Question

The wave function of a particle is

$\psi \left(x\right)=\left\{\begin{array}{l}\frac{b}{\left(1+x^2\right)}-1mm\le x<0mm\\ c\left(1+x^2\right)0mm\le x\le 1mm\end{array}\right.$

and zero elsewhere

Short answer

a) Suppose that the wave function is a continuous function

$b^2=c^2$

$0+{\int }_{-1}^0\frac{b^2}{{\left(1+x^2\right)}^2}dx+{\int }_0^1{c}^2{\left(1+x\right)}^{4}dx+0=1$

There fore the values of b and c are 0.382, 0.382

the probability that the particle will be found to the right of the origin is 90%

Step-by-step solution

Step 1:The wave function of a particle is 

$\psi \left(x\right)=\left\{\begin{array}{l}\frac{b}{\left(1+x^2\right)}-1mm\le x<0mm\\ c\left(1+x^2\right)0mm\le x\le 1mm\end{array}\right.$

and zero elsewhere

Sub part a) step 2: Suppose that the wave function is a continuous function

$$\begin{gathered} \underset{x\to 0^2}{\lim }\frac{b^2}{{\left(1+x^2\right)}^2}=\underset{x\to 0^2}{\lim }{c}^2{\left(1+x\right)}^4 \\ \frac{b^2}{{\left(1+0\right)}^2}=c^2{\left(1+0\right)}^2 \\ {b}^2=c^2 \end{gathered}$$

Sub part 2 (b) step 3:The graphs of the wave function, and the probability density function over the interval

$-2mm\le x\le 2mm$

Step 4:

Step 5:

Sub part (c) step 6:The probability that the particle will be found to the right of the origin, means that we have to

calculate the value of the probability of finding the particle in the interval $0mm\le x\le 1mm$

First of all, we have to calculate the values of b and c For the probability interpretation of $\psi \left(x\right)$to make sense, the wave function must satisfy the condition

$$\begin{gathered} {\int }_{-\infty }^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1 \\ {\int }_{-\infty }^{-1}{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_{-1}^0{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_0^1{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_1^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1 \\ 0+{\int }_{-1}^0\frac{b^2}{{\left(1+x^2\right)}^2}dx+{\int }_0^1{c}^2{\left(1+x\right)}^{4}dx+0=1 \end{gathered}$$

Step 7:

Since b=c then

$$\begin{gathered} {\int }_{-1}^0\frac{b^2}{{\left(1+x^2\right)}^2}dx+b^2{\int }_0^1{\left(1+x\right)}^{4}dx=1 \\ {b}^2{\int }_{-1}^0\frac{1}{\left(1+x^2\right)}dx+b^2{\int }_0^1{\left(1+x\right)}^{4}dx=1 \end{gathered}$$

Step 8:

$$\begin{gathered} b^2{\left[\frac{1}{2}\left(\frac{x}{x^2+1}+{\tan }^{-1}x\right)\right]}_{-1}^0+b^2{\int }_0^1\left(1+4x^3+6x^2+4x+x^4\right)dx=1 \\ \frac{b^2}{2}\left[0+0\left(\frac{-1}{2}-{\tan }^{-1}1\right)\right]+b^2\left[1+1+2+2+\frac{1}{5}-0\right]=1 \\ \frac{b^2}{2}\left[\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right]+b^2\left(6.2\right)=1 \\ \frac{b^2}{2}\left[\frac{1}{2}+\frac{\mathrm{\pi }}{4}\right]+6.2b^2=1 \\ {b}^2\left[6.45+0.3925\right]=1 \\ {b}^2=\frac{1}{6.8425} \\ =0.146 \\ b=0.382 \end{gathered}$$

There fore the values of b and c are 0.382, 0.382

Step 9:

Now, the probability that the particle will be found to the right of the origin is

$$\begin{gathered} p\left(0mm\le x\le 1mm\right)={\int }_0^1{\left|\psi \left(x\right)\right|}^{2}dx \\ ={\int }_0^1{\left[c{\left(1+x\right)}^2\right]}^{2}dx \\ ={\int }_0^{1}c{\left(1+x\right)}^{4}dx \\ =c^2{\int }_0^1\left(1+4x+6x^2+4x^3+x^4\right)dx \\ =\left(0.146\right){\int }_0^1\left(1+4x+6x^2+4x^3+x^4\right)dx \\ =\left(0.146\right){\left[x+\frac{4x^2}{2}+\frac{6x^3}{3}+\frac{4x^4}{4}+\frac{x^5}{5}\right]}_0^1 \\ =\left(0.146\right)\left(6.2\right)x100\% \\ =90.52\% \\ =91\% \end{gathered}$$