Q1P A hydrogen atom is placed in a (... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 9: Q1P (page 343)

A hydrogen atom is placed in a (time-dependent) electric $field E = E (t) \overset{铃�}{k .} calculate all four matrix elements H_{ij}^{,} of the perturbation \overset{铃�}{H^{,}} = eEz$ between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show ${thatH}_{ii}^{,} = 0$ for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is 鈥渁ccessible鈥� from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration鈥攁ssuming transitions to higher excited states can be ignored.

Short Answer

Expert verified

$H_{100, 210}^{,} = - 0.7449 e E a .$

Step by step solution

Concept of time varying electric field

Consider a hydrogen atom placed in a time-varying electric field:

$E = E (t) \overset{铃�}{z}$

Where E(t) is some arbitrary function of time. The spatial wave functions formula is given by:

$蠄_{n / m} = R_{nl} Y_{l}^{m}$

Calculating all for matrix elements and showing that Hii,=0

$蠄_{n / m} = R_{n |y| m} . From Tables 4.3 and 4.7 :$

$蠄_{100} = \frac{1}{\sqrt{{蟺补}^{3}}} e^{- r / a}; 蠄_{200} = \frac{1}{\sqrt{8^{'} {蟺补}^{3}}} (1 - \frac{r}{2 a}) e^{- r / 2 a} 蠄_{210} = \frac{1}{\sqrt{32 {蟺补}^{3}}} \frac{r}{a} e^{- r / 2 a} \cos 胃; 蠄_{21 卤 1} = 卤 \frac{1}{\sqrt{64 {蟺补}^{3}}} \frac{r}{a} e^{- r / 2 a} \sin 胃 e^{卤 i 蠒}$

$But r 肠辞蝉胃 = z and r {蝉颈苍胃别}^{卤颈蠒} = r \sin (肠辞蝉蠒卤 i 蝉颈苍蠒) = r 蝉颈苍胃肠辞蝉蠒卤 ir 蝉颈苍胃蝉颈苍蠒 . so {|蠄|}^{2 is} an even function of z in all cases, and hence 鈭� z {|蠄|}^{2} dxdydz, {soH}_{ii}^{,} = 0 moreover, 蠄_{100} is even in z, and so are 蠄_{200}, 蠄_{211}, and 蠄_{21 - 1}, so H_{ij}^{,} = 0 for all except H_{100, 210}^{,} = eE \frac{1}{\sqrt{{蟺补}^{3}}} \frac{1}{\sqrt{32 {蟺补}^{3}}} \frac{1}{a} 鈭� e^{- r / a} e^{- r / 2 a} z^{2} d^{3} r = \frac{eE}{4 \sqrt{2 {蟺补}^{4}}} 鈭� e^{- 3 r / 2 a} r^{2} \cos^{2} {胃谤}^{2} 蝉颈苍胃drd胃 . = \frac{eE}{4 \sqrt{2 {蟺补}^{4}}} {鈭�}_{0}^{鈭�} r^{4} e^{- 3 r / 2 a} dr {鈭�}_{0}^{蟺} \cos^{2} 胃蝉颈苍胃d胃 {鈭�}_{0}^{2 蟺} 诲蠒 = \frac{eE}{4 \sqrt{2 {蟺补}^{4}}} 4! {(\frac{2 a}{3})}^{5} \frac{2}{3} 2 蟺 = (\frac{2^{8}}{3^{5} \sqrt{2}}) or 0 : 7449 eEa .$

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Short Answer

Step by step solution

Concept of time varying electric field

Calculating all for matrix elements and showing that Hii,=0

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