Chapter 3, Problem 3.31P

To determine

The orthonormalization of the given functions.

Answer to Problem 3.31P

The orthonormalization of the given functions is given.

Explanation of Solution

Orthonormalization for the function $|{\psi }_1\rangle =1$ is given by,

    $\begin{array}{c}\langle {\psi }_1|{\psi }_1\rangle =1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left|{\psi }_1\right|}^{2}dx}=1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left(1\right)}^{2}dx}=1\\ A=\frac{1}{\sqrt{2}}\\ |{\psi }^{\text{'}}{}_1\rangle =\frac{1}{\sqrt{2}}\end{array}$                                                                                                

Orthonormalization for the function $|{\psi }_2\rangle =x$ is given by,

    $\begin{array}{c}\langle {\psi }^{\text{'}}{}_1|{\psi }_2\rangle =0\\ \langle {\psi }_2|{\psi }_2\rangle =1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left|{\psi }_2\right|}^{2}dx}=1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left(x\right)}^{2}dx}=1\\ A=\sqrt{\frac{3}{2}}\\ |{\psi }^{\text{'}}{}_2\rangle =\sqrt{\frac{3}{2}}x\end{array}$                                                                                                

Using problem A.4,

    $|{\psi }^{\text{'}}{}_3\rangle =x^2-\frac{1}{3}$                                         

Orthonormalization for the function $|{\psi }^{\text{'}}{}_3\rangle =x^2-\frac{1}{3}$ is given by,

    $\begin{array}{c}\langle {\psi }^{\text{'}}{}_3|{\psi }^{\text{'}}{}_3\rangle =1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left|{\psi }^{\text{'}}{}_3\right|}^{2}dx}=1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left(x^2-\frac{1}{3}\right)}^{2}dx}=1\\ A=\sqrt{\frac{45}{8}}\\ |{\psi }^{\text{'}}{}_3\rangle =\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\\ =\sqrt{\frac{5}{2}}\left(\frac{3x^2-1}{2}\right)\end{array}$                                                                                                 

Using problem A.4,

    $\begin{array}{c}|{\psi }^{\text{'}}{}_4\rangle =|{\psi }_4\rangle -\langle {\psi }^{\text{'}}{}_2|{\psi }_4\rangle |{\psi }^{\text{'}}{}_2\rangle \\ =x^3-\frac{3}{5}x\end{array}$                                         

Orthonormalization for the function $|{\psi }^{\text{'}}{}_4\rangle =x^3-\frac{3}{5}x$ is given by,

    $\begin{array}{c}\langle {\psi }^{\text{'}}{}_4|{\psi }^{\text{'}}{}_4\rangle =1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left|{\psi }^{\text{'}}{}_4\right|}^{2}dx}=1\\ {\displaystyle \underset{-1}{\overset{1}{\int }}{A}^2{\left(x^3-\frac{3}{5}x\right)}^{2}dx}=1\\ A=\sqrt{\frac{175}{8}}\\ |{\psi }^{\text{'}}{}_4\rangle =\sqrt{\frac{175}{8}}\left(x^3-\frac{3}{5}x\right)\\ =\sqrt{\frac{7}{2}}\left(\frac{5x^3-3x}{2}\right)\end{array}$

Conclusion:

Therefore, orthonormalization of the given functions is given.