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I have already read through the answers here.

So I understand that if the Hamiltonians commute, then they have the same eigenstates but not necessarily the same energy eigenvalues.

To formulate my question (skip to last paragraph if you prefer), we have the Hamiltonian $H_B$ with known lowest energy eigenstate $|\psi_{B_0}\rangle$ and corresponding energy eigenvalue $E_{B_0}$. Then our time-depenent Hamiltonian is

$$ H(t) = (1-\alpha(t))H_B + \alpha(t) H_C $$

where $H_C$ is the problem Hamiltonian with lowest energy eigenstate $|\psi_{C_0}\rangle$ and corresponding energy eigenvalue $E_{C_0}$, and $\alpha \in [0,1]$ is a time varying function that gradually moves from $0$ to $1$.

As I understand, if $H_C$ and $H_B$ don't commute, then $|\psi_{B_0}\rangle$ is an eigenstate of both of them, and therefore also of $H$. So the state won't evolve out of $|\psi_{B_0}\rangle$, and we aren't guaranteed that it is equal to $|\psi_{C_0}\rangle$. Another way people state it, which to me feels equivalent, is that there will be a crossover of energy bands and this degeneracy will cause the adiabatic speed limit to go to 0.

Here's my question. As I said we aren't guaranteed that $|\psi_{B_0}\rangle = |\psi_{C_0}\rangle$. But does that mean that we are sure $|\psi_{B_0}\rangle \neq |\psi_{C_0}\rangle$? Are there situations where $H_C$ and $H_B$ commute and we can know in advance that $|\psi_{B_0}\rangle = |\psi_{C_0}\rangle$ and can do QAA anyway?

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  • $\begingroup$ If you already knew that, what would you be trying to compute? $\endgroup$ – DaftWullie Jun 12 '20 at 15:21
  • $\begingroup$ btw, if $H_B$ and $H_C$ don't commute, it is not that $|\psi_{B_0}\rangle$ is an eigenstate of both. It's just that when you perform the adiabatic evolution, the ground state does change, and you stay in that ground state, so that, at the end of the evolution, you're in the ground state $|\psi_{C_0}\rangle$. $\endgroup$ – DaftWullie Jun 12 '20 at 15:22
  • $\begingroup$ Wait but I mean if $H_B$ and $H_C$ DO commute, then they are both diagonalizable in a common basis meaning they have the same eigenstates no? And regarding your first question, I was wondering if there are situations in which we know $|\psi_{B_0}\rangle = |\psi_{C_0}\rangle$ even if we don't necessarily know what they are. $\endgroup$ – Alexander Soare Jun 12 '20 at 15:55
  • $\begingroup$ Actually regarding your first question and what I just said... Yeah you must know $|\psi_{B_0}\rangle$, that's a premise of QAA. So yeah, makes sense. Thank you. $\endgroup$ – Alexander Soare Jun 12 '20 at 16:47

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