Representation of wavefunction as superposition of eigenstates

Question

In Quantum Physics it is postulated that, any general state $\psi$ can be represented as superposition of eigenstates with constant coefficients corresponding to any observable.

Say, I have all the eigenstates of energy, then I can represent wavefunction $\psi (x)$ as superposition of eigenstates of energy. But from this wavefunction I can calculate any other quantity, say probability density or expectation value of position. But the doubt is, since the eigenstates are energy eigenstates, then these eigenstates must have information about energy, then how their superposition can give us information about position of particle (as both are completely unrelated)?.

Is there any fundamental reason for this superposition, or it is just postulated like Newton's Laws?

Answer 1

This has a possibility of confusing you even more but I'll give it a go.

You seem to be comfortable with the idea that any state can be written as some $\psi(x)$ but not with the idea that any state can be written as some linear superposition of energy eigenstates. This is an internally inconsistent point of view. Let me elaborate.

What is $\psi(x)$? It's the probability amplitude of finding a given system in position $x$ upon measuring its position (let's ignore the fact that position eigenstates are not physically realizable, it's not super important to the relevant discussion). What this means is that you're expressing a state as a linear superposition of position eigenstate with the coefficients being the probability amplitudes. Explicitly, $$|\psi\rangle = \int dx |x\rangle\langle x | \psi\rangle = \int dx \psi(x) |x\rangle$$ So, describing a state as $\psi(x)$ is nothing but listing all the coefficients in the expansion of the given state in position eigenstates. So, now, I pose you the reverse question. Why should we be able to write an energy eigenstates, say, $|\phi_n\rangle$ as $\phi_n(x)$?

The reason is that the eigenstates of position operator forms a complete basis. Thus, if you've listed all the probability amplitudes of a given system to be found in a position eigenstate, i.e., $\psi(x)$ for all $x$ then you've described the state completely. The reason behind this claim is simply that the eigenbasis of any observable can be reached by a unitary transformation from the eigenbasis of some other observable (and vice versa). This ensures that probability amplitudes of finding the given state in different eigenstates of any observable can be constructed from the probability amplitudes of finding the given state in different eigenstates of some other observable (and vice versa).

This is the reason you can describe any state as an expansion in position basis with coefficients being the corresponding probability amplitudes. In particular, you can express any state as a list of probability amplitudes $\psi(E)$ of finding the system represented by $\psi$ in energy $E$ upon measuring its energy. This is no more strange or different than expressing any state as $\psi(x)$ which is nothing but the list of probabilities of finding the system represented by $\psi$ in a position $x$ upon measuring its position.

Answer 2

The two are disconnected. If you know you are preparing a state $\psi(x)=\alpha\psi_1(x)+\beta\psi_2(x)$, you know this presumably because you know that $E_1$ and $E_2$ are the only possible outcomes of energy, and you have managed to establish $\alpha$ and $\beta$ possibly through some additional measurements. That’s what your preparation tells you. This is prior information needed to construct your state. Moreover, the energy eigenstates have additional information beyond the energy: you can certainly compute $\langle x\rangle$ or $\langle x^2\rangle$ using those states.

Knowing this, you can then compute transition rates, average values etc of whatever observables are in your model using $\psi(x)$. The observables you measure are not constrained by the state that you prepared.