Question

For solutions of Klein-Gordon equation, the quantity,

$i{\psi }^{\ast }\frac{\partial }{\partial t}\psi -i\psi \frac{\partial }{\partial t}{\psi }^{\ast }$is interpreted as charge density. Show that for a positive-energy plane-wavesolution. It is a real constant, and for negative-energy solution.It is a negative of that constant.

Short answer

The quantity is a real constant for positive-energy plane-wave solution, and negative to that for negative-energy solution is proved.

Step-by-step solution

Given data

Charge density is $i{\psi }^{\ast }\frac{\partial }{\partial t}\psi -i\psi \frac{\partial }{\partial t}{\psi }^{\ast }$.

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,

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

Charge density is given as,

$\mathit{i}{\mathit{\psi }}^{\mathbf{\ast }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}\mathit{\psi }\mathbf{-}\mathit{i}\mathit{\psi }\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{t}}{\mathit{\psi }}^{\mathbf{\ast }}$

Calculation for positive wave function

The positive wave functions are given as follows:

${\Psi }_1(x,t)=A{e}^{ikx-i\omega t}$ ……(1)

${\Psi }_1^{*}(x,t)=A{e}^{-ikx+i\omega t}$ ……(2)

Differentiate the equation (1) partially with respect to$t$as:

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

………(3)

Differentiate the equation (2) partially with respect to $t$:

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

$\text{Charge\hspace{0.17em}density}=i{\Psi }_1^{*}\frac{\partial }{\partial t}{\Psi }_1-i{\Psi }_1\frac{\partial }{\partial t}{\Psi }_1^{*}$

Substitute $\frac{\partial }{\partial t}{\Psi }_1(x,t)=-i\omega {\Psi }_1(x,t)$ and$\frac{\partial }{\partial t}{\Psi }_1^{*}(x,t)=i\omega {\Psi }_1^{*}(x,t)$ in the equationand simplify as:

$\begin{array}{c}\text{Charge\hspace{0.17em}density}=i{\Psi }_1^{*}(-i\omega {\Psi }_1)-i{\Psi }_{1}i\omega {\Psi }_1^{*}(x,t)\\ =-i^2\omega {\Psi }_1^{*}{\Psi }_1-i^2\omega {\Psi }_1^{*}{\Psi }_1\\ =-i^2\omega \left(\left(A{e}^{-ikx+i\omega t}\right)\left(A{e}^{ikx-i\omega t}\right)\right){\Psi }_1-i^2\omega \left(\left(A{e}^{-ikx+i\omega t}\right)\left(A{e}^{ikx-i\omega t}\right)\right)\\ =2{\omega }^2{A}^2\end{array}$

Calculation for negative wave function 

The negative energy wave functions are:

${\mathit{\Psi }}_{\mathbf{1}}^{\mathbf{*}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{i}\mathbf{\omega }\mathbf{t}}$ ……(4)

${\mathit{\Psi }}_{\mathbf{1}}^{\mathbf{*}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{i}\mathbf{\omega }\mathbf{t}}$ ……(5)

Differentiate the equation (4) partially with respect to $t$ as:

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

Differentiate the equation (5) partially with respect to $t$as:

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

Substitute $\frac{\partial }{\partial t}{\Psi }_1(x,t)=i\omega {\Psi }_1(x,t)$and $\frac{\partial }{\partial t}{\Psi }_1^{*}(x,t)=-i\omega {\Psi }_1^{*}(x,t)$ in the equation and simplify as:

$\begin{array}{c}\text{Charge\hspace{0.17em}density}=i{\Psi }_1^{*}\left(i\omega {\Psi }_1\right)-i{\Psi }_1-i\omega {\Psi }_1^{*}(x,t)\\ =i^2\omega {\Psi }_1^{*}{\Psi }_1+i^2\omega {\Psi }_1^{*}{\Psi }_1\\ =i^2\omega \left(\left(A{e}^{ikx+i\omega t}\right)\left(A{e}^{-ikx-i\omega t}\right)\right){\Psi }_1+i^2\omega \left(\left(A{e}^{ikx+i\omega t}\right)\left(A{e}^{-ikx-i\omega t}\right)\right)\\ =-2{\omega }^2{A}^2\end{array}$

This is exactly the negative of the constant for this part.

The quantity is a real constant for positive-energy plane-wave solution, and negative to that for negative-energy solution is proved.