2
$\begingroup$

Suppose we have a finite dimensional Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$ and a Hamiltonian has the following form: $$H = H_A \otimes I \otimes I + I \otimes H_B \otimes I + |u\rangle \langle u| \otimes V,$$ where $|u\rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$.

Does there exist a state $|x\rangle|v\rangle$ such that $$(H_A \otimes I \otimes I + I \otimes H_B \otimes I + |u\rangle \langle u| \otimes V)|x\rangle|v\rangle = x |x\rangle |w\rangle?$$ In the above equation, we have $|x\rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$ and $|v\rangle, |w\rangle \in \mathcal{H}_C$.

The equation is true (we get a product state) whenever $|v\rangle$ is an eigenvector of $V$ and we get that $|v\rangle = |w\rangle$. However, I'm interested in a general case such that the RHS vector is a product state of systems AB and C.

$\endgroup$
2
  • 1
    $\begingroup$ Is there a reason why you're acting a Hamiltonian directly on the state rather than using its exponential to give you a unitary? $\endgroup$
    – DaftWullie
    May 3, 2023 at 8:25
  • $\begingroup$ @DaftWullie, at first, I was simply interested in an eigenvector of H. But then, I considered the case that is in my post. It seems a highly nontrivial problem. $\endgroup$
    – MonteNero
    May 3, 2023 at 18:36

1 Answer 1

2
$\begingroup$

Let's just do the special case of 3 qubits. Without loss of generality (up to unitary transformations, rescalings,...), $H_A$ and $H_B$ might as well be Pauli $Z$ matrices.

There are two possibilities for getting $|x\rangle$ on output, if $|x\rangle$ input.

  1. $|x\rangle$ is an eigenstate of $Z\otimes I+I\otimes Z$ (i.e. it has fixed number of 1s) of eigenvalue $\lambda$. In this case, we require $\langle u|x\rangle=0,1$. If 0, $|v\rangle=|w\rangle$. If 1, then $|w\rangle=\lambda|v\rangle+V|v\rangle$, and you have free choice of $V$. A non-trivial case might use something like $|x\rangle=(|01\rangle-|10\rangle)/\sqrt{2}$.

  2. If $|x\rangle$ is not an eigenstate of $Z\otimes I+I\otimes Z$, then $$ (Z\otimes I+I\otimes Z)|x\rangle=\lambda|x\rangle+\gamma |y\rangle. $$ ($|y\rangle$ is some state orthogonal to $|x\rangle$, $\gamma\neq 0$.) Then we require $$ |u\rangle(V|v\rangle)\langle u|x\rangle=(\alpha|x\rangle|-\delta |y\rangle)V|v\rangle. $$ In order to make sure we cancel off the non-x component, this requires $\delta V|v\rangle=\gamma|v\rangle$. Hence, this is the rather boring case of $|v\rangle$ being an eigenvector of $V$.

Although I made the simplifying assumption about the dimension of the Hilbert space, I don't immediately see anywhere that I've actually used it in the calculation, so I believe this is a general result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.