Let's say that I have some quantum system defined by Hamiltonian $\hat{H}$. The energy eigenstates of this Hamiltonian form a complete basis for the Hilbert space of all possible states corresponding to this quantum system.

Now, if I have some initial state $|\Psi(0)\rangle$, I can solve for $|\Psi(t)\rangle$ by finding the complex coefficients in the expansion of $|\Psi(0)\rangle$ in the energy basis, then add the usual time-dependence and append each of the coefficients to their corresponding eigenstate in a linear combination.

My question is, what if $|\Psi(0)\rangle$ is an unphysical state of our quantum system (does not obey the Schrödinger equation for this particular Hamiltonian)? Am I allowed to choose this as my initial state? If so, how do I know I can expand it in terms of energy eigenstates, as to my understanding, these form a complete basis for the Hilbert space of possible states for a given system?

Even more generally, is the Hilbert space of states for a given system restricted by the Hamiltonian itself?

  • 2
    $\begingroup$ The Schroedinger equation is $i\hbar\partial_t |\psi(t)\rangle = H|\psi(t)\rangle$. It describes the time evolution of the state. It does not restrict the state at a single point in time, $\psi(0)\rangle$, in any way -- what do you mean by "unphysical state"? $\endgroup$
    – Noiralef
    Commented Oct 1, 2019 at 3:11

1 Answer 1


The (time dependent) Schrodinger equation determines the time evolution of a state, so any single state ($| \Psi(0) \rangle$ for instance) can neither be said to obey or disobey the Schrodinger equation.

A function that takes in a time $t$ and outputs a state $| \Psi(t) \rangle$ can be said to obey the Schrodinger equation if $$ i \hbar \frac{d}{dt} |\Psi(t)\rangle = \hat{H} |\Psi(t)\rangle $$

So to answer your question any initial state is allowed. How one would experimentally prepare a given state is a different question though...


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