How to expand a wave function in terms of momentum eigenbasis without invoking the energy eigenbasis? So, I am trying to learn Quantum Mechanics on my own and I am using MIT OCW
$8.04$ lecture series and I was learning about the expansion of wavefunctions in terms of energy eigenfunctions. For some reason, the energy eigenfunctions evolve in a fairly easy and simple manner with a constant angular velocity phase, and thus constructing a general solution is as simple as latching on the phases to the energy eigenfunctions and summing them with appropriate coefficients. Now, we mostly avoid expanding wave functions in, say, momentum or position eigenbasis because time evolution of position or momentum eigenbasis is complicated. But how complicated? Can I, despite their complexity, expand a function in terms of its momentum eigenbasis? And I mean to do that without invoking the energy eigenfunctions i.e. first expanding the function in terms of energy eigenfunctions $\psi(x,t)=\sum_{n}c_n\phi_n(x)e^{-iE_nt/\hbar}$
and then taking a Fourier transform of the expanded wave function to calculate the expansion coefficients. $\tilde{\psi}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \psi(x,t)e^{-ikx} dx$
Instead, I would like to know how to expand a wave function in momentum eigenbasis starting from scratch and forgetting about the energy eigenfunctions completely.
Another thing that can be done is to find similarly the expansion in terms of position eigenbasis (since it can't be Fourier transformed to momentum eigenbasis).
 A: What you have done to get $\psi(x,t)$ is use the completeness relation for the energy eigenstates $\sum_n|\phi_n\rangle\langle\phi_n|=1$ so that
$$\psi(x,t)=\langle x|\psi(t)\rangle=\sum_n\langle x|\phi_n\rangle\langle\phi_n|\psi(t)\rangle=\sum_nc_n\phi_n(x)e^{-iE_nt/\hbar}$$
And then to find $\bar\psi(k,t)$ you used the completeness relation for the position basis
$\int|x\rangle\langle x|\,\text dx=1$ so that
$$\bar\psi(k,t)=\langle k|\psi(t)\rangle=\int\langle k|x\rangle\langle x|\psi(t)\rangle\,\text dx=\frac{1}{\sqrt{2\pi}}\int e^{-ikx}\psi(x,t)\,\text dx$$
which works since you already know $\psi(x,t)$ from above.
So this shows two places you can "opt out" of using energy eigenstates. You can either choose some other complete set of eigenstates $|\xi_m\rangle$ to determine $\psi(x,t)=\langle x|\psi(t)\rangle$, 
$$\psi(x,t)=\langle x|\psi(t)\rangle=\sum_m\langle x|\xi_m\rangle\langle\xi_m|\psi(t)\rangle=\sum_ma_m(t)\xi_m(x)$$
where you can keep your second step
$$\bar\psi(k,t)=\langle k|\psi(t)\rangle=\int\langle k|x\rangle\langle x|\psi(t)\rangle\,\text dx=\frac{1}{\sqrt{2\pi}}\int e^{-ikx}\psi(x,t)\,\text dx$$
or you can skip the first step by using some other complete set of eigenstates $|\chi_m\rangle$ to determine $\bar\psi(k,t)=\langle k|\psi(t)\rangle$
$$\bar\psi(k,t)=\langle k|\psi(t)\rangle=\sum_m\langle k|\chi_m\rangle\langle\chi_m|\psi(t)\rangle=\sum_mb_m(t)\bar\chi_m(k)$$
You could also use a continuous set of eigenstates for the above two cases.


Another thing that can be done is to find similiarly the expansion in terms of position eigenbasis

Note you already do this when you find $\psi(x,t)$ because your quantum state as expressed in terms of position eigenstates is in fact
$$|\psi(t)\rangle=\int\psi(x,t)|x\rangle\,\text dx$$
Just like how using what you (and I) describe above
$$|\psi(t)\rangle=\int\bar\psi(k,t)|k\rangle\,\text dk$$
