Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

To show,
(a) $Ψ_{1} (x, t) = A e^{i k x - i ω t}$ is the solution of both Klein-Gordon and the Schrodinger equations.
(b) $Ψ_{2} (x, t) = A e^{i k x} c o s ω t$ is the solution of both Klein-Gordon but not the Schrodinger equations.
(c) The $Ψ_{2}$ is a combination of positive and negative energy solutions of the Klein-Gordon equation.
(d) To compare the time dependence of $Ψ^{2}$ for $Ψ_{1}$ and $Ψ_{2}$ .

(a) The solution of both Klein-Gordon and the Schrodinger equations is $Ψ_{1} (x, t) = A e^{i k x - i ω t}$ .

(b) The solution of both Klein-Gordon but not the Schrodinger equations $Ψ_{2} (x, t) = A e^{i k x} \cos ω t$ .

(c) The first part $\frac{1}{2} A e^{i k x + i ω t}$ is the negative energy solution, and the second part $\frac{1}{2} A e^{i k x - i ω t}$ is the positive energy.

(d) The function ${| Ψ_{1} |}^{2}$ doesn't have the time dependence but ${| Ψ_{2} |}^{2}$ has the time dependence.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given functions are $Ψ_{1}$ and $Ψ_{2}$ .

Step 2: Concept of the Wave equation

The Klein-Gordon equation is given as,

$- c^{2} ℏ^{2} \frac{\partial^{2}}{\partial x^{2}} Ψ (x, t) + m^{2} c^{4} Ψ (x, t) = - ℏ^{2} \frac{\partial^{2}}{\partial t^{2}} Ψ (x, t)$

The Schrödinger Equation is given as,

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

Step 3: Take second order partial derivative of the Candidate wave function Ψ1(x,t)=Aeikx−iωt

(a)

Second order partial derivative of the wave function with respect to $x$ as:

$\frac{\partial^{2}}{d x^{2}} Ψ_{1} (x, t) = - k^{2} Ψ_{1} (x, t)$ ……(1)

Second order partial derivative of the wave function with respect $t$ as:

$\frac{\partial^{2}}{\partial t^{2}} Ψ_{1} (x, t) = - ω^{2} Ψ_{1} (x, t)$ …….(2)

After substitution equations (1) and (2) in the Klein-Gordon equation, obtain:

$c^{2} k^{2} ℏ^{2} Ψ_{1} (x, t) + m^{2} c^{4} Ψ_{1} (x, t) = ω^{2} ℏ^{2} Ψ_{1} (x, t)$

Substitute $p = k ℏ$ and $E = ω ℏ$ in the above equation and simplify as:

$c^{2} p^{2} + m^{2} c^{4} = E^{2}$

From special relativity, the wave function is proved to be a solution.

Step 4: Take first order partial derivative of Candidate function Ψ1(x,t)=Aeikx−iωt

Taking first order partial derivative of the Candidate wave function with respect to x as:

$\begin{matrix} \frac{\partial}{\partial t} Ψ_{1} (x, t) = \frac{\partial}{\partial t} (A e^{i k x - i ω t}) \\ = - i ω Ψ_{1} (x, t) \end{matrix}$

……(3)

After substitution equations (1) and (3) in the Schrödinger equation, obtain:

$\frac{ℏ^{2} k^{2}}{2 m} Ψ_{1} (x, t) = \frac{p^{2}}{2 m} Ψ_{1} (x, t)$

On the right, we have,

. $ℏ ω Ψ_{1} (x, t) = E Ψ_{1} (x, t)$

For nonrelativistic particles,

. $E = \frac{p^{2}}{2 m}$

The wave function is proved to be a solution.

$Ψ_{1} (x, t) = A e^{i k x - i ω t}$ is the solution of both Klein-Gordon and the Schrodinger equations.

Step 5: Take second order partial derivative of the Candidate wave functionΨ2(x,t)=Aeikxcosωt

(b)

The Candidate wave function is given as,

$Ψ_{2} (x, t) = A e^{i k x} \cos ω t$

Second order partial derivative of the Candidate wave function with respect to x as:

$\begin{matrix} \frac{\partial^{2}}{\partial x^{2}} Ψ_{2} (x, t) = A e^{i k x} t \cos ω t {(\pm i K)}^{2} \\ = - k^{2} Ψ_{2} (x, t) \end{matrix}$

Second order partial derivative of the Candidate wave function with respect t as:

$\begin{matrix} \frac{\partial^{2}}{\partial t^{2}} Ψ_{2} (x, t) = A e^{i k x} \cos ω t (- ω^{2}) \\ = - ω^{2} Ψ_{2} (x, t) \end{matrix}$

From special relativity, the wave function is proved to be a solution.

Similarly, takethe first order partial derivative of the Candidate wave function with respect to x as:

$\begin{matrix} \frac{\partial}{\partial t} Ψ_{2} (x, t) = A e^{i k x} \sin ω t (- ω) \\ \neq - ω Ψ_{2} (x, t) \end{matrix}$

The wave function is proved not to be a solution.

$Ψ_{2} (x, t) = A e^{i k x} \cos ω t$ is the solution of both Klein-Gordon but not the Schrodinger equations.

Step 6: Calculation to show Ψ2 is a combination of positive and negative energy solutions

(c)

The Candidate wave function is given, as shown below:

$\begin{matrix} Ψ_{2} (x, t) = A e^{i k x} \cos ω t \\ \cos θ = \frac{e^{i x} + e^{- i x}}{2} \end{matrix}$

The Candidate wave function is given as,

$Ψ_{2} (x, t) = A e^{i k x} \cos ω t$

After the expansion of $\cos ω t$ in exponential terms, obtain:

$\begin{array}{l} Ψ_{2} (x, t) = A e^{i k x} \cos ω t \\ Ψ_{2} (x, t) = A e^{i k x} \frac{e^{i ω t} + e^{- i a x}}{2} \\ Ψ_{2} (x, t) = \frac{1}{2} A e^{i k x + i ω t} + \frac{1}{2} A e^{i k x - i ω t} \end{array}$

The first part $\frac{1}{2} A e^{i k x + i ω t}$ is the negative energy solution, and the second part $\frac{1}{2} A e^{i k x - i ω t}$ is a positive energy.

Step 7: Comparison of the time dependence of Ψ2forΨ1 and Ψ2

(d)

The Candidate wave function is given as,

. $Ψ_{1} (x, t) = A e^{i k x - i ω t}$

The probability density is given as follows:

$\begin{array}{l} {| ψ_{1} (x, t) |}^{2} = {(A e^{i k x - i ω t})}^{2} \\ {| ψ_{1} (x, t) |}^{2} = A^{2} \end{array}$

For the first Candidate, since all the positive and time dependence is in terms of the exponential of an imaginary number,

${| ψ_{1} (x, t) |}^{2} = A^{2}$

So, it does not have time dependence.

The Candidate wave function is given as,

. $Ψ_{2} (x, t) = A e^{i k x} \cos ω t$

The probability density is given as below:

$\begin{array}{l} {| ψ_{2} (x, t) |}^{2} = {(A e^{i k x} \cos ω t)}^{2} \\ {| ψ_{2} (x, t) |}^{2} = A^{2} \cos^{2} ω t \end{array}$

So, it does have time dependence.

The function ${| Ψ_{1} |}^{2}$ doesn't have the time dependence but ${| Ψ_{2} |}^{2}$ has the time dependence.

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Concept of the Wave equation

Step 3: Take second order partial derivative of the Candidate wave function Ψ1(x,t)=Aeikx−iωt

Step 4: Take first order partial derivative of Candidate function Ψ1(x,t)=Aeikx−iωt

Step 5: Take second order partial derivative of the Candidate wave functionΨ2(x,t)=Aeikxcosωt

Step 6: Calculation to show Ψ2 is a combination of positive and negative energy solutions

Step 7: Comparison of the time dependence of Ψ2forΨ1 and Ψ2

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Modern Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Concept of the Wave equation

Step 3: Take second order partial derivative of the Candidate wave function Ψ1(x,t)=Aeikx−iωt

Step 4: Take first order partial derivative of Candidate function Ψ1(x,t)=Aeikx−iωt

Step 5: Take second order partial derivative of the Candidate wave functionΨ2(x,t)=Aeikxcosωt

Step 6: Calculation to show Ψ2 is a combination of positive and negative energy solutions

Step 7: Comparison of the time dependence of Ψ2forΨ1 and Ψ2

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