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I have studied the example of the simulation of LiH molecule in Learning Quantum Computation using Qiskit. It is available on this link: https://qiskit.org/textbook/ch-applications/vqe-molecules.html#VQE-Implementation-in-Qiskit . I don't understand the following cell:enter image description here

Why freeze_list and remove_list are initialized in this way? How does this procedure work?

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For the freeze list, as KAJ226 mentioned, it's the core orbital of LiH molecule. In this case it would be the 1$s$ orbital of Li and we only care about the 2$s$, 2$p_x$ orbital of Li, 1$s$ orbital of H, total 6 qubits considering the spin-degeneracy. It is revealed that frozen core approximation does not make any significant error in a quantitative manner.

I must emphasize that the remove list must be chosen with great care. So your question is why remove list is chosen to be [-3, -2]. Well, it's just the tutorial of qiskit and I don't know if you tried it or not but using such remove list on the VQE calculation of BeH2 molecule cause undesirable kink on the energy versus interatomic distance plot.

So the answer is clear. You could test several remove list on your system to choose the best remove list for qubit reduction. It's my hypothesis(not tested) but I think such orbital contained in remove_list would be Li 2$p_x$ orbital, since it is usually unoccupied in a ground state of LiH molecule.(So it does not contribute any bonding of LiH one)

Thankfully, there already is a code in qiskit chemistry tutorial that clearly shows how the choice of remove list affects on the BeH2 molecule VQE calculation. See link: https://github.com/qiskit-community/qiskit-community-tutorials/blob/master/chemistry/beh2_reductions.ipynb You could easily find that the choice of remove list dramatically contributes to the accuracy of result energy.

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The freeze list indicates frozen of the core orbital. If you execute: molecule.core_orbitals you will get the output: [0]. So that is why the freeze list initialize that way.

As for the other two orbitals, [-3,-2] in the remove_list, this have to do with some chemistry intuition or calculation to determine that it has no effect to the overall problem. Thus they can be removed without any consequences to the answer. This will save us to have to use additional qubits.

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  • $\begingroup$ Thanks. My doubt is about the correspondence between this numbers and the orbitals. Why the numbers -3 and -2? Does it depend on the particular mapping (parity in this case)? $\endgroup$ Mar 5 at 8:43
  • $\begingroup$ I don't think it depends on the mapping actually. $\endgroup$
    – KAJ226
    Mar 5 at 15:57

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