Question

To show,

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

(b) ${\mathit{\Psi }}_{\mathbf{2}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}$is the solution of both Klein-Gordon but not the Schrodinger equations.

(c) The${\mathit{\Psi }}_{\mathbf{2}}$ is a combination of positive and negative energy solutions of the Klein-Gordon equation.

(d) To compare the time dependence of${\mathit{\Psi }}^{\mathbf{2}}$ for ${\mathit{\Psi }}_{\mathbf{1}}$and ${\mathit{\Psi }}_{\mathbf{2}}$.

Short answer

(a) The solution of both Klein-Gordon and the Schrodinger equations is${\Psi }_1(x,t)=A{e}^{ikx-i\omega t}$.

(b) The solution of both Klein-Gordon but not the Schrodinger equations ${\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t$.

(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${\left|{\Psi }_1\right|}^2$ doesn't have the time dependence but ${\left|{\Psi }_2\right|}^2$ has the time dependence.

Step-by-step solution

Given data

The given functions are ${\Psi }_1$ and${\Psi }_2$ .

Concept of the Wave equation

The Klein-Gordon equation is given as,

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

.

The Schrödinger Equation is given as,

.

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

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:

$\frac{{\partial }^2}{d{x}^2}{\Psi }_1(x,t)=-k^2{\Psi }_1(x,t)$ ……(1)

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

$\frac{{\partial }^2}{\partial t^2}{\Psi }_1(x,t)=-{\omega }^2{\Psi }_1(x,t)$ …….(2)

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

$c^2{k}^2{\hslash }^2{\Psi }_1(x,t)+m^2{c}^4{\Psi }_1(x,t)={\omega }^2{\hslash }^2{\Psi }_1(x,t)$

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.

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:

$\begin{array}{c}\frac{\partial }{\partial t}{\Psi }_1(x,t)=\frac{\partial }{\partial t}\left(A{e}^{ikx-i\omega t}\right)\\ =-i\omega {\Psi }_1(x,t)\end{array}$

……(3)

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

$\frac{{\hslash }^2{k}^2}{2m}{\Psi }_1(x,t)=\frac{p^2}{2m}{\Psi }_1(x,t)$

On the right, we have,

.$\hslash \omega {\Psi }_1(x,t)=E{\Psi }_1(x,t)$

For nonrelativistic particles,

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

The wave function is proved to be a solution.

${\Psi }_1(x,t)=A{e}^{ikx-i\omega t}$is the solution of both Klein-Gordon and the Schrodinger equations.

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

(b)

The Candidate wave function is given as,

.

${\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t$

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

$\begin{array}{c}\frac{{\partial }^2}{\partial x^2}{\Psi }_2(x,t)=A{e}^{ikx}\text{t}\cos \omega t{\left(\pm iK\right)}^2\\ =-k^2{\Psi }_2(x,t)\end{array}$

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

$\begin{array}{c}\frac{{\partial }^2}{\partial t^2}{\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t(-{\omega }^2)\\ =-{\omega }^2{\Psi }_2(x,t)\end{array}$

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:

$\begin{array}{c}\frac{\partial }{\partial t}{\Psi }_2(x,t)=A{e}^{ikx}\sin \omega t(-\omega )\\ \ne -\omega {\Psi }_2(x,t)\end{array}$

The wave function is proved not to be a solution.

${\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t$is the solution of both Klein-Gordon but not the Schrodinger equations.

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

(c)

The Candidate wave function is given, as shown below:

$\begin{array}{c}{\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t\\ \cos \theta =\frac{e^{ix}+e^{-ix}}{2}\end{array}$

The Candidate wave function is given as,

${\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t$

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

$\begin{array}{l}{\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t\\ {\Psi }_2(x,t)=A{e}^{ikx}\frac{e^{i\omega t}+e^{-iax}}{2}\\ {\Psi }_2(x,t)=\frac{1}{2}A{e}^{ikx+i\omega t}+\frac{1}{2}A{e}^{ikx-i\omega t}\end{array}$

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.

Comparison of the time dependence of   Ψ2forΨ1   and  Ψ2 

(d)

The Candidate wave function is given as,

.${\Psi }_1(x,t)=A{e}^{ikx-i\omega t}$

The probability density is given as follows:

$\begin{array}{l}{\left|{\psi }_1(x,t)\right|}^2={\left(A{e}^{ikx-i\omega t}\right)}^2\\ {\left|{\psi }_1(x,t)\right|}^2=A^2\end{array}$

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

${\left|{\psi }_1(x,t)\right|}^2=A^2$

So, it does not have time dependence.

The Candidate wave function is given as,

.${\Psi }_2(x,t)=A{e}^{ikx}\cos \omega t$

The probability density is given as below:

$\begin{array}{l}{\left|{\psi }_2(x,t)\right|}^2={\left(A{e}^{ikx}\cos \omega t\right)}^2\\ {\left|{\psi }_2(x,t)\right|}^2=A^2{\cos }^2\omega t\end{array}$

So, it does have time dependence.

The function ${\left|{\Psi }_1\right|}^2$doesn't have the time dependence but${\left|{\Psi }_2\right|}^2$ has the time dependence.