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Q14E

Expert-verified
Found in: Page 556

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

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

(a) The solution of both Klein-Gordon and the Schrodinger equations is${\Psi }_{1}\left(x,t\right)=A{e}^{ikx-i\omega t}$ .

(b) The solution of both Klein-Gordon but not the Schrodinger equations ${\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t$ .

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

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

See the step by step solution

## Step 1: Given data

The given functions are ${\Psi }_{1}$ and${\Psi }_{2}$ .

## Step 2: Concept of the Wave equation

The Klein-Gordon equation is given as,

${\mathbf{-}}{{\mathbit{c}}}^{{\mathbf{2}}}{{\mathbit{\hslash }}}^{{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}{\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{+}}{{\mathbit{m}}}^{{\mathbf{2}}}{{\mathbit{c}}}^{{\mathbf{4}}}{\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{{\mathbit{\hslash }}}^{{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}{\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}$

.

The Schrödinger Equation is given as,

.

${\mathbf{-}}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}{\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{i}}{\mathbit{\hslash }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}{\mathbit{\Psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{t}}{\mathbf{\right)}}$

## 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:

$\frac{{\partial }^{2}}{d{x}^{2}}{\Psi }_{1}\left(x,t\right)=-{k}^{2}{\Psi }_{1}\left(x,t\right)$ ……(1)

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

$\frac{{\partial }^{2}}{\partial {t}^{2}}{\Psi }_{1}\left(x,t\right)=-{\omega }^{2}{\Psi }_{1}\left(x,t\right)$ …….(2)

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

${c}^{2}{k}^{2}{\hslash }^{2}{\Psi }_{1}\left(x,t\right)+{m}^{2}{c}^{4}{\Psi }_{1}\left(x,t\right)={\omega }^{2}{\hslash }^{2}{\Psi }_{1}\left(x,t\right)$

Substitute $p=k\hslash$ and$E=\omega \hslash$ 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:

$\begin{array}{c}\frac{\partial }{\partial t}{\Psi }_{1}\left(x,t\right)=\frac{\partial }{\partial t}\left(A{e}^{ikx-i\omega t}\right)\\ =-i\omega {\Psi }_{1}\left(x,t\right)\end{array}$

……(3)

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

$\frac{{\hslash }^{2}{k}^{2}}{2m}{\Psi }_{1}\left(x,t\right)=\frac{{p}^{2}}{2m}{\Psi }_{1}\left(x,t\right)$

On the right, we have,

.$\hslash \omega {\Psi }_{1}\left(x,t\right)=E{\Psi }_{1}\left(x,t\right)$

For nonrelativistic particles,

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

The wave function is proved to be a solution.

${\Psi }_{1}\left(x,t\right)=A{e}^{ikx-i\omega t}$ 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,

.

${\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t$

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

$\begin{array}{c}\frac{{\partial }^{2}}{\partial {x}^{2}}{\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\text{t}\mathrm{cos}\omega t{\left(±iK\right)}^{2}\\ =-{k}^{2}{\Psi }_{2}\left(x,t\right)\end{array}$

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

$\begin{array}{c}\frac{{\partial }^{2}}{\partial {t}^{2}}{\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t\left(-{\omega }^{2}\right)\\ =-{\omega }^{2}{\Psi }_{2}\left(x,t\right)\end{array}$

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:

$\begin{array}{c}\frac{\partial }{\partial t}{\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{sin}\omega t\left(-\omega \right)\\ \ne -\omega {\Psi }_{2}\left(x,t\right)\end{array}$

The wave function is proved not to be a solution.

${\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t$ 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:

$\begin{array}{c}{\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t\\ \mathrm{cos}\theta =\frac{{e}^{ix}+{e}^{-ix}}{2}\end{array}$

The Candidate wave function is given as,

${\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t$

After the expansion of $\mathrm{cos}\omega t$ in exponential terms, obtain:

$\begin{array}{l}{\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t\\ {\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\frac{{e}^{i\omega t}+{e}^{-iax}}{2}\\ {\Psi }_{2}\left(x,t\right)=\frac{1}{2}A{e}^{ikx+i\omega t}+\frac{1}{2}A{e}^{ikx-i\omega t}\end{array}$

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

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

(d)

The Candidate wave function is given as,

.${\Psi }_{1}\left(x,t\right)=A{e}^{ikx-i\omega t}$

The probability density is given as follows:

$\begin{array}{l}{|{\psi }_{1}\left(x,t\right)|}^{2}={\left(A{e}^{ikx-i\omega t}\right)}^{2}\\ {|{\psi }_{1}\left(x,t\right)|}^{2}={A}^{2}\end{array}$

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

${|{\psi }_{1}\left(x,t\right)|}^{2}={A}^{2}$

So, it does not have time dependence.

The Candidate wave function is given as,

.${\Psi }_{2}\left(x,t\right)=A{e}^{ikx}\mathrm{cos}\omega t$

The probability density is given as below:

$\begin{array}{l}{|{\psi }_{2}\left(x,t\right)|}^{2}={\left(A{e}^{ikx}\mathrm{cos}\omega t\right)}^{2}\\ {|{\psi }_{2}\left(x,t\right)|}^{2}={A}^{2}{\mathrm{cos}}^{2}\omega t\end{array}$

So, it does have time dependence.

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