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?

  • $\begingroup$ A quantum state can be an eigenstate of the Hamiltonian. More generally, a quantum state is a superposition of many eigenstates of the Hamiltonian. $\endgroup$
    – K_inverse
    Jan 7, 2019 at 11:31
  • $\begingroup$ Perfect, that was what I needed! $\endgroup$
    – CatoMaths
    Jan 7, 2019 at 13:24

1 Answer 1


You don't know in principle whether this is the full Hilbert space only from the Hamiltonian. It might be that there is an arbitrary (possibly infinite, or even continuous) amount of states with vanishing energy. A quantum system is only well defined, if you give its Hamiltonian and its Hilbert space.

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.

Also, note that your last sentence is not quite correct. The eigenvalues of the Hamiltonian are the possible results of energy measurements, not the possible energies of states. For nearly every state of the system it doesn't make sense to talk about its energy. Only the eigenstates of the Hamiltonian have definite energy. To superpositions of those eigenstates you can only assign energy expectation values which must not conincide with one of the eigenvalues.

  • $\begingroup$ 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 ? $\endgroup$
    – CatoMaths
    Jan 7, 2019 at 13:28

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