Let's break down the problem into two parts and solve each step-by-step.
### Part (i): Compute the Commutator [tex]\(\left[a, a^{+}\right]\)[/tex]
Given the operators:
[tex]\[ a = \sqrt{\frac{m \omega}{2 \hbar}}\left(x + \frac{i p}{m \omega}\right), \][/tex]
[tex]\[ a^+ = \sqrt{\frac{m \omega}{2 \hbar}}\left(x - \frac{i p}{m \omega}\right), \][/tex]
To compute the commutator [tex]\(\left[a, a^{+}\right]\)[/tex], we use the definition:
[tex]\[ [a, a^+] = a a^+ - a^+ a. \][/tex]
First, let's express [tex]\(a a^+\)[/tex] and [tex]\(a^+ a\)[/tex] explicitly.
[tex]\[ a a^+ = \left( \sqrt{\frac{m \omega}{2 \hbar}}\left(x + \frac{i p}{m \omega}\right)\right)
\left( \sqrt{\frac{m \omega}{2 \hbar}}\left(x - \frac{i p}{m \omega}\right)\right). \][/tex]
Simplify the expression:
[tex]\[ a a^+ = \frac{m \omega}{2 \hbar} \left( x^2 + \frac{i p}{m \omega}x - x \frac{i p}{m \omega} - \left( \frac{i p}{m \omega} \right)^2 \right). \][/tex]
Since [tex]\(xp - px = i \hbar\)[/tex],
[tex]\[ a a^+ = \frac{m \omega}{2 \hbar} \left( x^2 - \left(\frac{p^2}{m^2\omega^2}\right) + \left( \frac{i (xp - px)}{m \omega}\right)\right). \][/tex]
[tex]\[ a a^+ = \frac{m \omega}{2 \hbar} \left( x^2 - \left(\frac{p^2}{m^2\omega^2}\right) + \frac{i \hbar}{m \omega} \right). \][/tex]
Next, for [tex]\(a^+ a\)[/tex]:
[tex]\[ a^+ a = \left( \sqrt{\frac{m \omega}{2 \hbar}}\left(x - \frac{i p}{m \omega}\right)\right)
\left( \sqrt{\frac{m \omega}{2 \hbar}}\left(x + \frac{i p}{m \omega}\right) \right). \][/tex]
[tex]\[ a^+ a = \frac{m \omega}{2 \hbar} \left( x^2 + \frac{i p}{m \omega}x - x \frac{i p}{m \omega} - \left( \frac{i p}{m \omega} \right)^2 \right). \][/tex]
[tex]\[ a^+ a = \frac{m \omega}{2 \hbar} \left( x^2 - \left(\frac{p^2}{m^2\omega^2}\right) - \frac{i \hbar}{m \omega} \right). \][/tex]
Now, let's find their commutator:
[tex]\[ [a, a^+] = a a^+ - a^+ a. \][/tex]
[tex]\[ [a, a^+] = \frac{m \omega}{2 \hbar} \left( x^2 - \frac{p^2}{m^2 \omega^2} + \frac{i \hbar}{m \omega}\right) - \frac{m \omega}{2 \hbar} \left( x^2 - \frac{p^2}{m^2 \omega^2} - \frac{i \hbar}{m \omega} \right). \][/tex]
[tex]\[ [a, a^+] = \frac{m \omega}{2 \hbar} \left( \frac{2 i \hbar}{m \omega} \right). \][/tex]
[tex]\[ [a, a^+] = \frac{m \omega}{2 \hbar} \cdot \frac{2 i \hbar}{m \omega}. \][/tex]
[tex]\[ [a, a^+] = 1. \][/tex]
Therefore, [tex]\(\boxed{[a, a^+] = 1.}\)[/tex]
### Part (ii): Evaluate the matrix element [tex]\(\left\langle \psi_1 | A | \psi_2 \right\rangle\)[/tex]
Given:
[tex]\[ A = \begin{pmatrix} 1 & i \sqrt{3} & -i \\ -i \sqrt{3} & 0 & 3 \\ i & 3 & 1 \end{pmatrix}, \][/tex]
[tex]\[ \psi_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \][/tex]
[tex]\[ \psi_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}. \][/tex]
1. Compute [tex]\(\psi_1^\dagger\)[/tex] (the conjugate transpose of [tex]\(\psi_1\)[/tex]):
[tex]\[ \psi_1^\dagger = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right). \][/tex]
2. Compute [tex]\(A \psi_2\)[/tex]:
[tex]\[ A \psi_2 = \begin{pmatrix} 1 & i \sqrt{3} & -i \\ -i \sqrt{3} & 0 & 3 \\ i & 3 & 1 \end{pmatrix}
\cdot \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \][/tex]
[tex]\[ A \psi_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \cdot 1 + (i \sqrt{3}) \cdot (-1) + (-i) \cdot 0 \\
(-i \sqrt{3}) \cdot 1 + 0 \cdot (-1) + 3 \cdot 0 \\
i \cdot 1 + 3 \cdot (-1) + 1 \cdot 0 \end{pmatrix}, \][/tex]
[tex]\[ A \psi_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 - i \sqrt{3} \\ -i \sqrt{3} \\ i - 3 \end{pmatrix}. \][/tex]
3. Compute [tex]\(\left\langle \psi_1 | A | \psi_2 \right\rangle\)[/tex]:
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle = \psi_1^\dagger (A \psi_2), \][/tex]
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)
\cdot \frac{1}{\sqrt{2}} \begin{pmatrix} 1 - i \sqrt{3} \\ -i \sqrt{3} \\ i - 3 \end{pmatrix}, \][/tex]
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle = \frac{1}{\sqrt{6}}
\left( 1 - i \sqrt{3} - i \sqrt{3} + i - 3 \right), \][/tex]
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle = \frac{1}{\sqrt{6}}
\left( 1 - 3 -2 i \sqrt{3} + i \right), \][/tex]
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle = \frac{-2}{\sqrt{6}} + \frac{-2i \sqrt{3} + i}{\sqrt{6}}, \][/tex]
Evaluating the numerical results:
[tex]\[ \left\langle \psi_1 | A | \psi_2 \right\rangle \approx -0.816 + -1.006 i. \][/tex]
Therefore, the computed matrix element is:
[tex]\[
\boxed{\left\langle \psi_1 | A | \psi_2 \right\rangle = -0.816 + -1.006 i}.
\][/tex]
