I was reading this paper which introduces a mapping from a qubit Hamiltonian to an Ising model. Firstly, the first step of the mapping seems to assume that we know an eigenstate of the system (correct me here because it seems unlikely in practice). Below, is the mention of the first step : enter image description here

Secondly, such mapping seems extremely costly if we look at their complexity.

My question is: would this method be considered interesting for practice compared to classical methods and how can this be implemented on practical examples?

  • $\begingroup$ I am adding another link : docs.google.com/… $\endgroup$
    – cnada
    Nov 4, 2018 at 14:47
  • $\begingroup$ What is meant by "Classical Methods" later in your question? $\endgroup$ Nov 4, 2018 at 14:56
  • $\begingroup$ Any classical method/algorithms that people in the field use for such purpose. $\endgroup$
    – cnada
    Nov 4, 2018 at 15:00

1 Answer 1


In this paper, the authors use the Ising model to simulate the electronic structure Hamiltonian.
The electronic structure Hamiltonian after the Jordan-Wigner or Bravyi-Kitaev transformation (which the authors of this paper did use) has quadratic (and sometimes higher-order) terms containing $\sigma_x, \sigma_y$, and $\sigma_z$, but the Ising model does not have any quadratic terms containing $\sigma_x$ in any way.

It is possible to efficiently simulate any Hamiltonian using a Hamiltonian that has quadratic terms containing $\sigma_x$, as proven by Biamonte & Love. However, since the Ising Hamiltonian does not have such terms, which are required to simulate the electronic structure Hamiltonian efficiently, the method in the paper you mentioned, is not capable of efficiently finding the ground state of the electronic structure Hamiltonian.

  • $\begingroup$ What is your definition of efficiency in this kind of case? $\endgroup$
    – cnada
    Nov 5, 2018 at 18:35
  • $\begingroup$ Polynomial time $\endgroup$
    – user5019
    Nov 5, 2018 at 18:37
  • $\begingroup$ So correct me if I am misinterpreting, their mapping to an Ising Hamiltonian is certainly not efficient to get and that is what you mean by not efficient. But finding the ground state here is done on a quantum annealer, but what kind of complexity are we looking at in that case? Qubit efficiency right? $\endgroup$
    – cnada
    Nov 5, 2018 at 18:50
  • $\begingroup$ Their mapping to Ising is certainly efficient. It does not take long to do the mapping. The problem is that finding the ground state with a quantum annealer, will take exponentially long with the number of qubits. Therefore the number of qubits needs to be something like O(log(N)) for N basis functions in the electronic structure problem in order for the runtime (which is exponential in the number of qubits) to be polynomial in N. The problem is that they need more than N qubits, which is much more than O(log(N)). Sorry if it's not clear. $\endgroup$
    – user5019
    Nov 5, 2018 at 18:55
  • $\begingroup$ But in their mapping, the fact that the first part is based on an eigenstate, isn't it a bit problematic? $\endgroup$
    – cnada
    Nov 5, 2018 at 20:37

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.