# Why do the matrix elements of an operator correspond to the Fourier components of the observable in Heisenberg's Matrix Mechanics?

It is well-known that Heisenberg $$a$$ began developing his Matrix Mechanics by creating matrix components $$A(n,n-a,t)=A(n,n-a)e^{i\omega(n,n-a)t}$$ or $$A_{nm}(t)=A_{nm}e^{\frac{i}{\hbar}\omega(nm)t}$$ corresponding to the Fourier components of the classical $$A(n,t)=\sum_{a}A_a(n)e^{ia\omega(n)t}$$ But what corresponds to the "motion"of a classical particle is not the components of the observable's matrix, it's the particle's state vector, so I tried to find an explanation for this the other way around.

I think that, since a stationary state is time-independent, I should work with a superposition of energy levels close to $$n$$ $$|\psi\rangle=\sum_{k=0}^nc_{n-k} e^{-\frac{i}{\hbar}E_{n-k}t}|n-k\rangle$$ I think that the expectation value of the observable $$A$$ should have Fourier components corresponding to the classical motion folowing Ehrenfest's Theorem, so I take the expectation value $$\langle A\rangle=\langle\psi |\hat A|\psi\rangle=\sum_{j,k=0}^nc^*_{n-j}c_{n-k} e^{\frac{i}{\hbar}(E_{n-j}-E_{n-k})t}\langle n-j |\hat A|n-k\rangle$$ Replacing $$n-j=l$$ and $$n-k=m$$ $$\langle A\rangle=\langle\psi |\hat A|\psi\rangle=\sum_{l,m=0}^nc^*_{l}c_{m} e^{\frac{i}{\hbar}\omega(lm)t}\langle l |\hat A|m\rangle$$ This tells me that the matrix components (ignoring time dependence) $$A_{lm}$$ correspond to the 'Quantum Fourier Components' of the observable's expectation value, but multiplied by $$c_l^*c_m$$.