Question

Typically, the interaction potential depends only on the vector$\mathit{r}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\mathbf{}$between

the two particles. In that case the Schrodinger equation seperates, if we change variables from ${\mathit{r}}_{\mathbf{1}}\mathbf{,}{\mathit{r}}_{\mathbf{2}}$and$\mathit{R}\mathbf{=}\left(m_1{r}_1+m_2{r}_2\right)\mathit{I}\left(m_1+m_2\right)$(the center of mass).

(a)Show that

localid="1655976113066" $$\begin{gathered} {\mathbf{r}}_{\mathbf{1}}\mathbf{=}\mathbf{R}\mathbf{+}\mathbf{(}\mathbf{\mu }\mathbf{/}{\mathbf{m}}_{\mathbf{1}}\mathbf{)}\mathbf{r}\mathbf{,}\mathbf{}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{R}\mathbf{-}\mathbf{(}\mathbf{\mu }\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{r}\mathbf{,} \\ \mathbf{and}\mathbf{}{\mathbf{\nabla }}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{\mu }\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{}{\mathbf{\nabla }}_{\mathbf{R}}\mathbf{+}{\mathbf{\nabla }}_{\mathbf{r}\mathbf{,}}\mathbf{}{\mathbf{\nabla }}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{\mu }\mathbf{/}{\mathbf{m}}_{\mathbf{1}}\mathbf{)}\mathbf{}{\mathbf{\nabla }}_{\mathbf{R}}\mathbf{-}{\mathbf{\nabla }}_{\mathbf{r}\mathbf{,}}\mathbf{where} \end{gathered}$$

localid="1655976264171" $\mathbf{\mu }\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(5.15\right)\mathbf{.}$

is the reduced mass of the system

(b) Show that the (time-independent) Schrödinger equation (5.7) becomes

$$\begin{gathered} \mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{1}}}\mathbf{}{\mathbf{\nabla }}_{\mathbf{1}}^{\mathbf{2}}\mathbf{\psi }\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}{\mathbf{\nabla }}_{\mathbf{2}}^{\mathbf{2}}\mathbf{\psi }\mathbf{+}\mathbf{V\psi }\mathbf{=}\mathbf{E\psi } \\ \mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\left(m_1+m_2\right)}\mathbf{}{\mathbf{\nabla }}_{\mathbf{R}}^{\mathbf{2}}\mathbf{\psi }\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\mu }}\mathbf{}{\mathbf{\nabla }}_{\mathbf{r}}^{\mathbf{2}}\mathbf{\psi }\mathbf{+}\mathbf{V}\left(r\right)\mathbf{\psi }\mathbf{=}\mathbf{E\psi }\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(5.7\right)\mathbf{.} \end{gathered}$$

(c) Separate the variables, letting $\mathbf{\psi }\left(R,r\right)\mathbf{=}{\mathbf{\psi }}_{\mathbf{R}}\left(R\right){\mathbf{\psi }}_{\mathbf{r}}\left(r\right)$Note that ${\mathbf{\psi }}_{\mathbf{R}}$satisfies the

one-particle Schrödinger equation, with the total mass $\left(m_1+m_2\right)$in place of m, potential zero, and energy ${\mathbf{E}}_{\mathbf{R}}$while ${\mathbf{\psi }}_{\mathbf{r}}$satisfies the one-particle Schrödinger equation with the reduced mass in place of m, potential V(r) , and energy localid="1655977092786" ${\mathbf{E}}_r$. The total energy is the sum: $\mathit{E}\mathbf{=}{\mathit{E}}_{\mathbf{R}}\mathbf{+}{\mathit{E}}_{\mathbf{r}}$. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics; it reduces the two-body problem to an equivalent one-body problem.

Short answer

$\left(a\right)\mathbf{=}\left(\frac{m_2}{m_1+m_2}\right)\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{X}}\mathbf{-}\left(1\right)\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\frac{\mathbf{\mu }}{{\mathbf{m}}_{\mathbf{1}}}{\left({\nabla }_R\right)}_{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathit{s}\mathit{o}\mathbf{}{\mathbf{\nabla }}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{\mu }}{{\mathbf{m}}_{\mathbf{1}}}\mathbf{}{\mathbf{\nabla }}_{\mathbf{R}}\mathbf{-}{\mathbf{\nabla }}_{\mathbf{r}}\mathbf{}\mathbf{.}$

$\left(b\right)\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}{\mathbf{\nabla }}_{\mathbf{R}}^{\mathbf{2}}\mathit{\psi }\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\mu }}\mathbf{}{\mathbf{\nabla }}_{\mathbf{r}}^{\mathbf{2}}\mathit{\psi }\mathbf{+}\mathit{V}\left(r\right)\mathit{\psi }\mathbf{=}\mathit{E}\mathit{\psi }\mathbf{.}$

$\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\left(m_1+m_2\right)}\mathbf{}{\mathbf{\nabla }}^{\mathbf{2}}{\mathit{\psi }}_{\mathbf{R}}\mathbf{=}{\mathit{E}}_{\mathbf{R}}{\mathit{\psi }}_{\mathbf{R}\mathbf{,}}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\mu }}\mathbf{}{\mathbf{\nabla }}^{\mathbf{2}}{\mathit{\psi }}_{\mathbf{r}}\mathbf{+}\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}{\mathit{\psi }}_{\mathbf{r}}\mathbf{=}{\mathit{E}}_{\mathbf{r}}{\mathit{\psi }}_{\mathbf{r}}\mathbf{,}\mathit{w}\mathit{i}\mathit{t}\mathit{h}{\mathit{E}}_{\mathbf{R}}\mathbf{+}{\mathit{E}}_{\mathbf{r}}\mathbf{=}\mathit{E}\mathbf{.}$

Step-by-step solution

(a)Showingr1=R+(μ/m1)r, r2=R-(μ/m2)r,and ∇1=(μ/m2) ∇R+∇r,∇2=(μ/m1) ∇R-∇r, 

$$\begin{gathered} \left(m_1+m_2\right)R=m_1{r}_1+m_2{m}_2=m_1{r}_1+m_2\left(r_1-r\right)=\left(m_1+m_2\right)r_1-m_{2}r\Rightarrow \\ {r}_1=R+\frac{m_2}{m_1+m_2}r=R+\frac{\mu }{m_1}r \\ \left(m_1+m_2\right)R=m_1\left(r_2+r\right)+m_2{r}_2=\left(m_1+m_2\right)r_2+m_{1}r\Rightarrow r_2=R-\frac{m_1}{m_1{m}_2}r=R-\frac{\mu }{m_2}r. \end{gathered}$$

Let R = (X, Y,Z); r = (x,y, z).

$$\begin{gathered} {\left({\nabla }_1\right)}_x=\frac{\partial }{\partial x_1}=\frac{\partial X}{\partial x_1}\frac{\partial }{\partial X}+\frac{\partial x}{\partial x_1}\frac{\partial }{\partial x}. \\ =\left(\frac{m_1}{m_1+m_2}\right)\frac{\partial }{\partial X}+\left(1\right)\frac{\partial }{\partial x}=\frac{\mu }{m_2}{\left({\nabla }_R\right)}_X+{\left({\nabla }_r\right)}_X,so{\nabla }_1=\frac{\mu }{m_2}{\nabla }_R+{\nabla }_r. \\ {\left({\nabla }_2\right)}_x=\frac{\partial }{\partial x_2}=\frac{\partial X}{\partial x_2}\frac{\partial }{\partial X}+\frac{\partial x}{\partial x_2}\frac{\partial }{\partial x}. \\ =\left(\frac{m_2}{m_1+m_2}\right)\frac{\partial }{\partial X}-\left(1\right)\frac{\partial }{\partial x}=\frac{\mu }{m_1}{\left({\nabla }_R\right)}_X+{\left({\nabla }_r\right)}_X,so{\nabla }_2=\frac{\mu }{m_1}{\nabla }_R+{\nabla }_r. \end{gathered}$$

(b)Showing the (time-independent) Schrödinger equation

$$\begin{gathered} {\nabla }_1^2\psi ={\nabla }_1.\left({\nabla }_1\psi \right)={\nabla }_1.\left[\frac{\mu }{m_2}{\nabla }_R\psi +{\nabla }_r\psi \right]. \\ =\frac{\mu }{m_2}{\nabla }_R.\left(\frac{\mu }{m_2}{\nabla }_R\psi +{\nabla }_r\psi \right)+{\nabla }_r.\left(\frac{\mu }{m_2}{\nabla }_R\psi +{\nabla }_r\psi \right). \\ ={\left(\frac{\mu }{m^2}\right)}^2{\nabla }_R^2\psi +2\frac{\mu }{m_2}\left({\nabla }_r.{\nabla }_R\right)\psi +{\nabla }_r^2\psi . \\ Likewise,{\nabla }_2^2\psi =\left(\frac{\mu }{m_1}\right){\nabla }_R^2\psi -2\left({\nabla }_r.{\nabla }_R\right)+{\nabla }_r^2\psi . \end{gathered}$$

$$\begin{gathered} \therefore H\psi =-\frac{{\sout{h}}^2}{2m_1}{\nabla }_1^2\psi -\frac{{\sout{h}}^2}{2m_2}{\nabla }_2^2\psi +V\left(r_1,r_2\right)\psi . \\ =-\frac{{\sout{h}}^2}{2}\left(\frac{{\mu }^2}{m_1{m}_2}{\nabla }_R^2+\frac{2\mu }{m_1{m}_2}{\nabla }_r.{\nabla }_R+\frac{1}{m_1}{\nabla }_r^2+\frac{{\mu }^2}{m_1{m}_1^2}{\nabla }_R^2-\frac{2\mu }{m_2{m}_1}{\nabla }_r.{\nabla }_R+\frac{1}{m_2}{\nabla }_r^2\right)\psi \\ +V\left(r\right)\psi =-\frac{{\sout{h}}^2}{2}\left[\frac{{\mu }^2}{m_1{m}_2}\left(\frac{1}{m_2}+\frac{1}{m_2}\right){\nabla }_R^2+\left(\frac{1}{m_2}+\frac{1}{m_2}\right){\nabla }_R^2\right]\psi +V\left(r\right)\psi =E\psi . \\ But\left(\frac{1}{m_2}+\frac{1}{m_2}\right)=\frac{m_1+m_2}{m_1{m}_2}=\frac{1}{\mu },so\frac{{\mu }^2}{m_1{m}_2}\left(\frac{1}{m_2}+\frac{1}{m_2}\right)=\frac{{\mu }^2}{m_1{m}_2}=\frac{m_1{m}_2}{m_1{m}_2\left(m_1{m}_2\right)}=\frac{1}{m_1+m_2} \\ -\frac{{\sout{h}}^2}{2\left(m_1+m_2\right)}{\nabla }_R^2\psi -\frac{{\sout{h}}^2}{2\mu }{\nabla }_R^2\psi +V\left(r\right)\psi =E\psi . \end{gathered}$$

(c) Separating the variables, letting ψ(R,r)=ψR(R)ψr(r)

$$\begin{gathered} \mathrm{put}\mathrm{in}\psi ={\psi }_r\left(r\right){\psi }_R\left(R\right),anddivideby{\psi }_r{\psi }_R: \\ \left[-\frac{{\sout{h}}^2}{2\left(m_1+m_2\right)}\frac{1}{{\psi }_R}{\nabla }_R^2{\psi }_R\right]+\left[-\frac{{\sout{h}}^2}{2\mu }\frac{1}{{\psi }_r}{\nabla }_r^2{\psi }_r+V\left(r\right)\right]. \end{gathered}$$

The first term depends only on R, the second only on r, so each must be a constant; call

them $E_{R}and{E}_r$, respectively. Then:

$-\frac{{\sout{\mathrm{h}}}^2}{2({\mathrm{m}}_1+{\mathrm{m}}_2)}{\nabla }^2{\mathrm{\psi }}_{\mathrm{R}}={\mathrm{E}}_{\mathrm{R}}{\mathrm{\psi }}_{\mathrm{R},}-\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{\mu }}{\nabla }^2{\mathrm{\psi }}_{\mathrm{r}}+\mathrm{V}\left(\mathrm{r}\right){\mathrm{\psi }}_{\mathrm{r}}={\mathrm{E}}_{\mathrm{r}}{\mathrm{\psi }}_{\mathrm{r}},{\mathrm{withE}}_{\mathrm{R}}+{\mathrm{E}}_{\mathrm{r}}=\mathrm{E}.$