Chapter 5.1, Problem 5.1P

(a)

To determine

Show that $r_1=R+\left(\mu /m_1\right)r$, $r_2=R-\left(\mu /m_2\right)r$, and ${\nabla }_1=\left(\mu /m_2\right){\nabla }_R+{\nabla }_r$, ${\nabla }_2=\left(\mu /m_1\right){\nabla }_R-{\nabla }_r$.

Answer to Problem 5.1P

It is proved that $r_1=R+\left(\mu /m_1\right)r$, $r_2=R-\left(\mu /m_2\right)r$, and ${\nabla }_1=\left(\mu /m_2\right){\nabla }_R+{\nabla }_r$, ${\nabla }_2=\left(\mu /m_1\right){\nabla }_R-{\nabla }_r$.

Explanation of Solution

Given, the separation between the two particles is

    $r=r_1-r_2$        (I)

The position of the center of the mass is

    $R\equiv \frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$        (II)

The reduced mass of the system is

    $\mu =\frac{m_1{m}_2}{m_1+m_2}$        (III)

Substitute equation (I) in (II) and solve for $r_1$

  $\begin{array}{c}R=\frac{m_1{r}_1+m_2\left(r_1-r\right)}{m_1+m_2}\\ =\frac{m_1{r}_1+m_2{r}_1-m_{2}r}{m_1+m_2}\\ =\frac{\left(m_1+m_2\right)r_1-m_{2}r}{m_1+m_2}\\ =\frac{\left(m_1+m_2\right)r_1}{m_1+m_2}-\frac{m_{2}r}{m_1+m_2}\end{array}$

  $\begin{array}{c}=r_1-\left(\frac{m_2}{m_1+m_2}\right)r\\ R=r_1-\frac{\mu }{m_1}r\\ r_1=R+\frac{\mu }{m_1}r\end{array}$

Hence it is proved that $r_1=R+\frac{\mu }{m_1}r$.

Similarly substitute equation (I) in (II) and solve for $r_2$

  $\begin{array}{c}R=\frac{m_1\left(r_2+r\right)+m_2{r}_2}{m_1+m_2}\\ =\frac{m_1{r}_2+m_{1}r+m_2{r}_2}{m_1+m_2}\\ =\frac{\left(m_1+m_2\right)r_2+m_{1}r}{m_1+m_2}\\ =\frac{\left(m_1+m_2\right)r_2}{m_1+m_2}+\frac{m_{1}r}{m_1+m_2}\end{array}$

  $\begin{array}{c}=r_2+\left(\frac{m_1}{m_1+m_2}\right)r\\ R=r_2+\frac{\mu }{m_2}r\\ r_2=R-\frac{\mu }{m_2}r\end{array}$

Hence it is prove that $r_2=R-\frac{\mu }{m_2}r$.

Let the coordinates of $r=\left(x,y,z\right)$ and $R=\left(X,Y,Z\right)$. ${\nabla }_1$ and ${\nabla }_2$ indicates differentiation with respect to the coordinates of particle $1$ and particle $2$ respectively.

  ${\left({\nabla }_1\right)}_x=\frac{\partial X}{\partial x_1}\frac{\partial }{\partial X}+\frac{\partial x}{\partial x_1}\frac{\partial }{\partial x}$        (IV)

Equation (I) and (II) with respect to $x$

Substitute $\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ for $X$ and $x_1-x_2$ for $x$ in equation (IV) to solve for ${\left({\nabla }_1\right)}_x$

  $\begin{array}{c}{\left({\nabla }_1\right)}_x=\frac{\partial }{\partial x_1}\left(\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\right)\frac{\partial }{\partial X}+\frac{\partial }{\partial x_1}\left(x_1-x_2\right)\frac{\partial }{\partial x}\\ =\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial X}+\frac{\partial }{\partial x}\end{array}$

Similarly equation (I) and (II) with respect to $y$

  ${\left({\nabla }_1\right)}_y=\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial Y}+\frac{\partial }{\partial y}$

Similarly equation (I) and (II) with respect to $z$

  ${\left({\nabla }_1\right)}_z=\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial Z}+\frac{\partial }{\partial z}$

Adding,

  $\begin{array}{c}{\left({\nabla }_1\right)}_x+{\left({\nabla }_1\right)}_y{\left({\nabla }_1\right)}_z=\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial X}+\frac{\partial }{\partial x}+\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial Y}+\frac{\partial }{\partial y}+\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial Z}+\frac{\partial }{\partial z}\\ =\left(\frac{m_1}{m_1+m_2}\right)\left(\frac{\partial }{\partial X}+\frac{\partial }{\partial Y}+\frac{\partial }{\partial Z}\right)+\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)\\ {\nabla }_1=\frac{\mu }{m_2}{\nabla }_R+{\nabla }_r\end{array}$

Hence it is proved that ${\nabla }_1=\frac{\mu }{m_2}{\nabla }_R+{\nabla }_r$.

Similarly for particle $2$

  ${\left({\nabla }_2\right)}_x=\frac{\partial X}{\partial x_2}\frac{\partial }{\partial X}+\frac{\partial x}{\partial x_2}\frac{\partial }{\partial x}$        (IV)

Equation (I) and (II) with respect to $x$

Substitute $\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ for $X$ and $x_1-x_2$ for $x$ in equation (IV) to solve for ${\left({\nabla }_2\right)}_x$

  $\begin{array}{c}{\left({\nabla }_2\right)}_x=\frac{\partial }{\partial x_2}\left(\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\right)\frac{\partial }{\partial X}+\frac{\partial }{\partial x_2}\left(x_1-x_2\right)\frac{\partial }{\partial x}\\ =\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial X}-\frac{\partial }{\partial x}\end{array}$

Similarly equation (I) and (II) with respect to $y$

  ${\left({\nabla }_2\right)}_y=\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial Y}-\frac{\partial }{\partial y}$

Similarly equation (I) and (II) with respect to $z$

  ${\left({\nabla }_2\right)}_z=\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial Z}-\frac{\partial }{\partial z}$

Adding,

  $\begin{array}{c}{\left({\nabla }_2\right)}_x+{\left({\nabla }_2\right)}_y{\left({\nabla }_2\right)}_z=\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial X}-\frac{\partial }{\partial x}+\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial Y}-\frac{\partial }{\partial y}+\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial Z}-\frac{\partial }{\partial z}\\ =\left(\frac{m_2}{m_1+m_2}\right)\left(\frac{\partial }{\partial X}+\frac{\partial }{\partial Y}+\frac{\partial }{\partial Z}\right)-\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)\\ {\nabla }_2=\frac{\mu }{m_1}{\nabla }_R-{\nabla }_r\end{array}$

Hence it is proved that ${\nabla }_2=\frac{\mu }{m_1}{\nabla }_R-{\nabla }_r$.

Conclusion:

Thus, it is proved that $r_1=R+\left(\mu /m_1\right)r$, $r_2=R-\left(\mu /m_2\right)r$, and ${\nabla }_1=\left(\mu /m_2\right){\nabla }_R+{\nabla }_r$, ${\nabla }_2=\left(\mu /m_1\right){\nabla }_R-{\nabla }_r$.

(b)

To determine

Show that the Schrodinger equation becomes $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\left(r\right)\psi =E\psi$ .

Answer to Problem 5.1P

It is proved that the Schrodinger equation becomes $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\left(r\right)\psi =E\psi$ .

Explanation of Solution

Write the Schrodinger’s time-independent equation for two-particle system

    $-\frac{{\hslash }^2}{2m_1}{\nabla }_1^2\psi -\frac{{\hslash }^2}{2m_2}{\nabla }_2^2\psi +V\psi =E\psi$        (I)

From part (a), ${\nabla }_1=\frac{\mu }{m_2}{\nabla }_R+{\nabla }_r$ and ${\nabla }_2=\frac{\mu }{m_1}{\nabla }_R-{\nabla }_r$. Substituting in the above equation

$\begin{array}{c}-\frac{{\hslash }^2}{2m_1}{\left(\frac{\mu }{m_2}{\nabla }_R+{\nabla }_r\right)}^2\psi -\frac{{\hslash }^2}{2m_2}{\left(\frac{\mu }{m_1}{\nabla }_R-{\nabla }_r\right)}^2\psi +V\psi =E\psi \\ -\frac{{\hslash }^2}{2m_1}\left(\frac{{\mu }^2}{m_2^2}{\nabla }_R^2+{\nabla }_r^2+2\frac{\mu }{m_2}{\nabla }_R{\nabla }_r\right)\psi -\frac{{\hslash }^2}{2m_2}\left(\frac{{\mu }^2}{m_1^2}{\nabla }_R^2+{\nabla }_r^2-2\frac{\mu }{m_1}{\nabla }_R{\nabla }_r\right)\psi +V\psi =E\psi \\ -\frac{{\hslash }^2}{2}\left(\frac{{\mu }^2}{m_1{m}_2^2}{\nabla }_R^2+{\nabla }_r^2\left(\frac{1}{m_1}\right)+2\frac{\mu }{m_1{m}_2}{\nabla }_R{\nabla }_r\right)\psi -\frac{{\hslash }^2}{2}\left(\frac{{\mu }^2}{m_1^2{m}_2}{\nabla }_R^2+{\nabla }_r^2\left(\frac{1}{m_2}\right)-2\frac{\mu }{m_1{m}_2}{\nabla }_R{\nabla }_r\right)\psi +V\psi =E\psi \\ -\frac{{\hslash }^2}{2}\left(\left(\frac{{\mu }^2}{m_1{m}_2^2}+\frac{{\mu }^2}{m_1^2{m}_2}\right){\nabla }_R^2\psi +{\nabla }_r^2\psi \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right)+V\psi =E\psi \end{array}$

Since, from equation (III), $\frac{1}{\mu }=\frac{1}{m_1}+\frac{1}{m_2}$ and $\mu =\frac{m_1{m}_2}{m_1+m_2}$.

  $\begin{array}{c}-\frac{{\hslash }^2}{2}\left(\frac{{\mu }^2}{m_1{m}_2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right){\nabla }_R^2\psi +{\nabla }_r^2\psi \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right)+V\psi =E\psi \\ -\frac{{\hslash }^2}{2}\frac{{\mu }^2}{m_1{m}_2}\left(\frac{1}{\mu }\right){\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\psi =E\psi \\ -\frac{{\hslash }^2}{2}\frac{\mu }{m_1{m}_2}\left(\frac{m_1+m_2}{m_1+m_2}\right){\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\psi =E\psi \\ -\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\psi =E\psi \end{array}$

Hence it is proved that $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\psi =E\psi$.

Conclusion:

Thus, it is proved that the Schrodinger equation becomes $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\hslash }^2}{2\mu }{\nabla }_r^2\psi +V\left(r\right)\psi =E\psi$ .

(c)

To determine

The one-particle Schrodinger equation of ${\psi }_R$ with the total mass $\left(m_1+m_2\right)$ in place of $m$ and ${\psi }_r$ with the reduced mass in place of $m$. And solve for total energy.

Answer to Problem 5.1P

The one-particle Schrodinger equation of ${\psi }_R$ with the total mass $\left(m_1+m_2\right)$ in place of $m$ is $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2{\psi }_R=E_R{\psi }_R$ and ${\psi }_r$ with the reduced mass in place of $m$ is $-\frac{{\hslash }^2}{2\mu }{\nabla }_r^2{\psi }_r+V\left(r\right){\psi }_r=E{r}_{}{\psi }_r$. And the total energy is $E=-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}\left(\frac{1}{{\psi }_R}\right){\nabla }_R^2{\psi }_R+-\frac{{\hslash }^2}{2\mu }\left(\frac{1}{{\psi }_r}\right){\nabla }_r^2{\psi }_r+V\left(r\right)$.

Explanation of Solution

Write the one-particle Schrodinger equation

    $-\frac{{\hslash }^2}{2m}{\nabla }^2\psi +V\psi =E\psi$        (I)

The one-particle Schrodinger equation for ${\omega }_R$ with total mass $\left(m_1+m_2\right)$ in place of $m$, zero potential and energy $E_R$ can be written as

  $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2{\psi }_R=E_R{\psi }_R$

The one-particle Schrodinger equation for ${\omega }_r$ with the reduced mass in place of $m$, potential $V\left(r\right)$ and energy $E_r$ can be written as

  $-\frac{{\hslash }^2}{2\mu }{\nabla }_r^2{\psi }_r+V\left(r\right){\psi }_r=E{r}_{}{\psi }_r$

The total energy is

  $\begin{array}{l}E=E_r+E_R\\ E=-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}\left(\frac{1}{{\psi }_R}\right){\nabla }_R^2{\psi }_R+-\frac{{\hslash }^2}{2\mu }\left(\frac{1}{{\psi }_r}\right){\nabla }_r^2{\psi }_r+V\left(r\right)\end{array}$

Hence it is proved.

Conclusion:

The one-particle Schrodinger equation of ${\psi }_R$ with the total mass $\left(m_1+m_2\right)$ in place of $m$ is $-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}{\nabla }_R^2{\psi }_R=E_R{\psi }_R$ and ${\psi }_r$ with the reduced mass in place of $m$ is $-\frac{{\hslash }^2}{2\mu }{\nabla }_r^2{\psi }_r+V\left(r\right){\psi }_r=E{r}_{}{\psi }_r$. And the total energy is $E=-\frac{{\hslash }^2}{2\left(m_1+m_2\right)}\left(\frac{1}{{\psi }_R}\right){\nabla }_R^2{\psi }_R+-\frac{{\hslash }^2}{2\mu }\left(\frac{1}{{\psi }_r}\right){\nabla }_r^2{\psi }_r+V\left(r\right)$.