Q36P When an atom is placed in a unif... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q36P (page 289)

When an atom is placed in a uniform external electric field ,the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyse the Stark effect for the n=1 and n=2 states of hydrogen. Let the field point in the z direction, so the potential energy of the electron is
$H'_{s} = e E_{e x t} z = e E_{e x t} r c o s 胃$
Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.)
(a) Show that the ground state energy is not affected by this perturbation, in first order.
(b) The first excited state is 4-fold degenerate: $Y_{200}, Y_{211}, Y_{210}, Y_{200}, Y_{21 - 1}$ Using degenerate perturbation theory, determine the first order corrections to the energy. Into how many levels does $E_{2}$ split?
(c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment $(p_{e} = - er)$ in each of these "good" states.Notice that the results are independent of the applied field-evidently hydrogen in its first excited state can carry a permanent electric dipole moment.

Short Answer

Expert verified

${(a) E_{S}}^{1} = 0 (b) E_{2,} E_{2,} E_{2} + 3 a e E_{e x t} E_{2 -} 3 a e E_{e x t} (c) T h e E i g e n v e c t o r s : 蠄_{211'} 蠄_{21 - 1} \frac{1}{\sqrt{2}} (蠄_{200} + 蠄_{210}) $ . E x p e c t a t i o n v a l u e s : <\overset{鈬赌}{p_{e}}> - 0 <\overset{鈬赌}{p_{e}}> - 0 <\overset{鈬赌}{p_{e}}> - + 3 e a \hat{z}$

Step by step solution

Define the formula for wave function in ground state

Wave function of ground state in hydrogen is: $蠄_{100} = \frac{e^{- r / a}}{\sqrt{{蟺补}^{3}}}$

Correction in ground state ${E_{1}}^{1} = < 蠄_{100} | H_{S} | 蠄_{100} >$

Effect of perturbation on ground state energy

$100 > = \frac{1}{\sqrt{蟺 a^{3}}} e^{- r / a} . . . (4.80) {E^{1}}_{S} = <100 |H'| 100> e E_{e x t} \frac{1}{蟺 a^{3}} 鈭� e^{- 2 r / a} (r \cos 胃) r^{2} \sin 胃 d r d 胃 d 蠒 B u t t h e 胃 i n t e g r a l i s z e r o : {鈭�}_{0}^{蟺} \cos 胃 \sin 胃 d 胃 = \frac{\sin^{2}}{} {胃^{}}_{0}^{蟺} S o, = 0 {E^{1}}_{S} = 0$

first order corrections to the energy

From problem 4.11: $\{\begin{cases} \begin{matrix} \begin{matrix} 1 > = 蠄_{200 =} & \frac{1}{\sqrt{2 蟺 a^{2}}} \frac{1}{2 a} (1 - \frac{r}{2 a}) e^{- 2 / 2 a} \\ 2 > = 蠄_{211 =} & \frac{1}{\sqrt{蟺 a}} \frac{1}{8 a^{2}} r e^{- r / 2 a} \sin 胃 e^{i 蠒} \\ 3 > = 蠄_{210 =} & \frac{1}{\sqrt{2 蟺 a}} \frac{1}{4 a} r e^{- r / 2 a} \cos 胃 \\ 4 > = 蠄_{21 - 1 =} & \frac{1}{\sqrt{蟺 a}} \frac{1}{8 a^{2}} r e^{- r / 2 a} \sin 胃 e^{i 蠒} \end{matrix} \end{matrix} \end{cases} \begin{matrix} \end{matrix}$

$\begin{array}{r} \begin{matrix} <1 |H'_{s}| 1> = \{. . .\} {鈭�}_{0}^{蟺} & \cos 胃 \sin 胃 d 胃 = 0 \\ <2 |H'_{s}| 2> = \{. . .\} {鈭�}_{0}^{蟺} & \sin^{2} 胃 \cos 胃 \sin 胃 d 胃 = 0 \\ <3 |H'_{s}| 3> = \{. . .\} {鈭�}_{0}^{蟺} & \cos^{2} 胃 \sin 胃 d 胃 = 0 \\ <4 |H'_{s}| 4> = \{. . .\} {鈭�}_{0}^{蟺} & \sin^{2} 胃 \cos 胃 \sin 胃 d 胃 = 0 \begin{matrix} \end{matrix} \\ <1 |H'_{S}| 2> = \{. . .\} {鈭�}_{0}^{2 蟺} & e^{i 蠒} d 蠒 = 0 \\ <1 |H'_{S}| 4> = \{. . .\} {鈭�}_{0}^{2 蟺} & e^{i 蠒} d 蠒 = 0 \\ <2 |H'_{S}| 3> = \{. . .\} {鈭�}_{0}^{2 蟺} & e^{i 蠒} d 蠒 = 0 \\ <2 |H'_{S}| 4> = \{. . .\} {鈭�}_{0}^{2 蟺} & e^{i 蠒} d 蠒 = 0 \\ <2 |H'_{S}| 4> = \{. . .\} {鈭�}_{0}^{2 蟺} & e^{i 蠒} d 蠒 = 0 \end{matrix} \end{array}\}$

All matrix elements of are zero except $<1 |H'_{S}| 3>$ and $<3 |H'_{S}| 1>$ (which are complex conjugates, so only needs to be evaluated).
role="math" localid="1658313451117" $<1 |H'_{S}| 3> = e E_{e x t} \frac{1}{\sqrt{2 蟺 a}} \frac{1}{2 a} \frac{1}{\sqrt{2 蟺 a}} \frac{1}{4 a^{2}} 鈭� (1 - \frac{r}{2 a}) e^{- r / 2 a} \cos 胃 (r \cos 胃) r^{2} \sin 胃 d r d 胃 d 蠒 = <1 |H'_{S}| 3> = e E_{e x t} \frac{1}{\sqrt{2 蟺 a}} \frac{1}{2 a} \frac{1}{\sqrt{2 蟺 a}} \frac{1}{4 a^{2}} 鈭� (1 - \frac{r}{2 a}) e^{- r / 2 a} \cos 胃 (r \cos 胃) r^{2} \sin 胃 d r d 胃 d 蠒 = \frac{e E_{e x t}}{2 蟺 a 8 a^{3}} (2 蟺) [{鈭�}_{0}^{蟺} \cos^{2} 胃 \sin 胃 d 胃] {鈭�}_{0}^{鈭�} (1 - \frac{r}{2 a}) e^{- r / a} r^{4} d r = \frac{e E_{e x t}}{8 a^{4}} \frac{2}{3} \{{鈭�}_{0}^{鈭�} r^{4} e^{- r / a} d r - \frac{1}{2 a} {鈭�}_{0}^{鈭�} r^{5} e^{- r / a} d r\} = \frac{e E_{e x t}}{8 a^{4}} (4! a^{5} - \frac{1}{2 a} 5! a^{6}) = \frac{e E_{e x t}}{8 a^{4}} 24 a^{5} (1 - \frac{5}{2}) = e a E_{e x t} (- 3) = - 3 a e E_{e x t} W = - 3 a e E_{e x t} (\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$

There is need of eigenvalues of this matrix. The characteristic equation is:

$|\begin{matrix} - 位 & 0 & 1 & 0 \\ 0 & - 位 & 0 & 0 \\ 1 & 0 & - 位 & 0 \\ 0 & 0 & 0 & - 位 \end{matrix}| = - 位 |\begin{matrix} - 位 & 0 & 0 \\ 0 & - 位 & 0 \\ 0 & 0 & - 位 \end{matrix}| + |\begin{matrix} 0 & - 位 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 位 \end{matrix}| = - 位 {(- 位)}^{3} + (- 位^{2)} = 位^{2} (位^{2} - 10 = 0$

So, The eigenvalues are 0,01, and -1 , so the perturbed energies are

$E_{2,} E_{2,} E_{2} + 3 a e E_{e x t}, E_{2} - 3 a e E_{e x t}$

Obtain the electric dipole operator.

On the basis of Eigen vectors

$p {鈨�}_{e} = - e r 鈨� = - e r (s i n 鈦� 蠎 c o s 鈦� 蠁 x^+ s i n 鈦� 蠎 s i n 鈦� 蠁 y^+ c o s 鈦� 蠎 z^)$

$鉄� p {鈨�}_{e} {鉄�}_{4} = 鈭� 蠄_{(} 21 - 1)^{*} p {鈨�}_{e} 蠄_{(} 21 - 1) d^{3} r = - e \frac{1}{蟺 a} \frac{1}{8 a^{2}} 鈭� r^{2} e^{- r / a} \sin^{2} 蠎 e^{- i 蠁 + i 蠁} r (\sin 蠎 \cos 蠁 \hat{x} + \sin 蠎 \sin 蠁 \hat{y} + \cos 蠎 \hat{z)} = 0 {鈭�}_{0}^{2 蟺} \sin 蠁 d 蠁 = {鈭�}_{0}^{2 蟺} \cos 蠁 d 蠁 = 0 a n d {鈭�}_{0}^{蟺} \sin^{3} 蠎 \cos 蠎 d 蠎 = 0 A s = {<\overset{鈫�}{p_{e}}>}_{2} = 0 {<\overset{鈫�}{p_{e}}>}_{2} = - \frac{e}{2} 鈭� {(蠄_{200} + 蠄_{210})}^{2} r ((\sin 蠎 \cos 蠁 \hat{x} + \sin 蠎 \sin 蠁 \hat{y} + \cos 蠎 \hat{z)} r^{2} d r \sin 蠎 d 蠎 d 蠁 N o w i t e g t r a t e {鈭�}_{0}^{2 蟺} \sin 蠎 d 蠁 = {鈭�}_{0}^{2 蟺} \cos 蠁 d 蠁 = 0$

Here we have only Z-component

${<\overset{鈫�}{p_{e}}>}_{2} = - e 蟺 \hat{z} 鈭� ({蠄^{2}}_{200} + 2 蠄_{200 +} 蠄_{210 + 蠄^{2}} + {蠄^{2}}_{210}) r^{3} d r \sin 蠎 d 蠎 = - e 蟺 \hat{z} 鈭� [\frac{1}{2 蟺 a} \frac{1}{4 a^{2}} {(1 - \frac{r}{2 a})}^{2} e^{- r / a} + 2 \frac{1}{2 蟺 a} \frac{1}{8 a^{3}} (1 - \frac{r}{2 a}) r e^{- r / a} \cos 蠎 + \frac{1}{2 蟺 a} \frac{1}{16 a^{4}} r^{2} e^{- r / a} \cos^{2} 蠎] r^{3} d r \sin 蠎 d 蠎鈭� \sin 蠎 \cos 蠎 d 蠎 = 0 a n d 鈭� \sin 蠎 \cos^{3} 蠎 d 蠎 = 0 {<\overset{鈫�}{p_{e}}>}_{卤} = e 蟺 \hat{z} \frac{1}{8 a^{4}} {鈭�}_{0}^{鈭�} (1 - \frac{r}{2 a}) r^{4} e^{- r / a} d r {鈭�}_{0}^{蟺} \cos^{2} 蠎 \sin 蠎 d 蠎 = + \frac{e z}{8 a^{4}} \frac{2}{3} (a^{5} . 4! - \frac{1}{2 a} a^{6} 5!) {<\overset{鈫�}{p_{e}}>}_{卤} = 卤 3 e a \hat{z}$

Thus the Eigenvectors: $蠄_211, 蠄_(21 - 1), 1 / 鈭� 2 (蠄_200 + 蠄_210), 1 / 鈭� 2 (蠄_200 - 蠄_210) $ .$

Expectation values:

$鉄� p {鈨�}_{e} {鉄�}_{2} = 0 鉄� p {鈨�}_{e} {鉄�}_{4} = 0 鉄� p {鈨�}_{e} 鉄� = 卤 3 e a \hat{z}$

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Short Answer

Step by step solution

Define the formula for wave function in ground state

Effect of perturbation on ground state energy

first order corrections to the energy

Obtain the electric dipole operator.

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