# Hilbert space and Hamiltonians

Assume a system described by a Hamiltonian H, and assume that the eigenstates of H, $φ_i$(r) are integrable in absolute square. We say that these states belong to a Hilbert space (they can even form a base in that space).

But, is the opposite true? Let a system be described by a wave-function S(r, t), integrable in absolute square. Does that imply that the system behavior is also described by a Hamiltonian?

Remark: te evolution of a system does not always admit a Hamiltonian. E.g. if the evolution is non-unitary (or at least, if there is a Hamiltonian, it would take complex eigenvalues.) To be clear, I don't know if my system evolves unitarily or not. I just gave the example to show that the existence of a Hamiltonian is not guaranteed. Whatever I know of the function S(r, t) is that it belongs to a Hilbert space.

So the question is, absolute square-itegrability, ensures (as a sufficient condition) the existence of a Hamiltonian for the system?

An example: I expand S(r, t) in a quantum superposition of the eigenfunctions $φ_i$(r),

S(r, t) = $∑_i$ $C_i$(t) $φ_i$(r), with $C_i$(t) = $F_i$ (t) exp(-i$E_i$t/ħ).

Introducing this superposition in the Schrodinger equation with the Hamiltonian H, I obtain that iħ ∂S(r, t)/∂t is not equal with HS(r, t). But, could it be that another Hamiltonian H' may exist s.t. the Schrodinger eq. be satisfied?

• Comment to the question (v3): Given an arbitrary orthonormal basis for the Hilbert space, it is possible to find infinitely many self-adjoint operators $H$ that are diagonal in that basis. – Qmechanic Dec 6 '14 at 12:04
• If the evolution is non-unitary, you are not describing the entire system, but an open subsystem! Unitarity is one of the few tenets that no proper quantum theory breaks when considering closed systems. – ACuriousMind Dec 6 '14 at 13:01
• @Qmechanic: Yes, it seems simple. But something doesn't work. The discussed wave-function - let me call it S(r, t) - is tricky. Though abs. square-integrable, it is not an ordinary quantum-superposition of the base vectors ( $φ_i$(r) ) of the Hilbert space. What I have is S(r, t) = $∑_i$ $C_i$(t) $φ_i$(r). And the time-dependence of $C_i$(t) is not the ordinary dependence exp(-i$E_i$t/ħ), where $E_i$ are the energies of $φ_i$(r), but $C_i$(t) = exp[-i($E_i$ + Q)t/ħ], i.e. more complicated. – Sofia Dec 6 '14 at 15:00
• (continuation) By the way, it is a legitimate function in the Hilbert space. But introducing this superposition in the Schrodinger equation, the above time dependence makes trouble: what I obtain in the LHS (i.e. where I derivate by time) is not equal with what I get in the RHS. And I repeat, it's a legitimate function in the Hilbert space. – Sofia Dec 6 '14 at 15:07
• @Sofia Where do the $E_i$ come from? Some Hamiltonian I'd guess. The appearance of $Q$ merely shifts the whole spectrum by a constant. That has no physical consequence. The norm of the state $S$ is conserved during time-evolution, hence the dynamics is unitary. – Nephente Dec 6 '14 at 16:31

Any Hilbert space $\mathcal{H}$ with the notion of unitary time evolution also possesses the notion of Hamiltonian.

If $\mathcal{U}(t) : \mathcal{H} \to \mathcal{H}$ is the time evolution operator for every $t \in \mathbb{R}$, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element $H$ in the Lie algebra of linear operators. This generator is the Hamiltonian, and as the generator of a unitary operator, it is necessarily self-adjoint by Stone's theorem, so you get a Hamiltonian whose eigenvectors span the space.

Since non-unitary time-evolution comes into play if you are only considering a subspace of the full space of states (e.g. when you don't track all decay products for decaying systems), one can always get a unitary evolution by embedding the subsystem into "the whole system", find the Hamiltonian there, and then project it back onto the subsystem to get the Hamiltonian for the subsystem. But now, since time evolution was non-unitary here, it cannot be that this Hamiltonian is self-adjoint (since the exponential of self-adjoint operators is unitary), therefore we are forced to conclude that the eigenvectors of a Hamiltonian cannot span a subspace on which time evolution is non-unitary.

So, you cannot get a Hamiltonian both spanning the space and producing non-unitary time evolution, one of these must necessarily fail.

• Stone's theorem states that unitary evolution groups are generated by self-adjoint operators, not just hermitian ones, so there's no need to do any extension. Actually different extensions of hermitian operators yield different unitary evolution groups. – Mateus Sampaio Dec 6 '14 at 14:36
• @ACuriousMind: I saw that you answered and I read your comment, but I have to leave the computer now. I return much later. Let me say just in short. It is a disputed question whether the decay is unitary or not. So, I am not sure on this. In my question I only gave an example that existence of a Hamiltonian is not guaranteed. THAT'S ALL !!!. I don't know anything else about my system than the fact that its wave-function is abs. square-integrable. So, here comes my question. Abs. square integrability ensures the existence of a Hamiltonian? But, as I say, I'll return later. – Sofia Dec 6 '14 at 15:28
• @Sofia: I answer that, regardless of where the non-unitary evolution comes from: If time-evolution is non-unitary, you do not get a Hamitonian that simultaneously generates the time evolution and whose eigenvectors span the space on which the non-unitary evolution takes place. – ACuriousMind Dec 6 '14 at 15:31
• @Mateus: I am ignorant about the difference between a self-adjoint operator and a Hermitian one. I saw other people too claiming that there is a difference. What is this difference? But please, if possible, give me a simple explanation. What features has one, and doesn't have the other? What I learnt is that BOTH have the property of REAL eigenvalues. I hope that this is not wrong. – Sofia Dec 6 '14 at 20:36
• @Sofia: It is wrong, Hermitian not-self-adjoint operators can have complex eigenvalues. The difference exists only on infinite-dimensional Hilbert spaces, where the domains of definitions of linear operators need not necessarily be the whole space. An operator $A$ is Hermitian if it is the same as $A^\dagger$ on its domain of definition. It is self-adjoint if the domains of definition of $A$ and $A^\dagger$ coincide and they are the same on it. A nice overview of the issues with this difference is here. – ACuriousMind Dec 6 '14 at 20:41

No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian.

The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous idea is, that it is a space over the complex numbers.

The fact that a proper wavefunction should be $L^2$ reflects the interpretation as a probability-/ or charge-density.

Any (sensible) Hilbertspace admits a countable orthonormal basis. Pick a real number for each of those basis states and define

$$(H)_{mn} = \langle \psi_m,H\psi_n\rangle \equiv \epsilon_n$$

as matrix elements of an operator $H$. What kind of dynamics this "Hamiltonian" describes, depends on the choice of eigenvalues, e.g energies. You should also make sure, that the spectrum is not too pathological, e.g it should be bounded from below to be interpretable as a sensible physical spectrum.

Non-unitary time-evolutions come about in open quantum systems, where you forget about the environment which couples to the system of interest, but which is nevertheless there. Probability can "leak" into the environment, hence a non-unitary evolution. An example is e.g the Lindblatt-Master-Equation describing the Markovian limit of a system-environment coupling.

• So really, you are arguing that closed systems with a sensible probability density function should by definition be described by a Hamiltonian, whereas in an open system it is not the case. Am I understand your answer correctly? – Constandinos Damalas Dec 6 '14 at 13:11
• @PhotonicBoom I mean, you can't do with a Hilbertspace alone to describe a physical system. Some dynamical input is needed. One of the postulates of QM is that state evolution is generated by a Hamiltonian operator. The implicit assumption is, the system is closed. No coupling to an environment. Even with an "environment" the total system can still be considered closed and is thus governed by unitary time-evolution. It is only when one passes to an effective description of the subsystem by tracing out the environment, that non-unitary dynamics arise. – Nephente Dec 6 '14 at 16:41