Quantum Mechanics – Are Two Hamiltonians Producing the Same Evolution Up to a Phase Equivalent?

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Consider two Hamiltonian operators $H_1, H_2$ and the evolution of some fixed single-qubit state $|\psi\rangle$ and yielding:
$$
\exp(i H_1t)|\psi\rangle = |\phi\rangle \\
\exp(i H_2t)|\psi \rangle = e^{i\theta}|\phi \rangle
$$
So the evolved state picks up a global phase after the time evolution of $H_2$.

Questions:

  1. Do we consider $H_1$ equivalent to $H_2$ given that the probability amplitudes of the output states are equal? This is true because:
    $$||\phi\rangle|^2 = |e^{i\theta}|\phi \rangle|^2$$
    In this case, I would say that any Hamiltonians whose output states differ by global phases are experimentally equivalent.
  2. If instead of a single qubit we have $n$ qubits do the above hold true? By that I mean, can the Hamiltonian evolution yield global phases as above or there exist more generic such transformations?

Best Answer

Yes such Hamiltonians are considered equivalent. And, two different hamiltonians which only produce a global phase difference must differ by a real constant $H_2 = H_1 + C$.

You can have the same for $n$ qubits: a global - meaning, spin-independent - phase difference in the Hamiltonian can be neglected.

I believe that there are not generalizations for multiple qubits, either: it seems to me that a global phase change is the most general equivalent Hamiltonian of the kind you are looking at. If $H$ added a phase to each qubit in a product state, like

$$H_2 = (\theta_1\mathbb{I} \otimes \theta_2\mathbb{I} \otimes ...) +H_1$$

then these phases would add to just one global phase $e^{-it(\theta_1 + \theta_2 + ...)}$.

And if the modification $H_1 \to H_2$ instead added different phases to different terms in a superposition, the state would no longer be equivalent to the original state.

EDIT: Proof that the Hamiltonians must differ by a constant:

Defining our variables: $$U_1(t) = e^{-itH_1} \,\,\,\,\,\,\, U_1(t) = e^{-itH_2}$$

And by the premise, $U_2(t) |\psi \rangle = e^{-it \theta} U_1(t) |\psi \rangle $ for all times $t$ and all states $|\psi \rangle$. Note: You do need a $t$ here, or else you need to pick a fixed time. Then

$$e^{itH_2} |\psi \rangle = e^{i t \theta} e^{itH_1} |\psi \rangle $$

$$e^{itH_2} |\psi \rangle = e^{it(H_1+ \theta)} |\psi \rangle $$

Since this must be true for all states $|\psi \rangle$, the only possibility is if the operators are equal

$$e^{itH_2}=e^{it(H_1+ \theta)}$$

This implies that the Hamiltonians must differ only by a constant $\theta$.

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