Equation 2.70:
x=\sqrt{\frac{\hbar}{2 m \omega}}\left(\hat{a}_{+}+\hat{a}_{-}\right) ; \quad \hat{p}=i \sqrt{\frac{\hbar m \omega}{2}}\left(\hat{a}_{+}-\hat{a}_{-}\right) (2.70).
x=\sqrt{\frac{\hbar}{2 m \omega}}\left(a_{+}+a_{-}\right), \quad p=i \sqrt{\frac{\hbar m \omega}{2}}\left(a_{+}-a_{-}\right) ; Eq. 2.67 :
\hat{a}_{+} \psi_{n}=\sqrt{n+1} \psi_{n+1}, \quad \hat{a}_{-} \psi_{n}=\sqrt{n} \psi_{n-1} (2.67).
\left\{\begin{array}{l}a_{+}|n\rangle=\sqrt{n+1}|n+1\rangle \\ a_{-}|n\rangle=\sqrt{n}|n-1\rangle \end{array}\right..
\left\langle n|x| n^{\prime}\right\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left\langle n\left|\left(a_{+}+a_{-}\right)\right| n^{\prime}\right\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left[\sqrt{n^{\prime}+1}\left\langle n \mid n^{\prime}+1\right\rangle+\sqrt{n^{\prime}}\left\langle n \mid n^{\prime}-1\right\rangle\right]
=\sqrt{\frac{\hbar}{2 m \omega}}\left(\sqrt{n^{\prime}+1} \delta_{n, n^{\prime}+1}+\sqrt{n^{\prime}} \delta_{n, n^{\prime}-1}\right) = \sqrt{\frac{\hbar}{2 m \omega}}\left(\sqrt{n} \delta_{n^{\prime}, n-1}+\sqrt{n^{\prime}} \delta_{n, n^{\prime}-1}\right) .
\left\langle n|p| n^{\prime}\right\rangle = i \sqrt{\frac{m \hbar \omega}{2}}\left(\sqrt{n} \delta_{n^{\prime}, n-1}-\sqrt{n^{\prime}} \delta_{n, n^{\prime}-1}\right) .
Noting that n and n′ run from zero to infinity, the matrices are:
X=\sqrt{\frac{\hbar}{2 m \omega}}\left(\begin{array}{cccccc} 0 & \sqrt{1} & 0 & 0 & 0 & 0 \\ \sqrt{1} & 0 & \sqrt{2} & 0 & 0 & 0 \\ 0 & \sqrt{2} & 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 & \sqrt{4} & 0 \\ 0 & 0 & 0 & \sqrt{4} & 0 & \sqrt{5} \\ & & & \cdots & & \end{array}\right) ; P=i \sqrt{\frac{m \hbar \omega}{2}}\left(\begin{array}{cccccc} 0 & -\sqrt{1} & 0 & 0 & 0 & 0 \\ \sqrt{1} & 0 & -\sqrt{2} & 0 & 0 & 0 \\ 0 & \sqrt{2} & 0 & -\sqrt{3} & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 & -\sqrt{4} & 0 \\ 0 & 0 & 0 & \sqrt{4} & -\sqrt{5} \\ & & & \cdots & & \end{array}\right) .
Squaring these matrices:
X ^{2}=\frac{\hbar}{2 m \omega}\left(\begin{array}{cccccc} 1 & 0 & \sqrt{1 \cdot 2} & 0 & 0 & 0 \\ 0 & 3 & 0 & \sqrt{2 \cdot 3} & 0 & 0 \\ \sqrt{1 \cdot 2} & 0 & 5 & 0 & \sqrt{3 \cdot 4} & 0 \\ 0 & \sqrt{2 \cdot 3} & 0 & 7 & 0 & \sqrt{4 \cdot 5} \\ & \cdots & & \cdots \end{array}\right) .
P ^{2}=-\frac{m \hbar \omega}{2}\left(\begin{array}{cccccc} -1 & 0 & \sqrt{1 \cdot 2} & 0 & 0 & 0 \\ 0 & -3 & 0 & \sqrt{2 \cdot 3} & 0 & 0 \\ \sqrt{1 \cdot 2} & 0 & -5 & 0 & \sqrt{3 \cdot 4} & 0 \\ 0 & \sqrt{2 \cdot 3} & 0 & -7 & 0 & \sqrt{4 \cdot 5} \\ & & & \cdots & & \end{array}\right) .
So the Hamiltonian, in matrix form, is
H =\frac{1}{2 m} P ^{2}+\frac{m \omega^{2}}{2} X ^{2}
=-\frac{\hbar \omega}{4}\left(\begin{array}{cccccc} -1 & 0 & \sqrt{1 \cdot 2} & 0 & 0 & 0 \\ 0 & -3 & 0 & \sqrt{2 \cdot 3} & 0 & 0 \\ \sqrt{1 \cdot 2} & 0 & -5 & 0 & \sqrt{3 \cdot 4} & 0 \\ 0 & \sqrt{2 \cdot 3} & 0 & -7 & 0 & \sqrt{4 \cdot 5} \\ & & & \cdots & & \end{array}\right)
+\frac{\hbar \omega}{4}\left(\begin{array}{cccccc} 1 & 0 & \sqrt{1 \cdot 2} & 0 & 0 & 0 \\ 0 & 3 & 0 & \sqrt{2 \cdot 3} & 0 & 0 \\ \sqrt{1 \cdot 2} & 0 & 5 & 0 & \sqrt{3 \cdot 4} & 0 \\ 0 & \sqrt{2 \cdot 3} & 0 & 7 & 0 & \sqrt{4 \cdot 5} \\ & & & \cdots & & \end{array}\right) = \frac{\hbar \omega}{2}\left(\begin{array}{llllllll} 1 & 0 & 0 & 0 & \\ 0 & 3 & 0 & 0 & \\ 0 & 0 & 5 & 0 & \\ 0 & 0 & 0 & 7 & \\ & & & & \ddots \end{array}\right) .
It’s plainly diagonal, and the nonzero elements are H_{n n}=\left(n+\frac{1}{2}\right) \hbar \omega , as they should be.