20

GEORGE A. HAGEDORN

for some vectors b, c, and d. Generically 6, c, and d are linearly independent. By a rotation

of the coordinate system we may assume that only the first three components of 6, c, and d

are non-zero.

If 77(0) is a generic nuclear momentum vector, then we can rotate the first three coor-

dinate axes so that the projection of 7/(0) into the three dimensional subspace spanned by 6,

c, and d lies along the positive X\ axis.

At this point, the Xj coordinates for j 3 no longer play a role in the structure of

Ni(X). Furthermore, without altering the basic structure obtained so far, we still have the

freedom to rotate the X2 and X3 coordinate directions, and we can perform X-independent

complex rotations (unitary transformations) of the two dimensional space spanned by the

basic electronic wave functions ip\(X) and ip2(X). By doing both of these in a special way,

we claim that we may assume the following:

1. The first component of b is non-zero.

2. The first and third components of c are zero, but its second component is positive.

3. The first and second components of d are zero, but its third component is positive.

Thus, we may assume that N(X) has the form

N

(X) = ( blXl + ^ 2 + h^S c2X2 + id^Xs \

\ C2X2 - ^3X 3 -b\Xi - b2X2 - 63^3 /

To prove these claims we first do a unitary transformation of the span of ip\(X) and

ip2(X) so that when X2 = X3 = • • • = Xjy — 0, the matrix M\{X) is diagonal. Standard

one variable perturbation theory shows that this can always be done. Thus, we may assume

that c\ = d\ = 0.

Next, we show that we can do another unitary transformation of the span of ipi(X)

and ^2{X) so that c\ and d\ are unchanged, but c and d are transformed into perpendicular

vectors. The unitary transformation we use simply multiplies ip2(X) by a phase factor e .

This diagonal transformation leaves M\(X) diagonal when X2 = X3 = • • • = Xjy = 0, so c\

and d\ are not altered. However, the similarity transformation replaces

by

c2

h

d2\

dz) '

fc2 d2\

V C3 d3 J

fc2 d2\

V ^3 d3)

/cos 6

Vsin 6

— sin

cos t