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in two body problem we have $m_1$ and $m_2$ with positions $r_1$ and $r_2$

the position of the center of mass is $$ R = \frac{m_1r_1 + m_2r_2}{M} $$

where $$ {M} = {m_1 + m_2} $$

and the relative coordinates is $$ r = r_1 - r_2 $$

the reduced mass is $$ μ = \frac{m_1m_2}{m_1+m_2} $$

I found a relation in QM book by zettili

which is $$ \frac{1}{m_1}∇_{r_1}^2 + \frac{1}{m_2}∇_{r_2}^2 = \frac{1}{M}∇_R^2 + \frac{1}{μ} ∇_r^2$$

and I have no idea how did he get it. can someone explain to me the steps to get it

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1 Answer 1

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The classical Lagrangian for 2 free particles is $$ \mathcal L (\dot{\vec{r}} _1, \dot{\vec {r}} _2) = \frac{m_1}{2} \dot{\vec{r}} _1^2 + \frac{m_2}{2} \dot{\vec{r}} _2^2\quad.$$ Transforming it into center of mass coordinates yields $$\mathcal L (\dot{\vec{R}}, \dot{\vec {r}}) = \frac{M}{2} \dot{\vec{R}} ^2 + \frac{\mu}{2} \dot{\vec{r}} ^2\quad.$$ The canonical momenta therefore are $$\vec P _\text{CM} = M\dot{\vec R}\quad\text{ and } \quad \vec p _\text{rad} = \mu\dot{\vec r}\quad.$$ So the Legendre-Trafo of $\mathcal L $ yields $$\mathcal H(\vec P _\text{CM}, \vec p _\text{rad}) = \frac{\vec P _\text{CM} ^2}{2 M} + \frac{\vec p _\text{rad} ^2}{2 \mu}$$ Now changing to the quatum mechanical hamiltonian (using $\hbar = 1$) we get $$ \hat H = -\frac{\nabla^2 _R}{2M} - \frac{\nabla^2 _r}{2\mu}$$ Since the classical hamiltonian, before transforming in the center of mass frame, is $$ \mathcal H(\vec p_1, \vec p_2) = \frac{\vec p_1 ^2}{2m_1} + \frac{\vec p_2 ^2}{2m_2}\quad ,$$ which again using the operators yields the QM Operator $$ \hat H = -\frac{\nabla_{r_1} ^2}{2m_1} - \frac{\nabla_{r_2} ^2}{2m_2}\quad .$$ So we see that $$-\frac {\hat H} 2 = \frac{\nabla_{r_1} ^2}{m_1} + \frac{\nabla_{r_2} ^2}{m_2} = \frac{\nabla^2 _R}{M} + \frac{\nabla^2 _r}{\mu} $$

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