# Can a Hamiltonian of a tripartite system map an product state into a product state?

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.

• Is there a reason why you're acting a Hamiltonian directly on the state rather than using its exponential to give you a unitary? May 3, 2023 at 8:25
• @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. May 3, 2023 at 18:36

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 $$V$$.