Question

Question: Sometimes it is possible to solve Equation 6.10 directly, without having to expand in terms of the unperturbed wave functions (Equation 6.11). Here are two particularly nice examples.

(a) Stark effect in the ground state of hydrogen.

(i) Find the first-order correction to the ground state of hydrogen in the presence of a uniform external electric field${\mathit{E}}_{\mathbf{\text{ext}}}$ (the Stark effect-see Problem 6.36). Hint: Try a solution of the form

$\left(A+Br+C{r}^2\right){\mathit{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{a}}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$

your problem is to find the constants , and C that solve Equation 6.10.

(ii) Use Equation to determine the second-order correction to the ground state energy (the first-order correction is zero, as you found in Problem 6.36(a)). Answer:$\mathbf{-}\mathit{m}{\left(3a^{2}e{E}_{\text{ext}}/2\hslash \right)}^{\mathbf{2}}$ .

(b) If the proton had an electric dipole moment p the potential energy of the electron in hydrogen would be perturbed in the amount

$\mathit{H}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\frac{\mathbf{e}\mathbf{p}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}{\mathbf{4}\mathbf{\pi }\mathbf{̀}{\mathbf{o}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

(i) Solve Equation 6.10 for the first-order correction to the ground state wave function.

(ii) Show that the total electric dipole moment of the atom is (surprisingly) zero, to this order.

(iii) Use Equation 6.14 to determine the second-order correction to the ground state energy. What is the first-order correction?

Short answer

Answer

(a) (i) The first-order correction to the ground state of hydrogen for the given condition is${\psi }_1^1=-\beta r(2a+r)\cos \vartheta e^{-r/a},\beta =\frac{me{E}_{\text{ext}}}{2{\hslash }^2\sqrt{\pi a^3}}$.

(ii) Thesecond-order correction for the given condition is$E_1^2=-m{\left(\frac{3e{E}_{\text{ext}}{a}^2}{2\hslash }\right)}^2$.

(b) (i) The first-order correction to the ground state wave function is${\psi }_1^1=\frac{mep}{4\pi {\epsilon }_0{\hslash }^2\sqrt{\pi a^3}}\cos \vartheta e^{-r/a}.$

(ii) Total electric dipole moment of the atom vanishes.

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

Step-by-step solution

Definition of the wave function of Hamiltonian function 

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.

(a) Determination of the first order correction 

(i)

Consider equation 6.10.

$$\begin{gathered} \left(H^0-E_n^0\right){\psi }_n^1=-\left(H^{\text{'}}-E_n^1\right){\psi }_n^0 \\ {H}^0=-\frac{{\hslash }^2}{2m}{\nabla }^2-\frac{e^2}{4\pi {\epsilon }_{0}r} \\ =-\frac{{\hslash }^2}{2m}\left({\nabla }^2+\frac{2}{ar}\right) \end{gathered}$$

…(i)

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

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

Calculate first-order correction to wave function.

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

Rearrange the equation (i).

$$\begin{gathered} \left[-\frac{{\hslash }^2}{2m}\left({\nabla }^2+\frac{2}{ar}\right)+\frac{{\hslash }^2}{2m{a}^2}\right]{\psi }_1^1=-H\text{'}{\psi }_1^0 \\ {\nabla }^2{\psi }_1^1+\frac{2}{ar}{\psi }_1^1-\frac{1}{a^2}{\psi }_1^1=\frac{2m}{{\hslash }^2}H\text{'}{\psi }_1^0 \end{gathered}$$

Write the operator in spherical coordinates.

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

Use this equation.

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

Perform the simplification.

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

Here, $\lambda =\frac{2me{E}_{\text{ext}}}{{\hslash }^2\sqrt{\pi a^3}},f\text{'}=B+2Crandf=2C$

Further simplify the above expression.

$$\begin{gathered} -\frac{2B}{a}+4C-\frac{4Cr}{a}-\frac{2A}{r^2}-\lambda r=0 \\ \frac{2A}{r^2}-\lambda r=0 \\ A=0 \\ -\frac{4Cr}{a}-\lambda =0 \\ C=-\frac{a\lambda }{4} \\ =-\frac{me{E}_{\text{ext}}}{2{\hslash }^2\sqrt{\pi a^3}} \\ \frac{2B}{a}=4C \\ B=-\sqrt{\frac{a}{\pi }}\frac{me{E}_{\text{ext}}}{{\hslash }^2} \end{gathered}$$

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

$$\begin{gathered} {\psi }_1^1=r(B+Cr)\cos \theta e^{-r/a} \\ =-\beta r(2a+r)\cos \theta e^{-r/a} \end{gathered}$$

Thus, the first-order correction to wave function of ground state is${\psi }_1^1=-\beta r(2a+r)\cos \theta e^{-r/a}$, and$\beta =\frac{me{E}_{\text{ext}}}{2{\hslash }^2\sqrt{\pi a^3}}$.

(ii)

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

Further evaluate the expression.

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

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

 Step 3 : (b) Determination of the first order wave correction and the second order correction

Write the expression of Perturbation Hamiltonian.

$H^{\text{'}}=-\frac{ep\cos \theta }{4\pi {\epsilon }_0{r}^2}$

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

$$\begin{gathered} \cos \theta e^{-r/a}\left[f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-f\left(\frac{2}{r^2}+\frac{2}{ar}-\frac{1}{a^2}\right)\right]+\frac{2f\cos \theta e^{-r/a}}{ar}-\frac{f\cos \theta e^{-r/a}}{a^2}=\frac{2m}{{\hslash }^2}\left(-\frac{ep\cos \theta }{4\pi {\epsilon }_0{r}^2}\right)\frac{e^{-r/a}}{\sqrt{\pi a^3}} \\ f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{2f}{r^2}=-\frac{2mep}{4\pi {\epsilon }_0{\hslash }^2\sqrt{\pi a^3}}\frac{1}{r^2} \\ f\text{'}\text{'}+2f\text{'}\left(\frac{1}{r}-\frac{1}{a}\right)-\frac{2f}{r^2}=-\frac{2\alpha }{r^2} \\ f\left(r\right)=\alpha \\ =\frac{mep}{4\pi {\epsilon }_0{\hslash }^2\sqrt{\pi a^3}} \end{gathered}$$

Write the expression for the electric dipole.

${\psi }_1^1=\frac{mep}{4\pi {\epsilon }_0{\hslash }^2\sqrt{\pi a^3}}\cos \theta e^{-r/a}$

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 ${\psi }_1^1=\frac{mep}{4\pi {\epsilon }_0{\hslash }^2\sqrt{\pi a^3}}\cos \vartheta e^{-r/a}.$

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

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