[Physics] Ladder operators and the Hermitian adjoint

Question

We define $$\hat{a}=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}+i \frac{\hat{p}}{m w}\right)$$ $$\hat{a}^{\dagger}=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}-i \frac{\hat{p}}{m w}\right)$$
Lowering and raising operators respectively. Clearly, these are conjugates in the regular complex scalar sense but we label them and treat them as Hermitian adjoints (complex transpose). Since the differential operator can be represented by an infinitely large matrix, why do we not need to take the transpose part of it into account?

Answer 1

In the number basis of the harmonic oscillator (where the number operator energy is diagonal), your text must (and certainly Wikipedia does) have $$a^\dagger =\begin{pmatrix} 0 & 0 & 0 & \dots & 0 &\dots \\ \sqrt{1} & 0 & 0 & \dots & 0 & \dots\\ 0 & \sqrt{2} & 0 & \dots & 0 & \dots\\ 0 & 0 & \sqrt{3} & \dots & 0 & \dots\\ \vdots & \vdots & \vdots & \ddots & \vdots & \dots\\ 0 & 0 & 0 & \dots & \sqrt{n} &\dots & \\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots \end{pmatrix} $$ and $$a =\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\ 0 & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\ 0 & 0 & 0 & \sqrt{3} & \dots & 0 & \dots \\ 0 & 0 & 0 & 0 & \ddots & \vdots & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\ 0 & 0 & 0 & 0 & \dots & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, $$

both infinite-dimensional matrices and adjoint (actually transpose) of each other.

Adding them together gives you the hermitian (actually symmetric) $$ \hat x = \sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\ \sqrt{1} & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \dots & 0 & \dots \\ 0 & 0 & \sqrt{3} & 0 & \ddots & \vdots & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\ 0 & 0 & 0 & 0 & \sqrt{n} & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, $$ and it should be straightforward for you to construct the likewise hermitian (imaginary antisymmetric) $\hat p$.