Z2 symmetry in the hydrogen molecule when mapping to qubits

[Arne]

TL;DR Summary

When mapping a hamiltonian to qubits, one can reduce the number of qubits due to symmetries. How does an H2 molecule exhibit a Z2-symmetry?

Hello!

When using a Jordan-Wigner-mapping or parity-mapping to map the hydrogen molecule $\mathrm{H}_2$ with two electrons and 4 spin-orbitals to 4 qubits, it is possible to reduce the number of qubits down to two [1,2,3]. The reason is apparently that the molecule has a discrete $\mathbb{Z}_2$-symmetry. I'm wondering where this symmetry is present. Is it just that the molecule is mirror symmetric, since the two nuclei are the same? Or has it something to do with the spins, and we might flip the direction of all spins? Or is it rather that we maybe might occupy all unoccupied states and vice versa? In one reference I am reading it says the $\mathbb{Z}_2$-symmetry is due to fermionic particle conversation [3], but I don't understand where i can find the $\mathbb{Z}_2$-symmetry in this case. I'm sure some of these possibilities are related.

Thanks for your answers!

Regards.

[1] https://arxiv.org/abs/1208.5986 (describes Jordan-Wigner- and Parity mapping)
[2] https://arxiv.org/abs/1510.04048 (describes the reduction of qubits, but doesn't really talk about symmetries)
[3] https://arxiv.org/abs/1701.08213 (especially section VII, generalizes [2])
Edit: As it happens way to often to me, I asked the question when I had the feeling that I was really stuck, but a few moments after submitting somehow I actually get some better understanding of what the answer might be. Reference [3] clarifies, that non-relativistic molecular Hamiltonians conserve the number of particles with a fixed spin orientation, hence the number operator for spins with a certain orientation commutes with the Hamiltonian
$$[H,N_\uparrow] = [H,N_\downarrow] = 0 .$$
I'm not completely sure where the $\mathbb{Z}_2$-symmetry comes into play, but it might just be, that the physics are the same when flipping all the spins. Of course I would appreciate additional comments that might deepen my understanding into this matter :)

 

[eljose79]

Thank you for your question regarding the presence of a \mathbb{Z}_2-symmetry in the mapping of the hydrogen molecule to qubits. Your question is a very interesting one and it shows that you have a good understanding of the topic.

To answer your question, the \mathbb{Z}_2-symmetry in this case refers to the conservation of fermionic particles with a fixed spin orientation. This means that the Hamiltonian of the molecule commutes with the number operator for spins with a certain orientation. In other words, the physics remain the same even when we flip all the spins.

This symmetry is present due to the fact that the molecule is composed of two identical nuclei and two electrons with opposite spins. This leads to a mirror symmetry and the conservation of fermionic particles with a fixed spin orientation.

I hope this answers your question and helps deepen your understanding of the topic. If you have any further questions, please do not hesitate to ask.