Kreizhn
- 714
- 1
Hey everyone,
So we know that closed quantum systems evolve according to the Schrödinger equation. Now in open-loop quantum control, we essentially only ever deal with the operator version of this equation
[tex]i\hbar \frac{d}{dt} U(t) = H(t) U(t).[/tex]
That is, quantum control is often done in the Heisenberg picture rather than the Schrödinger equation.
My question is, why is this preferred? If the Hilbert space is finite dimensional, then the operator space has that dimension squared, and this isn't convenient to work with from a computational standpoint. I do have a hunch though.
Some times it's preferable to work with pure states, and sometimes it's preferable to work with density matrices. By considering the Schrödinger-operator equation above, it doesn't matter which we work in since by solving for U(t), pure state evolution is
[tex]|\psi(t)\rangle = U(t) |\psi(0)\rangle[/tex]
and density matrix evolution is
[tex]\rho(t) = U^\dagger(t) \rho(0) U(t)[/tex]
and so this allows us to "abstract" away the details of what states we're working with.
If there is a better explanation, I would love to hear it.
So we know that closed quantum systems evolve according to the Schrödinger equation. Now in open-loop quantum control, we essentially only ever deal with the operator version of this equation
[tex]i\hbar \frac{d}{dt} U(t) = H(t) U(t).[/tex]
That is, quantum control is often done in the Heisenberg picture rather than the Schrödinger equation.
My question is, why is this preferred? If the Hilbert space is finite dimensional, then the operator space has that dimension squared, and this isn't convenient to work with from a computational standpoint. I do have a hunch though.
Some times it's preferable to work with pure states, and sometimes it's preferable to work with density matrices. By considering the Schrödinger-operator equation above, it doesn't matter which we work in since by solving for U(t), pure state evolution is
[tex]|\psi(t)\rangle = U(t) |\psi(0)\rangle[/tex]
and density matrix evolution is
[tex]\rho(t) = U^\dagger(t) \rho(0) U(t)[/tex]
and so this allows us to "abstract" away the details of what states we're working with.
If there is a better explanation, I would love to hear it.