Physical meaning of the rows and columns matrix representation of operators in Quantum Mechanics When any operator is written in its matrix form, do the individual rows and columns of the matrices have any physical meaning to them? For example, the rows and columns of momentum and position operators $\textbf{x}$ and $\textbf{p}$ in Simple Harmonic Oscillator?
 A: Physical meaning? In simplified natural units,
$$ H = {1 \over 2} (P^2 + X^2) ~, $$
is manifestly diagonal,  I/2 + diag(0,1,2,3,...), for the standard matrix mechanics hermitian expressions
$$
\sqrt{2} X =  
\begin{bmatrix}
0 & \sqrt{1} & 0 & 0 & 0 & \cdots \\
\sqrt{1} & 0 & \sqrt{2} & 0 & 0 & \cdots \\
0 & \sqrt{2} & 0 & \sqrt{3} & 0 & \cdots \\
0 & 0 & \sqrt{3} & 0 & \sqrt{4} & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix},
 $$
and
$$
\sqrt{2} P  =  
\begin{bmatrix}
0 & -i\sqrt{1} & 0 & 0 & 0 & \cdots \\
i\sqrt{1} & 0 & -i\sqrt{2} & 0 & 0 & \cdots \\
0 & i\sqrt{2} & 0 & -i\sqrt{3} & 0 & \cdots \\
0 & 0 & i\sqrt{3} & 0 & -i\sqrt{4} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix},
  $$
as you have probably already checked, with eigenvalues 1/2+ n for H.
You then see that these matrices are extremely sparse,  flanking an empty diagonal, and you immediately  intuit their provenance  in creation and annihilator oscillator operators (which raise and lower states by one energy rung). So the first line of X tells you it only connects the first excited state to the ground state, etc...
That is, you see that X and P only connect states contiguous in energy, of great utility in perturbation theory. This is at the heart of their physical meaning in QM.
It is also straightforward to verify the basic commutation relation $[X,P]=iI$.
Finally, the Heisenberg time-evolving operators are trivial evolutes of the above matrix elements,
$$
X_{mn}(t) = X_{mn}  e^{i(m - n)t},\qquad P_{mn}(t) = P_{mn}  e^{i(m -n)t}~,
 $$
which should evoke for you Heisenberg's original epochal Umdeutung argument of 1925. The  phases of the non vanishing elements are only $\pm  t$, a single frequency!
$$
\sqrt{2} X (t)=  
\begin{bmatrix}
0 &  e^{-it} & 0 & 0 & 0 & \cdots \\
  e^{it}  & 0 & \sqrt{2}  e^{-it} & 0 & 0 & \cdots \\
0 & \sqrt{2}  e^{it} & 0 & \sqrt{3} e^{-it}  & 0 & \cdots \\
0 & 0 & \sqrt{3} e^{it}  & 0 & \sqrt{4} e^{-it}  & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix},
 \\      \sqrt{2} P (t) =  
\begin{bmatrix}
0 & -i e^{-it} & 0 & 0 & 0 & \cdots \\
i e^{-it} & 0 & -i\sqrt{2} e^{-it}& 0 & 0 & \cdots \\
0 & i\sqrt{2} e^{it} & 0 & -i\sqrt{3} e^{-it} & 0 & \cdots \\
0 & 0 & i\sqrt{3} e^{it} & 0 & -i\sqrt{4}  e^{-it} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}.
  $$
A: The matrix represents the operator on a basis set, usually of eigenfunctions of the Hamiltonian. They are a bookkeeping method to describe the effect of the operator on these eigenfunctions. Such matrices can be manipulated using linear algebra.
