# Are quantum mechanical orbits specified uniquely by a Hamiltonian and initial state?

EDIT: This is completely wrong, don't bother reading.

Consider a finite dimensional quantum mechanical system, say an $$N$$-qubit system so that $$\text{dim}(\mathcal{H})=2^{N}$$. Let's prepare our system in a simple product state, say $$|0\cdots 0\rangle \equiv |0\rangle$$. We can now pick out a Hamiltonian, an arbitrary Hermitian operator $$H$$, and evolve the state forward in time via $$|\psi(t)\rangle = e^{-iHt}|0\rangle$$ (again see edit below: $$H$$ actually needs to be non-integrable for this to be non-trivial).

Now, consider the set of quantum states which can be achieved by $$H$$, in other words $$\{|\psi\rangle\in\mathcal{H}| \exists t: |\psi\rangle = e^{-iHt}|0\rangle\}\equiv S(H)$$. Something that occurred to me is that if one can find a state $$|\psi\rangle$$ which is in both $$S(H)$$ and $$S(H')$$, it must be the case that $$H=H'$$ (up to rescaling). Proof:

$$e^{-iHt}|0\rangle = e^{-iH't'}|0\rangle \Rightarrow Ht = H't'$$

So indeed $$H\propto H'$$. This is somewhat surprising to me, since it implies that, for a given initial state, there is one and only one Hamiltonian which can take you to a particular final state. Naively this sounds like a strong result, and it certainly contradicted my initial intuition. This is particularly true because of results such as the Poincare recurrence theorem in the context of non-integrable Hamiltonian evolution. However I think this is precisely why the constraint cannot be very strong, since I should be able to find Hamiltonians which generate states arbitrarily close to any target state.

That said I would still like to develop some more intuition for this idea and thought some here could offer some input, and perhaps refute my argument somehow (though I don't see where I could have gone wrong). I somehow failed to notice this property of quantum mechanical orbits in the past.

EDIT: Some have graciously pointed out my carelessness in the problem statement. I am specifically interested in studying the case where $$H$$ is a non-integrable Hamiltonian, i.e., it has no local symmetries. A typical non-integrable Hamiltonian will not have a simple, unentangled state such as $$|0\rangle$$ as an eigenstate, so let us assume that is the case. If we want to be more precise about this, we can impose the condition of locality on $$H$$, and then it is known that for almost all (in the mathematically technical sense of the phrase) Hamiltonians, eigenvectors are not shared. That is, a state can be an eigenvector of at most one local Hamiltonian. This prevents us from doing something like finding compatible observables and using their simultaneous eigenvalues to find a degenerate point in the orbits they generate.

• Hey, if you really think no one should bother reading the question, you probably should delete it. – stafusa Feb 8 at 18:49
• I'm not allowed to by the website. – miggle Feb 8 at 21:01
• Ok. It might have to do with having an accepted answer already. Well, maybe you're being too harsh and it might end up being instructive for someone. :) – stafusa Feb 8 at 22:24

Consider $$\hat{H}_1 = \hat{S}_y$$ and $$\hat{H}_{2} = \hat{S}_z$$. Your initial state is $$|S_z = -1/2\rangle$$.

If you picture your initial state as the south pole of the Bloch sphere, $$\hat{H}_1$$ will rotate your state around the y-axis of the Bloch sphere. Let it rotate twice. This will bring the state back to the initial state (with positive sign). On the other hand, during the same time, the initial state is an eigenstate of $$\hat{H}_2$$ and stay where it is.

So we have two different Hamiltonians which can arrive at the same state after propagation in time.

• Sorry, your counterexample is perfectly correct but due to my carelessness it doesn't actually address my question fundamentally. I should have stated that I am thinking specifically about non-integrable Hamiltonians, i.e., those which have no symmetries other than energy conservation. I am also making the (physically reasonable) assumption that my initial product state is not an eigenstate of the Hamiltonian since such a case is obviously special. – miggle Feb 8 at 8:14
• I still don't understand your reasoning behind your claim $e^{-i\hat{H}t}|0\rangle = e^{-i\hat{H'}t'}|0\rangle \rightarrow \hat{H}t = \hat{H'}t'$. Since you imply that it has to do with quantum non-integrability, where do you use it? – IamAStudent Feb 8 at 8:27
• Upon further reflection I think I've completely messed up. I shouldn't post when tired. For the sake of closing this I'm picking your answer, thanks for the response. – miggle Feb 8 at 8:43

OP's argument is not valid.

Consider a vector $$\psi$$ that is eigenvector of: the self-adjoint operator $$A$$ with eigenvalue $$a$$, and of the operator $$B$$ with eigenvalue $$b$$. This is often possible if $$A$$ and $$B$$ commute, but also in cases where they do not. In any case, it is possible in some cases where $$A\neq \lambda B$$, $$\forall \lambda\in \mathbb{R}$$.

Now, clearly $$e^{-itA}\psi=e^{-isB}\psi$$ whenever $$ta=sb$$, i.e. when $$s=ta/b$$.

• Thanks for your response. I should have (and will state in an edit) that I am specifically interested in the case of non-integrable Hamiltonians, i.e., those without any symmetries other than global energy conservation. Such states will not (at least, with probability 0) have a simple state such as $|0\rangle$ as an eigenstate, so we can neglect that case. – miggle Feb 8 at 8:17