Q42P The (time-independent) momentum ... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 4: Q42P (page 191)

The (time-independent) momentum space wave function in three dimensions is defined by the natural generalization of Equation 3.54:
$桅 (p, t) = \frac{1}{\sqrt{2 蟺 h}} {鈭�}_{鈭�}^{鈭�} e^{- ipx / h} 蠄 (x, t) dx (3.54) . 蠒 (p) 鈮� \frac{1}{{(2 蟺 h)}^{3 / 2}} 鈭� e^{- i (p . r) I h} 蠄 (r) d^{3} r . (4.223) .$
(a)Find the momentum space wave function for the ground state of hydrogen (Equation 4.80). Hint: Use spherical coordinates, setting the polar axis along the direction of p. Do the 胃 integral first. Answer:
$蠄_{100} (r, 胃, 蠒) = \frac{1}{\sqrt{{蟺补}^{3}}} e^{- r / a} (4.80) . 蠒 (p) = \frac{1}{蟺} {(\frac{2 a}{h})}^{3 / 2} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2 .}} (4.224) .$
(b) Check that $桅 (p)$ is normalized.
(c) Use $桅 (p)$ to calculate $< p^{2} >$ , in the ground state of hydrogen.
(d) What is the expectation value of the kinetic energy in this state? Express your answer as a multiple of $E_{1}$ , and check that it is consistent with the virial theorem (Equation 4.218).
$< T > = - E_{n}; < V > = 2 E_{n} (4.218) .$

Short Answer

Expert verified

$(a) 蠒 (p) = \sqrt{\frac{2}{h}} \frac{1}{a^{3 / 2}} \frac{1}{蟺} \frac{2 {pa}^{3}}{h} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2}} = \frac{1}{蟺} {(\frac{2 a}{h})}^{3 / 2} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2 .}} (b) 鈭� {|蠒|}^{2} d^{3} p = \frac{32}{蟺} {(\frac{a}{h})}^{3} \frac{蟺}{32} {(\frac{h}{a})}^{3} . (c) <p^{2}> = \frac{4}{蟺} {(\frac{2 a}{h})}^{3} {(\frac{h}{a})}^{8} \frac{蟺}{32} {(\frac{h}{a})}^{- 3} = \frac{h^{2}}{a^{2}} . (d) <T> = \frac{1}{2 m} <p^{2}> = \frac{1}{2 m} \frac{h^{2}}{a^{2}} = \frac{h^{2}}{2 m} \frac{m^{2}}{h^{4}} {(\frac{e^{2}}{4 蟺 {鈭�}_{0}})}^{2} = \frac{m}{2 h^{2}} {(\frac{e^{2}}{4 蟺 {鈭�}^{0}})}^{2} = - E_{1}$

Step by step solution

(a) Finding the momentum of space wave function for the ground state of hydrogen.

$蠄 = \frac{1}{\sqrt{{蟺补}^{3}}} e^{- r / a} 鈬� 蠒 (p) = \frac{1}{{(2 蟺 h)}^{3 / 2}} \frac{1}{\sqrt{{蟺补}^{3}}} 鈭� e^{- ip . rI h} e^{- r / a} r^{2} 蝉颈苍胃诲谤诲胃诲蠒 . With axes as suggested, p . r = pr 肠辞蝉胃 . Doing the (trivial) 桅 integral : 蠒 (p) = \frac{2 蟺}{{(2 蟺补 h)}^{3 / 2}} \frac{1}{\sqrt{蟺}} {鈭�}_{0}^{鈭�} r^{2} e^{- r / a} [{鈭�}_{0}^{蟺} e^{- ipr 肠辞蝉胃 / h} \sin 胃诲胃] dr . {鈭�}_{0}^{蟺} e^{- ipr 肠辞蝉胃 / h} sin胃诲胃 = \frac{h}{ipr} {e^{- ipr 肠辞蝉胃 / h}}_{0}^{蟺} = \frac{h}{ipr} | (e^{ipr / h} - e^{^{ipr / h}}) = \frac{2 h}{pr} \sin (\frac{pr}{h}) . 蠒 (p) = \frac{1}{蟺 \sqrt{2}} \frac{1}{{(a h)}^{3 / 2}} \frac{2 h}{p} {鈭�}_{0}^{鈭�} {re}^{- r / a} \sin (\frac{pr}{h}) dr . {鈭�}_{0}^{鈭�} {re}^{- r / a} \sin (\frac{pr}{h}) dr = \frac{1}{2 i} [{鈭�}_{0}^{鈭�} {re}^{- r / a} e^{ipr / h} dr - {鈭�}_{0}^{鈭�} {re}^{- r / a} e^{ipr / h} dr] . = \frac{1}{2 i} [\frac{1}{{(1 / a - ip / h)}^{2}} - \frac{1}{{(1 / a + ip / h)}^{2}}] = \frac{1}{2 i} \frac{(2 ip / a h) 2}{{[{(1 / a)}^{2} + {(p / h)}^{2}]}^{2 .}} = \frac{(2 p / h) a^{3}}{{[1 + {(ap / h)}^{2}]}^{2 .}} 蠒 (p) = \sqrt{\frac{2}{h}} \frac{1}{a^{3 / 2}} \frac{1}{蟺辫} \frac{2 {pa}^{3}}{h} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2}} = \frac{1}{蟺} {(\frac{2 a}{h})}^{3 / 2} \frac{1}{{[1 + {(ap / h)}^{2}]}^{2 .}}$

Step2: (b) Checking that Φ(p) is normalized.

$鈭� {|蠒|}^{2} d^{3} p = 4 蟺 {鈭�}_{0}^{鈭�} p^{2} {|蠒|}^{2} dp = 4 蟺 \frac{1}{蟺^{2}} {(\frac{2 a}{h})}^{3} {鈭�}_{0}^{鈭�} \frac{p^{2}}{{[1 + {(ap / h)}^{2}]}^{4 .}} From math tables : {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(m + x^{2})}^{4}} dx = \frac{蟺}{32} m^{- 5 / 2}, so {鈭�}_{0}^{鈭�} \frac{p^{2}}{{[1 + {(ap / h)}^{2}]}^{4}} dp = {(\frac{h}{a})}^{8} \frac{蟺}{32} {(\frac{h}{a})}^{- 5} = \frac{蟺}{32} (\frac{h}{a}); 鈭� {|蠒|}^{2} d^{3} p = \frac{32}{蟺} {(\frac{a}{h})}^{3} \frac{蟺}{32} {(\frac{h}{a})}^{3} .$

(c) Calculating <p2>

$<p^{2}> = 鈭� p^{2} {|蠒|}^{2} d^{3} p = \frac{1}{蟺^{2}} {(\frac{2 a}{h})}^{3} 4 蟺 {鈭�}_{0}^{鈭�} \frac{p_{4}}{{[1 + {(ap / h)}^{2}]}^{4}} dp . From math tables : {鈭�}_{0}^{鈭�} \frac{x^{4}}{{[m + x^{2}]}^{4}} dx = (\frac{蟺}{32}) m^{- 3 / 2} . So <p^{2}> = \frac{4}{蟺} {(\frac{2 a}{h})}^{3} {(\frac{h}{a})}^{8} \frac{蟺}{32} {(\frac{h}{a})}^{- 3} = \frac{h^{2}}{a^{2}}'$

Step4: (d) Expressing the answer as a multiple of E1

$<T> = \frac{1}{2 m} <p^{2}> = \frac{1}{2 m} \frac{h^{2}}{a^{2}} = \frac{h^{2}}{2 m} \frac{m^{2}}{h^{4}} {(\frac{e^{2}}{4 蟺 {鈭�}_{0}})}^{2} = \frac{m}{2 h^{2}} {(\frac{e^{2}}{4 蟺 {鈭�}_{0}})}^{2} = - E_{1} . Which is consistent with Eq . 4.218 . <T> = - E_{n}; <V> = 2 E_{n} (4.218)$

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Short Answer

Step by step solution

(a) Finding the momentum of space wave function for the ground state of hydrogen.

Step2: (b) Checking that Φ(p) is normalized.

(c) Calculating <p2>

Step4: (d) Expressing the answer as a multiple of E1

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