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Expert-verified Found in: Page 556 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # 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$\frac{d}{dt}\int {\Psi }^{*}\left(x,t\right){\Psi }^{\text{'}}\left(x,t\right)dx=\int \left(\Psi \left(x,t\right)\frac{\partial }{\partial t}{\Psi }^{*}\left(x,t\right)+{\Psi }^{*}\left(x,t\right)\frac{\partial }{\partial t}\Psi \left(x,t\right)\right)dx$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 ${\psi }{\text{\hspace{0.17em}and\hspace{0.17em}}}\frac{\partial \psi }{\partial x}$are 0 when evaluated at .${±}{\infty }$(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 1: Given data

The given integral is$\frac{d}{dt}\int {\Psi }^{*}\left(x,t\right){\Psi }^{\text{'}}\left(x,t\right)dx=\int \left(\Psi \left(x,t\right)\frac{\partial }{\partial t}{\Psi }^{*}\left(x,t\right)+{\Psi }^{*}\left(x,t\right)\frac{\partial }{\partial t}\Psi \left(x,t\right)\right)dx$ .

## Step 2: Concept of the symmetric and asymmetric wave equation

The symmetric wave function is given as,

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

The Anti-symmetric wave function is given as,

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

## 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.

$\begin{array}{c}\frac{d}{dt}\int {\Psi }^{*}\Psi dx=\int \left(\Psi \frac{\partial }{\partial t}{\Psi }^{*}+{\Psi }^{*}\frac{\partial }{\partial t}\Psi \right)dx\\ =\int \left(-\Psi \frac{i{\hslash }^{2}}{2m}\frac{{\partial }^{2}}{\partial {x}^{2}}{\Psi }^{*}+{\Psi }^{*}\frac{i{\hslash }^{2}}{2m}\frac{{\partial }^{2}}{\partial {x}^{2}}\Psi \right)dx\\ =\frac{i{\hslash }^{2}}{2m}\int -\Psi \frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}{\Psi }^{*}\right)dx+\frac{i{\hslash }^{2}}{2m}\int {\Psi }^{*}\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\Psi \right)dx\\ =\frac{i{\hslash }^{2}}{2m}\left(-\Psi \frac{\partial }{\partial x}{\Psi }^{*}+\int \frac{\partial }{\partial x}{\Psi }^{*}\left(\frac{\partial }{\partial x}\Psi \right)dx\right)+\frac{i{\hslash }^{2}}{2m}\left(\Psi *\frac{\partial }{dx}\Psi +\int \frac{\partial }{\partial x}\Psi \left(\frac{\partial }{\partial x}{\Psi }^{*}\right)dx\right)\end{array}$

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

.$\frac{d}{dt}\int {\Psi }^{*}\Psi dx=0$

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. ### Want to see more solutions like these? 