Consider a quantum system that is prepared at some initial state $| \text{in}\rangle$ and we are able to measure the output state $| \text{out}\rangle$. Furthermore assume this evolution of the state is governed by a Hamiltonian without any control signals and let us consider for simplicity that the system is closed (such simulations can be easily run on Qutip).

Essentially: $$ H:| \text{in}\rangle \mapsto | \text{out}\rangle $$ This can easily be visualized by considering a qubit. In the Bloch sphere this evolution corresponds to a curve whose initial point is the input state and the final point is the output state.

Question: Given fixed $|\text{in}\rangle,|\text{out}\rangle$ is there a unique Hamiltonian that maps $|\text{in}\rangle \to |\text{out}\rangle$?

Thinking of paths, we know that there are many inequivalent paths from one state to another. Nevertheless, nature will choose the one that minimizes the path integral. This amounts into minimizing the Lagrangian. Furthermore, thinking of paths again, one can find symmetries on the system which are redundant.

Thinking of Hamiltonians, is there something equivalent? Specifically:

  • Is it known what are the symmetries in the space of Hamiltonians that results in equivalent evolution $|\text{in}\rangle \to|\text{out}\rangle$?

Naively thinking, I can think of some toric action (a phase if you will) multiplying any Hamiltonian that should produce the same class of evolution and that can be normalized (choosing phase to be equal to unity).

Layman's terms: Is the evolution $|\text{in}\rangle \to |\text{out}\rangle$ be given by a unique Hamiltonian and what is actually known about it regardless of the answer?

  • $\begingroup$ Do you allow piecewise-continuous evolutions? What about relative phase difference between final states? $\endgroup$ Sep 18, 2021 at 12:44
  • $\begingroup$ I don't follow this comment. The initial and final states are completely different. A phase difference only would correspond to the same state. $\endgroup$
    – Marion
    Sep 18, 2021 at 17:24
  • $\begingroup$ When you ask for "symmetries" in the space of Hamiltonians, aren't you essentially asking for a class of Hamiltonians with a specific eigenspace decomposition? E.g. "I only want to consider Hamiltonians that leave invariant the space $\{|1\rangle,|4\rangle\}$ for some at-least-four-dimensional underlying space". I think the answer strongly depends on how much symmetry you impose, as you'd probably expect. But something not clear from the question is the time: are you asking for $H,H'$ such that $e^{iHt}|\psi\rangle=e^{iH't}|\psi\rangle$ for all times $t$? This changes the answer a lot $\endgroup$
    – glS
    Sep 20, 2021 at 8:50
  • $\begingroup$ This paper discusses the related problem of finding $H$ such that $U=e^{iH}$ for a given $U$. This is thus a stronger condition: you investigate classes of Hamiltonians sending all input states to the same output. Time is fixed. The role of symmetry in the answer is made explicitly obvious, as well as the nontriviality of the solutions. E.g. Eq (5) and Eq (22) give classes of Hamiltonian implementing the Toffoli gate; note that many of the solutions break the symmetries of the Toffoli during the dynamics, but recover it at $t=1$. $\endgroup$
    – glS
    Sep 20, 2021 at 8:53
  • $\begingroup$ Right, so your final equation is what I am asking but only for $t=t_\text{final}$. So I am asking if $e^{iHt}|\psi \rangle|_{t=t_\text{final}} = e^{iH't}|\psi \rangle|_{t=t_\text{final}} $ implies that up to some "symmetry" $H$ is equivalent to $H'$. $\endgroup$
    – Marion
    Sep 20, 2021 at 8:53

1 Answer 1


This is too long for a comment but consider the following situation. Suppose $\vert \uparrow\rangle$ and $\vert\downarrow\rangle$ are you initial and final states.

Then $e^{i\pi \sigma_y/2}\vert\uparrow\rangle=\vert \downarrow\rangle$ while $e^{i\pi \sigma_x/2}\vert\uparrow\rangle =i\vert\downarrow\rangle$, so you get the same final state up to a phase.

If you allow for piecewise Hamiltonian, then you can always rotate away the phase $i$ in $i\vert\downarrow\rangle$ using $e^{i\pi\sigma_z/2}$.

If you do allow final states up to a phase, there are clearly more solutions since you can go from the North Pole to the South Pole of the Bloch sphere along any line of constant latitude.

If you do allow piecewise Hamiltonian, you can find even more solutions. Pictorially, you can imagine going from the North Pole to the equator, travel some distance along the equator, and then from the equator to the South Pole.

You might have to include piecewise Hamiltonians if you have constraints on your Hamiltonians: there are some groups which are not simply connected so you might more than one transformation to reach a final state from an initial state. An example would be an $\mathfrak{su}(1,1)$ evolution, where the exponential map does not cover the whole group with a single exponential. Details can be found in

Chiribella G, D’Ariano GM, Perinotti P. Applications of the group SU (1, 1) for quantum computation and tomography. Laser physics. 2006 Nov;16(11):1572-81.

or the arXiv version.

  • $\begingroup$ Thanks a lot for the answer. I guess that I would like to cut phases down since given an experiment the output is "agnostic" about the overall phase so we would only read/estimate the state as it is given: In this case the complex $i$ makes a difference as to the numerical (not physical) output. But even if we allow phases, then there must be a family of Hamiltonians parametrized by some variable right? So I guess I am asking the equivalence classes classifications of such Hamiltonians. $\endgroup$
    – Marion
    Sep 18, 2021 at 17:29
  • $\begingroup$ still not quite precise. Do you allow piecewise Hamiltonian evolution? Do you specify input AND output? Do you have restrictions on your Hamiltonians? (for instance, you cannot go from $\ell=2,m=2$ to $\ell=2,m=1$ states using a single Hamiltonian constructed from angular momentum operators alone... $\endgroup$ Sep 19, 2021 at 2:14
  • $\begingroup$ I am not sure what is $\ell,m$ here but let's say I don't allow piecewise Hamiltonians. It seems from your answer that the evolution of the state is unique up to a complex number so one can say it is "conformally unique" since $i$ scales the result in the complex plane (actually rotates it). And this is the result of a toric action. But is the state $|n\rangle$ and $i|n\rangle$ indistinguishable? $\endgroup$
    – Marion
    Sep 20, 2021 at 7:50
  • $\begingroup$ $\ell,m$ are angular momentum quantum number… $i$ has magnitude 1 so it’s hardly a scaling, and since all states differing by an overall phase are equivalent, this isn’t an issue unless you have grounds to keep track of this overall phase. $\endgroup$ Sep 20, 2021 at 11:21
  • $\begingroup$ Ok, but one might not have angular momentum or such as a parameter. Depends on what is the system considering, right? For a superconducting qubit I guess there is not much point to talk about angular momentum. But point is that in the Bloch sphere, am I correct that any two states related by a $U(1)$ rotation (that is a phase) are then physically equivalent? $\endgroup$
    – Marion
    Sep 20, 2021 at 11:49

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