A particle is described by the wave function

$\mathbf{\text{ψ(x)=}}\frac{\sqrt{\mathbf{\text{2/Π}}}}{{\mathbf{\text{x}}}^{\mathbf{\text{2}}}\mathbf{\text{-x+1.25}}}$

(a) Show that the normalization constant$\sqrt{\mathbf{2}\mathbf{/}\mathbf{\Pi }}$is correct.

(b) A measurement of the position of the particle is to be made. At what location is it most probable that the particle would be found?

(c) What is the probability per unit length of finding the particle at this location?

(a) The wave function is given in the question which is

$\Psi \left(x\right)=\frac{\sqrt{2/\Pi }}{x^2-x+1.25}$

Here we can see that the probability to find the particle somewhere is 1.

So, we can write it as:

$\begin{array}{l}\underset{-\infty }{\overset{\infty }{\int }}\Psi \left(x\right){\Psi }^0\left(x\right)dx=\underset{-\infty }{\overset{\infty }{\int }}\frac{\sqrt{2/\Pi }}{x^2-x+1.25}\frac{\sqrt{2/\Pi }}{x^2-x+1.25}dx\\ =\frac{2}{\Pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{{(x^2-x+1.25)}^2}dx\\ =\frac{2}{\Pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{{({(x-0.5)}^2+1)}^2}dx\end{array}$

Now,

By using trigonometry transformation,$x=0.5+\tan \theta$

The equation can be written as:

$\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{{({(x-0.5)}^2+1)}^2}dx=\underset{-\Pi /2}{\overset{\Pi /2}{\int }}\frac{1}{{({\tan }^2\theta +1)}^2}{\sec }^2\theta d\theta$

$\begin{array}{c}=\underset{-\Pi /2}{\overset{\Pi /2}{\int }}\frac{1}{{\sec }^2\theta }d\theta \\ =\underset{-\Pi /2}{\overset{\Pi /2}{\int }}{\cos }^2\theta d\theta \\ =\frac{\Pi }{2}\end{array}$

Here,

$\begin{array}{c}\underset{-\Pi /2}{\overset{\Pi /2}{\int }}{\cos }^2\theta d\theta =\underset{-\Pi /2}{\overset{\Pi /2}{\int }}\frac{1+\cos 2\theta }{2}d\theta \\ =\frac{1}{2}\Pi +\underset{-\Pi /2}{\overset{\Pi /2}{\int }}\frac{\cos 2\theta }{2}d\theta \\ =\frac{\Pi }{2}+0\\ =\frac{\Pi }{2}\end{array}$

The probability is.$\frac{2}{\Pi }\frac{\Pi }{2}=1$ The normalization is right,

Therefore the normalization constant will be $\sqrt{\frac{2}{\Pi }}$.

Now by finding the maximum for this we will use this equation,

$\rho =\frac{2}{\pi }\frac{1}{{(x^2-x+1.25)}^2}$

Now by Differentiating above equation with respect to $x$,

$\begin{array}{c}\frac{d\rho }{dx}=\frac{2}{\pi }\frac{-2(x^2-x+1.25)(2x-1)}{{(x^2-x+1.25)}^4}\\ =\frac{-4(2x-1)}{\pi {(x^2-x+1.25)}^3}\end{array}$

The derivative will be$0$when it’s maximum, so

$\begin{array}{c}\frac{-4(2x-1)}{\pi {(x^2-x+1.25)}^3}=0\\ \therefore 2x-1=0\\ 2x=1\\ x=\frac{1}{2}\end{array}$

Therefore, the most probable location of particle is .$x=\frac{1}{2}$

Now for the probability per unit length:

$\begin{array}{l}\rho \left(0.5\right)=\frac{2}{\pi }\frac{1}{{({0.5}^2-0.5+1.25)}^2}\\ =\frac{2}{\pi }\end{array}$

Therefore, the probability per unit length for the particle at the location is$=\frac{2}{\pi }$.$=\frac{2}{\pi }$