2
$\begingroup$

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 $|v\rangle$.

$\endgroup$
2
  • $\begingroup$ Just to clarify the setting: you know $H$ and can implement unitary evolutions $e^{-iHt}$? You know that $|v\rangle$ is an eigenstate. Do you know (i) its eigenvalue? (ii) is that eigenvalue unique? (iii) what $|v\rangle$ itself actually is? $\endgroup$
    – DaftWullie
    Commented Oct 10, 2022 at 10:09
  • $\begingroup$ @DaftWullie I've added some clarifications. Hope this helps. Thanks! $\endgroup$
    – MonteNero
    Commented Oct 11, 2022 at 5:46

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ @ DaftWullie, thanks for the answer. Could you clarify what you mean by "the trick now is to apply a controlled-controlled-...-controlled-phase gate on the extra register that applies −1 on the value of |n⟩ corresponding to the one eigenvector you want"? How does it look mathematically? Also, by "extra register" do you mean a brand-new register? $\endgroup$
    – MonteNero
    Commented Oct 12, 2022 at 20:39
  • $\begingroup$ Also, is it possible to do get $I + (e^{-\textrm{i} s t}-1) |v \rangle \langle v|$ instead of $(I-2|v\rangle\langle v|)$? $\endgroup$
    – MonteNero
    Commented Oct 12, 2022 at 20:57
  • $\begingroup$ Yes, the brand-new register. $\endgroup$
    – DaftWullie
    Commented Oct 13, 2022 at 15:05
  • $\begingroup$ Sure, again you apply a controled-controlled-...-controlled phase, where this time the phase is $e^{-ist}$ if the target state is $|n\rangle$ and 2 if it's snything else. $\endgroup$
    – DaftWullie
    Commented Oct 13, 2022 at 15:06
  • $\begingroup$ I'm not sure how you want it to look mathematically, as you clearly know how to write these things: $I-2|n\rangle\langle n|$ $\endgroup$
    – DaftWullie
    Commented Oct 13, 2022 at 15:06
0
$\begingroup$

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.

enter image description here

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.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. But you left the most relevant part about $V$ unanswered and just wrote about generalized Grover diffusion. I also considered that venue but then it all boiled down to expressing $|v\rangle$ or $V$. $\endgroup$
    – MonteNero
    Commented Sep 10, 2022 at 15:49

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.