Subatomic Physics

358 The Weak Interaction
vanishes: Scalars transform like Y0, vectors like Y1, andtheintegral � d 3 xY ∗
0 Y1Y0
vanishes. The only term left under the integral is V0, and the matrix element takes
on the form
〈β|Hw|α〉 = 1
√ 2 GF Vudu ∗ e (V0 − A0)u¯ν〈1〉, (11.64)
where 〈1〉 is the symbol used in nuclear physics for the integral
�
〈1〉 =
d 3 xψ ∗ 0 + β V0ψ o + α. (11.65)
The recoil energy imparted to the decaying nucleus is very small so that the nuclear
matrix element 〈1〉 can be computed nonrelativistically; the result is
〈1〉 = √ 2, (11.66)
if the states β and α have the same isospin and are part of the same multiplet.
• To verify Eq. (11.66), we use the nonrelativistic operator V0 = 1 from
Eq. (11.40) so that
�
〈1〉 = d 3 xψ ∗ 0 + βψ0 + α.
A new problem arises here: the wave functions ψβ and ψα belong to different isobars
and hence are orthogonal. As written, the integral vanishes. The solution to the
problem is simple if the isospin formalism is introduced. The states in 14 Oand 14 N
belong to the same I = 1 isospin multiplet, with I3 values of 1 and 0, respectively.
They have the same spatial wave function so that the total wave functions can be
written
14 O: ψα = ψ0(x)Φ1,1
14 N: ψβ = ψ0(x)Φ1,0.
where, Φ1,1 and Φ1,0 denote the normalized isospin functions. The weak current
changes 14 Ointo 14 N; it lowers the I3 value by one unit. This lowering is expressed
by the operator I−, given in Eq. (8.26). In the isospin formalism the complete
matrix element 〈1〉 thus becomes
�
〈1〉 =
The isospin part is evaluated with Eq. (8.27):
d 3 xψ ∗ 0 (x)ψ0(x)Φ ∗ 1,0 I−Φ1,1.
Φ ∗ 1,0I−Φ1,1 = √ 2Φ ∗ 1,0Φ1,0 = √ 2.
The spatial wave function is normalized to 1 so that the final result, 〈1〉 = √ 2,
verifies Eq. (11.66). •