# What is the physical meaning of Hamiltonian eigenstates for a single particle?

Let us assume we have one 2-dimensional quantum system with a Hamiltonian

$$H = \sum_{n=1}^2 n \omega \mid n\rangle\langle n\mid$$

Do I understand it correctly when I assume that the eigenstates of this Hamiltonian $$\mid 1\rangle$$ and $$\mid 2 \rangle$$ are the 2 possible states the quantum system can be in ? (or some superposition between the 2 like $$\mid \psi \rangle=\alpha \mid 1\rangle+\beta\mid 2\rangle)$$.

And thus the corresponding eigenvalues of the eigenstates are the energies the system can have in a given state?

• A quantum state can be an eigenstate of the Hamiltonian. More generally, a quantum state is a superposition of many eigenstates of the Hamiltonian. – K_inverse Jan 7 at 11:31
• Perfect, that was what I needed! – CatoMaths Jan 7 at 13:24

If you are only given this Hamiltonian, it might be informally implied that the Hilbert space is spanned by $$|1\rangle$$ and $$|2\rangle$$, but there is no formal way of proving this. But, if you know that the Hilbert space is two-dimensional and that $$|1\rangle$$ and $$|2\rangle$$ are linearly independent, then they necessarily span the entire space.
• Ah yes. So only if the hilbert space (coincides?) with the dimensions of the Hamiltonian I know that the entire space is spanned by it. Hm, I see. So the eigenvalues are the results I can expect from a measurement. And of course this makes only sense when we are talking about the eigenstates. But when talking about superpositions it makes no sense as it is just a mixture of prob. to get either result $E_1$ or $E_2$. But then again this makes sense because a superposition itself is NOT an eigenstate of the Hamiltonian, no ? – CatoMaths Jan 7 at 13:28