Q40P Sometimes it is possible to solv... [FREE SOLUTION]

Chapter 6: Q40P (page 291)

Sometimes it is possible to solve Equation 6.10 directly, without having to expand ${蠄^{1}}_{n}$ 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 (the Stark effect-see Problem 6.36). Hint: Try a solution of the form
$(A + B r + C r^{2}) e^{- r / a} c o s 胃$
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: $- m {(3 a^{2} e E_{ext} / 2 h)}^{2}$
(b) If the proton had an electric dipole moment p the potential energy of the electron in hydrogen would be perturbed in the amount
$H' = - \frac{别辫肠辞蝉胃}{4 蟺 \tilde{0_{0} r^{2}}}$
(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

Expert verified

(a) (i) The first -order correction to the ground state of hydrogen for the given condition is $蠄_{1}^{1} = - 尾 r (2 a + r) c o s 鈦� 蠎 e^{(} - r / a), 鈥� 鈥� 鈥� 尾 = (m e E_{"} e x t ") / (2 {鈩�}^{2} 鈭� (蟺 a^{3})$ .

(ii) The second-order correction for the given condition is $E_{1}^{2} = - m {(\frac{(3 e E_{"} e x t " a^{2})}{2 h})}^{2}$ .

(b) (i) The first-order correction to the ground state wave function is

${蠄_{1}}^{1} = \frac{m e p}{4 蟺蔚_{0} {鈩�}^{2} \sqrt{蟺 a^{3}}} \cos 蠎 e^{- r / a}$

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

(iii) The second-order correction to the ground state energy $E_{1}^{2} = \frac{4}{3} {(\frac{p}{e a})}^{2} E_{1}$

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.

$(H^{0} - E_{n}^{0}) 蠄_{n}^{1} = - (H - E_{n}^{1}) 蠄_{n}^{0} H^{0} = - \frac{h^{2}}{2 m} {鈭�}^{2} - \frac{e^{2}}{4 蟺蔚_{0} r} = - \frac{h^{2}}{2 m} ({鈭�}^{2} + \frac{2}{a r})$

鈥�(颈)

Ground state, n=1, and for $H' = e E_{e x t} r \cos 胃$ , first-order correction to ground state energy $E_{1}^{1}$ diminishes.

Now, the non-perturbed ground state energy $E_{1}^{0} = - h^{2} / (2 m a^{2})$ 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.

$蠄_{1}^{1} = (A + B r + C r^{2}) \cos 胃 e^{- r / a} = f (r) \cos 胃 e^{^{- r / a}}$

Rearrange the equation (i).

$[- \frac{h^{2}}{2 m} ({鈭�}^{2} + \frac{2}{a r}) + \frac{h^{2}}{2 m a^{2}}] 蠄_{1}^{1} = - H' 蠄_{1}^{0} {鈭�}^{2} 蠄_{1}^{1} + \frac{2}{a r} 蠄_{1}^{1} - \frac{1}{a^{2}} 蠄_{1}^{1} = \frac{2 m}{h^{2}} H' 蠄_{1}^{0}$

Write the operator in spherical coordinates.

${鈭�}^{2} 蠄_{1}^{1} = \frac{1}{r^{2}} \frac{d}{d r} [r^{2} \frac{d}{d r} (f (r) \cos 胃 e^{- r / a})] + \frac{1}{r^{2} \sin 胃} \frac{d}{d 胃} [\sin 胃 \frac{d}{d 胃} (f (r) \cos 胃 e^{- r / a})] = \frac{\cos 胃}{r^{2}} \frac{d}{d r} (r^{2} f' e^{- r / a} - \frac{r^{2} f}{a} e^{- r / a}) + \frac{f e^{- r / a}}{r^{2} \sin 胃} \frac{d}{d 胃} (- \sin^{2} 胃) = \frac{\cos 胃}{r^{2}} (2 r f' e^{- r / a} + r^{2} f " e^{- r / a} - - \frac{r^{2} f}{a} e^{- r / a} - \frac{r^{2} f}{a} e^{- r / a} - \frac{r^{2} f}{a} e^{- r / a} + \frac{r^{2} f}{a^{2}} e^{- r / a}) - \frac{2 r f e^{- r / a}}{r^{2}} \cos 胃 = \frac{\cos 胃 e^{- r / a}}{r^{2}} [r^{2} f " + 2 r f' (1 - \frac{r}{a}) - f (2 + \frac{2 r}{a} - \frac{r^{2}}{a^{2}})] = \cos 胃 e^{- r / a} [f " + 2 f (\frac{1}{r} - \frac{1}{a}) - f (\frac{2}{r^{2}} + \frac{2}{a r} - \frac{1}{a^{2}})]$

Use this equation.

$\cos 胃 e^{- r / a} [f " + 2 f' (\frac{1}{r} - \frac{1}{a}) - f (\frac{2}{r^{2}} + \frac{2}{a r} - \frac{1}{a^{2}})] + \frac{2 f \cos 胃 e^{- r / a}}{a r} = \frac{2 m}{h^{2}} e E_{e x t} r \cos 胃 \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}}$

Perform the simplification.

$f " + 2 f' (\frac{1}{r} - \frac{1}{a}) - \frac{f}{r^{2}} = 位 r f (r) = A + B r + C r^{2} 2 C + 2 (B = 2 C r) (\frac{1}{r} - \frac{1}{a}) - \frac{2 A}{r^{2}} - \frac{2 A}{r} - 2 C - 位 r = 0$

Here, $位 = \frac{2 m e E_{e x t}}{h^{2 \sqrt{蟺 a^{3}}}}, f' = B + 2 C r, a n d f " = 2 C$

Further simplify the above expression.

$- \frac{2 B}{a} + 4 C - \frac{4 C r}{a} - \frac{2 A}{r^{2}} - 位 r = 0 \frac{2 A}{r^{2}} - 位 r = 0 A = 0 \frac{4 C r}{a} - 位 = 0 C = - \frac{a 位}{4} = - \frac{m e E_{e x t}}{2 h^{2} \sqrt{蟺 a^{3}}} \frac{2 B}{a} = - 4 C B = - \sqrt{\frac{a}{蟺}} \frac{m e E_{e x t}}{h^{2}}$

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

$蠄_{1}^{1} = r (B + C r) \cos 胃 e^{- r / a} = - 尾 r (2 a + r) \cos 胃 e^{- r / a}$

Thus, the first-order correction to wave function of ground state is .

$蠄_{1}^{1} = = - 尾 r (2 a + r) \cos 胃 e^{- r / a}, a n d 尾 = \frac{m e E_{e x t}}{2 h^{2} \sqrt{蟺 a^{3}}}$

(ii)

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

$E_{1}^{1} = <蠄_{1}^{0} |H'| 蠄_{1}^{1}> - O . <蠄_{1}^{0} 蠄_{1}^{1}> = <蠄_{1}^{0} |H'| 蠄_{1}^{1}> = 鈭� \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}} e E_{e x t} r \cos 胃 (- 尾 r) (2 a + r) \cos 胃 e^{- r / a} r^{2} d r \sin 胃 d 胃 d 蠁 = - \frac{m {(e E_{e x t})}^{2}}{a^{2} h^{2}} {鈭�}_{0}^{鈭�} r^{4} (r + 2 a) e^{- r / a {鈭�}_{0}^{蟺}} \cos^{2} 胃 \sin 胃 d 胃$

Further evaluate the expression.

$E_{1}^{2} = - \frac{m {(e E_{e x t})}^{2}}{a^{2} h^{2}} [5! {(\frac{a}{2})}^{6} + 2 a 路 4! {(\frac{a}{2})}^{5}] 路 \frac{2}{3} = - m (\frac{3 e E_{e x t} a^{2}}{2 h})$

Therefore, the second-order correction for the given condition is .

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

Write the expression of Perturbation Hamiltonian.

$H^{'} = - \frac{e p \cos 胃}{4 蟺蔚_{0} r^{2}} \cos 胃 e^{- r / a} [f " + 2 f' (\frac{1}{r} - \frac{1}{a}) - f (\frac{2}{r^{2}} + \frac{2}{a r} + \frac{1}{a^{2}})] + \frac{2 f \cos 胃 e^{- r / a}}{a r} - \frac{f \cos 胃 e^{- r / a}}{a^{3}} = \frac{2 m}{h^{2}} (- \frac{e p \cos 胃}{4 蟺蔚_{0} r^{2}}) \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}} f' + 2 f' (\frac{1}{r} - \frac{1}{a}) - \frac{2 f}{r^{2}} = - \frac{2 m e p}{4 蟺蔚_{0} h^{2} \sqrt{蟺 a^{3}}} \frac{1}{r^{2}} f' + 2 f' (\frac{1}{r} - \frac{1}{a}) - \frac{2 f}{r^{2}} = - \frac{2 a}{r^{2}} f (r) = 伪 = \frac{2 m e p}{4 蟺蔚_{0} h^{2} \sqrt{蟺 a^{3}}}$

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

$= 蠄_{1}^{1} \frac{m e p}{4 蟺蔚_{0} h^{2} \sqrt{蟺 a^{3}}} \cos 胃 e^{- r / a}$

Write the expression for the electric dipole.

$<p_{e}> = - <e r \cos 胃> = - e <蠄_{1}^{0} + 蠄_{1}^{1} |r \cos 胃| 蠄_{1}^{0} + 蠄_{1}^{1}> = [<蠄_{1}^{0} |r \cos 胃| 蠄_{1}^{0}> + 2 <蠄_{1}^{0} |r \cos 胃| 蠄_{1}^{1}> + <蠄_{1}^{1} |r \cos 胃| 蠄_{1}^{1}>]$

Write the third order of the term.

$<p_{e}> = - 2 <蠄_{1}^{0} |r \cos 胃| 蠄_{1}^{1}> = - 2 e 鈭� \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}} r \cos 胃伪 \cos 蠎^{- r / a} r^{2} d r \sin 胃 d 胃 d 蠁 = - \frac{4 蟺 e 伪}{\sqrt{蟺 a^{3}}} {鈭�}_{0}^{鈭�} r^{3} e^{- 2 / a} d r {鈭�}_{0}^{蟺} \cos^{2} 胃 \sin 胃 d 胃 = - \frac{4 蟺 e 伪}{\sqrt{蟺 a^{3}}} 路 3! {(\frac{a}{2})}^{4} 路 \frac{2}{3} = - \frac{m e^{2} a}{4 蟺蔚_{0} h^{2}} 路 p = - p$

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

$E_{1}^{2} = <蠄_{1}^{0} |H| 蠄_{1}^{1}> = 鈭� \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}} (- \frac{e p \cos 胃}{4 蟺蔚_{0} r^{2}}) 伪 \cos 胃 e^{- r / a} r^{2} d r \sin 胃 d 胃 d 蠁 = - 2 m \frac{{(e p)}^{2}}{{(4 蟺蔚_{0})}^{2} h^{2}} 路 (\frac{a}{2}) 路 \frac{2}{3} = \frac{4}{3} (- \frac{m e^{4}}{2 {(4 蟺蔚_{0})}^{2} h^{2}}) \frac{p^{2}}{e^{2} a^{2}} = \frac{4}{3} {(\frac{p}{e a})}^{2} E_{1}$

(i) Thus, the first-order correction to the ground state wave function is

$蠄_{1}^{1} = \frac{m e p}{4 蟺蔚_{0} {鈩�}^{2} 鈭� (蟺 a^{3}} \cos 蠎 e^{- r / a}$

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

(iii) Thus, the second-order correction to the ground state energy is $E_{1}^{2} = \frac{4}{3} {(\frac{p}{e a})}^{2} E_{1}$

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Short Answer

Step by step solution

Definition of the wave function of Hamiltonian function

(a) Determination of the first order correction

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

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