
126 L. D. Faddeev and 0. A. Yakubovskij
Therefore. we can write
(7)
J=e.in = -
(j T m) (j ± m + 1) ej.r1at
This formula enables us for any unit basis vector to construct new
unit basis vectors e j,7 ± 1 satisfying all the requirements The minus
sign in front of the square root is writ ten for convenience.
We have not yet explained what values the numbers j and m can
take. We start from the equalities
(8)
J_ ejmi = 0,
J+ejr2 = 0
Multiplying these equalities by J +. and by J_ and using (5), we get
that,
(J2J+J3)jmi=0.
(J2-JJ.3)jm2=0,
3
or
J(j + 1) _ M2 + mi = 0,
.
j(j +1) -rn -m2 =0
From (9) we get at, once that (mi + M2) (MI - m2
- 1) = 0. We take
the solution ml = -m2 of this equation, since m2 > rnl. Further,
rn2 - rn z = 2rn2 is an integer or zero. Therefore, rrm2 can take the
values 0, 1/2. 1, 3/2, .... Finally, we see from (0) that j can be taken
to be m2. This means that the eigenvalues of 'I 2 have the form
j (j + 1), where j = 0, 1/2, 1, 3/2, .
. ,
and for a given j the eigenvalues
m of J3 ruri through the (2j + 1) values -j, + 1, ... , j - 1. j. The
riuinbers j and in are simultaneously either integers or half-integers.
We emphasize once more that these properties of the spectrum of the
operators J2 and J have been obtained using only the commutation
relations.
To finish the proof it remains for us to see that the eigerivectors
constructed form a basis in £. This follows from the irreducibility of
the representation. Indeed, the subspace E` spanned by the vectors
e37 n with m = -j, -j + 1,. .. , j will be invariant with respect to the
operators Jk, k = 1, 2, 3, and therefore must, coincide with e, and the
dimension of the representation is n = 2j + 1. The formulas (1) and
(7) show that for this choice of basis the matrices Jk coincide with
the matrices Mk.