Chapter 1: Q7P (page 18)

Calculate d〈p〉/dt. Answer:
$\frac{d }{d x} = <- \frac{\partial V}{\partial x}>$

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

Short Answer

Expert verified

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

Schrodinger equation and its complex conjugate:

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

And according to Ehrenfest’s theorem,

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

Step by step solution

Determining the expectation value of x

Finding the expectation values.

$<x> = \frac{\int_{- \infty}^{\infty} x {|ψ_{(x . t)}|}^{2} d x}{\int_{- \infty}^{\infty} {|ψ_{(x . t)}|}^{2} d x} <x> = \int_{- \infty}^{\infty} x {|ψ_{(x . t)}|}^{2} d x <x> = \int_{- \infty}^{\infty} ψ_{(x . t)} {ψ^{*}}_{(x . t)} d x <x> = \int_{- \infty}^{\infty} ψ_{(x . t)} {ψ^{*}}_{(x . t)} d x$

Differentiating both the sides with respect to t,

$\frac{d (x)}{d t} = \int_{- \infty}^{\infty} ψ_{(x . t)} {ψ {(x)}^{*}}_{(x . t)} d x \frac{d (x)}{d t} = \int_{- \infty}^{\infty} \times \frac{\partial ψ}{\partial t} [ψ_{(x . t)} {ψ^{*}}_{(x . t)}] d x \frac{d (x)}{d t} = \int_{- \infty}^{\infty} \times (\frac{\partial ψ}{\partial t} ψ + ψ^{*} \frac{\partial ψ}{\partial t}) d x$

Now, substituting the Schrodinger equation for the time derivatives

$\int_{- \infty}^{\infty} X (- \frac{i h}{2 m} \frac{\partial^{2} ψ^{*}}{\partial x} ψ + \frac{i}{h} V ψ^{*} ψ + \frac{i h}{2 m} ψ^{*} \frac{\partial^{2} ψ}{\partial x^{2}} - \frac{i}{h} V ψ^{*} ψ) d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} \int_{- \infty}^{\infty} \times (ψ^{*} \frac{\partial^{2} ψ}{\partial x^{2}} - \frac{\partial^{2} ψ^{*}}{\partial x^{2}} ψ) d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} \int_{- \infty}^{\infty} \times [(\frac{\partial ψ^{*}}{\partial x} \frac{\partial ψ}{\partial x^{2}} + ψ^{*} \frac{\partial^{2} ψ^{2}}{\partial x^{2}}) - (\frac{\partial^{2} ψ^{2}}{\partial x^{2}} ψ + \frac{\partial ψ^{*}}{\partial x} \frac{\partial ψ}{\partial x})] d x$

role="math" localid="1657786380532" $\frac{d <x>}{d t} = \frac{i h}{2 m} \int_{- \infty}^{\infty} \times [\frac{\partial}{\partial x} (ψ^{*} \frac{\partial ψ}{\partial x}) - \frac{\partial}{\partial x} (\frac{\partial ψ^{*}}{\partial x} ψ)] d x$

$\frac{d <x>}{d t} = \frac{i h}{2 m} \int_{- \infty}^{\infty} \times [\frac{\partial}{\partial x} (ψ^{*} \frac{\partial ψ}{\partial x}) - \frac{\partial}{\partial x} (\frac{\partial ψ^{*}}{\partial x} ψ)] d x \frac{d <x>}{d t} = \frac{i h}{2 m} \times (ψ^{*} ψ_{- \infty}^{\infty} - \int_{- \infty}^{\infty} ψ^{*} \frac{\partial ψ}{\partial x} d x - \int_{- \infty}^{\infty} ψ^{*} \frac{\partial ψ}{\partial x} d x) \frac{d <x>}{d t} = \frac{i h}{2 m} \int_{- \infty}^{\infty} ψ^{*} \frac{\partial ψ}{\partial x} d x$

Now, multiplying both sides by m

$m \frac{d <x>}{d t} = - i h \int_{- \infty}^{\infty} ψ^{*} \frac{\partial ψ}{\partial x} d x$

And using equation (3)

$ = - i h \int_{- \infty}^{\infty} {ψ^{*}}^{\infty} \frac{\partial ψ}{\partial x} d x = \int_{- \infty}^{\infty} ψ^{*} (- i h \frac{\partial}{\partial x}) ψ 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} \int_{- \infty}^{\infty} ψ^{*} \frac{\partial ψ}{\partial x} d x \frac{d <p>}{d t} = - i h \int_{- \infty}^{\infty} (ψ^{*} \frac{\partial ψ}{\partial x}) d x

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

We get,

$\frac{d }{d t} = - i h \int_{- \infty}^{\infty} [(\frac{\partial ψ^{*}}{\partial t}) \frac{\partial ψ}{\partial t} + ψ^{*} \frac{\partial}{\partial x} (\frac{\partial ψ}{\partial t})] d x \frac{d }{d t} = - i h \int_{- \infty}^{\infty} [(- \frac{i h}{2 m} \frac{\partial^{2} ψ}{\partial x^{2}} + \frac{i}{h} V ψ^{*}) \frac{\partial ψ}{\partial x} + ψ^{*} \frac{\partial}{\partial x} (\frac{i h}{2 m} \frac{\partial^{2} ψ}{\partial x^{2}} + \frac{i}{h} V ψ)] d x$
$\frac{d }{d t} = - i h \int_{- \infty}^{\infty} [- \frac{i h}{2 m} \frac{\partial^{2} ψ}{\partial x^{2}} \frac{\partial ψ}{\partial x} + \frac{i}{h} V ψ^{*} \frac{\partial ψ}{\partial x} \frac{i h}{2 m} ψ^{*} \frac{\partial^{3} ψ}{\partial x^{3}} - \frac{i}{h} \frac{\partial V}{\partial x} ψ^{*} ψ - \frac{i}{h} V ψ^{*} \frac{\partial ψ}{\partial x}] \frac{d }{d t} = - i h - (- \frac{i h}{2 m} \frac{\partial ψ^{*}}{\partial x} \frac{\partial ψ}{\partial x} - \int_{- \infty}^{\infty} \frac{\partial ψ}{\partial x} \frac{\partial^{2} ψ}{\partial x^{2}} d x) + \int_{- \infty}^{\infty} [\frac{i h}{2 m} ψ \frac{\partial^{3} ψ}{\partial x^{3}} - \frac{i}{h} \frac{\partial V}{\partial x} ψ^{*} ψ] d x$

localid="1657947924145" $\frac{d }{d t} = - i h \int_{- \infty}^{\infty} [- \frac{i h}{2 m} ψ^{*} \frac{\partial^{3} ψ}{\partial x} + \frac{i h}{2 m} ψ^{*} \frac{\partial^{3}}{\partial x^{3}} - \frac{i}{h} \frac{\partial V}{\partial X} ψ^{*} ψ] d x \frac{d }{d t} = i^{2} \int_{- \infty}^{\infty} [\frac{\partial V}{\partial X} ψ^{*} ψ] d x$

$\frac{d }{d t} = i^{2} \int_{- \infty}^{\infty} [\frac{\partial V}{\partial x} ψ^{*} ψ] d x$

$\frac{d }{d t} = - \int_{- \infty}^{\infty} [\frac{\partial v}{\partial x} ψ^{*} ψ] d x \frac{d }{d t} = \int_{- \infty}^{\infty} [ψ^{*} (- \frac{\partial V}{\partial x}) ψ] d x \frac{d }{d t} = <- \frac{\partial V}{\partial x}>$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Calculate d〈p〉/dt. Answer:
$\frac{d <p>}{d x} = <- \frac{\partial V}{\partial x}>$

This is an instance of Ehrenfest’s 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

Solid State Physics

Radiation

Waves Physics

Work Energy and Power

Turning Points in Physics

Further Mechanics and Thermal Physics

Study anywhere. Anytime. Across all devices.

Company

Product

Help

Calculate d〈p〉/dt. Answer:dpdx=-∂V∂x This is an instance of Ehrenfest’s 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

Solid State Physics

Radiation

Waves Physics

Work Energy and Power

Turning Points in Physics

Further Mechanics and Thermal Physics

Study anywhere. Anytime. Across all devices.

Calculate d〈p〉/dt. Answer:
$\frac{d <p>}{d x} = <- \frac{\partial V}{\partial x}>$

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