How to find a circuit for a unitary operator $e^{-i s |v \rangle \langle v| t }$?

Question

Let $|v \rangle$ be an eigenstate of an $n$-qubit and $2$-local Hamiltonian
$$H = \sum_{i=1}^n \left (X_i + a_i Z_i \right) + \sum_{(i,j)} b_{i,j} Z_i Z_j,$$
where $\sigma_i = I \otimes \cdots \otimes \sigma \otimes \cdots \otimes I$ for $\sigma \in \{X,Z\}$ and $a_i,b_i, h_{i,j} \in \mathbb{R}$.

I would like to know how to do a circuit implementation of $$ U(t) = e^{-\textrm{i} s |v \rangle \langle v| t },$$ where $s,t \in \mathbb{R}$.

If it helps, $U(t)$ can also be written as $$U(t) = e^{-\textrm{i} s |v \rangle \langle v| t } = I + (e^{-\textrm{i} s t}-1) |v \rangle \langle v|.$$

Edit:
My initial idea was to express $|v \rangle \langle v|$ as a linear combination of $(3^2-1)\binom{n}{2}$ $n$-qubit Pauli matrices that have a form $P_{i,j} = I \otimes \sigma_i \otimes \cdots \otimes \sigma_j \otimes I$ where $\sigma_i, \sigma_j \in \{I, X, Z\}$. Then we can write: \begin{align*} |v\rangle \langle v| &= \sum_{i,j} c_{i,j} P_{i,j}. \\ e^{-\textrm{i} s |v \rangle \langle v| t} &= e^{-\textrm{i} s \left (\sum_{i,j} c_{i,j} P_{i,j} \right)t}. \end{align*}

Since most Pauli terms $P_{i,j}$ don't commute, we could apply the first-order Trotterization and get: \begin{align*} e^{-\textrm{i} s |v \rangle \langle v| t} = e^{-\textrm{i} s \left (\sum_{i,j} c_{i,j} P_{i,j} \right) t } \approx \prod_{i,j} e^{-\textrm{i} s c_{i,j} P_{i,j}t}. \end{align*} This is a product of $(3^2-1)\binom{n}{2}$ gates $RZ$, $RZZ$ and $RX$.

I think this is too many gates to approximate the original unitary matrix! I also assumed that $P_{i,j}$ are made of $I, X$ and $Z$ and that if $\sigma_i= X$ then $\sigma_j \neq X$.

Some clarifications:
$H$ is known, and it is possible to implement an approximate evolution given by a trotterized version of $e^{-iHt}$. I guess we could assume that the associated eigenvalue of $|v\rangle$ is known and that it is unique. So far, I don't have an exact expression for $|v\rangle$.

Answer 1

If you know the eigenvalue associated with the eigenvector, then by far the best thing to do is run a phase estimation protocol using $e^{-iHt}$. There's probably an optimum choice of $t$ depending on what you know about the spectrum, but as a bare minimum, if all your eigenvalues are between 0 and $\Lambda$, then you set $t=2\pi/\Lambda$ to maximally spread out the eigenvalues without being affected by the $2\pi$ modulus that arises from the exponentiation.

To convey what happens, let $|\lambda_n\rangle$ be eigenvectors of $H$ with eigenvalue $\lambda_n$. Any initial state may be decomposed as $$ |\psi\rangle=\sum_n\alpha_n|\lambda_n\rangle. $$ To apply phase estimation, you introduce a second register of $t$ qubits, initially in $|0\rangle$. After phase estimation you have, at least to a good approximation, $$ |\psi\rangle=\sum_n\alpha_n|\lambda_n\rangle|n\rangle. $$ where by $|n\rangle$ I mean the $t$-bit representation of the eigenvalue $\lambda_nt/(2\pi)$. The trick now is to apply a controlled-controlled-...-controlled-phase gate on the extra register that applies $-1$ on the value of $|n\rangle$ corresponding to the one eigenvector you want. The trick being that while we apply $$ I\otimes(I-2|n\rangle\langle n|), $$ it has exactly the same effect as the operation $$ (I-2|v\rangle\langle v|)\otimes I. $$ Finally, you have to apply the inverse of the phase estimation to undo the entanglement between the original register and the extra one.

Answer 2

This operation $U(t) = e^{-i s t|v\rangle \langle v|}$ that you wish to implement resembles the Grover diffusion operator if we take $|v \rangle = (H |0\rangle)^{\otimes n}$ and $st = \pi$. Hence, in the spirit of its generalization, given a unitary $V$ such that $V |0 \rangle^{\otimes n} = |v \rangle$, we can implement $U(t)$ via the following $(n+1)$-qubit circuit, where an ancillary qubit in state $|0\rangle$ is added for convenience.

In words, ignoring the $V$ and $V^{\dagger}$ subcircuits for the moment, we have a $R_z(2st)$ gate acting on a qubit in state $|0\rangle$ if and only if all $n$ qubits in the main register are in state $|0\rangle$. Hence, only the state $|0\rangle^{\otimes n}$ will be applied a phase shift of $e^{-i st}$ per the phase kickback effect. Now, once we sandwich this n-controlled-$R_z$ gate between $V^{\dagger}$ and $V$, the special state that is applied the phase shift is no longer $|0\rangle^{\otimes n}$ but instead $V|0\rangle^{\otimes n} \equiv |v\rangle$. That is precisely what $U(t)$ amounts to.

The n-controlled-$R_z$ gate can be decomposed into elementary gates following, e.g., this reference. The only missing piece is therefore the implementation of the $n$-qubit operation $V$ that prepares the eigenstate $|v\rangle$ of the given Hamiltonian. In the absence of the longitudinal field (i.e., setting $a_i = 0$), this model is exactly solvable by mapping the local spins-$\frac{1}{2}$ to fermions via the Jordan-Wigner transformation, which yields a quadratic Hamiltonian for which all eigenstates are just Slater determinants, which can be prepared on a quantum computer using the method discussed, e.g., here. For the general case where the longitudinal field is present, I suppose one has to resort to some ansatz to find the eigenstates of the model.