Question

Although the Schrödinger equation for helium itself cannot be solved exactly, there exist “helium-like” systems that do admit exact solutions. A simple example is “rubber-band helium,” in which the Coulomb forces are replaced by Hooke’s law forces:

$\mathit{H}\mathbf{=}\mathbf{-}\frac{{\mathbf{ħ}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\mathbf{(}{\mathbf{\nabla }}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{\nabla }}_{\mathbf{2}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{\omega }}^{\mathbf{2}}{\left|r_1-r_1\right|}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{78}\mathbf{)}\mathbf{.}$

(a) Show that the change of variables from

${\mathit{r}}_{\mathbf{1}}\mathbf{,}{\mathit{r}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathit{r}\mathbf{1}\mathbf{,}\mathit{r}\mathbf{2}\mathbf{,}\mathit{t}\mathit{o}\mathbf{}\mathit{u}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(r_1+r_2\right)\mathbf{,}\mathit{v}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left(r_1+r_2\right)$ (8.79).

turns the Hamiltonian into two independent three-dimensional harmonic oscillators:

$\mathit{H}\mathbf{=}\left[-\frac{ħ}{2m}{\nabla }_u^2+\frac{1}{2}m{\omega }^2{u}^2\right]\mathbf{+}\left[-\frac{ħ}{2m}{\nabla }_u^2+\frac{1}{2}\left(1-\lambda \right)m{\omega }^2{u}^2\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{80}\mathbf{)}$

(b) What is the exact ground state energy for this system?

(c) If we didn’t know the exact solution, we might be inclined to apply the method of Section 7.2 to the Hamiltonian in its original form (Equation 7.78). Do so (but don’t bother with shielding). How does your result compare with the exact answer? Answer:$\left(H\right)\mathbf{=}\mathbf{3}\mathit{ħ}\mathit{\omega }\left(1-\lambda /4\right)\mathit{a}$.

Short answer

(a) Required equation is proved.

(b)${\mathbf{E}}_{\mathbf{gs}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{ħ\omega }\left(1+\sqrt{1-\lambda }\right)$

(c)${\mathit{E}}_{\mathbf{g}\mathbf{s}}\mathbf{\approx }\frac{\mathbf{3}}{\mathbf{2}}\mathit{ħ}\mathit{\omega }\left(1+1-\frac{1}{2}ħ\right)\mathbf{=}\mathbf{3}\mathit{ħ}\mathit{\omega }\left(1-\frac{\lambda }{4}\right)$

Step-by-step solution

(a)Showing the changes of variables

$$\begin{gathered} r_1=\frac{1}{\sqrt{2}}(u+v);r_2=\frac{1}{\sqrt{2}}(u-v);r_1^2+r_2^2=\frac{1}{2}(u^2+2u\cdot v+v^2+u^2-2u\cdot v+v^2)=u^2+v^2 \\ ({\nabla }_1^2+{\nabla }_2^2)f(r_1,r_2)=\left(\frac{{\partial }^{2}f}{\partial x_1^2}+\frac{{\partial }^{2}f}{\partial y_1^2}+\frac{{\partial }^{2}f}{\partial z_1^2}\frac{{\partial }^{2}f}{\partial x_2^2}+\frac{{\partial }^{2}f}{\partial y_2^2}+\frac{{\partial }^{2}f}{\partial z_2^2}\right) \\ \frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial u_x}\frac{\partial u_x}{\partial x_1}+\frac{\partial f}{\partial v_x}\frac{\partial v_x}{\partial x_1}=\frac{1}{\sqrt{2}}\left(\frac{\partial f}{\partial u_x}+\frac{\partial f}{\partial v_x}\right);\frac{\partial f}{\partial x_2}=\frac{\partial f}{\partial u_x}\frac{\partial u_x}{\partial x_2}+\frac{\partial f}{\partial v_x} \\ \frac{\partial f}{\partial x_1}=\frac{1}{\sqrt{2}}\frac{\partial }{\partial x_1}\left(\frac{\partial f}{\partial u_1}+\frac{\partial f}{\partial v_x}\right)=\frac{1}{\sqrt{2}}\left(\frac{{\partial }^{2}f}{\partial u_x^2}\frac{\partial u_x}{\partial x_1}+\frac{{\partial }^{2}f}{\partial u_x\partial v_x}\frac{\partial v_x}{\partial x_1}+\frac{{\partial }^{2}f}{\partial u_x\partial u_x}\right) \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\left(\frac{{\partial }^{2}f}{\partial u_x^2}+2\frac{{\partial }^{2}f}{\partial u_x\partial v_x}+\frac{{\partial }^{2}f}{\partial v_x^2}\right). \\ \frac{{\partial }^2}{\partial x_2^2}=\frac{1}{\sqrt{2}}\frac{\partial }{\partial x_2}\left(\frac{\partial f}{\partial u_x}-\frac{\partial f}{\partial v_x}\right)=\frac{1}{\sqrt{2}}\left(\frac{{\partial }^{2}f}{\partial u_x^2}\frac{\partial u_x}{\partial x_2}+\frac{{\partial }^{2}f}{\partial u_x\partial v_x}\frac{\partial v_x}{\partial x_2}-\frac{{\partial }^{2}f}{\partial v_2\partial u_x}\frac{\partial u_x}{\partial x_2}-\frac{{\partial }^{2}f}{\partial {v^2}_x}\frac{\partial v_x}{\partial x_2}\right). \\ =\frac{1}{2}\left(\frac{{\partial }^{2}f}{\partial u_x^2}-2\frac{{\partial }^{2}f}{\partial u_x\partial v_x}+\frac{{\partial }^{2}f}{\partial v_x^2}\right). \\ So\left(\frac{{\partial }^{2}f}{\partial x_1^2}+\frac{{\partial }^{2}f}{\partial x_2^2}\right)=\left(\frac{{\partial }^{2}f}{\partial u_x^2}+\frac{{\partial }^{2}f}{\partial v_x^2}\right),\mathrm{and}\mathrm{likewise}\mathrm{for}\mathrm{y}\mathrm{and}\mathrm{z}:\nabla 12+\nabla 22=\nabla \mathrm{u}2+\nabla \mathrm{v}2. \\ \mathrm{H}=\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\left({\nabla }_{\mathrm{u}}^2+{\nabla }_{\mathrm{v}}^2\right)+\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{u}}^2+{\mathrm{v}}^2\right)-\frac{\mathrm{\lambda }}{4}{\mathrm{m\omega }}^{2}2{\mathrm{v}}^2\mathrm{H} \\ =\left[-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}{\nabla }_{\mathrm{u}}^2+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{u}}^2\right]+\left[\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}{\nabla }_{\mathrm{v}}^2+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{u}}^2-\frac{1}{2}{\mathrm{\lambda m\omega }}^2{\mathrm{v}}^2\right] \end{gathered}$$

(b) The exact ground state energy 

The energy is

$\frac{3}{2}ħ\omega \left(\mathrm{for}\mathrm{the}\mathrm{u}\mathrm{part}\right)\mathrm{and}\frac{3}{2}ħ\omega \sqrt{1-\mathrm{\lambda }}\left(\mathrm{for}\mathrm{the}\mathrm{v}\mathrm{part}\right):E_{gs}=\frac{3}{2}ħ\omega \left(1+\sqrt{1-\lambda }\right).$

(c) The ground state for one dimension oscillator 

The ground state for a one-dimensional oscillator is

localid="1655995511764"

$$\begin{gathered} 3ħ\omega \left(1-\frac{\lambda }{4}\right)>\frac{3}{2}ħ\omega \left(1+\sqrt{1-\lambda }\right)\iff 2-\frac{\lambda }{2}>1+\sqrt{1-\lambda }\iff 1-\frac{\lambda }{2}>\sqrt{1-\lambda }\iff 1-\lambda -\frac{{\lambda }^2}{4}>1-\lambda . \\ \mathrm{It}\mathrm{checks}.\mathrm{In}\mathrm{fact},\mathrm{expanding}\mathrm{the}\mathrm{exact}\mathrm{answer}\mathrm{in}\mathrm{powers}\mathrm{of} \\ \mathrm{\lambda },{\mathrm{E}}_{\mathrm{gs}}\approx \frac{3}{2}\mathrm{ħ\omega }\left(1+1-\frac{1}{2}\mathrm{\lambda }\right)=3\mathrm{ħ\omega }\left(1-\frac{\mathrm{\lambda }}{4}\right),\mathrm{we}\mathrm{recover}\mathrm{the}\mathrm{variation}\mathrm{result}. \end{gathered}$$