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.

  • 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


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.


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