Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

Answers without the blur.

Just sign up for free and you're in.

Short Answer

In non-relavistic quantum mechanics, governed by the Schrodinger equation, the probability of finding a particle does not change with time.
(a)
Prove it, Begin with the time derivative of the total probability
$\frac{d}{d t} \int Ψ^{} (x, t) Ψ^{'} (x, t) d x = \int (Ψ (x, t) \frac{\partial}{\partial t} Ψ^{} (x, t) + Ψ^{*} (x, t) \frac{\partial}{\partial t} Ψ (x, t)) d x$
Then use the Schrodinger equation to eliminate the partial time derivatives, integrate by parts, and show that the result is zero. Assume that the particle is well localised, so that $ψ and \frac{\partial ψ}{\partial x}$ are 0 when evaluated at . $\pm \infty$
(b) Does this procedure lead to the same conclusion if Wave function obeyKlein-Gordon rather than Shrodinger equation? Why and why not?

(a) The integrals from the two integrals by parts cancel each other, and the result is $0$ .

(b) It will not work if the wave function obeys Klein-Gordon equations.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given integral is $\frac{d}{d t} \int Ψ^{*} (x, t) Ψ^{'} (x, t) d x = \int (Ψ (x, t) \frac{\partial}{\partial t} Ψ^{*} (x, t) + Ψ^{*} (x, t) \frac{\partial}{\partial t} Ψ (x, t)) d x$ .

Step 2: Concept of the symmetric and asymmetric wave equation

The symmetric wave function is given as,

$\frac{\partial}{\partial t} Ψ (x, t) = \frac{i ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} Ψ (x, t)$

The Anti-symmetric wave function is given as,

$\frac{\partial}{\partial t} Ψ^{*} (x, t) = - \frac{i ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} Ψ^{*} (x, t)$

Step 3: Proof of the integrals from the two integrals by parts cancel each other

(a)

The time derivative of the probability density function is given, as shown below.

$\begin{matrix} \frac{d}{d t} \int Ψ^{*} Ψ d x = \int (Ψ \frac{\partial}{\partial t} Ψ^{*} + Ψ^{*} \frac{\partial}{\partial t} Ψ) d x \\ = \int (- Ψ \frac{i ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} Ψ^{*} + Ψ^{*} \frac{i ℏ^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} Ψ) d x \\ = \frac{i ℏ^{2}}{2 m} \int - Ψ \frac{\partial}{\partial x} (\frac{\partial}{\partial x} Ψ^{*}) d x + \frac{i ℏ^{2}}{2 m} \int Ψ^{*} \frac{\partial}{\partial x} (\frac{\partial}{\partial x} Ψ) d x \\ = \frac{i ℏ^{2}}{2 m} (- Ψ \frac{\partial}{\partial x} Ψ^{*} + \int \frac{\partial}{\partial x} Ψ^{*} (\frac{\partial}{\partial x} Ψ) d x) + \frac{i ℏ^{2}}{2 m} (Ψ * \frac{\partial}{d x} Ψ + \int \frac{\partial}{\partial x} Ψ (\frac{\partial}{\partial x} Ψ^{*}) d x) \end{matrix}$

The integrals from the two integrals by parts, and we obtain:

. $\frac{d}{d t} \int Ψ^{*} Ψ d x = 0$

The integrals from the two integrals by parts cancel each other, and the result is $0$ .

Step 4: Explanation of wave function obeying Klein-Gordon equation lead to the same result or not

(b)

It will not work if the wave function obeys the Klein-Gordon equation.

Because the first-time derivative of the wave function does not appear in the equations.

So we cannot trade them into space derivatives.

Find Study Materials for

Create Study Materials

Select your language

Modern Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Concept of the symmetric and asymmetric wave equation

Step 3: Proof of the integrals from the two integrals by parts cancel each other

Step 4: Explanation of wave function obeying Klein-Gordon equation lead to the same result or not

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Modern Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Concept of the symmetric and asymmetric wave equation

Step 3: Proof of the integrals from the two integrals by parts cancel each other

Step 4: Explanation of wave function obeying Klein-Gordon equation lead to the same result or not

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics