Q43P In Problem 2.21 you analyzed the... [FREE SOLUTION]

Chapter 2: Q43P (page 86)

In Problem 2.21 you analyzed the stationary gaussian free particle wave packet. Now solve the same problem for the traveling gaussian wave packet, starting with the initial wave function. $蠄 (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} 露$

Short Answer

Expert verified

$a) 鈫� The - value - A - is - A = {(\frac{2 a}{蟺})}^{1 / 4} - 露 b) 鈫� 蠄 (x, t) = {(\frac{2 a}{蟺})}^{1 / 4} \frac{e \frac{- {ax}^{2} + i (lx - {魔濒}^{2} t / 2 m)}{\sqrt{1 + 2 颈魔补迟 / m}}}{\sqrt{1 + 2 颈魔补迟 / m}} Where, K = \frac{1}{\sqrt{2 蟺}} {(\frac{1}{2 蟺补})}^{1 / 4} 露 c) 鈫� {|蠄 (x, t)|}^{2} = \sqrt{\frac{2}{蟺}} {蝇别}^{- 2 蝇^{2 {(x - 胃濒濒 2 a)}^{2}}} 露 d) 鈫� <x> = \frac{魔濒}{m} t, = 魔濒, <x^{2}> = \frac{1}{4 蠔^{2}} + {(\frac{魔颈迟}{m})}^{2}, <p^{2}> = 魔^{2} (a + /^{2}), 蟽_{x} = \frac{1}{2 蝇}, 蟽_{p} = 魔 \sqrt{a} . 露 e) 鈫� The - uncertainty - principle - holds - 露露露露$

Step by step solution

Normalizing and finding A¶

$露 a) 露$

The initial wave function, $露$

$露蠄 (x, 0) = A e^{- a x^{2}} 露露$

Converting .this .equation.to. a. traveling. wave,. so .we. multiply .the .stationary .wave .function. by .a. factor .of .localid="1658142498331" width="49" style="max-width: none;" $e^{i l x}, 露$ ,

$露蠄 (x, 0) = A e^{- a x^{2}} e^{i l x} 露露 Now, 露露 {|A|}^{2} {鈭�}_{- 鈭�}^{鈭�} e^{- 2 {ax}^{2}} dx = 1 露露 The . integral . comes . out . to . be . \sqrt{蟺 / 2 a .} 露露 Therefore, 露露 A = {(\frac{2 a}{蟺})}^{1 / 4} 露$

$T h e . v a l u e . o f . A . i s . A = {(\frac{2 a}{蟺})}^{1 / 4} . 露露 b) 露露 F o l l o w i n g . t h e . s a m e . p r o c e d u r e . a s . i n . t h e . s t a t i o n a r y . c a s e, 露露蠄 (x, t) = \frac{1}{\sqrt{2 蟺}} {鈭�}_{- 鈭�}^{鈭�} 蠒 (k) e^{ikx} e^{- {颈魔办}^{2} t / 2 m} d k 露露 W h e r e, 露露$

$蠒 (k) = \frac{1}{\sqrt{2 蟺}} {鈭�}_{- 鈭�}^{鈭�} 蠄 (x, o) e^{- i k x} d k 露蠒 (k) = \frac{1}{\sqrt{2 蟺}} {(\frac{2 a}{蟺})}^{1 / 4} {鈭�}_{- 鈭�}^{鈭�} e^{- a x^{2} - i k x + i l x} d k 露蠒 (k) = {(\frac{1}{2 蟺 a})}^{1 / 4} e^{- {(k - l)}^{2} / 4 a} 露露 Thus, . 蠒 (k) = {(\frac{1}{2 蟺补})}^{1 / 4} e^{- {(k - l)}^{2} / 4 a} 露露 Now . 露蠄 (x, t) = K {鈭�}_{- 鈭�}^{鈭�} e^{- / 2 / 4 a} \exp^{[(\frac{1}{4 a} + i \frac{魔迟}{2 m}) k^{2} - (ix + \frac{1}{2 a}) k]} dk 露$

$蠄 (x, t) = {Ke}^{- l^{2} / 4 a} \sqrt{\frac{蟺}{(\frac{1}{4 a} + i \frac{魔迟}{2 m})}} e^{\frac{{(ix + / / 2 a)}^{2}}{4 (\frac{1}{4 a} + i \frac{魔迟}{2 m})}} The value calculated is 蠄 (x, t) = {(\frac{2 a}{蟺})}^{1 / 4} \frac{e^{\frac{- {ax}^{2} + i (ix - {魔濒}^{2} t / 2 m)}{\sqrt{1 + 2 颈魔补迟 / m}}}}{\sqrt{1 + 2 颈魔补迟 / m}} Where, K = \frac{1}{\sqrt{2 蟺}} {(\frac{1}{2 蟺补})}^{1 / 4} .$

Finding |ψ|2

${|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + \frac{4 魔^{2} a^{2} t^{2}}{m^{2}}}} e x p \{a [\frac{{(i x + / / 2 a)}^{2}}{1 + 2 i 魔 a t / m} + \frac{{(i x + / / 2 a)}^{2}}{1 + 2 i 魔 a t / m}]\} Let 胃 = 2 魔补迟 / m {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} e^{- l^{2} l 2 a} \exp \{a [\frac{{(ix + / / 2 a)}^{2}}{1 + 胃} + \frac{{(ix + / / 2 a)}^{2}}{(1 - 胃)}]\}$

Expanding the term in the square brackets,

$[T] = \frac{1}{1 + 胃^{2}} [(1 - i 胃) {(i x + \frac{l}{2 a})}^{2} + (1 + i 胃) {(- i x + \frac{1}{2 a})}^{2}] [T] = \frac{1}{1 + 胃^{2}} [(- x^{2} + \frac{i x l}{a} + \frac{l^{2}}{4 a^{2}}) + (- x^{2} + \frac{i x l}{a} + \frac{l^{2}}{4 a^{2}}) + i 胃 (x^{2} + \frac{i x l}{a} + \frac{l^{2}}{4 a^{2}}) + i 胃 (- x^{2} + \frac{i x l}{a} + \frac{l^{2}}{4 a^{2}})] [T] = \frac{1}{1 + 胃^{2}} [- 2 x^{2} + \frac{l^{2}}{2 a^{2}} + 2 胃 \frac{x l}{a}] [T] = \frac{1}{1 + 胃^{2}} [- 2 x^{2} + 2 胃 \frac{x l}{a} - \frac{胃^{2} l^{2}}{2 a^{2}} + \frac{1^{2}}{2 a^{2}}] [T] = \frac{- 2}{1 + 胃^{2}} {(x - \frac{胃 l}{2 a})}^{2} + \frac{1^{2}}{2 a^{2}}$

$Hence, {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} e^{- l^{2} l 2 a} \exp \{\frac{- 2 a}{1 + 胃^{2}} {(x - \frac{胃濒}{2 a})}^{2} + \frac{l^{2}}{2 a^{2}}\} {|蠄|}^{2} = \sqrt{\frac{2 a}{蟺}} \frac{1}{\sqrt{1 + 胃^{2}}} \exp \{\frac{- 2 a}{1 + 胃^{2}} {(x - \frac{胃濒}{2 a})}^{2}\} Using, 蝇 = {(\frac{a}{1 + {(2 魔补迟 / m)}^{2}})}^{1 / 2} 蝇 = {(\frac{a}{1 + 胃^{2}})}^{1 / 2} Therefore, {|蠄 (x, t)|}^{2} = \sqrt{\frac{2 a}{蟺}} {蝇别}^{- 2 蝇^{2} {(x - 胃濒濒 2 a)}^{2}} .$

Calculating the expectation values of x

$<x> = {鈭�}_{- 鈭�}^{鈭�} x {|蠄 (x, t)|}^{2} d x <x> = {鈭�}_{- 鈭�}^{鈭�} \sqrt{\frac{2}{蟺}} 蝇 x e^{- 2 蝇^{2} {(x - 胃 l l 2 a)}^{2}} d x L e t y = x - 胃 l l 2 a = x - v t, a n d t h u s, x = y + v t w h e r e, v = 魔 l l m . W e g e t, <x> = {鈭�}_{- 鈭�}^{鈭�} (y + v t) \sqrt{\frac{2}{蟺}} 蝇 x e^{- 2 蝇^{2} y^{2}} d y$

Since, we have two integrals, the first one is zero as the function is odd, and the integration of an odd function from $- 鈭�$ to $鈭�$ is zero, and the second integration is 1 by normalization, so:

role="math" localid="1658204074614" $<x> = v t <x> = \frac{魔 l}{m} t$

Calculating the expectation value of momentum

$ = m \frac{d <x>}{d t} = 魔 l$

Calculating the expectation value of x2

$<x^{2}> = {鈭�}_{- 鈭�}^{鈭�} {(y + v t)}^{2} \sqrt{\frac{2}{蟺}} 蝇 e^{- 2 蝇^{2} y^{2}} d y$

Since ${(y + v t)}^{2} = y^{2} + 2 y v t + v^{2} t^{2}$ , to find the values of the first integral, we used the integral calculator, the second integral is zero since it鈥檚 an odd function, and the third integral is one by normalization, so the value of this integral would be,

$<x^{2}> = \frac{1}{4 蝇^{2}} + 0 + v^{2} t^{2} <x^{2}> = \frac{1}{4 蝇^{2}} + {(\frac{魔 l t}{m})}^{2}$

Calculating the expectation value of <p2>

$<p^{2}> = - 魔^{2} {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{d^{2} 蠄}{d x} d x$

Since,

$\frac{d 蠄}{d x} = \frac{2 i a (i x + \frac{l}{2 a})}{(1 + i 胃)} 蠄 \frac{d 蠄}{d x} = [\frac{- 4 a^{2} (i x + / / 2 a)}{1 + i 胃^{2}} - \frac{2 a}{1 + i 胃}] 蠄$

Therefore,

$<p^{2}> = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} {鈭�}_{- 鈭�}^{鈭�} [{(i x + \frac{l}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] {|蠄|}^{2} d x <p^{2}> = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} {鈭�}_{- 鈭�}^{鈭�} [- {(y + v t - \frac{i l}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] {|蠄|}^{2} d y <p^{2}> = \frac{4 a^{2} 魔^{2}}{{(1 + i 胃)}^{2}} {鈭�}_{- 鈭�}^{鈭�} [- \frac{1}{4 蝇^{2}} + 0 - {(v t - \frac{i l}{2 a})}^{2} + \frac{(1 + i 胃)}{2 a}] <p^{2}> = \frac{a 魔^{2}}{(1 + i 胃)} [(1 + i 胃) (1 + \frac{l^{2}}{a})] <p^{2}> = 魔^{2} (a + l^{2})$

Calculating the standard deviation

$蟽_{x} = \sqrt{<x^{2}> - <x>} 蟽_{x} = \sqrt{\frac{1}{4 蝇^{2}} + {(\frac{魔 l t}{m})}^{2} - {(\frac{魔 l t}{m})}^{2}} 蟽_{x} = \frac{1}{2 蝇} And, 蟽_{p} = \sqrt{<p^{2}> - } 蟽_{p} = \sqrt{魔^{2} a + 魔^{2} l^{2} - 魔^{2} l^{2}} 蟽_{p} = 魔 \sqrt{a} Thus, <x> = \frac{魔濒}{m} t, = 魔濒, <x^{2}> = \frac{1}{4 蝇^{2}} + {(\frac{魔濒t}{m})}^{2}, <p^{2}> = 魔^{2} (a + l^{2}), 蟽_{x} = \frac{1}{2 蝇}, 蟽_{p} = 魔 \sqrt{a}$

Checking for the uncertainty principle

(e)

$蟽_{x} 蟽_{p} = \frac{魔 \sqrt{a}}{2 蝇} Putting the value of 蝇 as : 蟽_{x} 蟽_{p} = {(\frac{1 + {(2 魔补迟 / m)}^{2}}{a})}^{1 / 2} \frac{魔 \sqrt{a}}{2} 蟽_{x} 蟽_{p} = \frac{魔}{2} \sqrt{1 + \frac{{(2 魔补迟)}^{2}}{m^{2}}} 蟽_{x} 蟽_{p} 鈮� \frac{魔^{2}}{2}$

Hence, the uncertainty principle holds.

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In Problem 2.21 you analyzed the stationary gaussian free particle wave packet. Now solve the same problem for the traveling gaussian wave packet, starting with the initial wave function. $蠄 (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} 露$

Short Answer

Step by step solution

Normalizing and finding A¶

Finding |ψ|2

Calculating the expectation values of x

Calculating the expectation value of momentum

Calculating the expectation value of x2

Calculating the expectation value of <p2>

Calculating the standard deviation

Checking for the uncertainty principle

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Most popular questions from this chapter

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91影视

In Problem 2.21 you analyzed the stationary gaussian free particle wave packet. Now solve the same problem for the traveling gaussian wave packet, starting with the initial wave function.蠄(x,0)=Ae-ax2eilx露

Short Answer

Step by step solution

Normalizing and finding A¶

Finding |ψ|2

Calculating the expectation values of x

Calculating the expectation value of momentum

Calculating the expectation value of x2

Calculating the expectation value of <p2>

Calculating the standard deviation

Checking for the uncertainty principle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physics of Motion

Electromagnetism

Nuclear Physics

Mechanics and Materials

Solid State Physics

Electricity and Magnetism

Study anywhere. Anytime. Across all devices.

In Problem 2.21 you analyzed the stationary gaussian free particle wave packet. Now solve the same problem for the traveling gaussian wave packet, starting with the initial wave function. $蠄 (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} 露$