0
$\begingroup$

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?

$\endgroup$
1
  • 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
    Oct 1, 2019 at 3:11

1 Answer 1

2
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.