Q12P (a) Generalize聽Problem 7.2, usi... [FREE SOLUTION]

Chapter 7: Q12P (page 309)

(a) GeneralizeProblem 7.2, using the trial wave function
$蠄 (x) = \frac{A}{{(x^{2} + b^{2})}^{n}},$
For arbitrary n. Partial answer: The best value of b is given by
localid="1658300238725" $b^{2} = \frac{h}{m 蝇} {[\frac{n (4 n - 1) (4 n - 3)}{2 (2 n + 1)}]}^{1 / 2}$
(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form
$蠄 (x) = \frac{B x}{{(x^{2} + b^{2})}^{n}} .$
Partial answer: The best value of b is given by
localid="1658300555415" $b^{2} = \frac{h}{m 蝇} {[\frac{n (4 n - 5) (4 n - 3)}{2 (2 n + 1)}]}^{1 / 2} .$
(c) Notice that the bounds approach the exact energies as n 鈫掆垶. Why is that? Hint: Plot the trial wave functions for n = 2 , n = 3 , and n = 4, and compare them with the true wave functions (Equations 2.59 and 2.62). To do it analytically, start with the identity
$e^{z} = \underset{n 鈫� 鈭�}{l i m} {(1 + \frac{z}{n})}^{n} 蠄_{0} (x) = {(\frac{m 蝇}{蟺 h})}^{1 / 4} e^{- \frac{m 蝇}{2 h} x^{2}}$ $(2.59) .$
$蠄_{1} (x) = A_{1} {\hat{a}}_{+} 蠄_{0} = \frac{A_{1}}{\sqrt{2 h m 蝇}} (- h \frac{d}{d x} + m 蝇 x) {(\frac{m 蝇}{蟺 h})}^{1 / 4} e^{- \frac{m 蝇}{2 h} x^{2}} 蠄_{1} (x) = A_{1} {(\frac{m 蝇}{蟺 h})}^{1 / 4} \sqrt{\frac{2 m 蝇}{h}} x e^{- \frac{m 蝇}{2 h} x^{2}} (2.62) .$

Short Answer

Expert verified

$(a) <H_{m i n}> \frac{1}{2} h 蝇 \sqrt{\frac{8 n^{2} - 2 n}{8 n^{2} - 2 n - 3}} > \frac{1}{2} h 蝇 (b) <H_{m i n}> \frac{3}{2} h 蝇 \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 3}} > \frac{3}{2} h 蝇 (c) 蠄_{0} (x) = {(\frac{m 蝇}{蟺 h})}^{1 / 4 n} e^{- \frac{m 蝇}{2 h} x^{2}}$

Step by step solution

Given.

We will need the following integral repeatedly:

$蠄 (x) = \frac{A}{{(x^{2} + b^{2})}^{n}}, 蠄 (x) = \frac{B x}{{(x^{2} + b^{2})}^{n}} .$

(a) Generalizing using the trial wave function

$1 = {鈭�}_{- 鈭�}^{鈭�} {|蠄|}^{2} d x = 2 {|A|}^{2} {鈭�}_{0}^{鈭�} \frac{1}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{{|A|}^{2}}{b^{4 n - 1}} \frac{螕 (\frac{1}{2}) 螕 (\frac{4 n - 1}{2})}{螕 (2 n)} 鈬� A = \sqrt{\frac{b^{4 n - 1} (2 n)}{螕 (\frac{1}{2}) 螕 (\frac{4 n - 1}{2})}}$

$<T> = - \frac{h^{2}}{2 m} {鈭�}_{- 鈭�}^{鈭�} 蠄 \frac{d^{2} x}{d x^{2}} d x = - \frac{h^{2}}{2 m} A^{2} {鈭�}_{- 鈭�}^{鈭�} \frac{1}{{(x^{2} + b^{2})}^{n}} \frac{d}{d x} [\frac{- 2 n x}{{(x^{2} + b^{2})}^{n + 1}}] d x$

$= \frac{n h^{2}}{m} A^{2} {鈭�}_{- 鈭�}^{鈭�} \frac{1}{{(x^{2} + b^{2})}^{n}} [\frac{1}{{(x^{2} + b^{2})}^{n + 1}} - \frac{2 (n + 1) x^{2}}{{(x^{2} + b^{2})}^{n + 2}}] d x = \frac{2 n h^{2}}{m} A^{2} [{鈭�}_{0}^{鈭�} \frac{1}{{(x^{2} + b^{2})}^{2 n + 1}} d x - 2 (n + 1) {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 2}} d x]$

role="math" localid="1658310221074" $= \frac{2 n h^{2}}{m} \frac{b^{4 n - 1} 螕 (2 n)}{螕 (\frac{1}{2}) 螕 (\frac{4 n - 1}{2})} [\frac{1}{2 b^{4 n + 1}} \frac{螕 (\frac{1}{2}) 螕 (\frac{4 n + 1}{2})}{螕 (2 n + 1)} - \frac{2 (n + 1)}{2 b^{4 n - 1}} \frac{螕 (\frac{1}{2}) 螕 (\frac{4 n + 1}{2})}{螕 (2 n + 2)}] = \frac{h^{2}}{4 m b^{2}} \frac{n (4 n - 1)}{(2 n + 1)} . <V> = \frac{1}{2} m 蝇^{2} {鈭�}_{- 鈭�}^{鈭�} 蠄^{2} x^{2} d x = \frac{1}{2} m 蝇^{2} 2 A^{2} {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n}} d x . = m 蝇^{2} \frac{b^{4 n - 1} 螕 (2 n)}{螕 (\frac{1}{2}) 螕 (\frac{4 n - 1}{2})} \frac{1}{2 b^{4 n - 3}} \frac{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})}{螕 (2 n)} = \frac{m 蝇^{2} b^{2}}{2 (4 n - 3)} . <H> = \frac{h^{2}}{4 m b^{2}} \frac{n (4 n - 1)}{(2 n + 1)} + \frac{m 蝇^{2} b^{2}}{(4 n - 3)}; \frac{鈭� <H>}{鈭� b} = - \frac{h^{2}}{2 m b^{3}} \frac{n (4 n - 1)}{(2 n + 1)} + \frac{m 蝇^{2} b^{2}}{(4 n - 3)} = 0$

$鈬� b = \sqrt{\frac{h}{m 蝇}} {[\frac{n (4 n - 1) (4 n - 3)}{2 (2 n + 1)}]}^{1 / 4} . {<H>}_{m i n} = \frac{h^{2}}{4 m} \frac{n (4 n - 1)}{(2 n + 1)} \frac{m 蝇}{h} \sqrt{\frac{2 (2 n + 1)}{n (4 n - 1) (4 n - 3)}} + \frac{m 蝇^{2}}{2 (4 n - 3)} \frac{h}{m 蝇} \sqrt{\frac{n (4 n - 1) (4 n - 3)}{2 (2 n + 1)} .} = \frac{1}{2} h 蝇 \sqrt{\frac{2 n (4 n - 1)}{(2 n + 1) (4 n - 3)}} = \frac{1}{2} h 蝇 \sqrt{\frac{8 n^{2} - 2 n}{8 n^{2} - 2 n - 3}} > \frac{1}{2} h 蝇$

(b) Finding the least upper bound on the first excited state

$<H> = \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} + \frac{3}{2} \frac{m 蝇^{2} b^{2}}{(4 n - 5)}; \frac{鈭� <H>}{鈭� b} = - \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} + \frac{m 蝇^{2} b^{2}}{(4 n - 5)} = 0 b = \sqrt{\frac{h}{m 蝇}} {[\frac{n (4 n - 3) (4 n - 5)}{2 (2 n + 1)}]}^{1 / 4} . <H> = \frac{3 h^{2}}{4 m} \frac{n (4 n - 3)}{(2 n + 1)} \frac{m 蝇}{h} \sqrt{\frac{2 (2 n + 1)}{n (4 n - 3) (4 n - 5)}} + \frac{3}{2} \frac{m 蝇^{2}}{(4 n - 5)} \frac{h}{m 蝇} \sqrt{\frac{n (4 n - 3) (4 n - 5)}{2 (2 n + 1)}} . = \frac{3}{2} h 蝇 \sqrt{\frac{2 n (4 n - 3)}{(2 n + 1) (4 n - 5)}} = \frac{3}{2} h 蝇 \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 5}} > \frac{3}{2} h 蝇 = \frac{3}{2} h 蝇 \sqrt{\frac{2 n (4 n - 3)}{(2 n + 1) (4 n - 5)}} = \frac{3}{2} h 蝇 \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 5}} > \frac{3}{2} h 蝇 .$ $1 = 2 {|B|}^{2} {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{{|B|}^{2}}{b^{4 n - 3}} \frac{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})}{螕 (2 n)} 鈬� B = \sqrt{\frac{b^{4 n - 3} 螕 (2 n)}{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})}} <T> = \frac{h^{2}}{2 m} B^{2} {鈭�}_{- 鈭�}^{鈭�} \frac{x}{{(x^{2} + b^{2})}^{n}} \frac{d}{d x} [\frac{x}{{(x^{2} + b^{2})}^{n}} - \frac{2 n x^{2}}{{(x^{2} + b^{2})}^{n + 1}}] d x = - \frac{h^{2} B^{2}}{2 m} {鈭�}_{- 鈭�}^{鈭�} \frac{x}{{(x^{2} + b^{2})}^{n}} [\frac{- 2 n x}{{(x^{2} + b^{2})}^{n + 1}} - \frac{4 n x}{{(x^{2} + b^{2})}^{n + 1}} + \frac{4 n (n + 1) x^{3}}{{(x^{2} + b^{2})}^{n + 2}}] d x = \frac{4 n h^{2} B^{2}}{2 m} [3 {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 1}} d x - 2 (n + 1) {鈭�}_{0}^{鈭�} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 2}} d x] = \frac{2 n h^{2}}{m} \frac{b^{4 n - 3} 螕 (2 n)}{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})} [\frac{3}{2 b^{4 n - 1}} \frac{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})}{螕 (2 n + 1)} - \frac{2 (2 n + 1)}{2 b^{4 n - 1}} \frac{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})}{螕 (2 n + 2)}] = \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} .$

$<V> = \frac{1}{2} m 蝇^{2} 2 B^{2} {鈭�}_{0}^{鈭�} \frac{x^{4}}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{1}{2} m 蝇^{2} \frac{b^{4 n - 3} 螕 (2 n)}{螕 (\frac{3}{2}) 螕 (\frac{4 n - 3}{2})} \frac{2}{2 b^{4 n - 5}} \frac{螕 (\frac{5}{2}) 螕 (\frac{4 n - 5}{2})}{螕 (2 n)} = \frac{3}{2} \frac{m 蝇^{2} b^{2}}{(4 n - 5)} .$

Step4: (c) The bounds approach the exact energy as n→∞

As $n 鈫� 鈭�,$ 蠄 becomes more and more Gaussian鈥�.

Analytically, for large n $b 鈮� \sqrt{\frac{h}{m 蝇}} {(\frac{n . 4 n . 4 n}{2.2 n})}^{1 / 4} = \sqrt{\frac{2 n h}{m 蝇}},$ so

${(x^{2} + b^{2})}^{n} = b^{2 n} (1 + x)$ ²b²n鈮�b2n1+m蝇x22hnn鈫�b2nem蝇x2/h

Meanwhile, using Sterling鈥檚 approximation in the form $螕 (z + 1) 鈮� z^{z} e^{- z} :$

$A^{2} = \frac{b^{4 n - 1} 螕 (2 n)}{螕 (\frac{1}{2}) 螕 (2 n - \frac{1}{2})} 鈮� \frac{b^{4 n - 1}}{\sqrt{蟺}} \frac{{(2 n - 1)}^{2 n - 1} e^{- (2 n - 1)}}{{(2 n - \frac{3}{2})}^{2 n - 3 / 2} e^{- (2 n - 3 / 2)}} 鈮� \frac{b^{4 n - 1}}{\sqrt{蟺}} \frac{1}{\sqrt{e}} (\frac{2 n - 1}{2 n - \frac{3}{2}})$

localid="1658317073153" $But (\frac{1 - \frac{1}{2 n}}{1 - \frac{3}{4 n}}) 鈮� (1 - \frac{1}{2 n}) (1 - \frac{3}{4 n}) 鈮� 1 + \frac{3}{4 n} - \frac{1}{2 n} = 1 + \frac{1}{4 n}; s o {(\frac{2 n - 1}{2 n - \frac{3}{2}})}^{2 n - 1} 鈮� {[{(1 + \frac{1}{4 n})}^{n}]}^{2} \frac{1}{1 + 1 / 4 n} 鈫� {(e^{1 / 4})}^{2} = \sqrt{e} . = \frac{b^{4 n - 1}}{\sqrt{蟺别}} \sqrt{e} \sqrt{2 n} = \sqrt{\frac{2 n}{蟺}} b^{4 n - 1} 鈬� A 鈮� {(\frac{2 n}{蟺})}^{1 / 4} b^{2 n - 1 / 2}$

So,

$蠄鈮� {(\frac{2 n}{蟺})}^{1 / 4} b^{2 n - 1 / 2} \frac{1}{b^{2 n}} e^{- m 蝇 x^{2} / 2 h} = {(\frac{2 n}{蟺})}^{1 / 4} {(\frac{m 蝇}{2 n h})}^{1 / 4} e^{- m 蝇 x^{2} / 2 h} = {(\frac{m 蝇}{蟺丑})}^{1 / 4} e^{- m 蝇 x^{2} / 2 h}$

Which is precisely the ground state of the harmonic oscillator (Eq. 2.60). So it鈥檚 no accident that we get the exact energies, in the limit n 鈫� 鈭�.

$蠄_{0} (x) = {(\frac{m 蝇}{蟺 h})}^{1 / 4} e^{- \frac{m 蝇}{2 h} x^{2}} .$

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Short Answer

Step by step solution

Given.

(a) Generalizing using the trial wave function

(b) Finding the least upper bound on the first excited state

Step4: (c) The bounds approach the exact energy as n→∞

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