Subatomic Physics

14.2. The Pion–Nucleon Interaction—Survey 429
states in terms of eigenstates of I and I3, denoted by |I,I3〉. Starting with the
|π + p〉 = � �
� 3 3
(7)
2 , 2 state and applying twice the isospin-lowering operator:
|π + � �
�
p〉 = �
3 3
� ,
2 2
|π − � � � � � �
1 �
p〉 = �
3
2 �
3 � , −1 − �
1
2 2 3 � , −1
(14.8)
2 2
|π 0 � � � � � �
2 �
n〉 = �
3
1 �
3 � , −1 + �
1
2 2 3 � , −1 .
2 2
To describe pion–nucleon scattering, a scattering operator S is introduced. The
operator S is not as frightening as it usually appears to the beginner, and all we
have to know about it are two properties. (1) The scattering amplitude f for a
collision ab → cd is proportional to the matrix element of S,
f ∝〈cd|S|ab〉.
The cross section is related to f by Eq. (6.2), or dσ/dΩ =|f| 2 . (2) The pion–nucleon
force is strong and assumed to be charge-independent. Thus the Hamiltonian HπN
must commute with the isospin operator,
[HπN, � I]=0.
Since pion–nucleon scattering occurs through the pion–nucleon force as shown in
Fig. 14.4, the scattering operator can be constructed from HπN. It therefore must
also commute with � I,
and with I 2 ,
[S, � I ]=0, (14.9)
[S, I 2 ]=0. (14.10)
Thus, if |I,I3〉 is an eigenstate of I 2 with eigenvalues I(I +1), so is S|I,I3〉. Consequently,
the state S|I,I3〉 is orthogonal to the state |I ′ ,I ′ 3〉, andthematrixelement
〈I ′ ,I ′ 3 |S|I,I3〉 vanishes unless I ′ = I,I ′ 3 = I3. Moreover, S does not depend on
I3, as is indicated by Eq. (14.9); the matrix element is independent of I3 and can
simply be written as 〈I|S|I〉. With the abbreviations
f 1/2 =
7 Merzbacher, Section 17.6; see also problem 15.7.
� � � �
1 1
3 3
|S| , f3/2 = |S|
2 2
2 2