Imagine two situations. In one, there are two qubits that are next to each other, that is, they have non-zero coupling terms in their Hamiltonian, and thus suffer from cross-talk and energy can leak from one to the other. In the other situation, the same qubits are separated, unconnected.

Question: How would you describe the Hilbert spaces of these two situations? What mathematical properties would the space of the coupled qubits have that the separated qubits wouldn't have?


1 Answer 1


According to the Composite system postulate:

"The Hilbert space of a composite system is the Hilbert space tensor product of the state spaces associated with the component systems. For a non-relativistic system consisting of a finite number of distinguishable particles, the component systems are the individual particles."

So the Hilbert spaces are the same in either case $\mathbb{C^2}\otimes \mathbb{C^2}$. What is different is not the Hilbert space, but rather the trajectories of states in these spaces, as they will depend on the unitary dynamics being driven by either a coupled or uncoupled Hamiltonian.

  • $\begingroup$ Thanks. Can you say (or mathematically formulate) anything about those trajectories? That's what I'm actually focused on --- I know it's the same Hilbert space. $\endgroup$
    – psitae
    Commented Apr 8, 2022 at 18:12
  • $\begingroup$ it's going to depend greatly on the form of the hamiltonian. But if all the qubit 1 terms in your hamiltonian commute with all your qubit 2 terms, so that $H=H_{q1}+H_{q2}$ where $[H_{q1},H_{q2}]=0$, then your unitary $U(t)=\exp(iHt/\hbar)$ will factor into two commuting unitaries $U(t)=U_1(t)U_2(t)=U_2(t)U_1(t)$ with $U_1(t)=\exp(iH_{q1}t/\hbar)$ and $U_2(t)=\exp(iH_{q2}t/\hbar)$. Otherwise its tricky... $\endgroup$
    – Condo
    Commented Apr 8, 2022 at 18:26

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