Is there any useful application to estimating the expectation value for an Ising model without magnetic field?

Question

In the same line of thoughts as this post, I am trying to understand better in which cases quantum computers could be useful to simulate materials under some constraints on what the quantum computer can do.

I consider a class of Hamiltonians among which we can find the Ising models without external magnetic field:

$$H=\sum_{i=1}^{\textrm{Poly}(n)} c_i P^{(i)}_Z \tag{1}.$$

$P^{(i)}_Z$ is an n-Pauli operator which is a tensor product of $\mathbb{I}$ and $Z$. $\textrm{Poly}(n)$ means that my sum contains a polynomial number of terms as a function of the number of qubits $n$. The Ising model in the absence of an external magnetic field, being written $H=\sum_{\langle i,j\rangle} J_{ij} Z_i Z_j$ is then one particular choice for $H$.

My question:

Let's assume that I am able to evaluate $\langle \psi | H | \psi \rangle$ efficiently, for any entangled state $|\psi\rangle$. Is there anything useful I could access in terms of physics with that?

More context

I know that one "hot question" about the Ising model is to be able to find its ground state. However this ground state in the absence of magnetic field is necessarily a tensor product (so it is not very useful to consider an entangled state in this case). Also, to find it, a brute-force approach would be to try exponentially many ansatz (hence even if $\langle \psi | H | \psi \rangle$ is "easy" to evaluate it is not enough to do something interesting).

That being said, it could be that in some cases it is interesting to study the properties of this Ising model with initial entangled state. I don't have any specific idea in mind but I don't find a reason why such cases would necessarily be un-interesting.

I hope it provides a little bit more context behind my question.

Answer 1

"I know that one "hot question" about the Ising model is to be able to find its ground state. However this ground state in the absence of magnetic field is necessarily a tensor product (so it is not very useful to consider an entangled state in this case)."

This is not only true for the ground state. All eigenstates of your Eq. 1 can be written as a tensor product of single spin states.

"Also, to find it, a brute-force approach would be to try exponentially many ansatze (hence even if ⟨ψ|H|ψ⟩ is "easy" to evaluate it is not enough to do something interesting)."

It's true that you can find the ground state by trying exponentially many ansatze and seeing which one gives you the lowest energy. This doesn't mean that finding the ground state is not enough to do something interesting. Finding the ground state of an Ising model can solve many problems such as factoring integers and finding Ramsey numbers along with many others. Perhaps you mean that simply being able to calculate ⟨ψ|H|ψ⟩ efficiently is not enough to find the ground state efficiently, which is correct because there's an infinite number of possible states |ψ⟩ in the Hilbert space. However, quantum computers can already calculate the ground state of H efficiently, without calculating ⟨ψ|H|ψ⟩ for all possible |ψ⟩ by brute force. Therefore, it would seem that by saying "constraints on what the quantum computer can do" you meant that the quantum calculator can do ⟨ψ|H|ψ⟩ efficiently but nothing else directly (i.e. it can't find the ground state efficiently).

In that case, it's still useful to be able to calculate ⟨ψ|H|ψ⟩ when |ψ⟩ is an arbitrary tensor product of single spin states (i.e. |ψ⟩ is not necessarily the ground state). This would tell you the energy of the ↑↓↓↑↑↑↓↑ configuration of electron spins, under the given Hamiltonian, for example. Beyond the realm of physics, there's also many applications of QUBO in classical computing, and being able to evaluate the energy for a particular configuration of {0,1}ⁿ could be helpful in many QUBO applications, assuming your the quantum calculator can evaluate ⟨ψ|H|ψ⟩ efficiently with respect to the number of spins, rather than with respect to the number of states in the Hilbert space.

"That being said, it could be that in some cases it is interesting to study the properties of this Ising model with initial entangled state."

If the state is entangled, then it is not an eigenstate of Eq. 1, so it's unlikely that it would have many useful applications in physics. The following code shows this for $Z_1Z_2$ acting on a Bell state, and can easily be verified here:

z1 = kron([1 0 ; 0 -1],eye(2)); z2 = kron(eye(2),[1 0 ; 0 -1]);
H = z1*z2;
psi = (1/sqrt(2))*[1; 0 ; -1 ; 0 ];
H*psi;
[H*psi psi]

The result is a matrix whose first column is H|ψ⟩ and the second column is |ψ⟩ and they are not scalar multiples of each other, so |ψ⟩ is not an eigenstate of H.

Most of physics concerns states that are eigenstates of their Hamiltonian, so applications of your quantum calculator are limited by that. States that are not eigenstates of their Hamiltonian can still exist (maybe not for long!) and they are sometimes called "resonance states", which are very unstable. Perhaps there's applications there, or elsewhere, but likely not anything worth a large investment (even fully-fledged "universal" quantum computers will be very limited in terms of practical applications that are useful to society).