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

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

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

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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)

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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

Kinematics Physics

Further Mechanics and Thermal Physics

Nuclear Physics

Physical Quantities And Units

Fundamentals of Physics

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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