Eq. (2.4.7) of Weinberg, "Lectures on QM" from eq. (2.4.6) Could anyone please tell me the proof of (2.4.7), using (2.4.6)?
Substituting (2.4.7) into the right equation of (2.4.4), we get:
$$P_1 = -i \hbar \left( \frac{\partial}{\partial x_{1e}} + \frac{\partial}{\partial x_{1N}} \right).$$
Also, substituting the right equation of (2.4.3) into the RHS of the right equation of (2.4.7), we get:
$$(-i \hbar \bf \nabla _{X} \rm )_1 = - i \hbar \frac{\partial}{\partial X_1} = - i \hbar \frac{\partial}{\frac{m_e \partial x_{1e} + m_N \partial x_{1N}}{m_e + m_N}} = - i \hbar (m_e + m_N) \frac{\partial}{m_e \partial x_{1e} + m_N \partial x_{1N}}.$$
Why we can say these are equal?

 A: When we say two operators $A_1$ and $A_2$ are equal, we mean that for any wave function $\psi$, we have
$$A_1\psi = A_2\psi.$$
I will answer your question in the one-dimension case. The generalization to three-dimension case is straightforward.
Suppose the wave function for the system is $\psi(x_{\rm e}, x_{\rm n})$. Then 
$$\hat{p}\psi(x_{\rm e}, x_{\rm n}) = \frac{-i\hbar}{m_{\rm e}+ m_{\rm n}}\left(m_{\rm n}\nabla_{\rm e} - m_{\rm e}\nabla_{\rm n}\right)\psi(x_{\rm e}, x_{\rm n})$$
On the other hand, we have
$$-i\hbar\nabla_{x}\psi(x_{\rm e}, x_{\rm n}) = -i\hbar\left(\nabla_{\rm e}\psi\times\frac{\partial x_e}{\partial x} + \nabla_{\rm n}\psi\times\frac{\partial x_n}{\partial x}\right)$$
From 2.4.3, we can get
$$x_{\rm n} = X - \frac{m_{\rm e}}{m_{\rm e}+m_{\rm n}}x$$
$$x_{\rm e} = X + \frac{m_{\rm n}}{m_{\rm e}+m_{\rm n}}x$$
Put it into the equation above, we can prove that
$$\hat{p}\psi(x_{\rm e}, x_{\rm n}) = -i\hbar\nabla_{x}\psi(x_{\rm e}, x_{\rm n}) $$
So, we can prove
$$\hat{p} = -i\hbar\nabla_{x}$$
To prove 
$$\hat{P} = -i\hbar\nabla_{X}$$
is similar and you can do it yourself.
