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. $ψ (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} ¶$

Short Answer

Expert verified

$a) \to The - value - A - is - A = {(\frac{2 a}{π})}^{1 / 4} - ¶ b) \to ψ (x, t) = {(\frac{2 a}{π})}^{1 / 4} \frac{e \frac{- {ax}^{2} + i (lx - {ħl}^{2} t / 2 m)}{\sqrt{1 + 2 iħat / m}}}{\sqrt{1 + 2 iħat / m}} Where, K = \frac{1}{\sqrt{2 π}} {(\frac{1}{2 πa})}^{1 / 4} ¶ c) \to {|ψ (x, t)|}^{2} = \sqrt{\frac{2}{π}} {ωe}^{- 2 ω^{2 {(x - θll 2 a)}^{2}}} ¶ d) \to <x> = \frac{ħl}{m} t, = ħl, <x^{2}> = \frac{1}{4 ϖ^{2}} + {(\frac{ħit}{m})}^{2}, <p^{2}> = ħ^{2} (a + /^{2}), σ_{x} = \frac{1}{2 ω}, σ_{p} = ħ \sqrt{a} . ¶ e) \to The - uncertainty - principle - holds - ¶ ¶ ¶ ¶$

Step by step solution

Normalizing and finding A¶

$¶ a) ¶$

The initial wave function, $¶$

$¶ ψ (x, 0) = A e^{- a x^{2}} ¶ ¶$

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;" $e^{i l x}, ¶$ ,

$¶ ψ (x, 0) = A e^{- a x^{2}} e^{i l x} ¶ ¶ Now, ¶ ¶ {|A|}^{2} \int_{- \infty}^{\infty} e^{- 2 {ax}^{2}} dx = 1 ¶ ¶ The . integral . comes . out . to . be . \sqrt{π / 2 a .} ¶ ¶ Therefore, ¶ ¶ A = {(\frac{2 a}{π})}^{1 / 4} ¶$

$T h e . v a l u e . o f . A . i s . A = {(\frac{2 a}{π})}^{1 / 4} . ¶ ¶ b) ¶ ¶ F o l l o w i n g . t h e . s a m e . p r o c e d u r e . a s . i n . t h e . s t a t i o n a r y . c a s e, ¶ ¶ ψ (x, t) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} ϕ (k) e^{ikx} e^{- {iħk}^{2} t / 2 m} d k ¶ ¶ W h e r e, ¶ ¶$

$ϕ (k) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} ψ (x, o) e^{- i k x} d k ¶ ϕ (k) = \frac{1}{\sqrt{2 π}} {(\frac{2 a}{π})}^{1 / 4} \int_{- \infty}^{\infty} e^{- a x^{2} - i k x + i l x} d k ¶ ϕ (k) = {(\frac{1}{2 π a})}^{1 / 4} e^{- {(k - l)}^{2} / 4 a} ¶ ¶ Thus, . ϕ (k) = {(\frac{1}{2 πa})}^{1 / 4} e^{- {(k - l)}^{2} / 4 a} ¶ ¶ Now . ¶ ψ (x, t) = K \int_{- \infty}^{\infty} e^{- / 2 / 4 a} \exp^{[(\frac{1}{4 a} + i \frac{ħt}{2 m}) k^{2} - (ix + \frac{1}{2 a}) k]} dk ¶$

$ψ (x, t) = {Ke}^{- l^{2} / 4 a} \sqrt{\frac{π}{(\frac{1}{4 a} + i \frac{ħt}{2 m})}} e^{\frac{{(ix + / / 2 a)}^{2}}{4 (\frac{1}{4 a} + i \frac{ħt}{2 m})}} The value calculated is ψ (x, t) = {(\frac{2 a}{π})}^{1 / 4} \frac{e^{\frac{- {ax}^{2} + i (ix - {ħl}^{2} t / 2 m)}{\sqrt{1 + 2 iħat / m}}}}{\sqrt{1 + 2 iħat / m}} Where, K = \frac{1}{\sqrt{2 π}} {(\frac{1}{2 πa})}^{1 / 4} .$

Finding |ψ|2

${|ψ|}^{2} = \sqrt{\frac{2 a}{π}} \frac{1}{\sqrt{1 + \frac{4 ħ^{2} a^{2} t^{2}}{m^{2}}}} e x p \{a [\frac{{(i x + / / 2 a)}^{2}}{1 + 2 i ħ a t / m} + \frac{{(i x + / / 2 a)}^{2}}{1 + 2 i ħ a t / m}]\} Let θ = 2 ħat / m {|ψ|}^{2} = \sqrt{\frac{2 a}{π}} \frac{1}{\sqrt{1 + θ^{2}}} e^{- l^{2} l 2 a} \exp \{a [\frac{{(ix + / / 2 a)}^{2}}{1 + θ} + \frac{{(ix + / / 2 a)}^{2}}{(1 - θ)}]\}$

Expanding the term in the square brackets,

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

$Hence, {|ψ|}^{2} = \sqrt{\frac{2 a}{π}} \frac{1}{\sqrt{1 + θ^{2}}} e^{- l^{2} l 2 a} \exp \{\frac{- 2 a}{1 + θ^{2}} {(x - \frac{θl}{2 a})}^{2} + \frac{l^{2}}{2 a^{2}}\} {|ψ|}^{2} = \sqrt{\frac{2 a}{π}} \frac{1}{\sqrt{1 + θ^{2}}} \exp \{\frac{- 2 a}{1 + θ^{2}} {(x - \frac{θl}{2 a})}^{2}\} Using, ω = {(\frac{a}{1 + {(2 ħat / m)}^{2}})}^{1 / 2} ω = {(\frac{a}{1 + θ^{2}})}^{1 / 2} Therefore, {|ψ (x, t)|}^{2} = \sqrt{\frac{2 a}{π}} {ωe}^{- 2 ω^{2} {(x - θll 2 a)}^{2}} .$

Calculating the expectation values of x

$<x> = \int_{- \infty}^{\infty} x {|ψ (x, t)|}^{2} d x <x> = \int_{- \infty}^{\infty} \sqrt{\frac{2}{π}} ω x e^{- 2 ω^{2} {(x - θ l l 2 a)}^{2}} d x L e t y = x - θ l l 2 a = x - v t, a n d t h u s, x = y + v t w h e r e, v = ħ l l m . W e g e t, <x> = \int_{- \infty}^{\infty} (y + v t) \sqrt{\frac{2}{π}} ω x e^{- 2 ω^{2} y^{2}} d y$

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

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

Calculating the expectation value of momentum

$ = m \frac{d <x>}{d t} = ħ l$

Calculating the expectation value of x2

$<x^{2}> = \int_{- \infty}^{\infty} {(y + v t)}^{2} \sqrt{\frac{2}{π}} ω e^{- 2 ω^{2} y^{2}} d y$

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’s an odd function, and the third integral is one by normalization, so the value of this integral would be,

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

Calculating the expectation value of <p2>

$<p^{2}> = - ħ^{2} \int_{- \infty}^{\infty} ψ^{*} \frac{d^{2} ψ}{d x} d x$

Since,

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

Therefore,

$<p^{2}> = \frac{4 a^{2} ħ^{2}}{{(1 + i θ)}^{2}} \int_{- \infty}^{\infty} [{(i x + \frac{l}{2 a})}^{2} + \frac{(1 + i θ)}{2 a}] {|ψ|}^{2} d x <p^{2}> = \frac{4 a^{2} ħ^{2}}{{(1 + i θ)}^{2}} \int_{- \infty}^{\infty} [- {(y + v t - \frac{i l}{2 a})}^{2} + \frac{(1 + i θ)}{2 a}] {|ψ|}^{2} d y <p^{2}> = \frac{4 a^{2} ħ^{2}}{{(1 + i θ)}^{2}} \int_{- \infty}^{\infty} [- \frac{1}{4 ω^{2}} + 0 - {(v t - \frac{i l}{2 a})}^{2} + \frac{(1 + i θ)}{2 a}] <p^{2}> = \frac{a ħ^{2}}{(1 + i θ)} [(1 + i θ) (1 + \frac{l^{2}}{a})] <p^{2}> = ħ^{2} (a + l^{2})$

Calculating the standard deviation

$σ_{x} = \sqrt{<x^{2}> - <x>} σ_{x} = \sqrt{\frac{1}{4 ω^{2}} + {(\frac{ħ l t}{m})}^{2} - {(\frac{ħ l t}{m})}^{2}} σ_{x} = \frac{1}{2 ω} And, σ_{p} = \sqrt{<p^{2}> - } σ_{p} = \sqrt{ħ^{2} a + ħ^{2} l^{2} - ħ^{2} l^{2}} σ_{p} = ħ \sqrt{a} Thus, <x> = \frac{ħl}{m} t, = ħl, <x^{2}> = \frac{1}{4 ω^{2}} + {(\frac{ħlt}{m})}^{2}, <p^{2}> = ħ^{2} (a + l^{2}), σ_{x} = \frac{1}{2 ω}, σ_{p} = ħ \sqrt{a}$

Checking for the uncertainty principle

(e)

$σ_{x} σ_{p} = \frac{ħ \sqrt{a}}{2 ω} Putting the value of ω as : σ_{x} σ_{p} = {(\frac{1 + {(2 ħat / m)}^{2}}{a})}^{1 / 2} \frac{ħ \sqrt{a}}{2} σ_{x} σ_{p} = \frac{ħ}{2} \sqrt{1 + \frac{{(2 ħat)}^{2}}{m^{2}}} σ_{x} σ_{p} \geq \frac{ħ^{2}}{2}$

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. $ψ (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} ¶$

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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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.ψ(x,0)=Ae-ax2eilx¶

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

Turning Points in Physics

Electromagnetism

Fluids

Oscillations

Linear Momentum

Magnetism and Electromagnetic Induction

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. $ψ (x, 0) = {Ae}^{- {ax}^{2}} e^{ilx} ¶$