2
$\begingroup$

Let's say I am given a Hamiltonian

\begin{equation} H = \sum_{i = 1}^{m} H_{i}, \end{equation}

where $H$ acts on $n$-qubits, and each $H_{i}$ acts non-trivially on at most $k$ qubits. The eigenvalues of $H$ are between $0$ and $1$. As can be seen, $H$ is a $k$-local Hamiltonian. Now, let's say I am given any quantum state $|\psi\rangle$ over $n$ qubits. I want to implement the POVM

\begin{equation} \{H_{i}, \mathbb{1} - H_{i}\}. \end{equation}

  1. How do I implement this POVM using a fixed universal gate set and the ability to measure in the standard basis? What is the unitary that I have to apply before measuring in the standard basis and how much error can I tolerate?
  2. What is the guarantee this implementation is efficient?
  3. Is there any rule regarding when implementing such POVMs is efficient?
$\endgroup$
3
$\begingroup$

What is the guarantee this implementation is efficient? Is there any rule regarding when implementing such POVMs is efficient?

The implementation of such a gate will only depend on the parameter $k$ (which I assume you mean to be fixed), not $n$. Since efficiency is generally phrased in terms of scaling with $n$, and you have no dependence on that, it is efficient.

How do I implement this POVM using a fixed universal gate set and the ability to measure in the standard basis? What is the unitary that I have to apply before measuring in the standard basis

Let $H_i=UDU^\dagger$, where $D$ is diagonal (with entries between 0 and 1 on the diagonal) and $U$ is a unitary. Apply $U^\dagger$ to the appropriate set of qubits. This now reduces you to the problem of performing the measurement $\{D,1-D\}$.

You'll need to introduce a single ancilla qubit, prepared in the $|0\rangle$ state. It is this ancilla that you will measure in the computational basis, with the two outcomes corresponding to the two different measurement operators. But before that, we need to construct a unitary between the original system (S) and the ancilla (A). Let $D=\sum_id_i|i\rangle\langle i|$, and let $V|i\rangle_S|0\rangle_A=\sqrt{d_i}|i\rangle|0\rangle+\sqrt{1-d_i}|i\rangle|1\rangle$. You can decompose this unitary via standard techniques. Apply $V$, and measure the ancilla.

To see that this works, let your input state be $|\psi\rangle=U\sum_i\alpha_i|i\rangle$. You sould get the measurement outcome with probaility $$ \langle\psi|H_i|\psi\rangle=\sum_i|\alpha_i|^2d_i. $$ This is what we need to check that we get. So, our simulation first applies $U^\dagger$, so we have $$ \sum_i\alpha_i|i\rangle_S|0\rangle_A. $$ We apply $V$ to prepare $$ |\Psi\rangle=\sum_i\alpha_i|i\rangle_S(\sqrt{d_i}|0\rangle_A+\sqrt{1-d_i}|1\rangle_A). $$ We calculate the probability of the 0 outcome: $$ \langle\Psi| 1_S\otimes|0\rangle\langle 0|_A|\Psi\rangle=\sum_i|\alpha_i|^2d_i, $$ as required.

Note that I've not worried about the state after the measurement because you've only specified a POVM, which immediately implies you're only interested in the measurement probability, not the output state.

and how much error can I tolerate?

This depends on what you mean, and is probably an entirely separate question to do justice to.

$\endgroup$
6
  • $\begingroup$ Regarding the error, what I meant is, when I am decomposing the unitaries with gates from my universal gate set (comprising of H, S, T, and the CNOT gate, let's say), what is the trade-off between the error incurred and the size of the circuit? $\endgroup$ – BlackHat18 Oct 12 '20 at 9:56
  • $\begingroup$ Also, what is the cost (in terms of circuit zie) of implementing the unitary $U$? Why do we assume it is not high? $\endgroup$ – BlackHat18 Oct 12 '20 at 10:09
  • $\begingroup$ Because everything only acts on $k$ qubits. It might b exponentially large in $k$, but if $k$ i fixed as $n$ scales, we don't care about that. $\endgroup$ – DaftWullie Oct 12 '20 at 10:14
  • $\begingroup$ Regarding the issue of error, why is this different to any standard "build a unitary" task, where we know the optimal solution and how the error scales (run time is $\log(1/\epsilon)$ for each single qubit gate) $\endgroup$ – DaftWullie Oct 12 '20 at 10:15
  • $\begingroup$ Thanks! It's clear now. One thing though, in the analysis, I am not quite sure why you used the state $|\psi\rangle=U\sum_i\alpha_i|i\rangle$ for your analysis. Won't the analysis work for states like $|\psi\rangle=\sum_i\alpha_i|i\rangle$? $\endgroup$ – BlackHat18 Oct 12 '20 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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