Question

Question: In a crystal, the electric field of neighbouring ions perturbs the energy levels of an atom. As a crude model, imagine that a hydrogen atom is surrounded by three pairs of point charges, as shown in Figure 6.15. (Spin is irrelevant to this problem, so ignore it.)

(a) Assuming that $\mathit{r}\mathbf{\ll }{\mathit{d}}_{\mathbf{1}}\mathbf{,}\mathit{r}\mathbf{\ll }{\mathit{d}}_{\mathbf{2}}\mathbf{,}\mathit{r}\mathbf{\ll }{\mathit{d}}_{\mathbf{3}}$show that

$\mathit{H}\mathbf{\text{'}}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{+}\mathbf{3}\mathbf{(}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}\mathbf{-}\mathbf{(}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{)}{\mathit{r}}^{\mathbf{2}}$

where

$\mathbf{}{\mathit{\beta }}_{\mathbf{i}}\mathbf{\equiv }\mathbf{-}\frac{\mathbf{e}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{i}}}{{\mathbf{d}}_{\mathbf{i}}^{\mathbf{3}}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathbf{2}\mathbf{(}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{d}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{d}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}{\mathbf{d}}_{\mathbf{3}}^{\mathbf{2}}\mathbf{)}$

(b) Find the lowest-order correction to the ground state energy.

(c) Calculate the first-order corrections to the energy of the first excited states Into how many levels does this four-fold degenerate system split,

(i) in the case of cubic symmetry${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}{\mathit{\beta }}_{\mathbf{3}}\mathbf{;}$, (ii) in the case of tetragonal symmetry${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{3}}\mathbf{;}$, (iii) in the general case of orthorhombic symmetry (all three different)?

Short answer

Answer

(a) The given expression is verified.

(b) The lowest-order correction to the ground state energy is$V_0$.

(c) (i) If ${\beta }_1={\beta }_2={\beta }_3,$ then $E_1=E_2=E_3=E_4=V_0$ one level of degeneracy is equal to 4.

(ii) If ${\beta }_1={\beta }_2\ne {\beta }_3,$ three level

(iii). if ${\beta }_1\ne {\beta }_2\ne {\beta }_3,$ no level and then all states have a different energy, and there is no degeneracy.

Step-by-step solution

 Step 1: Definition of the ground state energy.

A quantum mechanical system's ground state is its stationary, lowest energy state; this energy is often referred to as the system's zero-point energy.

(a) Verification of the expression 

Consider that interaction between electron and charges at$x=\pm d$.

$V=-\frac{eq}{4\pi {\epsilon }_0}\left(\frac{1}{\sqrt{{(x+d)}^2+y^2+z^2}}+\frac{1}{\sqrt{{(x-d)}^2+y^2+z^2}}\right)$

Simplify the expression,$\sqrt{{(x+d)}^2+y^2+z^2}$.

$$\begin{gathered} {\left[{(x\pm d)}^2+y^2+z^2\right]}^{-1/2}={\left[x^2\pm 2xd+d^2+y^2+z^2\right]}^{-1/2} \\ {\left[{(x\pm d)}^2+y^2+z^2\right]}^{-1/2}={\left[d^2\pm 2xd+r^2\right]}^{-1/2} \\ {\left[{(x\pm d)}^2+y^2+z^2\right]}^{-1/2}=\frac{1}{d}{\left[1\pm \frac{2x}{d}+\frac{r^2}{d^2}\right]}^{-1/2} \\ \approx \frac{1}{d}\left(1\mp \frac{x}{d}-\frac{r^2}{2d^2}+\frac{3x^2}{2d^2}\right) \\ =\frac{1}{d}\left(1\mp \frac{x}{d}+\frac{3x^2-r^2}{2d^2}\right) \end{gathered}$$

Substitute the above value in .

$$\begin{gathered} V=-\frac{eq}{4\pi {\epsilon }_{0}d}\left(1-\frac{x}{d}+\frac{3x^2-r^2}{2d^2}+1+\frac{x}{d}+\frac{3x^2-r^2}{2d^2}\right) \\ =-\frac{eq}{4\pi {\epsilon }_{0}d}\left(2+\frac{3x^2-r^2}{d^2}\right) \end{gathered}$$

Substitute $\beta$ for $-\frac{eq}{4\pi {\epsilon }_0{d}^3}$ in the above expression.

For all six charges,

$$\begin{gathered} H\text{'}=2\left({\beta }_1{d}_1^2+{\beta }_2{d}_2^2+{\beta }_3{d}_3^2\right)+3\left({\beta }_1{x}^2+{\beta }_2{y}^2+{\beta }_3{z}^2\right)-r^2\left({\beta }_1+{\beta }_2+{\beta }_3\right) \\ =V_0+3\left({\beta }_1{x}^2+{\beta }_2{y}^2+{\beta }_3{z}^2\right)-r^2\left({\beta }_1+{\beta }_2+{\beta }_3\right) \end{gathered}$$

Thus, the given expression is verified.

(b) Determination of the lowest order correction

Perform the correction on ground state energy.

The first term is equal to $V_0$ as the wave function is normalized. Write the value of $r^2$.

The function is spherically symmetric. Write the value of .

Write the value of the lowest order correction to the ground state.

Thus, the lowest-order correction to the ground state energy is v₀ .

(c) Determination of the first order corrections of energy when n=2

Write the expression for the wave function.

$$\begin{gathered} {\psi }_{200}=\frac{1}{\sqrt{2\pi a}}\frac{1}{2a}\left(1-\frac{r}{2a}\right)e^{-r/\left(2a\right)} \\ {\psi }_{21\pm 1}=\mp \frac{1}{\sqrt{\pi a}}\frac{1}{8a^2}r{e}^{-r/\left(2a\right)}\sin \theta e^{\pm i\phi } \\ {\psi }_{210}=\frac{1}{\sqrt{2\pi a}}\frac{1}{4a^2}r{e}^{-r/\left(2a\right)}\cos \theta \end{gathered}$$

Construct the perturbation matrix.

When n = 2, I = 1 then .

Calculate the value of $x^2,y^2,and{z}^2$

role="math" localid="1659007692072"

Determinethe off-diagonal element.

Construct the perturbation matrix.

$W=\left(\begin{array}{cccc}{V}_0& 0& 0& 0\\ 0& V_0-12a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)& 0& 0\\ 0& 0& V_0+6a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)& 18a^2\left(-{\beta }_1+{\beta }_2\right)\\ 0& 0& 18a^2\left(-{\beta }_1+{\beta }_2\right)& V_0+6a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)\end{array}\right)$

Determine the eigenvalues andseparately diagonalize.

$$\begin{gathered} \left(V_0-E\right)\left[V_0-12a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)-E\right]=0 \\ {E}_1=V_0,E_2=V_0-12a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right) \\ {V}_0+6a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)-E18a^2\left({\beta }_2-{\beta }_1\right) \\ 18a^2\left({\beta }_2-{\beta }_1\right)V_0+6a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)-E \\ {\left[V_0+6a^2\left({\beta }_1+{\beta }_2-2{\beta }_3\right)-E\right]}^2-{\left[18a^2\left({\beta }_2-{\beta }_1\right)\right]}^2=0 \end{gathered}$$

Thus,

i) If ${\beta }_1={\beta }_2={\beta }_3,$} then: $E_1=E_2=E_3=E_4=V_0$

one level of degeneracy =4

ii) If ${\beta }_1={\beta }_2\ne {\beta }_3,$

iii). if ${\beta }_1\ne {\beta }_2\ne {\beta }_3,$then all states have different energy, and there is no degeneracy