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Are eigenfunctions of observables solutions to the time-dependent Schrödinger equation? Or is this not necessarily the case? From what I had been reading they are not necessarily solutions to Schrödinger, but their associated eigenvalues are possible measurements results. Not only that, the eigenfunctions and their linear combinations are possible initial wave functions (when evaluated at some time $t$) since apparently any function in Hilbert space is a possible initial wave function, but that does not mean the eigenfunctions are solutions for all $t$. Not all members in Hilbert space satisfy the time-dependent Schrödinger equation.

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  • $\begingroup$ No. Why would you think so? $\endgroup$ Commented Mar 8, 2018 at 23:26

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The eigenfunctions of the Hamiltonian, i.e. energy observable, are stationary solutions to the Schrödinger equation, since the latter is $i\,\hbar\,\partial_t\,\psi = \hat{H}\,\psi$, so if we have an eigenfunction $\psi_H$ of $\hat{H}$, a solution of the Schrödinger equation is $e^{i\,E\,t/\hbar} \psi_H$, where $E$ is the energy eigenvalue.

I think this is where you may be getting confused: something like your statement is true for the energy observable, but not for observables in general.

Possibly reinforcing your confusion is the fact that in some important cases, the eigenfunctions of other observables have the same behavior. This happens whenever the observable in question commutes with the Hamiltonian, such as the momentum observable for a free particle.

Moreover, the eigenfunctions of $\hat{H}$, for most if not all Hamiltonians of interest, span the whole Hilbert space, so that you can build any Hilbert space member as a superposition of them.

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  • $\begingroup$ Ok, thanks I got it. I just want to confirm something else. Even though eigen functions of observables may not be solutions, they are assumed to be complete, this means they can span any member of Hilbert space, which means they can span any solution to time dependent Schrodinger equation by choosing the appropriate constants, right? $\endgroup$ Commented Mar 9, 2018 at 1:26
  • $\begingroup$ @angelleonardo For many observables, that is true. To answer that question in full generality calls for full appreciation of the relevant spectral theorem; in general one has to deal with self adjoint operators with no eigenfunctions in the Hilbert space (for example, the momentum operator). One can retrieve a generalized version of the notion of eigenfunction that retrieves completeness in most cases of importance; see my answer here. This Wiki page is also of interest. $\endgroup$ Commented Mar 9, 2018 at 3:01

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