In the paper https://arxiv.org/abs/1909.05820 the authors introduce several Hamiltonians. For example they define $$ H_G = A^\dagger \left( \mathbb{I} - \vert b \rangle \langle b \vert \right) A $$ in equation (4), where $A$ is a complex-weighted sum of unitary matrices and $\vert b \rangle$ a quantum state prepared with the unitary $U$ (that will appear later in the question): $U\vert 0 \rangle = \vert b \rangle$.

This Hamiltonian is depicted as the "global" Hamiltonian and is defined such as its expectation value when the system is in the state $\vert x \rangle$ is $$ \langle x \vert H \vert x \rangle = \langle x \vert \left[ A^\dagger \left( \mathbb{I} - \vert b \rangle \langle b \vert \right) A \right]\vert x \rangle = \langle \psi \vert \left( \mathbb{I} - \vert b \rangle \langle b \vert \right) \vert \psi \rangle $$ where $\vert \psi \rangle = A \vert x \rangle$.

I may be wrong, but I think this expectation value can be interpreted as the "part" of $\vert \psi \rangle$ that is in the subspace orthogonal to $\vert b \rangle$.

The authors also define a "local" Hamiltonian: $$ H_L = A^\dagger U \left( \mathbb{I} - \frac{1}{n} \sum_{j=1}^n \vert 0_j \rangle \langle 0_j \vert \otimes \mathbb{I}_{\bar{j}} \right) U^\dagger A $$ where $U$ is a unitary matrix "preparing" $\vert b \rangle$ and $\mathbb{I}_\bar{j}$ is the identity operation on all qubits except the $j$-th one.

My question is: is there a interpretation of the formula for $H_L$ as I wrote above for $H_G$? I particularly need help for the interpretation of the sum, I would tend to say it is a projection on all the states that have their $j$-th qubit in the $\vert 0 \rangle$ state, but I am not sure.

EDIT: after more research, this notebook gave me a re-formulation of one of the terms: $$ \frac{1}{n} \sum_{j=1}^n \vert 0_j \rangle \langle 0_j \vert \otimes \mathbb{I}_\bar{j} = P = \frac{1}{2} \left( \mathbb{I} + \frac{1}{n}\sum_{j=1}^n Z_j \right) $$ where $Z_j$ is the $\sigma_z$ Pauli matrix applied to the $j$-th qubit, qubit indexes starting with $0$.

NOTE: the question is open, i.e. a possible answer could be "I am pretty sure there is no easy interpretation of that thing".

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    $\begingroup$ As DaftWullie pointed out in the comments to my deleted answer, $I - P$ isn't a projection operator indeed. However, I do believe that there should be a simple interpretation of that term in terms of a restricted/local Hamiltonians (it's a big deal in perturbation theory). For instance. $I - P$ acts non-trivially only on the subspace in which the $j$-th qubit is $|0\rangle$. Nonetheless, it is evident why when $|\psi\rangle \propto |b\rangle$ the energy (cost) function vanishes. $\endgroup$ – Sanchayan Dutta Dec 11 '19 at 15:21

To start with, let's look at $$\mathbb I - \frac{1}{n} \sum_{j=1}^n \vert 0_j \rangle \langle 0_j \vert \otimes \mathbb{I}_\bar{j} =\mathbb I - P.$$

We can rewrite $$\mathbb I = \frac{1}{n}\sum_{j=1}^n\mathbb I = \frac{1}{n}\sum_{j=1}^n\left(\left|0_j\rangle\langle 0_j\right| + \left|1_j\rangle\langle 1_j\right|\right)\otimes \mathbb I_\bar j$$ to get $$\mathbb I - P = \frac{1}{n}\sum_{j=1}^n\left|1_j\rangle\langle 1_j\right|\otimes \mathbb I_{\bar j}.$$

This is a normalised sum of projectors $$\mathbb I - P =\frac{1}{n}\sum_{j=1}^n\mathbb P_{1,j}$$ where $\mathbb P_{1, j}$ projects the $j^{\text{th}}$ qubit into state $\left|1\right>$. That is, $\mathbb I - P$ gives the proportion of qubits in state $\left|1\right>$, or more accurately, the expectation value of $\mathbb I-P$ with respect to some state gives the expected proportion of qubits to return $'1'$ when that state is measured. Equivalently, the expectation value of $P$ is the expected proportion of qubits to return $'0'$ when measured.

As $U\left|0\right> = \left|b\right>$, $U\left(\mathbb I - P\right)U^\dagger$ does with $\left|b\right>$ what $\mathbb I - P$ does with $\left|0\right>$.

Of course, the difficulty now is that $\left|b\right>$ can presumably be an entangled state, so explaining this in terms of "expectation value of the proportion of qubits returning $'1'$ upon measurement" isn't quite so straightforward.

However, asking the question 'are you measuring the state $\left|b\right>$?', $\left<x|H_L|x\right>$ is the expectation value of the proportion of qubits that would return 'no'.

An alternative way of thinking about it that might make more sense in this context is by realising that if $A\left|x\right> = \left|b\right> = U\left|0\right>$, $U^\dagger A\left|x\right> = \left|0\right>$ and so we arrive back at the expectation value of the proportion of qubits returning $'1'$ upon measurement of $U^\dagger A\left|x\right>$.

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    $\begingroup$ Very nice answer! It took me a little bit of time to understand it, but it seems like it solves my problem. The conclusions are coherent with the context and fit nicely, I am happy with this! Thanks a lot! $\endgroup$ – Adrien Suau Dec 17 '19 at 15:57

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