91影视

In quantum physics, a wave function represents the quantum state of a particle as a function of spin, time, momentum, and location.

Additionally, it depends on the degrees of freedom that correspond to the broadest group of observables that can coexist.

(i)

Consider equation 6.10.

$$\begin{gathered} \left(H^0-E_n^0\right){蠄}_n^1=-\left(H^{\text{'}}-E_n^1\right){蠄}_n^0 \\ {H}^0=-\frac{{\mathrm{鈩�}}^2}{2m}{鈭�}^2-\frac{e^2}{4蟺{蔚}_{0}r} \\ =-\frac{{\mathrm{鈩�}}^2}{2m}\left({鈭�}^2+\frac{2}{ar}\right) \end{gathered}$$

鈥�(颈)

Ground state, n = 1 , and for$H^{\text{'}}=e{E}_{\text{ext}}r\cos 胃$, first-order correction to ground state energy$E_1^1$diminishes.

Now, the non-perturbed ground state energy$E_1^0=-{\mathrm{鈩�}}^2/\left(2m{a}^2\right)$and non-perturbed wave function for ground state of hydrogen is${蠄}_1^0=e^{-r/a}/\sqrt{蟺a^3}$.

Calculate first-order correction to wave function.

$$\begin{gathered} {蠄}_1^1=\left(A+Br+C{r}^2\right)\cos 胃e^{-r/a} \\ =f\left(r\right)\cos 胃e^{-r/a} \end{gathered}$$

Rearrange the equation (i).

$$\begin{gathered} \left[-\frac{{\mathrm{鈩�}}^2}{2m}\left({鈭�}^2+\frac{2}{ar}\right)+\frac{{\mathrm{鈩�}}^2}{2m{a}^2}\right]{蠄}_1^1=-H\text{'}{蠄}_1^0 \\ {鈭�}^2{蠄}_1^1+\frac{2}{ar}{蠄}_1^1-\frac{1}{a^2}{蠄}_1^1=\frac{2m}{{\mathrm{鈩�}}^2}H\text{'}{蠄}_1^0 \end{gathered}$$

Write the operator in spherical coordinates.

$$\begin{gathered} {鈭�}^2{蠄}_1^1=\frac{1}{r^2}\frac{d}{dr}\left[r^2\frac{d}{dr}\left(f\left(r\right)\cos 胃e^{-r/a}\right)\right]+\frac{1}{r^2\sin 胃}\frac{d}{d胃}\left[\sin 胃\frac{d}{d胃}\left(f\left(r\right)\cos 胃e^{-r/a}\right)\right] \\ =\frac{\cos 胃}{r^2}\frac{d}{dr}\left(r^{2}f\text{'}{e}^{-r/a}-\frac{r^{2}f}{a}{e}^{-r/a}\right)+\frac{f{e}^{-r/a}}{r^2\sin 胃}\frac{d}{d胃}\left(-{\sin }^2胃\right) \\ =\frac{\cos 胃}{r^2}\left(2rf\text{'}{e}^{-r/a}+r^{2}f{e}^{-r/a}-\frac{r^2{f}^{\text{'}}}{a}{e}^{-r/a}-\frac{2rf}{a}{e}^{-r/a}-\frac{r^{2}f\text{'}}{a}{e}^{-r/a}+\frac{r^{2}f}{a^2}{e}^{-r/a}\right)-\frac{2f{e}^{-r/a}}{r^2}\cos 胃 \\ =\frac{\cos 胃e^{-r/a}}{r^2}\left[r^{2}f+2rf\text{'}\left(1-\frac{r}{a}\right)-f\left(2+\frac{2r}{a}-\frac{r^2}{a^2}\right)\right] \\ =\cos 胃e^{-r/a}\left[f+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-f\left(\frac{2}{r^2}+\frac{2}{ar}-\frac{1}{a^2}\right)\right] \end{gathered}$$

Use this equation.

$\cos 胃e^{-r/a}\left[f+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-f\left(\frac{2}{r^2}+\frac{2}{ar}-\frac{1}{a^2}\right)\right]+\frac{2f\cos 胃e^{-r/a}}{ar}-\frac{f\cos 胃e^{-r/a}}{a^2}=\frac{2m}{{\mathrm{鈩�}}^2}e{E}_{\text{ext}}r\cos 胃\frac{e^{-r/a}}{\sqrt{蟺a^3}}$

Perform the simplification.

$$\begin{gathered} f+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{f}{r^2}=位r \\ f\left(r\right)=A+Br+C{r}^2 \\ 2C+2\left(B+2Cr\right)\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{2A}{r^2}-\frac{2B}{r}-2C-位r=0 \end{gathered}$$

Here, $位=\frac{2me{E}_{\text{ext}}}{{\mathrm{鈩�}}^2\sqrt{蟺a^3}},f\text{'}=B+2Crandf=2C$

Further simplify the above expression.

$$\begin{gathered} -\frac{2B}{a}+4C-\frac{4Cr}{a}-\frac{2A}{r^2}-位r=0 \\ \frac{2A}{r^2}-位r=0 \\ A=0 \\ -\frac{4Cr}{a}-位=0 \\ C=-\frac{a位}{4} \\ =-\frac{me{E}_{\text{ext}}}{2{\mathrm{鈩�}}^2\sqrt{蟺a^3}} \\ \frac{2B}{a}=4C \\ B=-\sqrt{\frac{a}{蟺}}\frac{me{E}_{\text{ext}}}{{\mathrm{鈩�}}^2} \end{gathered}$$

Write the first-order correction to wave function of ground state.

$$\begin{gathered} {蠄}_1^1=r(B+Cr)\cos 胃e^{-r/a} \\ =-尾r(2a+r)\cos 胃e^{-r/a} \end{gathered}$$

Thus, the first-order correction to wave function of ground state is${蠄}_1^1=-尾r(2a+r)\cos 胃e^{-r/a}$, and$尾=\frac{me{E}_{\text{ext}}}{2{\mathrm{鈩�}}^2\sqrt{蟺a^3}}$.

(ii)

Secondly find second-order correction of ground state energy, using equation 6.17.

Further evaluate the expression.

$$\begin{gathered} E_1^2=-\frac{m{\left(e{E}_{\text{ext}}\right)}^2}{a^2{\mathrm{鈩�}}^2}\left[5!{\left(\frac{a}{2}\right)}^6+2a路4!{\left(\frac{a}{2}\right)}^5\right]路\frac{2}{3} \\ =-m{\left(\frac{3e{E}_{\text{ext}}{a}^2}{2\mathrm{鈩�}}\right)}^2 \end{gathered}$$

Therefore, the second-order correction for the given condition is $E_1^2=-m{\left(\frac{3e{E}_{\text{ext}}{a}^2}{2\mathrm{鈩�}}\right)}^2$.

Write the expression of Perturbation Hamiltonian.

$H^{\text{'}}=-\frac{ep\cos 胃}{4蟺{蔚}_0{r}^2}$

Write the first-order correction to wave function of ground state energy.

$$\begin{gathered} \cos 胃e^{-r/a}\left[f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-f\left(\frac{2}{r^2}+\frac{2}{ar}-\frac{1}{a^2}\right)\right]+\frac{2f\cos 胃e^{-r/a}}{ar}-\frac{f\cos 胃e^{-r/a}}{a^2}=\frac{2m}{{\mathrm{鈩�}}^2}\left(-\frac{ep\cos 胃}{4蟺{蔚}_0{r}^2}\right)\frac{e^{-r/a}}{\sqrt{蟺a^3}} \\ f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{2f}{r^2}=-\frac{2mep}{4蟺{蔚}_0{\mathrm{鈩�}}^2\sqrt{蟺a^3}}\frac{1}{r^2} \\ f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{2f}{r^2}=-\frac{2伪}{r^2} \\ f\left(r\right)=伪 \\ =\frac{mep}{4蟺{蔚}_0{\mathrm{鈩�}}^2\sqrt{蟺a^3}} \end{gathered}$$

Write the expression for the electric dipole.

${蠄}_1^1=\frac{mep}{4蟺{蔚}_0{\mathrm{鈩�}}^2\sqrt{蟺a^3}}\cos 胃e^{-r/a}$

Write the third order of the term.

Write the expressions for second-order correction to the ground state energy.

(i) Thus,the first-order correction to the ground state wave function is ${蠄}_1^1=\frac{mep}{4蟺{蔚}_0{\mathrm{鈩�}}^2\sqrt{蟺a^3}}\cos 蠎e^{-r/a}.$

(ii) Thus, the total electric dipole moment of the atom vanishes.

(iii) Thus, the second-order correction to the ground state energy is$E_1^2=\frac{4}{3}{\left(\frac{p}{ea}\right)}^2{E}_1.$