Q7P Calculate d銆坧銆�/dt. Answer:d... [FREE SOLUTION]

Chapter 1: Q7P (page 18)

Calculate d銆坧銆�/dt. Answer:
$\frac{d }{d x} = <- \frac{鈭� V}{鈭� x}>$

This is an instance of Ehrenfest鈥檚 theorem, which asserts that expectation values obey the classical laws

Short Answer

Expert verified

$\frac{d }{d x} = <- \frac{鈭� V}{鈭� x}>$

Schrodinger equation and its complex conjugate:

role="math" localid="1657778797745" $\frac{鈭� 蠄}{鈭� t} = \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} - \frac{i}{h} V (x, t) 蠄 (x, t) (1) \frac{鈭� 蠄^{*}}{鈭� t} = - \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄^{*}}{鈭� x^{2}} + \frac{i}{h} V (x, t) 蠄 (x, t) (2)$

And according to Ehrenfest鈥檚 theorem,

$ = m <v> = m \frac{d <x>}{d t}$

Step by step solution

Determining the expectation value of x

Finding the expectation values.

$<x> = \frac{{鈭�}_{- 鈭�}^{鈭�} x {|蠄_{(x . t)}|}^{2} d x}{{鈭�}_{- 鈭�}^{鈭�} {|蠄_{(x . t)}|}^{2} d x} <x> = {鈭�}_{- 鈭�}^{鈭�} x {|蠄_{(x . t)}|}^{2} d x <x> = {鈭�}_{- 鈭�}^{鈭�} 蠄_{(x . t)} {蠄^{*}}_{(x . t)} d x <x> = {鈭�}_{- 鈭�}^{鈭�} 蠄_{(x . t)} {蠄^{*}}_{(x . t)} d x$

Differentiating both the sides with respect to t,

$\frac{d (x)}{d t} = {鈭�}_{- 鈭�}^{鈭�} 蠄_{(x . t)} {蠄 {(x)}^{*}}_{(x . t)} d x \frac{d (x)}{d t} = {鈭�}_{- 鈭�}^{鈭�} 脳 \frac{鈭� 蠄}{鈭� t} [蠄_{(x . t)} {蠄^{*}}_{(x . t)}] d x \frac{d (x)}{d t} = {鈭�}_{- 鈭�}^{鈭�} 脳 (\frac{鈭� 蠄}{鈭� t} 蠄 + 蠄^{*} \frac{鈭� 蠄}{鈭� t}) d x$

Now, substituting the Schrodinger equation for the time derivatives

${鈭�}_{- 鈭�}^{鈭�} X (- \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄^{*}}{鈭� x} 蠄 + \frac{i}{h} V 蠄^{*} 蠄 + \frac{i h}{2 m} 蠄^{*} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} - \frac{i}{h} V 蠄^{*} 蠄) d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} {鈭�}_{- 鈭�}^{鈭�} 脳 (蠄^{*} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} - \frac{{鈭�}^{2} 蠄^{*}}{鈭� x^{2}} 蠄) d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} {鈭�}_{- 鈭�}^{鈭�} 脳 [(\frac{鈭� 蠄^{*}}{鈭� x} \frac{鈭� 蠄}{鈭� x^{2}} + 蠄^{*} \frac{{鈭�}^{2} 蠄^{2}}{鈭� x^{2}}) - (\frac{{鈭�}^{2} 蠄^{2}}{鈭� x^{2}} 蠄 + \frac{鈭� 蠄^{*}}{鈭� x} \frac{鈭� 蠄}{鈭� x})] d x$

role="math" localid="1657786380532" $\frac{d <x>}{d t} = \frac{i h}{2 m} {鈭�}_{- 鈭�}^{鈭�} 脳 [\frac{鈭�}{鈭� x} (蠄^{*} \frac{鈭� 蠄}{鈭� x}) - \frac{鈭�}{鈭� x} (\frac{鈭� 蠄^{*}}{鈭� x} 蠄)] d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} {鈭�}_{- 鈭�}^{鈭�} 脳 [\frac{鈭�}{鈭� x} (蠄^{*} \frac{鈭� 蠄}{鈭� x}) - \frac{鈭�}{鈭� x} (\frac{鈭� 蠄^{*}}{鈭� x} 蠄)] d x \frac{d <x>}{d t} = \frac{i h}{2 m} 脳 (蠄^{*} 蠄_{- 鈭�}^{鈭�} - {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{鈭� 蠄}{鈭� x} d x - {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{鈭� 蠄}{鈭� x} d x) \frac{d <x>}{d t} = \frac{i h}{2 m} {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{鈭� 蠄}{鈭� x} d x$

Now, multiplying both sides by m

$m \frac{d <x>}{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{鈭� 蠄}{鈭� x} d x$

And using equation (3)

$ = - i h {鈭�}_{- 鈭�}^{鈭�} {蠄^{*}}^{鈭�} \frac{鈭� 蠄}{鈭� x} d x = {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} (- i h \frac{鈭�}{鈭� x}) 蠄 d x$

Differentiating both the sides with respect to t

We get the desired value,

\frac{d <p>}{d t} = - i h \frac{d}{d t} {鈭�}_{- 鈭�}^{鈭�} 蠄^{*} \frac{鈭� 蠄}{鈭� x} d x \frac{d <p>}{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} (蠄^{*} \frac{鈭� 蠄}{鈭� x}) d x

Using Clairaut’s theorem and substituting equation (1) and (2)

We get,

$\frac{d }{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} [(\frac{鈭� 蠄^{*}}{鈭� t}) \frac{鈭� 蠄}{鈭� t} + 蠄^{*} \frac{鈭�}{鈭� x} (\frac{鈭� 蠄}{鈭� t})] d x \frac{d }{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} [(- \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} + \frac{i}{h} V 蠄^{*}) \frac{鈭� 蠄}{鈭� x} + 蠄^{*} \frac{鈭�}{鈭� x} (\frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} + \frac{i}{h} V 蠄)] d x$
$\frac{d }{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} [- \frac{i h}{2 m} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} \frac{鈭� 蠄}{鈭� x} + \frac{i}{h} V 蠄^{*} \frac{鈭� 蠄}{鈭� x} \frac{i h}{2 m} 蠄^{*} \frac{{鈭�}^{3} 蠄}{鈭� x^{3}} - \frac{i}{h} \frac{鈭� V}{鈭� x} 蠄^{*} 蠄 - \frac{i}{h} V 蠄^{*} \frac{鈭� 蠄}{鈭� x}] \frac{d }{d t} = - i h - (- \frac{i h}{2 m} \frac{鈭� 蠄^{*}}{鈭� x} \frac{鈭� 蠄}{鈭� x} - {鈭�}_{- 鈭�}^{鈭�} \frac{鈭� 蠄}{鈭� x} \frac{{鈭�}^{2} 蠄}{鈭� x^{2}} d x) + {鈭�}_{- 鈭�}^{鈭�} [\frac{i h}{2 m} 蠄 \frac{{鈭�}^{3} 蠄}{鈭� x^{3}} - \frac{i}{h} \frac{鈭� V}{鈭� x} 蠄^{*} 蠄] d x$

localid="1657947924145" $\frac{d }{d t} = - i h {鈭�}_{- 鈭�}^{鈭�} [- \frac{i h}{2 m} 蠄^{*} \frac{{鈭�}^{3} 蠄}{鈭� x} + \frac{i h}{2 m} 蠄^{*} \frac{{鈭�}^{3}}{鈭� x^{3}} - \frac{i}{h} \frac{鈭� V}{鈭� X} 蠄^{*} 蠄] d x \frac{d }{d t} = i^{2} {鈭�}_{- 鈭�}^{鈭�} [\frac{鈭� V}{鈭� X} 蠄^{*} 蠄] d x$

$\frac{d }{d t} = i^{2} {鈭�}_{- 鈭�}^{鈭�} [\frac{鈭� V}{鈭� x} 蠄^{*} 蠄] d x$

$\frac{d }{d t} = - {鈭�}_{- 鈭�}^{鈭�} [\frac{鈭� v}{鈭� x} 蠄^{*} 蠄] d x \frac{d }{d t} = {鈭�}_{- 鈭�}^{鈭�} [蠄^{*} (- \frac{鈭� V}{鈭� x}) 蠄] d x \frac{d }{d t} = <- \frac{鈭� V}{鈭� x}>$

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Calculate d銆坧銆�/dt. Answer:
$\frac{d <p>}{d x} = <- \frac{鈭� V}{鈭� x}>$

This is an instance of Ehrenfest鈥檚 theorem, which asserts that expectation values obey the classical laws

Short Answer

Step by step solution

Determining the expectation value of x

Differentiating both the sides with respect to t,

Now, substituting the Schrodinger equation for the time derivatives

Differentiating both the sides with respect to t

Using Clairaut’s theorem and substituting equation (1) and (2)

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Most popular questions from this chapter

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91影视

Calculate d銆坧銆�/dt. Answer:dpdx=-鈭�V鈭�x This is an instance of Ehrenfest鈥檚 theorem, which asserts that expectation values obey the classical laws

Short Answer

Step by step solution

Determining the expectation value of x

Differentiating both the sides with respect to t,

Now, substituting the Schrodinger equation for the time derivatives

Differentiating both the sides with respect to t

Using Clairaut’s theorem and substituting equation (1) and (2)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Work Energy and Power

Physical Quantities And Units

Astrophysics

Nuclear Physics

Mechanics and Materials

Study anywhere. Anytime. Across all devices.

Calculate d銆坧銆�/dt. Answer:
$\frac{d <p>}{d x} = <- \frac{鈭� V}{鈭� x}>$

This is an instance of Ehrenfest鈥檚 theorem, which asserts that expectation values obey the classical laws