Parity measure of a fermionic state

DMRG and Numerical Methods
1

Hi!,
I am trying to measure the parity of a fermion state, however first of all I have the following problem:

If I start with (for example) with:

N = 100 #number of sites
Npart=N+1 #number of particles

sites = siteinds("Electron", N;
                conserve_qns = false,
                conserve_sz = false,
                conserve_nf = false,
                conserve_nfparity = true,
                qnname_sz = "Sz",
                qnname_nf = "Nf",
                qnname_nfparity = "NfParity")

state = ["Emp" for n in 1:N]
p = Npart
for i in N:-1:1
  if p > i
    state[i] = "UpDn"
    p -= 2
  elseif p > 0
    state[i] = (isodd(i) ? "Up" : "Dn")
    p -= 1
  end
end

psi0 = productMPS(sites,state)

and measure the number of particles of psi0, the result is Npart = 101

But when i do a DMRG calculation,

energy, psi = dmrg(H, psi0, sweeps; observer=obs, outputlevel=outputlevel)

and measure the number of particles,

Number_p = sum(expect(psi,"Ntot"))

the result is Number_p = 100.00000000506148.

As the Hamiltonian does not conserve the number of particles I would not expect to find Number_p =101, but yes an odd number of particles because I’m imposing the conservation of the parity.
I’m not sure if my problem is conceptual or computational,

From already thank you very much!

2

Welcome to the new forum. Your question is a really interesting one & I would not say the answer is obvious.

I think I can only guess, really. My best guess is based on the fact that, first of all sum(expect(psi,"Ntot")) is only computing the average number of particles. So the distribution of particle number could have a spread to it. Like it could be the case, hypothetically, there’s an equal probability of either 99 particles or of 101 particles. Then the average would be 100 but the state would never fluctuate into an even-parity subspace.

One thing you haven’t said is what Hamiltonian psi the ground state of. I wonder what would happen if you added a chemical potential? You might be able to get the ground state to have other average particle numbers besides 100, and also some fractional values perhaps.

The only way I can think of to really understand this fully is to study the wavefunction in more detail. Conceptually one could imagine projecting psi onto different particle numbers, and in this way work out the distribution over particle number. But it’s not so easy to make the operator that projects onto fixed particle number. There may be a good trick to do it but it’s not obvious to me.

3

Thanks for the answer Miles!

I get your point, I hadn’t really thought of it that way but it must be correct. The Hamiltonian is the next:

H0 = \sum_{j} \left[\left(t \textbf{c}_{j}^{\dag}\textbf{c}_{j+1} - \frac{\mu}{2} \textbf{c}_j^{\dag} \textbf{c}_{j} + \Delta c_{j,\uparrow}^{\dag}c_{j+1,\downarrow}^{\dag} +i\alpha \textbf{c}_j^{\dag}\sigma_2\textbf{c}_{j+1} + h.c. \right) + J \textbf{c}_j^{\dag}\sigma_3\textbf{c}_j \right]

I am going to continue meditating what you told me and if I find a way that works I will share it here.

Greetings!

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