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.


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 Dec 11 '18 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)? $\endgroup$ – Nelimee Dec 11 '18 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 Dec 11 '18 at 13:45
  • $\begingroup$ Ok that solved my problem! Thanks for the quick clarification :) $\endgroup$ – Nelimee Dec 11 '18 at 13:54

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