On operator and momentum space Might be the question is not up to the mark and caused due to some misunderstanding. Please still give me the idea where I was wrong.
I am studying quantum mechanics and came across the fact that we can use the Fourier transforms to define the wavefunction in momentum space.
What caught my attention was that the momentum operator in momentum space acts like a multiplication.
My question is can we define the Hamiltonian as just a multiplication by moving to a possible "space"? (I think that it's not possible as energy is not a vector like position and momentum).
So in that possible situation what will the time evolution be like? Will it be multiplicative?
 A: The basic idea of the Fourier transform is that of a transformation between two different bases in a vector space:

*

*In the position representation, the basis is that of states with well-defined positions $\phi_x$ (i.e. Dirac delta functions), and the state is written as $\psi = \int \psi(x)\phi_x \mathrm dx$, i.e. as a linear combination of the basis functions, with coefficients $\psi(x)$.

*In the momentum representation, the basis functions $\chi_p$ have well-defined momentum, and we write $\psi = \int \tilde \psi(p) \chi_p \mathrm dp$ in the same way.

*The coefficients of the two representations are related to each other by a linear transformation, $$\psi(x) = \int e^{ixp} \tilde \psi(p)\mathrm dp,$$ which is analogous to a change-of-basis matrix when working in finite dimensions.

*The property you note, that $\hat p$ acts on the momentum-basis components multiplicatively, i.e. $\tilde \psi(p) \mapsto p \, \tilde \psi(p)$, basically says that $\hat p$ is diagonal on the momentum basis.

In that language, you're asking "for an operator $A$, is there a basis where it is diagonal?", and the answer is yes, so long as the operator is hermitian; this result is known as the Spectral Theorem.
For the specific example of the hamiltonian, this is very much the case, and the states involved are known as the energy eigenstates of the system. They will be critically important once you start solving the Schrödinger equation (normally, via separation of variables), precisely because once you've found them the time evolution is multiplicative.
A: In essence, the answer is yes. You do it by going to the "energy-space", i.e., you use the energy eigenbasis to span your Hilbert space. Let's say the kets forming the energy eigenbasis are $\vert E \rangle$ then the wavefunction of a state $\vert \psi\rangle$ in this basis would be given by $\psi(E) \equiv \langle E \vert \psi \rangle$. Just like $\psi(x)$ means the probability amplitude of finding the particle in a state $\vert x\rangle$ upon the measurement of the position operator, $\psi(E)$ gives the probability amplitude of finding the system in a state $\vert E\rangle$ upon the measurement of the energy operator (i.e., the Hamiltonian).
Now, the action of the Hamiltonian in the energy basis could be deduced from the following observation: $$\langle E\vert \hat{H} \vert \psi\rangle = \sum_{E'}\langle E\vert \hat{H}\vert E'\rangle\langle E'\vert \psi\rangle=\sum_{E'}E'\delta_{EE'}\psi(E')=E~\psi(E)$$ Thus, in the energy-space, the action of the Hamiltonian on the wavefunction would indeed be multiplicative, as expected. As for the time evolution, a similar calculation with the exponentiated Hamiltonian would tell you that $\psi(E,t)=e^{-iEt}\psi(E,0)$. This is simply reflective of the fact that each eigenstate of the Hamiltonian only picks up a phase during the time evolution.

As @EmilioPisanty explains in their more comprehensive answer, you can go to the space spanned by the eigenbasis of any operator so long as the operator is a Hermitian operator (reminder: all observables are Hermitian). And thus, the translations produced by the said operator (such as translations in space by the momentum operator and translations in time by the Hamiltonian) would be produced by a multiplicative phase factor in this new basis in which the said operator is diagonalized.
