Asymmetric angular-momentum distribution of electrons in axial channeling in bicrystals

Asymmetric angular-momentum distribution of electrons in axial channeling in bicrystals

Nuclear Instruments and Methods in Physics Research B48 (1990) 263-265 North-Holland 263 ASYMMETRIC ANGULAR-MOMENTUM DISTRIBUTION IN AXIAL CHANNELIN...

316KB Sizes 0 Downloads 40 Views

Nuclear Instruments and Methods in Physics Research B48 (1990) 263-265 North-Holland

263

ASYMMETRIC ANGULAR-MOMENTUM DISTRIBUTION IN AXIAL CHANNELING IN BICRYSTALS

OF ELECTRONS

V.V. BELOSHITSKY I. V. Kurchatov Institute of Atomic Energy, Moscow, 123182, USSR

A.H. HOKONOV and M.H. HOKONOV Kabardino - Balkarian State University, Nalchik, 360004, USSR

Populations of positive and negative angular-momentum projections of electrons channeling in a bicrystal are shown to be different. The absolute value of this difference is about 10%.

As is well known, an electron channeled along an atomic row has, besides the transverse energy E, , another integral of motion - the angular-momentum projection M, on the row direction. A sign-asymmetric angular-momentum projection distribution is of interest as, for this condition, the possibility of polarization of the electron spin due to spin-dependent radiation in the atomic row potential has recently been predicted [l]. The estimates of radiative self-polarization given in ref. [l] show that the degree of self-polarization in axial electron channeling for a single electron is - 20% if #,Y .=E1 and - 70% if 4,~ B 1, where I/J, is a critical channeling angle. An asymmetric distribution over angular momenta of channeling electrons does not arise for common experimental conditions when a collimated beam of particles bombards a single-crystal target. The cause of this lies in the practical uniformity of the beam over the transverse section in the vicinity of the atomic row. This uniformity means that the distribution function f. of the incident particles does not depend on the transverse coordinates hr, = {x, JJ}. Let the incident beam be ideally collimated so that

fo(Pd =

+)s(p, -Pox)G(P,-Poyh

(1)

where S, is the transverse section area per row. The distribution function of electrons over transverse energy E, and angular momentum it4, can be obtained by a convolution of the microcanonical distribution with the incident distribution function fo:

F(E, 9 4)

E, -

where U(r,) is the continuum potential of the atomic string and M the electron’s relativistic mass. Substituting f. from eq. (1) into eq. (2) we obtain

where 1c, is the polar angle of the radius vector p:,, P& + P& and rl o is defined by the relation E,

= 2

+ U(rlO)

The angles & in eq. (3) obey the condition sin(o,

- &) = M,/P.,~,,,

0168-583X/90/$03.50 (North-Holland)

XP,, +YP,),

0 Elsevier Science Publishers B.V.

(2)

(5)

where a, is the polar angle of p I ,,. For each sign of the momentum projection M,, eq. (5) has two roots, which leads to the summation over i = 1, 2 in eq. (3). The roots of eq. (5) are shown in fig. 1. The independence of the continuum potential of polar angle leads to equal weights in eq. (3) of particles with opposite initial radius vectors rI in = rI a and rI in = - rI ,, (and opposite angular momenta, correspondingly) at any incidence angle of the beam. Thus we obtain F(E,

1 M,) =F(E,,

-M,) 1

2rLo

= Sl

2 - u(r,))+fz-

=

I U’(r,o)

I {_

.

(6)

The situation is different when a beam of electrons transmitted through an auxiliary thin crystal is used (see the scheme in fig. 2a). Such a compound system yields a redistribution of particles over transverse energies and IV. CHANNELING

RADIATION

264

V. K Beloshitsky et al. /Asymmetric

angular-momentum distribution

To obtain the distribution over angular momenta and transverse energies in the second crystal we again use eq. (2) with corresponding fc(r, , pA> at the front surface of the second crystal. The function fo(r, , pi can be expressed &rough the distribution function at the rear surface of the first crystal. The phase spaces of the first and the second blocks are related by the equations P; =pJ. +4* i r; = rL +b,

Fig. 1. ~IIus~ation of the conditions for the compensation of the angular mom~fa of chilling electrons at different stiffs -b and tilts q of the crystals. The numbers 1,2 and 1, 2 indicate the transverse momenta perpendicular to the afomic row of the second crystal. The compensation is fulfilled at: (a) b = 0% #==O, (b) &==bXe,, B=O, (c) b=O, B=B.,e,, but if is not realized at (d) 6 = bxe,, B = t9,,ey.

(71

where primed coordinates refer to the first crystal , whereas 4 and b are constant vectors determined by the relative tilt angle and the shift of the blocks respectively. The vector g assumes the value q = (E/c)@, where 8 is the tilt angle. From the invariance of the phase space voIume dr dp it follows that fs( r, , p i ) = g( r: , p:>. The function g( r; , p>) is supposed to obey the equilib~um dist~bution

where T is the period of the transverse motion and F,, F, are the distributions of the bound and free electrons,

a

respectively. The distribution (8) is symmetric in the p: . Therefore the ~gul~-momentum distribution is also symmetric if there is only a shift of the blocks without any tilt {see fig. lb). However, if there is also a tilt of the crystals (see fig. Id), then there is no compensation of angular momenta of opposite signs. Indeed, the tilt is equivalent to a subtraction of the vector q from p: and therefore different absolute values of angular momentum are gained at the same entrance point rLin for different signs of p; : (9)

%=[f+~hX(P~-4)1.~

As follows from eq. (9& the compensation will also occur if the entrance points ri, = rL,-, and rI in = -rlo will contribute to the angle-momentum dist~bution

b Fig. 2. The relative disposition of the atomic rows of the crystals in the parallei and transverse planes.

angular momenta in the second single crystal. Then there is a possibility of populating the vicinity of the points r&e and -rla with different weigbts owing to the substantial inhomogeneity of the beam over the space coordinates on the rear surface of the first crystal PI.

Tabte 1 The asynmefry of the angle-momentum distribution .$ at $Q, = 0, A = 0.6% vs shift b, and tilt B,,. The number of bound electrons is given in brackets.

4/d 0.1 0.2 0.3

0.08

0.13

0.20

0.08 (0.61) 0.14 (0.56) 0.17 (0.47)

0.11 (0.59) 0.1s (0.53) 0.20 (0.44)

0.12 (0.53) 0.18 (0.48) 0.17 (0.40)

V. V. Beloshitsky et al. / Asymmetric

angular-momentum

distribution

Table 2 Same as table 1 but for A = 0.15~~ bx /d

@,I?&

0.1

0.2 0.3

0.08

0.13

0.06 (0.72) 0.18 (0.78) 0.20 (0.67)

0.07 (0.72) 0.20 (0.70) 0.25 (0.62)

with equal weights. This will take place if there is no shift of the second crystal (see fig. lc). We introduce the asymmetry parameter (N+ N-)/N+ - N-) = 5, where N * are the number of electrons with positive and negative angular-momentum projections M, respectively. Calculations were made for 1 GeV electrons channeled in a Si bicrystal along the (111) axis. The results are presented in tables l-3, where b = bXe, and q = (E/c)Be,,. The directions of the axes x, y are shown in fig. 2b. The incidence angle at the first crystal is supposed to be zero and the angular divergence of the beam is A = 0.64$= (tables 1, 3) and A = O.lS#, (table 2) where $c is the critical angle for channeling and d = 2.217 A is the spacing of the atomic rows. The number Nch of electrons bound to the axis is given in brackets. If only free (unbound) electrons are included into g(r’, p’) then the asymmetry parameter is equal to zero, as the electrons with EL > 0 enter the second block with opposite values of M, with equal probabilities. Therefore, the asymmetry increases with the number of channeling electrons NC,,. In table 3 the calculation of the asymmetry including multiple scattering in the first block is shown. The distribution function F(E; , Mz‘) has been obtained as a numerical solution of the Fokker-Planck equation [2] by the method developed in ref. [3]. Despite the small thickness of the first block the distribution function is substantially changed by the multiple scattering, which makes it more smooth at EL = 0 as compared to the initial distribution. This causes a reduction of the asymmetry parameter (at small tilt angles BY< 0.3~~) because just the region in the vicinity of El = 0 gives the main contribution to the integral (2) though the channeling fraction of the beam has increased by 15%. The sign of the asymmetry changes with increasing tilt angle 0, at BY> 0.5& This can be understood from fig. 3 where the functions f( M,) = jF( E, , At__)dE, are shown for

0.2 *

0

0.1 0.2 0.3 0.4 c.5 0.6 0.7

0.8

N l , rrldt.

0.9

units

Fig. 3. The distribution of electrons over the angular momenta f( M,) and f( - M,) (the solid and dashed lines) at A = 0.6&, &, = 0, b, = 0.13d, I = 1 pm, and different tilt angles 19,. positive and negative values of M, at different values of B,,. It is seen that the crossing point of the curves f( M,) and f(-M,) shifts to the small angular-momentum band and to large absolute values 1E, 1 when 0,, increases. At large 1E, 1 electrons are localized near atomic rows of the second crystal. Then the entrance points from the small region between the atomic rows of the first and the second crystal become important, and at B,, > 0.5$1~ the number of electrons with p; = I -pl +ql outweighs that with p; = (pl +ql. At small eY the situation is different. The maximum value of asymmetry for the total beam, 5, = .$‘N,, is of about 0.05 at small B,,( = 0.2&), though we did not optimize the shift parameter 6. In conclusion, it has been shown that by means of the bicrystal system it is possible to obtain a polarized electron beam. The degree of self-polarization is about a few percent, which is important for some experimental investigations. References [l] V.G. Bagrov, I.M. Temov and B.V. Holomay, Sov. Phys.

JETP 59 (1984) 622. [2] V.V. Beloshitsky and M.A. Kumakhov, Sov. Phys. JETP 55 (1982) 265. [3] V.I. Telegin and M.H. Hokonov, Sov. Phys. JETP 56 (1982) 142.

Table 3 Same as table 1 but for b, = 0.13d and including multiple scattering in the first crystal of thickness I = 1 pm.

w+c

0.1

0.2

0.3

0.5

0.6

0.7

0.8

1.0

6

0.08 0.58

0.09 0.54

0.07 0.47

- 0.06 0.32

- 0.09 0.25

-0.15 0.19

-0.25 0.15

-- 0.47 0.10

%

IV. CHANNELING

RADIATION