Circumventing linear birefringence in ferromagnetic crystals
A. Sterl and K. 0. Thielheim
Institut fur Reine und Angewandte Kernphysik, University of Kiel, Olshausenstrasse 40-60, 2300 Kiel,Federal Republic of Germany
Received April 6, 1983; accepted August 8, 1983
Utilization of the Faraday effect in room-temperature ferromagnetic crystals transparent in the visible, such asFeBO 3 or FeF3 , is complicated by the dominant linear birefringence present in such crystals. We devise a methodto overcome this obstacle by stacking of properly oriented platelets. To take into account the reflections withinsuch a stack, we develop a new matrix method based on the well-known Jones calculus and a paper by Teitler andHenvis [J. Opt. Soc. Am. 60, 830 (1970)].
1. INTRODUCTION
For applications in optical information-processing devicesmade of ferromagnetic materials it is desirable to rotate theplane of linearly polarized visible light.1'2 There are someferromagnetic materials, such as FeBO3, FeF3, and LuFeO 3,1 '2
that are transparent in the visible, but their utilization iscomplicated by the fact that they possess linear birefringencein addition to the circular birefringence. The combined oc-currence of the two sorts of birefringence causes an oscillationof the angle of rotation with increasing layer thickness ratherthan the linear growth that would be produced by pure cir-cular birefringence.14
One way of circumventing the effect of the linear birefrin-gence may be the stacking of properly oriented platelets,1 atechnique that is examined in this paper. To describe thepropagation of light through those platelets, we use the well-known Jones method.k7 Because reflections at the boundarybetween two platelets might affect the behavior of the stack,we have to take them into account. To do so we use a methoddeveloped by Teitler and Henvis8 that describes the reflectionand transmission of a slab of (anisotropic) material betweentwo isotropic half-spaces. We extend that method to a stackof slabs and relate it to the Jones method.
2. APPLICATION OF JONES METHOD TOMATERIALS POSSESSING BOTH LINEAR ANDCIRCULAR BIREFRINGENCE
Jones MethodThe Jones method describes the propagation of a completelypolarized electromagnetic wave in some medium by calcu-lating the dependence of the (transverse) electric-field vectorE = (E., EY) on the propagation length z. The medium islocally characterized by a 2 X 2 matrix A', defined by (di: =
d/dz)
dzE(z) = AV(z) E(z). (2.1)
A' is called the differential matrix. 5- 7' 9 For homogeneousmaterials, i.e., A' independent of z, Eq. (2.1) can be integratedto give
E(z) = exp(JVz) - E(0) = A(z) -E(0). (2.2)
At is the Jones matrix. It characterizes a special sample ofthickness z of a given material. A' can be expressed as a sumof the eight fundamental 0 matrices6'9 forming a basis ofC2X2:
8A'= Z Mio.
i=1(2.3)
The coefficients as describe the amount of the eight basicoptical properties of the material, among which are linear andcircular birefringence.
As an example, consider a material such as FeBO3 or FeF3.A' takes the form
(2.4)-A I1-i(n + g) - JI
where ij is the average phase constant, g is the linear bire-fringence (i.e., half of the difference between the two mainphase constants), y is the circular birefringence (half of thedifference between the phase constants of left- and right-handed circularly polarized light), and x is the absorptioncoefficient, which will be neglected in what follows.
To evaluate a function f with a matrix as the argument, suchas the exponential function of Eq. (2.2), we use a method de-scribed by Azzam.10 For a general 2 X 2 matrix,
A (a, a41
a3 a2)
Thus we get
fQ(A) =-2W
XA+f(Xj)
{a3Vf(XI)
where
- A I (X2 ); a4V(XI) - f(X2 ))- f (X2)]; A'f(X 2 ) - A -f(X1 )l
Linear and Circular BirefringenceApplication of Eqs. (2.2) and (2.6) with A' from Eq. (2.4) yields(cf. Ref. 3)
03 = [(y - k)2 + g 2]1 /2. (2.13)
If g = k, i.e., the twist rate equals the rotation rate of the planeof polarization caused by the circular birefringence, Eq. (2.12)reduces to
(cos (d + i g sin (3d
At(d) = (9 sin/3d
- A sin (3d 1
cos f3d - i gg sin (3d /(3
where
( = (g2 + g2)112.
At(d) = S(gd) (ei d) e -igd-ind (2.14)
Obviously, if E - (1, 0) [orE i- (0, 1)], the effect of a sampledescribed by Eq. (2.14) is [besides an uninteresting absolute
(2.8) phase retardation of -(-q ± g)d) to rotate the plane of polar-ization by an angle a = gd. Thus we managed to find amethod to circumvent the effect of the linear birefrin-
(2.9) gence-at least theoretically.
It can easily be seen that a sample described by Eq. (2.8)does not act as a simple rotator. A more detailed investigationshows that the effect of such a sample is to rotate the plane ofpolarization back and forth for some number of degrees withincreasing sample thickness but that there is no linear growthof the rotation angle a.
To understand the physical reasons for this behavior,consider light polarized parallel to the optic axis of linear bi-refringence (birefringent axis). Then, in a purely birefringentmedium, transmission without change of polarization statewould be allowed. Circular birefringence then causes arotation away from that axis, leading to a splitting up of thebeam into two components polarized perpendicularly to eachother and traveling with different phase velocities. These twocomponents again get rotated, split up, and so on.1' 2
This description suggests a possible method of circum-venting the effect of linear birefringence: If the whole me-dium is twisted along the z direction in such a way that thebirefringent axis and the (rotating) plane of polarization ofthe light always remain parallel, there should be no splittingup but a pure rotation of the plane of polarization, increasinglinearly with path length.
Jones provides a formula for the M matrix of a linearlytwisted medium,6 namely,
At(d) = S(kd) expl(A'o -k9(r/2)]dj
= SM(kd) exp(JA'd), (2.10)
where k is the twist rate (in degrees per centimeter, perhaps),A'o is the N matrix of the untwisted material, and & is therotation matrix:
SW = (cos a -sin a. (2.11)\sin a cos I
Using A' from Eq. (2.4) as A'o and applying Eq. (2.6), weget
Stacked PlateletsNow, for FeBO 3 , g = 23000/cm. 2 Even if it were possible toobtain a twist rate of such an amount, it would result in analteration of the optic properties of the material because ofstress and strain, so that Eq. (2.14) would no longer be valid.The question arises whether it is possible to reach the desiredeffect, not by continuously twisting the material but by takingN slabs of thickness h = diN and stacking them with each slabrotated by an angle aN = a/N = gd/N = gh with respect tothe one preceding it. For such a stack, the M matrix wouldbe (cf. Ref. 5)
At = S(N aN)Ato(h)QV(-NaN) ... &(aN)Ato(h)&(-aN)
= 9(NaN)[Ato(h)&(-OaN)]N
= &(gd)jexp(A'oh) exp[-k9(r/2)h]}N, (2.15)
where AO is the M matrix of a nonrotated slab and A'o is thecorresponding N matrix. The equation
S(-aN) = exp[-kS(r/2)h] (2.16)
can easily be verified by use of Eq. (2.6).Now, as long as A'oh and S(vr/2)h do not commute (and, in
fact, they usually do not), the expressions given for At by Eqs.(2.10) and (2.15) are not equal. Only as h approaches zero(and N infinity) are they approximately equal. The conditionat which both expressions coincide for finite h remains to beexamined.
Let us therefore look at Eq. (2.15). Here Ato(h) is given byEq. (2.8). If h meets the requirement Oh = nir, n e 91, AOreduces to the unit matrix 1 (or its negative) and A be-comes
At = &(NaN)l1ei?7h&(-HaN)]N
= &(NaN)&(-NaN)e-Ind
= le-ind, (2.17)
Jt(d) = &(kdd) AB A-) e,wnere
A- = cos O'd ± i - sin ,'d,
B = - sk sin /'d,'3'
and
(2.12) i.e., the stack under consideration has no influence on thepolarization of the light. The twisted medium [Eq. (2.12)]does not show this effect that occurs with a period of P1 =2-7rlg as a function of layer thickness h.
Now let us try to write the product of the two exponentialsof Eq. (2.15) as one exponential, i.e.,
[exp(JV0h)exp(-kcS(ir/2)hjN = (exp A)N = exp(N X A).
(2.18)
The Baker-Campbell-Hausdorff formulall1 2 gives for A
The first term of this expression alone obviously leads to Eq.(2.10). Taking into account the second term (still a poorapproximation) and applying Eq. (2.6) yields (neglectingterms of order g2h2 )
exp(N X A) fe igd
\igh sin gdigih sin gd\ (2.20)
e -igd)
The off-diagonal elements show that there should be an-other periodicity in the behavior of the considered stack oflayers as a function of h, occurring with a period of P 2 =
2rr/(Ng). Approximation (2.20) clearly shows that approxi-mation (2.19) for A is very rough, because the periodicity Piis not represented.
Numerical ResultsWe have evaluated Eq. (2.15) numerically using the values g= 2300'/cm and g = -2 X 1050 /cm given by Wolfe et al. 2 forFeBO3 . The results are shown in Fig. 1, where we put
d = Nh, gd = NaN = 7r/4, N = 16. (2.21)
(For other values of N, the same behavior is shown qualita-
I (%)110,
100
90
80
70
60
50
40
6/0
56
48
40
32
24
16
a
0.64 0.60 096 1.12 1.28 1.44 1.60 1.76 1.92 2.08
13 lo-,3 H(a)
0,64 0.60 0.96 1.12 1.28 1.44 1.60 1.76 1.92 208
.10ocm H
(b)
Fig. 1. Dependence of (a) I and (b) a on h, the thickness of one layerin a stack of N = 16 layers. p is considered to be variable (see thetext). The two periodicities P1 and P2 , arising from fh = nw and gd= mr, n, m a S, respectively, can be seen clearly.
l %1
0 10 20 3 40 50 60 70 80 90 100
N(b)
Fig. 2. (a) I and (b) a for different numbers of layers N. 4 and d areheld constant, so that h varies as dIN. At peaks (I) and (®, Oih -and Oh - 27r, respectively.
tively.) Thus g is variable, leading to a constant (expected)rotation angle of a = 7r/4 for all values of h.
In Fig. 1, l and a, the degree of linear polarizations and therotation angle of the plane of polarization, respectively, oflinearly polarized light incident upon the stack, are plottedas a function of layer thickness h. The two periodicities P 1and P 2 can be seen clearly. [A computation for other valuesof h revealed that there is no other dependence on h than thetwo periodicities. This is not in accord with expression (2.20),again showing the inaccuracy of the approximation, Eq.(2.19).]
When sin(^h) = 0, a = 0 and I = 1, in agreement with Eq.(2.17). The periodicity P2 then leads to I << I and a #d r/4in the close neighborhood of points with Oh = ni7r. So if wewant to approximate the continuously twisted medium by ourstack of thin layers, we must not choose a value of h with Oh
nw7r, because this will not lead to a good approximation. Thebest approximation is met when
Oh = (2n + 1)(r/2), nE 91. (2.22)
This leads to an understanding of Fig. 2, in which we fixedg and d (as in a real stack) and varied N, the number of layers,and accordingly h, their thickness. At the two peaks in Fig.2(a), where 1 deviates from 1 = 1, coinciding with a changefrom a < 7r/4 to a > r/4 [Fig. 2(b)], h (= d/N) takes valuesleading to Oh -ir and 2wr, respectively. As is shown above,
(a) N
-N=16
A. Sterl and K. 0. Thielheim
48 J. Opt. Soc. Am. A/Vol. 1, No. 1/January 1984
at such points the stack does not represent the twisted me-dium.
To meet the requirement of Eq. (2.22) one has to choose N= 15, N = 9, N = 6, or N > 40. Figure 2 shows that this leadsto 1 1 and a - r/4, as desired.
3. REFLECTIONS IN MULTILAYERSTRUCTURES
Extension of Jones's MethodAs we have shown, Jones's method predicts the circumventingof linear birefringence by stacking of thin layers, rotated withrespect to one another by an angle aN. However, this methoddoes not take into account the reflections taking place at theboundaries between two adjacent layers. We now try to finda method that accounts for this effect and then apply it to thestack.
Jones's Eq. (2.1) describes light traveling in the positive zdirection. If we want to describe light traveling in the nega-tive z direction, a minus must be added on the right-hand sideof Eq. (2.1)14 So there are two equations for light travelingin both directions.
Differentiating Jones's Eq. (2.1) again yields (with JA beingconstant)
d, 2E(z) = ind 2 E(z) = N 2E(z) =N. E(z). (3.1)
This is the same equation for both directions, but, unfortu-nately, it is of second order. To reduce the order, we use theequations
dE.(z) = -iwfH,(z),dzEy(z) = iwftHx(z), (3.2)
following from the Maxwell equation rot E = - =-fH, fbeing the scalar magnetic permeability, and a harmonictime-dependence eiwt.
Inserting Eqs. (3.2) into (3.1), we get
dF(z) = W- F(z), (3.3)
where
,X21 +1 = 0IT*(11 TJ* )e/i(
iw~l'(A'*l')Ji W' ,(T
Inserting this together with the definition
N*1'= = a2
and the identity
1'Js = A'2
(3.7)
(3.8)
(3.9)
into the well-known power series of the exponential function,we get for L
[ = cosh(Vz)
£ W(l'-') sinh(A'z)/iwftict(1'A' a~) sinh(NA'z)
cosh(NAz) I(3.10)
This equation contains only functions of 2 X 2 matrices thatcan easily be evaluated by use of Eq. (2.6).
Teitler and Henvis8 use Eq. (3.3) as a starting point, too, butthey do not relate L and I to A', so that they do not connectthis approach to the Jones method.
Slab between Two Half-SpacesWe next consider a layer of some (anisotropic) materialstratified in the x, y plane between two isotropic half-spaces(Fig. 3). A wave F+ coming from medium (® is perpendicularlyincident upon the layer, giving rise to a reflected wave F_ inthe same medium, a field Ft0 t in the layer, and a transmittedfield Ft. in medium ®. Ftot(O) and Ft0 t(d) are related by theL matrix of the layer:
Ftt(d) = L X Ftot(0). (3.11)
At the boundaries, the tangential components of E and H (i.e.,the components of F) must be continuous, yielding
F+ + F_ = Ftot(0)
and
(3.12)Ftot(d) = Ft.
Combining Eqs. (3.11) and (3.12) leads to
r (F+ + F-) = Ft. (3.13)
0l;,
F = (E., EY , H.,, HY)T,
N,, = O icon I'\
1 1 0 )
and
w* = (3 - 2 )-n1 -n4 (3.4)
Here Ti are the elements of N, and 0 is the null matrix. Inanalogy to Jones's Eq. (2.1), Eq. (3.3) can be integrated togive
(3.6) giving rise to the reflected wave F_, the field Ftot in the layer, and atransmitted field Ft.
A. Sterl and K. 0. Thielheim
Vol. 1, No. 1/January 1984/J. Opt. Soc. Am. A 49
Expanding F+, F-, and Ft into the eigenvectors of the corre-sponding media AT) and ® [eigenvectors are solutions of Eq.(3.3) of the form fe7z, leading to (I - -y1)f = 0] and insertinginto Eq. (3.13) yields eight linear algebraic equations for theeight elements of the reflection and transmission matrices Rand T, defined by
E_ = RE+, Et = T- E+. (3.14)
Thus T corresponds to the Jones matrix A. B = (r/ik) andT = (tik) are given by Teitler and Henvis 8 :
ril = - (b ap+ -a, bP+),
r2 = (bsas+ -a-b+)
matrices of a stack of layers are obtained by replacing L byLtot in Eqs. (3.15) and (3.16). Yi and T then represent allinternal reflections.
Now we apply this method to the stack of layers alreadyexamined in Section 2. As was shown by Jones, 5 the N matrixJVo transforms into
N = S[a(z)]J1oV[-a(z)] (3.23)
when the sample is twisted by an angle a(z) along the z axis.Straightforward calculation shows that No transforms like
N = V2[(z)1R0Y2[-a(z)], (3.24)
where
e2 =( ). (3.25)
r2l = N-(-bp-ap+ + aP~bpt),
r22 = I (-bp-as+ + ap bs+);
In the special case a(z) = a = constant, it follows from Eqs.(3.5) and (3.24) that
have been used. Here ni is the index of refraction of mediumi and c is the speed of light.
Thus, to get the transmission matrix T of a given materialcharacterized by the differential matrix JV, one has to calcu-late L from Eq. (3.10) and R from Eq. (3.15) and then insertthe results into Eq. (3.16).
Stack between Two Half-SpacesNow we want to examine not one layer but a stack of N layerswith thicknesses hi, i = 1, ... , N, between two isotropichalf-spaces. Then the resulting fields Fi (a) and Fi (h) at thetwo boundaries of layer i are related by
Fo(h) = LiFiM, (3.20)
AC being the L matrix of the ith layer. From the principle ofcontinuity of the tangential components of E and H, it caneasily be shown that
Ltot(F+ + F-) = Ft, (3.21)1tot = 1 N X 1 N-1, X ... X L1. (3.22)
As Eq. (3.21) equals Eq. (3.13), the reflection and transmission
(3.26)
Thus the stack of layers rotated with respect to one anotherby an angle ajN is described-in formal analogy to Eq.(2.15)-by
1 tot = 8(NaN)[Lo(h)82(-aN)]N.To see whether the reflections within the stack alter the resultsgiven in Section 2, we calculated 1tot and T from Eq. (3.16)numerically and compared it with the Jones matrix Al of Eq.(2.15). We found that two sorts of reflection must be distin-guished: (1) reflections taking place at the boundaries be-tween the stack and the surrounding media (media (I) and (2)in Fig. 3) and (2) reflections at the boundaries between twolayers.
The type (2) reflections obviously have no influence (i.e.,they can be neglected); this can be understood when one rec-ognizes that the indices of refraction of adjacent layers arenearly the same. The type (1) reflections, however, have aninfluence on the behavior of the stack, being dependent on theindices of refraction of media (D and (M. When they equal the(average) index of refraction of FeBO 3, no reflections occur,and T = At. When they differ from that of FeBO3 , the re-flections result in losses, i.e., I tikI < j mikI, and in an additionalphase lag between the diagonal and the off-diagonal elementsof T, which did not occur in At. Thus the reflections can beneglected if the indices of refraction of surrounding mediaequal those of FeBO3. The result of Section 2, that the stackshows no linear birefringence and acts as a simple rotator, stillholds.
4. CONCLUSIONS
We have arrived at the following two important results:
(1) We have shown that it is possible to circumvent theeffect of linear birefringence in a material possessing bothlinear and circular birefringence by stacking thin layers of thatmaterial rotated with respect to one another by an angleOaN.
(2) We have devised a method that can describe the re-flections within a stack of layers. It is based on the Jonescalculus, and its formulas are formally equal to those of theJones method [cf., e.g., Eqs. (3.23) and (3.24) or Eqs. (2.1) and(3.3)].
(3.27)
A. Sterl and K. 0. Thielheim
-C = 82(0142(-a).
50 J. Opt. Soc. Am. A/Vol. 1, No. 1/January 1984
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