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Scattering of slow neutrons by molecules
J. Nucl. Energy.PartA:
ReactorScience. l%l.
Vol. 13.Pp. 128to 132. Pcr~amoaPressLtd. Printedin NorthernIreland
SCATTERING
OF SLOW NEUTRONS
BY MOLECULES
A. RAHMAN* Tata Institute of Fundamental Research, Bombay, India (Received 18 March 1960)
Abstract-The
scattering of neutrons from a rotating rigid molecule has been analysed for incident neutron
energy of the same order of magnitude mations.
as the rotational
level separations,
without making any approxi-
When the energy of the incident neutron is large, the method of ZEMACH and GLAUBER(1956) of using the rotator as a model for a spherical top molecule, has heen justified. The inclusion of zero-point vibrations have been considered briefly.
1. INTRODUCTION
WHEN the energy of a neutron incident on a molecule
is of the same order of magnitude as the separation of the rotational energy levels of the molecule, it is necessary to calculate the partial cross-section for scattering from each rotational level without making any approximations. The formalism of ZEMACH and GLAUBER covers the whole neutron energy range but only for spherically symmetric molecules; in extending this formalism to molecules of arbitrary shape, KRIEGER and NELKIN (1957) were able to justify the mass-tensor approximation of SACHS and TELLER (1941); however this could be done only for an incident energy of the neutron which is large compared to the separation of the rotational energy levels. Following the method of GORYUNOV (1957), we shall consider the scattering of slow neutrons from a rotating and vibrating molecule; the energy of the incident neutron will be assumed to be much lower than the lowest vibrational level so that only the zero-point vibrations of the molecule need be considered. In another paper we shall give the results for the water molecule, using the formulae given here. 2. BORN APPROXIMATION FOR THE SCATTERING CROSS-SECTION In the system of co-ordinates in which the centre of mass of the molecule of mass M and the neutron (mass m) is considered as fixed, the differential scattering cross-section per unit solid angle is given by
da -= dfA
where fi k,, ti k are the initial and final momenta of the neutron; y is the product of the rotational, vibrational and spin functions of the molecule and the spin function of the neutron; tin = ti+, - hn is the momentum transfer; r,, Y = 1,2, . . . , N are the position vectors, with respect to the molecular centre of mass of the N nuclei forming the molecule. Let the energy of the states Yinitial and yfinal be Ei and E,. The energy condition is then
(A2(M + m)/2mM)(k~ - k2) = Ef - Ei. The spin-dependent bound scattering the Yth nucleus is B,, and is given by (2S, + 1) B, = (S, + l)b,+ + S,b, + 2(b,+ - b,)s . S,
(2)
where S, is the spin operator of the Yth nucleus with spin S,; s is the neutron spin operator; b,* are the scattering lengths for the compound spin S, f 4; To get the differential cross-section from a particular rotational and vibrational state of the molecule we require an averaging over all initial states degenerate with this particular state and a summation over all the final states. Now the wave function y occurring in (1) can be broken up into three parts; one depends on the spin of the neutron, one on the total spin state of the molecule and one on the space co-ordinates rr of the N nuclei of the molecule. It is easy to show that the processes of averaging over the initial and summing over the final spin states of the neutron give the following: -=-
* Present address : Argonne National Laboratory, Lemont,
Illinois,
length for
U.S.A.
128
2 A,A A~~‘) + CL&‘)s
(1’)
Scattering of slow neiJtrons by molecules
in which (vr’) = Wls
=
(vi
lexp
(yf
IS,
(ix exp
??
dl
0%
W&f
lexp
?? rJ(
yi>(yf
I%
0% exp
??
rJl
vi)*
wi)*
(ix*rd)l
where y now is a product of the spin function of the molecule and its rotational and vibrational state functions. B, are defined in terms of A, and C, by B, = A, + 2C, s S,, so that
129
as in KHUBCHANDANI and RAHMAN(1960). Denoting the rotational quantum numbers of the initial and final states byj, e, m and J, E, M respectively, writing I’,,,@, Y) = 2/[(& + 1)/4dQ,- da, B, r) and using the integral (2n+ 1) j KM,-= D;,+ D-&-k dr = 8r2(- l)“+“‘C(.7j n; M - m 0)
??
x C(Jjn;K-kk)
(4)
we get
(2% + 1) A, = (S, + l)b,+ + S&b,; (2S, + 1) C” = b,c - b,.
(2’)
We shall not consider the spin dependent parts of the matrix elements in (1’). We shall only be concerned with matrix elements of exp (ix rv) between the initial and final rotational and vibrational functions of the molecule.
(~~t.,ti~~il exp 6% . r) I vrot..initd = (- 1)“1x d(2j + 1) 2 (i)“j,(scr)
M ---:, O)G~;,#3, +)
x C(Jjn;
??
(5)
where G&A& 4) = 4/[4~(2J + 1)/(2n +
x B&_, Bi,-k
Y&e,
$) C(Jjn;
~)l~~KW)g E; ‘- k Z) (6)
We note that G does not depend on m, M. More important still is the fact that the G’s are geometric constants for the particular nucleus (0,#) of the molecule and are independent of the incident andjnal neutron energy and momentum.
In evaluating (1’) we shall require products of the type (lexp (ix - r,)l)( exp (in - r,,)()* + C.C.which will have to be summed over M and averaged over m ; let the result of this be denoted by (YV’); we shall use the relation
FIG. 1 .-Eulerian angles a, /?, y specifying the principal axes A, B, C of the molecule and co-ordinates r, 0, + of an atom with respect to axes A, B, C.
,FV C(Jjn; I.
3. THE MATRIX ELEMENT OF exp(ix ‘r) FOR A RIGID MOLECULE
(ix
. r) =
4~ L: (i)njn(wr) 41
Yn,,
y) r,,,(e,
N-
Oz
(3) Using (3) we can reduce a typical term of the matrix element occurring in (1’). Since we are considering the molecule as rigid, only the rotational functions will appear in y and these can be expanded in terms of the symmetric top eigenfunctions Dfn,a (a, /3, r); for this expansion we shall use the notation t We shall use throughout ROSE (1957). 3
the notations
and conventions
O)C(Jjn’;
M - m 0) = a,,,.
and write
The Eulerian angles a, p, y for fixing the principal axes A, B, C of the molecule are shown in Fig. 1. Let r(r, 0, 4) be the co-ordinates of any nucleus P in the system A, B, C as shown in the figure. 2 is the direction of X. Now exp (ix - r) can be expressed as followst in terms of a, p, y (ROSE, 1957): exp
M -m
as in
Gj$,(e,
4) = G$, cos 4) + i G&,(0,
sin 4) (7)
where: G$$;,(O,:;
4) =
x v’bl(2n +
d/w $1 1
(--l)= B& B:,_, k.K.1 l)lP,,,(W(Jj n; K - k 1)(;31# (8)
It is easy to show that 2/2j + 1 Gg;n(B, 2; 4) = &Z/W + 1 G&(0,
;y; 4) (8’) Equation (8) allows us to use only real magnitudes by writing k’> = 2 ~j&r.)j,(xr,~) x {G$,@,, cos +,)G&,(e”,, cos A.,) $ G$Jeu, sin +,)G&@., sin &}
(9)
$ The factor d[4R/(Zn +- l)] in (8) goes with P,,,1(8); (see definition of P.,@) in ROSE (1957) Appendix III).
130
A. RAHMAN
The )I = 0 term in (9) is simply 2j,(~r,)j,(~r,,)6~,~6e,& since from (6) G$,(6, 4) = ~j,,6,,. Hence terms with n = 0 contribute to elastic scattering only. If an atom of the pair Y,V’ is situated at the origin we get
where 0,.,,,is the angle between the vectors r, and rV,. For an atom, say v’, situated at the origin we again get equations (IO) as for an asymmetric molecule. Example of CH,
In other words, collision of the neutron with an atom situated at the centre of mass of the molecule cannot bring about a change of angular momentum and hence of the energy of the molecule. 4. ROTATIONAL SCATTERING FROM A SYMMETRIC TOP MOLECULE In this case the summation (8) over k, K disappears. The energy depends only on k2 and K2 so that, for example, if k # 0 the states Dj+,, are energetically degenerate. Hence in getting (YV’)we have to perform an additional averaging and summing. Instead of (5) we have (-l)nr+-%‘2j
,L
x G&,@, G;&;,,(O, c$) = d/w -I- l\/4rr/(2n
4)
K - k)
(11)
To get {YV’)we average over m and sum over M as before; in addition we average over ik (when k # 0) and sum over &K. This gives for a transition from state j, k to state J, K (k, K being taken positive)
Pn,~+lc(%,) cos (K + k)(& - 4,s)
(12)
5. ROTATIONAL SCATTERING FROM A SPHERICAL TOP MOLECULE In this case we can further use the symmetry to put 4. =; $,, = 0 and 0, = 0. The averaging has to be done over all the (2j + 1) values of k and summation over all the (W + 1) values of K. This gives in an analogous way, for a transition from state j to final state J, (j+J)
[2(W + 1)/(2j + l)]
+ ;j4%r,
fj:(Kr) . . .)
(14)
where ‘c’ indicates that the choice of the terms inside the brackets is from 1j - J( to (j + J).
In the formalism of ZEMACHand GLAUBER(1956) the motion of a spherical top is considered as that of a rotator; in that case the co-ordinate y disappears and Dj__,r,_-kare restricted now to simply Di_),X,owhich is the same as Y,,,. Then, for a transition from j to J, we get instead of (13) @‘> = 2 ; (2n + l)jn(Kr”)jn(KrV*)1/[477/(2n +
I>1
Now C(jn J; 00) # 0 only if j + n + J is even; otherwise it is 0. It is an easy matter to calculate expression (15). In the formalism of ZEMACHand GLAUBER(1956) this expression is summed over all the final states J to get the so-called static approximation. It is easy to see that summing over J gives
2 z\ (212+ 1) j&r,)
n
(13)
j&r,,)
2/[4d(2n
+
111cdt.d
and this is the expansion of 2jo(Kr,,,), where r,,. is the distance between v and v’. This coincides with the result of ZEMACHand GLAUBER. We shall come to this again in Section 9 below. But it must be remarked that (15) is only an approximation to the correct (YV’)for a spherically symmetric molecule given by (12). This approximation amounts to the replacement of C(jn J; 0 0) by z/(W + l)/ 2/(2j + 1)(2n + 1) which is not in fact the value of C(jnJ;
2 s=ij-J/
x jn(Krv)j,(Krvt) 1/[4~/(2n + 1)i ~~,~(6.,,)
-
x ~,,&,JC( jn J; W2 (15)
Symmetry around the axis of y requires that this be a function of (4, - &) and not of $, and q$, separately.
=
;jlw - fj((Kr) +Zj3yKI)
+ 1)(4rr/2n + 1)
x C(Jj n ; K Q2 + (1 - S,,s) P, ,,-,(%) P,,K--k(&) x cos (K - k)(4, - &)C(Jj n; K -k)2}
(Y1”j
f&=/J-JI
6. SPHERICAL TOP MOLECULE SIMULATED BY A ROTATOR
+ 1)
XY n,~-@, &C(Jin;
x {P,,,:(Q
[2(2J + 1)/(2j + l)] ji!
M - m0)
-t-+1~(i)“j,(Kr)C(.@?;
(rv’) = 2 zj,(Kr,)j,(Kr,.)(W
For every pair of protons we have cos 8,,. = - l/3 ; if r is the distance C-H we get for every pair of protons
00).
For an atom, say v’, situated at the origin, rv, = 0, and we again get equations (10).
131
Scattering of slow neutrons by molecules
Example of CH, Every pair of protons will contribute, instead of (14), 2 xj&cr) II
d4n/(2n
+ 1) P,,,(cos 0 = -l/3) x C( j 17J; 00)2
where r is the distance C-H. 7. INCLUSION OF ZERO-POINT
VIBRATIONS
The position vector r, of a nucleus of mass m, can be written as r, = py + u, where u, is the displacement from py due to the vibrations. Let e(vln) denote the direction of displacement of mv for the nth mode of vibration. If [, denotes the nth mode so that 2T = 1 in2 and 2V = 2 ~~~~~~~~~ we can write l/m, u, =
8. TOTAL SCATTERING CROSS-SECTION Whereas (9) and (10) give the contribution of a pair Y,V’to the differential scattering cross-section for a chosen pair of initial and final states of the molecule, we can get the total differential scattering from a particular initial state by summing over all the final states which satisfy the energy-conservation condition. To get the total scattering from a particular initial state of the molecule we have to integrate over all scattering angles. This requires the evaluation of integrals of the type
j dd(‘%)
This can easily be shown to be
2 t, e(vln). Therm ”
exp (ix
??
(27&%,2)
rJ = exp (ix - pJ exp (i 2 6J~)5,)
the limits being:
n
where b,(jt) = K - e(vln)/l/m,. As has been shown by POPE (1952) the diagonal matrix element of the exponential containing the E,, for the zero-point state of vibration is exp (-I:
n
bn2(y)/4a,)
where
a,L = w&i.
Hence we require the rotational matrix elements of exp (in - pv) exp (-2
bn2(v)/4a,) = exp (ix - p,,) r8
x {I - 1 (K - e(vln))2/4a, m,}. Now e(vln) has components iA( e&yin), ec(vln) in the directions A, B, C of the principal axes. Thus x
* e(a+z)
=
K{-e,(vIn)
sin p cos y + eB(y]n)
x sin fi sin y + ec(+) ~(w*e(+z))2/4a,m, n
= +
Denoting
cos fi}
K2(Sin2~cos2y~(eA(Y~n))2/4a,m, n
similar terms).
2\ eA(vln) eg(vln)/4anmv
(16) by
(AB),
for
j&Y,) j,(Kr,*).
x1 =
r,lk
-
I2x in(x) j,(xrJv,,) s 21 k,]
;
- 2 sin B cos /? cos y (AC), $ sin2 @sin2 y (BB), -I- 2 sin B cos ,8 sin y (BC), + cos2 /3 (CC),) All the expressions in j3,r occurring in (16) can be expressed as products or sums of products of any two of %, 0; &l so that integration (4) can again be performed ‘quite easily leading to a rather long expression involving a sum of products of six ClebschGordon coefficients.
(17)
x2 = r,(k _L k”).
9. THE ‘HIGH’ ENERGY APPROXIMATION When the energy of the incident neutron (=:=VkOz/ 2m) is large compared to the separations of the rotational levels it is possible to make certain simplifying approximations. In this case the energy conservation condition cannot prevent any transitions so that a summation over all final states is possible. Further k = k, so that K effectively depends only on the incident momentum k, and the angle of scattering. Thus in (9), (12) and (13) the (vv’) can be summed over all the final states and this summation can disregard the products jn(~rv)jl,(Krv,) from what has been said above. In the general case for example, we get the total differential cross-section from state (j,e) by summing (vv’) from (9) over all J,E, giving 2 t: j&%)jn(Kr,$;
1{G$,;,,(%, cos $,)G$,~
x (i,,,, cos $,,) + G&,(0,,, sin +,)G$,(B,.,
sin +,.,) (18)
example, wenget for right-hand side of (16) K2(sin2t!?cos2 y (A A), - 2 sin2 B sin y cos y (AB),
d.y
The integration required for getting the total scattering cross-section then has the limits 0, 2k,r, instead of those given in the previous section. The effect of summing over all final states has already been commented upon in the case of a rotator, in Section 6. Also in the case of a spherical top molecule treated as such and not as a rotator we get some simplification on summing over the final states. In fact (13) gives, on summing over J, 2 2 (2n + l)j,cKrl)jn(Kr,,) n
1/4~1(2n -t 1) ~,,,oOL)
132
A. RAHMAN
which is the expansion of 2j,(~r,,), where r,,., is the distance between Y and Y’. We are thus led to the two rather important results; firstly that in the ‘high’ energy approximation for rotational scattering from a rigid spherical top molecule the reduction of the top to a rotator is justified; secondly that the scattering is independent of the state in which the molecule is found before scattering. It is interesting to note that though it has not been found possible to prove this result in the case of a nonspherical top, nevertheless, detailed computation on the water molecule, a symmetric top, has shown that this independence is still true even though not exactly. These results will be presented in another paper.
Ackno~Ie&entm&-The author gratefully acknowledges that this work and a computation on the water molecule which will be reported in another paper, were produced while the author was receiving a fellowship from the Comite d’Etude et d’Exploitation des Calculateurs Electroniques at Brussels. REFERENCES GORYWOV A. F. (1957) J. Nucl. Energy 2, 109. KHUBCHANDANI P. G. and RAHMAN A. (1960) Ann. Sot. Sci.
Bruxclles 74, 35. KRIEGERT. J. and NELKIN M. S. (1957) Whys. Rev. 106,290. POPEN. K. (1952) Conud. J. Phys. 30,597.
Rests M. E. (1957) Elementary Theory of Angular Momentum, John Wiley, New York. SACHSR. G. and TELLERE. (1941) Phys. Rev. 60,18. ZEMACHA. C. and GLAUBERR. J. (1956) Whys. Rev. 101, 118, 129.
ReactorScience. l%l.
Vol. 13.Pp. 128to 132. Pcr~amoaPressLtd. Printedin NorthernIreland
SCATTERING
OF SLOW NEUTRONS
BY MOLECULES
A. RAHMAN* Tata Institute of Fundamental Research, Bombay, India (Received 18 March 1960)
Abstract-The
scattering of neutrons from a rotating rigid molecule has been analysed for incident neutron
energy of the same order of magnitude mations.
as the rotational
level separations,
without making any approxi-
When the energy of the incident neutron is large, the method of ZEMACH and GLAUBER(1956) of using the rotator as a model for a spherical top molecule, has heen justified. The inclusion of zero-point vibrations have been considered briefly.
1. INTRODUCTION
WHEN the energy of a neutron incident on a molecule
is of the same order of magnitude as the separation of the rotational energy levels of the molecule, it is necessary to calculate the partial cross-section for scattering from each rotational level without making any approximations. The formalism of ZEMACH and GLAUBER covers the whole neutron energy range but only for spherically symmetric molecules; in extending this formalism to molecules of arbitrary shape, KRIEGER and NELKIN (1957) were able to justify the mass-tensor approximation of SACHS and TELLER (1941); however this could be done only for an incident energy of the neutron which is large compared to the separation of the rotational energy levels. Following the method of GORYUNOV (1957), we shall consider the scattering of slow neutrons from a rotating and vibrating molecule; the energy of the incident neutron will be assumed to be much lower than the lowest vibrational level so that only the zero-point vibrations of the molecule need be considered. In another paper we shall give the results for the water molecule, using the formulae given here. 2. BORN APPROXIMATION FOR THE SCATTERING CROSS-SECTION In the system of co-ordinates in which the centre of mass of the molecule of mass M and the neutron (mass m) is considered as fixed, the differential scattering cross-section per unit solid angle is given by
da -= dfA
where fi k,, ti k are the initial and final momenta of the neutron; y is the product of the rotational, vibrational and spin functions of the molecule and the spin function of the neutron; tin = ti+, - hn is the momentum transfer; r,, Y = 1,2, . . . , N are the position vectors, with respect to the molecular centre of mass of the N nuclei forming the molecule. Let the energy of the states Yinitial and yfinal be Ei and E,. The energy condition is then
(A2(M + m)/2mM)(k~ - k2) = Ef - Ei. The spin-dependent bound scattering the Yth nucleus is B,, and is given by (2S, + 1) B, = (S, + l)b,+ + S,b, + 2(b,+ - b,)s . S,
(2)
where S, is the spin operator of the Yth nucleus with spin S,; s is the neutron spin operator; b,* are the scattering lengths for the compound spin S, f 4; To get the differential cross-section from a particular rotational and vibrational state of the molecule we require an averaging over all initial states degenerate with this particular state and a summation over all the final states. Now the wave function y occurring in (1) can be broken up into three parts; one depends on the spin of the neutron, one on the total spin state of the molecule and one on the space co-ordinates rr of the N nuclei of the molecule. It is easy to show that the processes of averaging over the initial and summing over the final spin states of the neutron give the following: -=-
* Present address : Argonne National Laboratory, Lemont,
Illinois,
length for
U.S.A.
128
2 A,A A~~‘) + CL&‘)s
(1’)
Scattering of slow neiJtrons by molecules
in which (vr’) = Wls
=
(vi
lexp
(yf
IS,
(ix exp
??
dl
0%
W&f
lexp
?? rJ(
yi>(yf
I%
0% exp
??
rJl
vi)*
wi)*
(ix*rd)l
where y now is a product of the spin function of the molecule and its rotational and vibrational state functions. B, are defined in terms of A, and C, by B, = A, + 2C, s S,, so that
129
as in KHUBCHANDANI and RAHMAN(1960). Denoting the rotational quantum numbers of the initial and final states byj, e, m and J, E, M respectively, writing I’,,,@, Y) = 2/[(& + 1)/4dQ,- da, B, r) and using the integral (2n+ 1) j KM,-= D;,+ D-&-k dr = 8r2(- l)“+“‘C(.7j n; M - m 0)
??
x C(Jjn;K-kk)
(4)
we get
(2% + 1) A, = (S, + l)b,+ + S&b,; (2S, + 1) C” = b,c - b,.
(2’)
We shall not consider the spin dependent parts of the matrix elements in (1’). We shall only be concerned with matrix elements of exp (ix rv) between the initial and final rotational and vibrational functions of the molecule.
(~~t.,ti~~il exp 6% . r) I vrot..initd = (- 1)“1x d(2j + 1) 2 (i)“j,(scr)
M ---:, O)G~;,#3, +)
x C(Jjn;
??
(5)
where G&A& 4) = 4/[4~(2J + 1)/(2n +
x B&_, Bi,-k
Y&e,
$) C(Jjn;
~)l~~KW)g E; ‘- k Z) (6)
We note that G does not depend on m, M. More important still is the fact that the G’s are geometric constants for the particular nucleus (0,#) of the molecule and are independent of the incident andjnal neutron energy and momentum.
In evaluating (1’) we shall require products of the type (lexp (ix - r,)l)( exp (in - r,,)()* + C.C.which will have to be summed over M and averaged over m ; let the result of this be denoted by (YV’); we shall use the relation
FIG. 1 .-Eulerian angles a, /?, y specifying the principal axes A, B, C of the molecule and co-ordinates r, 0, + of an atom with respect to axes A, B, C.
,FV C(Jjn; I.
3. THE MATRIX ELEMENT OF exp(ix ‘r) FOR A RIGID MOLECULE
(ix
. r) =
4~ L: (i)njn(wr) 41
Yn,,
y) r,,,(e,
N-
Oz
(3) Using (3) we can reduce a typical term of the matrix element occurring in (1’). Since we are considering the molecule as rigid, only the rotational functions will appear in y and these can be expanded in terms of the symmetric top eigenfunctions Dfn,a (a, /3, r); for this expansion we shall use the notation t We shall use throughout ROSE (1957). 3
the notations
and conventions
O)C(Jjn’;
M - m 0) = a,,,.
and write
The Eulerian angles a, p, y for fixing the principal axes A, B, C of the molecule are shown in Fig. 1. Let r(r, 0, 4) be the co-ordinates of any nucleus P in the system A, B, C as shown in the figure. 2 is the direction of X. Now exp (ix - r) can be expressed as followst in terms of a, p, y (ROSE, 1957): exp
M -m
as in
Gj$,(e,
4) = G$, cos 4) + i G&,(0,
sin 4) (7)
where: G$$;,(O,:;
4) =
x v’bl(2n +
d/w $1 1
(--l)= B& B:,_, k.K.1 l)lP,,,(W(Jj n; K - k 1)(;31# (8)
It is easy to show that 2/2j + 1 Gg;n(B, 2; 4) = &Z/W + 1 G&(0,
;y; 4) (8’) Equation (8) allows us to use only real magnitudes by writing k’> = 2 ~j&r.)j,(xr,~) x {G$,@,, cos +,)G&,(e”,, cos A.,) $ G$Jeu, sin +,)G&@., sin &}
(9)
$ The factor d[4R/(Zn +- l)] in (8) goes with P,,,1(8); (see definition of P.,@) in ROSE (1957) Appendix III).
130
A. RAHMAN
The )I = 0 term in (9) is simply 2j,(~r,)j,(~r,,)6~,~6e,& since from (6) G$,(6, 4) = ~j,,6,,. Hence terms with n = 0 contribute to elastic scattering only. If an atom of the pair Y,V’ is situated at the origin we get
where 0,.,,,is the angle between the vectors r, and rV,. For an atom, say v’, situated at the origin we again get equations (IO) as for an asymmetric molecule. Example of CH,
In other words, collision of the neutron with an atom situated at the centre of mass of the molecule cannot bring about a change of angular momentum and hence of the energy of the molecule. 4. ROTATIONAL SCATTERING FROM A SYMMETRIC TOP MOLECULE In this case the summation (8) over k, K disappears. The energy depends only on k2 and K2 so that, for example, if k # 0 the states Dj+,, are energetically degenerate. Hence in getting (YV’)we have to perform an additional averaging and summing. Instead of (5) we have (-l)nr+-%‘2j
,L
x G&,@, G;&;,,(O, c$) = d/w -I- l\/4rr/(2n
4)
K - k)
(11)
To get {YV’)we average over m and sum over M as before; in addition we average over ik (when k # 0) and sum over &K. This gives for a transition from state j, k to state J, K (k, K being taken positive)
Pn,~+lc(%,) cos (K + k)(& - 4,s)
(12)
5. ROTATIONAL SCATTERING FROM A SPHERICAL TOP MOLECULE In this case we can further use the symmetry to put 4. =; $,, = 0 and 0, = 0. The averaging has to be done over all the (2j + 1) values of k and summation over all the (W + 1) values of K. This gives in an analogous way, for a transition from state j to final state J, (j+J)
[2(W + 1)/(2j + l)]
+ ;j4%r,
fj:(Kr) . . .)
(14)
where ‘c’ indicates that the choice of the terms inside the brackets is from 1j - J( to (j + J).
In the formalism of ZEMACHand GLAUBER(1956) the motion of a spherical top is considered as that of a rotator; in that case the co-ordinate y disappears and Dj__,r,_-kare restricted now to simply Di_),X,owhich is the same as Y,,,. Then, for a transition from j to J, we get instead of (13) @‘> = 2 ; (2n + l)jn(Kr”)jn(KrV*)1/[477/(2n +
I>1
Now C(jn J; 00) # 0 only if j + n + J is even; otherwise it is 0. It is an easy matter to calculate expression (15). In the formalism of ZEMACHand GLAUBER(1956) this expression is summed over all the final states J to get the so-called static approximation. It is easy to see that summing over J gives
2 z\ (212+ 1) j&r,)
n
(13)
j&r,,)
2/[4d(2n
+
111cdt.d
and this is the expansion of 2jo(Kr,,,), where r,,. is the distance between v and v’. This coincides with the result of ZEMACHand GLAUBER. We shall come to this again in Section 9 below. But it must be remarked that (15) is only an approximation to the correct (YV’)for a spherically symmetric molecule given by (12). This approximation amounts to the replacement of C(jn J; 0 0) by z/(W + l)/ 2/(2j + 1)(2n + 1) which is not in fact the value of C(jnJ;
2 s=ij-J/
x jn(Krv)j,(Krvt) 1/[4~/(2n + 1)i ~~,~(6.,,)
-
x ~,,&,JC( jn J; W2 (15)
Symmetry around the axis of y requires that this be a function of (4, - &) and not of $, and q$, separately.
=
;jlw - fj((Kr) +Zj3yKI)
+ 1)(4rr/2n + 1)
x C(Jj n ; K Q2 + (1 - S,,s) P, ,,-,(%) P,,K--k(&) x cos (K - k)(4, - &)C(Jj n; K -k)2}
(Y1”j
f&=/J-JI
6. SPHERICAL TOP MOLECULE SIMULATED BY A ROTATOR
+ 1)
XY n,~-@, &C(Jin;
x {P,,,:(Q
[2(2J + 1)/(2j + l)] ji!
M - m0)
-t-+1~(i)“j,(Kr)C(.@?;
(rv’) = 2 zj,(Kr,)j,(Kr,.)(W
For every pair of protons we have cos 8,,. = - l/3 ; if r is the distance C-H we get for every pair of protons
00).
For an atom, say v’, situated at the origin, rv, = 0, and we again get equations (10).
131
Scattering of slow neutrons by molecules
Example of CH, Every pair of protons will contribute, instead of (14), 2 xj&cr) II
d4n/(2n
+ 1) P,,,(cos 0 = -l/3) x C( j 17J; 00)2
where r is the distance C-H. 7. INCLUSION OF ZERO-POINT
VIBRATIONS
The position vector r, of a nucleus of mass m, can be written as r, = py + u, where u, is the displacement from py due to the vibrations. Let e(vln) denote the direction of displacement of mv for the nth mode of vibration. If [, denotes the nth mode so that 2T = 1 in2 and 2V = 2 ~~~~~~~~~ we can write l/m, u, =
8. TOTAL SCATTERING CROSS-SECTION Whereas (9) and (10) give the contribution of a pair Y,V’to the differential scattering cross-section for a chosen pair of initial and final states of the molecule, we can get the total differential scattering from a particular initial state by summing over all the final states which satisfy the energy-conservation condition. To get the total scattering from a particular initial state of the molecule we have to integrate over all scattering angles. This requires the evaluation of integrals of the type
j dd(‘%)
This can easily be shown to be
2 t, e(vln). Therm ”
exp (ix
??
(27&%,2)
rJ = exp (ix - pJ exp (i 2 6J~)5,)
the limits being:
n
where b,(jt) = K - e(vln)/l/m,. As has been shown by POPE (1952) the diagonal matrix element of the exponential containing the E,, for the zero-point state of vibration is exp (-I:
n
bn2(y)/4a,)
where
a,L = w&i.
Hence we require the rotational matrix elements of exp (in - pv) exp (-2
bn2(v)/4a,) = exp (ix - p,,) r8
x {I - 1 (K - e(vln))2/4a, m,}. Now e(vln) has components iA( e&yin), ec(vln) in the directions A, B, C of the principal axes. Thus x
* e(a+z)
=
K{-e,(vIn)
sin p cos y + eB(y]n)
x sin fi sin y + ec(+) ~(w*e(+z))2/4a,m, n
= +
Denoting
cos fi}
K2(Sin2~cos2y~(eA(Y~n))2/4a,m, n
similar terms).
2\ eA(vln) eg(vln)/4anmv
(16) by
(AB),
for
j&Y,) j,(Kr,*).
x1 =
r,lk
-
I2x in(x) j,(xrJv,,) s 21 k,]
;
- 2 sin B cos /? cos y (AC), $ sin2 @sin2 y (BB), -I- 2 sin B cos ,8 sin y (BC), + cos2 /3 (CC),) All the expressions in j3,r occurring in (16) can be expressed as products or sums of products of any two of %, 0; &l so that integration (4) can again be performed ‘quite easily leading to a rather long expression involving a sum of products of six ClebschGordon coefficients.
(17)
x2 = r,(k _L k”).
9. THE ‘HIGH’ ENERGY APPROXIMATION When the energy of the incident neutron (=:=VkOz/ 2m) is large compared to the separations of the rotational levels it is possible to make certain simplifying approximations. In this case the energy conservation condition cannot prevent any transitions so that a summation over all final states is possible. Further k = k, so that K effectively depends only on the incident momentum k, and the angle of scattering. Thus in (9), (12) and (13) the (vv’) can be summed over all the final states and this summation can disregard the products jn(~rv)jl,(Krv,) from what has been said above. In the general case for example, we get the total differential cross-section from state (j,e) by summing (vv’) from (9) over all J,E, giving 2 t: j&%)jn(Kr,$;
1{G$,;,,(%, cos $,)G$,~
x (i,,,, cos $,,) + G&,(0,,, sin +,)G$,(B,.,
sin +,.,) (18)
example, wenget for right-hand side of (16) K2(sin2t!?cos2 y (A A), - 2 sin2 B sin y cos y (AB),
d.y
The integration required for getting the total scattering cross-section then has the limits 0, 2k,r, instead of those given in the previous section. The effect of summing over all final states has already been commented upon in the case of a rotator, in Section 6. Also in the case of a spherical top molecule treated as such and not as a rotator we get some simplification on summing over the final states. In fact (13) gives, on summing over J, 2 2 (2n + l)j,cKrl)jn(Kr,,) n
1/4~1(2n -t 1) ~,,,oOL)
132
A. RAHMAN
which is the expansion of 2j,(~r,,), where r,,., is the distance between Y and Y’. We are thus led to the two rather important results; firstly that in the ‘high’ energy approximation for rotational scattering from a rigid spherical top molecule the reduction of the top to a rotator is justified; secondly that the scattering is independent of the state in which the molecule is found before scattering. It is interesting to note that though it has not been found possible to prove this result in the case of a nonspherical top, nevertheless, detailed computation on the water molecule, a symmetric top, has shown that this independence is still true even though not exactly. These results will be presented in another paper.
Ackno~Ie&entm&-The author gratefully acknowledges that this work and a computation on the water molecule which will be reported in another paper, were produced while the author was receiving a fellowship from the Comite d’Etude et d’Exploitation des Calculateurs Electroniques at Brussels. REFERENCES GORYWOV A. F. (1957) J. Nucl. Energy 2, 109. KHUBCHANDANI P. G. and RAHMAN A. (1960) Ann. Sot. Sci.
Bruxclles 74, 35. KRIEGERT. J. and NELKIN M. S. (1957) Whys. Rev. 106,290. POPEN. K. (1952) Conud. J. Phys. 30,597.
Rests M. E. (1957) Elementary Theory of Angular Momentum, John Wiley, New York. SACHSR. G. and TELLERE. (1941) Phys. Rev. 60,18. ZEMACHA. C. and GLAUBERR. J. (1956) Whys. Rev. 101, 118, 129.