In this research paper, the authors introduce a new algorithm to perform Hamiltonian simulation.

The beginning of their abstract is

Given a Hermitian operator $\hat{H} = \langle G\vert \hat{U} \vert G\rangle$ that is the projection of an oracle $\hat{U}$ by state $\vert G\rangle$ created with oracle $\hat{G}$, the problem of Hamiltonian simulation is approximating the time evolution operator $e^{-i\hat{H}t}$ at time $t$ with error $\epsilon$.

In the article:

  • $\hat{G}$ and $\hat{U}$ are called "oracles".
  • $\hat{H}$ is an Hermitian operator in $\mathbb{C}^{2^n} \times \mathbb{C}^{2^n}$.
  • $\vert G \rangle \in \mathbb{C}^d$ (legend of Table 1).

My question is the following: what means $\hat{H} = \langle G\vert \hat{U} \vert G\rangle$? More precisely, I do not understand what $\langle G\vert \hat{U} \vert G\rangle$ represents when $\hat{U}$ is an oracle and $\vert G \rangle$ a quantum state.


1 Answer 1


You want to start by being careful with the sizes of the operators. $\hat U$ acts on $q$ qubits, and $\hat H$ acts on $n<q$ qubits. I believe that $|G\rangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use subscripts to denote which qubits the different operators and states act upon:

$$ \hat H_A=(\langle G|_B\otimes\mathbb{I}_A)\hat U_{AB}(|G\rangle_B\otimes\mathbb{I}_A) $$

  • $\begingroup$ @Nelimee I'm not sure if this is sufficient to resolve your confusion? Or is there something more that you're asking? $\endgroup$
    – DaftWullie
    Commented Dec 11, 2018 at 13:24
  • $\begingroup$ I am still trying to understand your answer but the sizes of the operators were definitely one of the points I missed! About your answer, what does $\vert G \rangle_B \otimes \mathbb{I}_A$ represent? A tensor product between a quantum state (a vector) and an operator (a matrix)? $\endgroup$ Commented Dec 11, 2018 at 13:39
  • $\begingroup$ Yes, exactly. Where, of course, you should think of a vector as a matrix where one of the dimensions is just 1. $\endgroup$
    – DaftWullie
    Commented Dec 11, 2018 at 13:45
  • $\begingroup$ Ok that solved my problem! Thanks for the quick clarification :) $\endgroup$ Commented Dec 11, 2018 at 13:54

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.