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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?

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Qmechanic
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ynn
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1 Answers1

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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.

Eric Yang
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