_5\right] u\left( {\mbox{\boldmath $p $}}\right) .  
\label{alambda0}$$

with $p_d=p+k$, $p_x=p^{\prime }-k$. Using Dirac equation, we rewrite (\[alambda0\]) as

$$T\left( \lambda \right) =-ig^2\tau _c\bar u\left( {\mbox{\boldmath $p'$}}\right) 
\left[ \frac{\not{k}}{p_d^2-m^2}+\frac{\not{k}}{p_x^2-m^2}+\frac \lambda
{gf_\pi }\right] \gamma _5u\left( {\mbox{\boldmath $p $}}\right) .
\label{alambda}$$

It is worth pointing out that, as shown in [@Mae+97], this result does not depend on our choice of the $PS$ coupling in non-linear Lagrangian and it would be the same if the $PV$ scheme were adopted.

This amplitude contains positive frequency states, which do not enter in the construction of the proper $\pi NN$ kernel. In order to isolate these contributions, we write the nucleon propagator as $$\frac{\not p+m}{p^2-m^2}=\frac 1{2E}\left[ \frac 1{p^0-E}\sum_su^s\left( {%
\mbox{\boldmath $p $}}\right) \bar u^s\left( {\mbox{\boldmath $p $}}\right) +%
\frac 1{p^0+E}\sum_sv^s\left( -{\mbox{\boldmath $p $}}\right) \bar v^s\left(
-{\mbox{\boldmath $p $}}\right) \right] ,$$ where $E=\sqrt{{\mbox{\boldmath $p $}}^{\ 2}+m^2}$. Thus $$\left[ \frac{\not p+m}{p^2-m^2}\right] _{(+)}=\frac 1{2E}\left[ \frac{\not
p+m}{p^0-E}-\gamma ^0\right]$$ and the positive energy contribution is $$\begin{aligned}
T_{\left( +\right) } &=&-ig^2\tau _c\bar u\left( {\mbox{\boldmath $p $}}%
^{\prime }\right) \left[ \frac{\not k}{2E_d\left( p_d^0-E_d\right) }+\frac{%
\not k}{2E_x\left( p_x^0-E_x\right) }\right.  \nonumber \\
&&\left. -\left( \frac 1{2E_d}-\frac 1{2E_x}\right) \gamma ^0\right] \gamma
_5u\left( {\mbox{\boldmath $p $}}\right) ,  \label{Tmais}\end{aligned}$$

with $\tilde p_i=\left( E_i,\vec p_i\right) $ for $i=d,x$.

Subtracting (\[Tmais\]) 