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

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

is the reduced mass of the system

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

$$\begin{gathered} \mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{1}}}\mathbf{}{\mathbf{∇}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{ψ}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}{\mathbf{∇}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{ψ}\mathbf{+}\mathbf{³Õψ}\mathbf{=}\mathbf{·¡Ïˆ} \\ \mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\left(m_1+m_2\right)}\mathbf{}{\mathbf{∇}}_{\mathbf{R}}^{\mathbf{2}}\mathbf{ψ}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{μ}}\mathbf{}{\mathbf{∇}}_{\mathbf{r}}^{\mathbf{2}}\mathbf{ψ}\mathbf{+}\mathbf{V}\left(r\right)\mathbf{ψ}\mathbf{=}\mathbf{·¡Ïˆ}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(5.7\right)\mathbf{.} \end{gathered}$$

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

Step by step solution

01

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

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

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

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

02

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

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

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

03

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

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

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:

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

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