# What is the physical meaning of the eigenstates of an operator in quantum mechanics?

Let us suppose that we have an Hamiltonian that describes a quantum system. If one would like to know all of the possible values that the energy of the system described by that hamiltonian, one has to solve

$$\hat{H}\psi_m = E\psi_m$$

Where $$\hat{H}$$ is the hamiltonian operator and $$E$$ is the eigenvalue for the eigenstate $$\psi_m$$. My question is, if I denote $$\Psi$$ by

$$\Psi = \sum_m c_m \psi_m$$

where the sum is on all the eigenstates, do I get the "famous" $$\Psi$$ that has the property that $$\lvert\Psi\rvert^2$$ is the probability to find the particle(s) 0f the system in some position or the other? And do the $$\psi_m$$ also have this property?

This expansion is useful because to get the time evolution of the system, you just do: $$\Psi(t)=\sum c_n\psi_ne^{-iE_nt/\hbar}$$
• Great! And will the square norm of $\Psi$ and any of the $\psi_m$ give the same information with regards to position? May 6, 2022 at 18:10
• Also, could you clarify what you mean by "position basis"? I am diagonalizing H, so wouldn't the $\psi_m$ form the basis of the energy operator basis? May 6, 2022 at 18:13
• @peppece By position basis, I mean that we may for example express $\psi(x)=Ae^{-x^2/2}$, rather than expressing it in some other basis like $\psi(p)=Ae^{-p^2/2}$. The probabilistic interpretation holds in either case. May 6, 2022 at 18:26