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Q16E

Expert-verifiedFound in: Page 556

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**

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

**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 ** **.${\pm}{\infty}$**

**(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.

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

**The symmetric wave function is given as,**

** **

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

** **

**The Anti-symmetric wave function is given as,**

** **

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

(a)

The time derivative of the probability density function is given, as shown below.

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

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 $0$ .

(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.

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