …eld have been neglected, so
the equations to be derived should be compared with Eqs. (12) in the limit that
! 0.
The Heisenberg equations of motion for the positive frequency components of the oscillator,
obtained using
p_ =
1
[p; H] ;
ih
r_ =
1
[r; H] ;
ih
(24)
are
eB0
e2 B02
e2 B0 Ey (Z) exp [i (nkZ
py+
x+
2mo
4mo
4imo !
2 2
2
eB0
e B0
e B0 Ey (Z) exp [i (nkZ
p_y+ = mo ! 20 y+ +
px+
y+ +
2mo
4mo
4imo !
px+ eEx (Z) exp [i (nkZ !t)] eB0
x_ + =
+
y+ ;
mo
2imo !
2mo
py+ eEy (Z) exp [i (nkZ !t)] eB0
y_ + =
+
+
x+ ;
mo
2imo !
2mo
p_x+ =
mo ! 20 x+
!t)]
!t)]
;
(25a)
;
(25b)
(25c)
(25d)
from which it follows that
eEx (Z) exp [i (nkZ !t)] eB0
y_ +
2mo
2mo
eEx (Z) exp [i (nkZ !t)] eB0
y_ +
= ! 20 x+
2mo
mo
p_y+ eEy (Z) exp [i (nkZ !t)] eB0
y•+ =
+
x_ +
mo
2mo
2mo
eEy (Z) exp [i (nkZ !t)] eB0 x_ +
= ! 20 y+
+
:
2mo
mo
x•+ =
p_x+
mo
7
(26a)
(26b)
These equations agree with Eqs. (12) in the limit
! 0. If I had used Hamilton’s equations
for the classical calculation instead of the Lorentz force equation, the classical and quantum
calculations would have been identical.
The calculation in this section illustrates the power of using the Heisenberg equations
of motion. One obtains the equations of motion for the relevant operators without having
to resort to a solution of Schrödinger’s equation. Of course, life was simple here, owing to
the fact that I considered the quantum system to be an (intrinsically linear) oscillator. Had
I tried to calculate the response of an atom to the applied …elds, I would not be able to
write closed form di¤erential equations of the type (26), since the atom acts as a non-linear
device.9
IV.
SCHRÖDINGER PICTURE CALCULATION
The two lowest lying electronic state manifolds of the oscillator are shown in Fig. 1.
The ground state has angular momentum L = 0 and the …rst excited state L = 1: Under
the assumption that the external …elds are weak, these are the only levels that need be
considered.10 Moreover, since the external …elds are weak, one can drop any terms that vary
as E02 or B02 in the Hamiltonian (22). In this limit the approximate Hamiltonian for the
oscillator is
1
e [px Ex (Z) + py Ey (Z)]
p2
+ mo ! 20 r2 +
exp [i (nkZ !t)] + adj.
2mo 2
2imo !
eB0
e2 B0 [yEx (Z) xEy (Z)]
[px y py x]
exp [i (nkZ !t)] + adj.
2mo
2mo
2imo !
eEx (Z) (px eyB0 =2)
= H0 +
exp [i (nkZ !t)] + adj.
2imo !
eEy (Z) (py + exB0 =2)
eLz B0
+
exp [i (nkZ !t)] + adj. +
2imo !
2mo
H=
where Lz = (xpy
(27a)
(27b)
ypx ) is the z component of angular momentum and
H0 =
p2
1
+ mo ! 20 r2
2mo 2
(28)
is the Hamiltonian in the absence of any applied …elds. Note that I had to keep terms
that vary as Ex B0 or Ey B0 , originating from the 2AE AB cross term in the A2 term in the
Hamiltonian [see Eqs. (23)]. This term contributes to the Faraday rotation in lowest order
8
and its inclusion is the price one must pay for using the eA p=mo rather than the er E
interaction Hamiltonian for the optical …eld. I return to this point in the Sec. V.
In this section, I want to calculate hxi and hyi from Schrödinger’s equation. As in Sec. II,
I assume that the …eld is x polarized initially and propagates for a small distance in which
the Faraday rotation angle is small, allowing one to replace Ex (Z) by E0 and to neglect the
Ey (Z)B0 term in in Eq. (27b). As such the e¤ective Hamiltonian for the problem is
H = H0 +
e [E0 (px
e E0 (px
eyB0 =2) + py Ey (Z)]
exp [i (nkZ !t)]
2imo !
eyB0 =2) + py Ey (Z)
exp [ i (nkZ !t)] .
2imo !
(29)
Since the external …elds are weak, a perturbative approach can be used.10
In a perturbation theory limit, one can work with state amplitudes rather than density
matrix elements. The state vector for the oscillator can be written as
j (t)i = a0 (t) jL = 0i +
X
m= 1;1
am (t) jL = 1; mi ;
where a0 (t) is the ground state amplitude and am (t) (m =
(30)
1) is the excited state amplitude
for sublevel m (the m = 0 excited state sublevel is not excited since the optical electric …eld
has no z component). In terms of these state amplitudes, the expectation values hxi and
hyi for the state vector (30) are given by
hx(t)i =
hy(t)i =
X
x0m am (t)a0 (t) + c.c.;
(31a)
y0m am (t)a0 (t) + c.c.;
(31b)
m= 1;1
X
m= 1;1
where
x0m = hL = 0j x jL = 1; mi ; y0m = hL = 0j y jL = 1; mi
(32)
are matrix elements between the jL = 1; mi excited state and the ground state.
To calculate the state amplitudes, one uses the time-dependent Schrödinger equation
with the Hamiltonian (29), along with the fact that the energy levels in the absence of the
optical …eld are given by
EL=0 = h! 0 =2;
EL=1;m = 3h! 0 =2 + mehB0 =2mo :
9
(33a)
(33b)
If the oscillator is prepared in its ground state at t = 0, then, in lowest order perturbation
theory, the ground state amplitude is given by
(34)
a0 (t) = exp ( i! 0 t=2) ;
while the time evolution of the excited state amplitudes is governed by
a_ m =
i! m am
am
e
[E0 (px eyB0 =2) + py Ey (Z)]m0 exp [i (nkZ !t)] a0 (t)
2mo h!
e
+
E (px eyB0 =2) + py Ey (Z) m0 exp [ i (nkZ !t)] a0 (t);
2mo h! 0
(35)
where
!m =
EL=1;m
3! 0 emB0
=
+
h
2
2mo
(36)
and I have included radiative damping to allow for a steady-state solution.
The steady-state solution of Eq. (35), with a0 (t) given by Eq. (34), is
am (t) =
+
e [E0 (px
e E0 (px
eyB0 =2) + py Ey (Z)]m0 exp [i (nkZ !t)] i!0 t=2
e
2mo h!
+ i (! m0 !)
eyB0 =2) + py Ey (Z) m0 exp [ i (nkZ !t)] i!0 t=2
e
;
2mo h!
+ i (! m0 + !)
(37)
where
! m0 = ! m
! 0 =2 = ! 0 +
emB0
2mo
(38)
is an optical transition frequency. Keeping terms to zeroth order in B0 for hx+ i, …rst order
in B0 for hy+ i and neglecting the back action of …eld component Ey (Z) on x+ ; one can use
Eqs. (31), (37), and (34) to show that the positive frequency components [those varying as
exp ( i!t)] of the displacement are given by
hx+ (Z; t)i =
hy+ (Z; t)i =
X eE0 exp [i (nkZ
2mo h!
m= 1;1
X e exp [i (nkZ
2mo h!
m= 1;1
n
y0m py Ey (Z) + px E0
!t)]
x0m (px )m0
+ i (! m0 !)
y0m [py Ey (Z) + px E0
+ i (! m0
o 9
=
eyE0 B0 =2
!t)]
m0
;
i (! m0 + !)
10
[x0m (px )m0 ]
i (! m0 + !)
; (39a)
eyE0 B0 =2]m0
!)
(39b)
Matrix elements for the 3-D oscillator are
r
h
1
x0m =
( m;1
m; 1 ) ;
2 mo ! 0
r
i
h
y0m = ym0 =
( m;1 + m; 1 ) ;
2 mo ! 0
ip
(px )m0 =
hmo ! 0 ( m;1
m; 1 ) ;
2
1p
hmo ! 0 ( m;1 + m; 1 ) ;
(py )m0 =
2
where
i;j
(40a)
(40b)
(40c)
(40d)
is a Kronecker delta. Note that the matrix elements depend on ! 0 and not ! m0
since they are taken with respect to the unperturbed basis. Combining Eqs. (39) and (40),
one …nds
X
hx+ (Z; t)i =
m=
hy+ (Z; t)i =
eE0 ei(nkZ !t)
4mo (! 2m0 ! 2 )
1;1
eE0 ei(nkZ !t)
;
2mo (! 20 ! 2 )
X eEy (Z)ei(nkZ !t)
4mo (! 2m0 ! 2 )
m= 1;1
X ieE0 ei(nkZ !t) ! m0
(
2
2)
4m
!
(!
!
o
m0
m= 1;1
where I have taken the limit that
X
! m0
(! 2m0
m= 1;1
2
!2)
eB0
6 ! 0 + 2mo
=4
eB0
! 0 + 2m
o
(
m;1
eB0
2mo
!0
eB0
2mo
2
+
eB0
;
2mo ! 0
(41b)
! 0: Using Eq. (38), one can show that
m; 1 )
!0
m; 1 )
m;1
(41a)
2
+
+
eB0
2mo ! 0
! 10
! 210
3
eB0
7
5
mo (! 20 ! 2 )
!
!2
!
2
10
!2
10
2eB0 ! 2
mo (! 20
+
eB0
mo (! 20 ! 2 )
(42)
2;
!2)
as a consequence, Eq. (41b) reduces to
hy+ (Z; t)i
eEy (Z)
e2 B0 E0
(i!)
+
2
2
2
2
2mo (! 0 ! )
2mo (! 0 ! 2 )2
ei(nkZ
!t)
:
(43)
Equations (41a) and (43) are in agreement with Eqs. (14).
The Schrödinger approach has led to a result that is consistent with the classical result.
Moreover, it can help us to understand the breakdown of the reasoning given in Sec. I. To
understand what went wrong, consider a
+
circularly polarized …eld having electric …eld
vector,
E (Z; t; +) =
E0 x
^ + i^
y i(n+ kZ
p
e
2
2
11
!t)
+ c.c.,
(44)
with E0 real, that drives transitions between the L = 0 ground state and the L = 1 excited
state. Using the Hamiltonian (see Sec. V below)
H = H0 + er E(Z; t; +) +
with the state vector (30) and a0 (t)
eLz B0
;
2mo
(45)
exp ( i! 0 t=2), one …nds that the excited state
amplitudes evolve as
a_ m =
i! m am
i
E0 e
am
i
E0 ei(n+ kZ
p
2h 2
!t)
[x + iy]m0 exp ( i! 0 t=2)
i(n+ kZ !t)
p
2 2
[x
(46)
iy]m0 exp ( i! 0 t=2) ;
where I have included radiative decay. The needed matrix elements are given by Eqs. (40a)
and (40b), and one obtains
a_ m =
i! m am
am
ieE0
r
1
ei(n+ kZ
8hmo ! 0
!t)
m;1
e
i(n+ kZ !t)
m; 1
exp ( i! 0 t=2) :
(47)
From Eq. (47), it is clear that, for the Hamiltonian (45), the only e¤ect of the magnetic
…eld is to shift the energy of the m =
1 excited state sublevels. As a consequence, the
changes the index of refraction associated with the
+
and
components of the electric
…eld result solely from the magnetic-…eld-induced level shifts, and it seems that arguments
of the type presented in the Sec. I lead to an inconsistent value for the Verdet constant.
However, this conclusion is valid only if
circularly polarized …elds drive only
m=
1
transitions on absorption
In the resonance or rotating-wave approximation (RWA), the last term in Eq. (47) can be
dropped. In the RWA, it is correct to say that a
+
circularly polarized excites only
m=1
transitions on absorption. If the last term in Eq. (47) is included, however, this statement is
no longer true. Thus the breakdown in the reasoning in Sec. I is that a circularly polarized
…eld drives both
m=
1 transitions if the resonance approximation is not made. Thus,
there are contributions to the index of refraction of
+
radiation from both the
m=
1
transitions.
Equations (31) and (47) can be used to calculate n+ . The steady state solution of Eq.
(47) is
am (Z; t) =
ieE0
r
1
8hmo ! 0
ei(n+ kZ !t) m;1
+ i (! m0 !)
12
e
i(n+ kZ !t)
m; 1
+ i (! m0 + !)
exp ( i! 0 t=2) :
(48)
To obtain an expression for n+ in terms of am (Z; t); I …rst write the polarization as
P (Z; t) = Px+ x
^ + iPy+ y
^ ei(n+ kZ
!t)
+ c.c.
x
^ + i^
y
x
^ i^
y i(n+ kZ
e
= P+ (+) p
+ P+ ( ) p
2
2
!t)
+ c.c.,
(49)
where the positive frequency component of the circular components of the polarization is
given by
iPy+
hx+ i i hy+ i
p
= Ne
:
(50)
2
2
By combining Eqs. (31), (40a), (40b), (48) and (50), one can show that P+ ( ) = 0 (the
P+ ( ) =
+
P x+
p
polarized incident …eld induces only a
P+ (+) =
where the limit
+
component of the polarization) and
1
N e2 E0
4mo ! 0 ! + !
1
;
! +!
! 0 has been taken (recall that ! = !
10
(51)
= !0
eB0 =2mo ). Finally,
since the polarization and …eld amplitudes appearing in Eqs. (49) and (44) are related by
P+ (+) = n2+
1
(52)
0 E0 =2;
it follows that
n2+
Similarly, for a
1=
N e2
1
2mo 0 ! 0 ! + !
1
:
! +!
(53)
1
;
!+ + !
(54)
circularly polarized …eld, one …nds
n2
1=
N e2
1
2mo 0 ! 0 !
!
implying that
n
n+
N e3
!B0
;
2
nmo 0 (! 20 ! 2 )2
(55)
in agreement with Eq. (1). Note that the index of refraction for each circularly polarized
…eld in Eqs. (53) and (54) contain contributions from both
V.
m=
1 transitions.
DISCUSSION
I have presented three di¤erent calculation of the Faraday rotation, a classical calculation using the Lorentz force equation, a quantum calculation using a Heisenberg operator
approach, and a quantum calculation using a Schrödinger wave function approach. All three
methods yield consistent results. An oversimpli…ed picture in which one assigns an index of
13
refraction for
circularly polarized radiation resulting only from
m=
1 transitions on
absorption breaks down when non-RWA terms are included in the calculation.
Finally, I would like to return to the question as to the correct form of the Hamiltonian
in weak …elds. One might think that using a Hamiltonian of the form
H1 = H0 +
ep AE (Z; t) eLz B0
+
;
mo
2mo
(56)
with neglect of the A2 term would lead to correct results. However we have seen that this
is not the case. It is essential to include the 2AE AB cross term to arrive at the correct
expression for the Faraday rotation. This is true for both the quantum calculation and a
classical calculation based on Hamilton’s equations. On the other hand, if one starts from
the Hamiltonian
H2 = H0 + er E(Z; t) +
eLz B0
;
2mo
(57)
the correct Faraday rotation is obtained without any problem. This is another example
of where it is advantageous to use the er E rather than the ep A=mo form of the interaction potential for interactions of atoms or oscillators with optical …elds in the dipole
approximation.11
Of course, the Hamiltonians (22) and (57) must lead to the same expectation values for
any physical observables since they are related by a unitary transformation.11,12 However, the
dynamics of the state evolution for each Hamiltonian can be quite di¤erent. For example, it
follows from Eq. (35) that, for the Hamiltonian (27b), the magnetic …eld shifts the excited
state sublevels, the optical …eld couples the ground and excited states, and the combined
action of the optical and magnetic …elds couples the ground and excited states. On the other
hand, for the Hamiltonian (57), the magnetic …eld shifts the excited state sublevels and the
optical …eld couples the ground and excited states, but there is no cross term involving the
combined action of the optical and magnetic …elds. As a consequence, it is easier to solve for
the system dynamics and to interpret the roles of the …elds if one uses the er E Hamiltonian
(57).
The calculation carried out for the oscillators would be identical to that for atoms having
ground state angular momentum L = 0 and excited state angular momentum L = 1; as long
as the …elds are weak. In contrast to the oscillator problem, however, there are nonlinear
e¤ects for atoms that enter as the optical …eld strength is increased.6 Moreover, for ground
states in which there is magnetic state degeneracy, …ne, or hyper…ne structure, optical
14
pumping can modify the Faraday rotation.7 For atoms, it is not possible to get a closed from
di¤erential equation for the Heisenberg operator r(t) as it was for the oscillator. Instead,
using either a Heisenberg or Schrödinger approach, one must calculate the expectation value
of the dipole moment operator of the atoms.
VI.
ACKNOWLEDGMENTS
The motivation for this calculation came from a request from Carl Akerlof at the University of Michigan to provide a theory write-up for an advanced undergraduate laboratory on
Faraday rotation. I am also happy to acknowledge helpful discussions with Peter Milonni
and I bene…tted from helpful comments of the referees.
1
See, for example, A. D. Buckingham and P. J. Stephens, "Magnetic Optical Activity," Annu.
Rev. Phys. Chem. 17, 399-432 (1966), and references therein.
2
A. Sommerfeld, Optics (Academic Press, New York, 1964) pp. 101-106. Sommerfeld uses n
for left circularly polarized radiation and n+ for right circularly polarized radiation, but I have
interchanged his notation since n corresponds to
direction, where
radiation induced
m=
radiation for a …eld propagating in the ẑ
1 transitions on absorption (m is the magnetic
quantum number.)
3
F. J. Loe- er, "A Faraday rotation experiment for the undergraduate physics laboratory," Amer.
J. Phys. 51, 661-663 (1983); F. L. Pedrotti and P. Bandettini, "Faraday rotation in the undergraduate advanced laboratory," Amer. J. Phys. 58, 542-545 (1990).
4
B. Rossi, Optics (Addison-Wesley, Reading MA, 1957) pp. 427-430.
5
See, for example, D. M. Bishop and S. M. Cybulski, "Magnetic optical rotation in H2 and D2 ,"
J. Chem Phys. 93, 590-599 (1990); M. Krykunov, A. Banerjee, T. Ziegler, and J. Autschbach,
"Calculation of Verdet constants with time-dependent density functional theory: Implementation and results for small molecules," J. Chem Phys. 112, 074105 (2005).
6
See, for example, F. Schuller, M. J. D. MacPherson, and D. N. Stacey, "Saturation and collisional
e¤ects on magnetic optical rotation," Physica 147C, 321-331 (1987); D. Budker, D. J. Orlando,
and V. Yashchuk, "Nonlinear laser spectroscopy and magneto-optics," Amer. J. Phys. 67, 584-
15
592 (1999).
7
F. Schuller, M. J. D. MacPherson, and D. N. Stacey, "Magneto-optical rotation in an atomic
vapor," Optics Comm. 71, 61-64 (1989); F. Schuller, D. N. Stacey, R. B. Warrington, and K.
P. Zetie, "Theory of Faraday rotation produced by atoms near a Jg =
1
2
! Jg =
1
2
transition,"
J. Phys. B 28, 3783-3790 (1995) (this paper also includes collisional and saturation e¤ects).
8
For successive intervals, one can rede…ne the direction of polarization as the x direction and
then calculate the rotation in the next small interval. The treatment is strictly valid only for
eB0 !=2mo
9
! 20
!2 :
P. R. Berman, "Two-Level Approximation in Atomic Systems," Amer. J. Phys. 42, 992-997
(1974).
10
Since the equations of motion for the displacement operator of a harmonic oscillator contain
a source term that is linear in the external electric …eld, a perturbative calculation of the
Heisenberg operator r(t) yields a result that is correct to all orders in the electric …eld amplitude.
11
See, for example, J. R. Ackerhalt and P. W. Milonni, "Interaction Hamiltonian of quantum
optics," J. Opt. Soc. Amer. B 1, 116-120 (1984).
12
The Hamiltonians are related via the unitary transformation H 0 = U HU y with U =
exp[ier AE (Z; t)=h]; where H is given by Eq. (22), AE (Z; t) by Eq. (23b), and H 0 =
H2 + e2 B02 (x2 + y 2 )=8mo is equal to H2 [Eq. (57)], plus a term corresponding to the quadratic
Zeeman shift.
16