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The Hamiltonian for a 1D Hubbard model reads $$H= -t \sum_i c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_{i} + U\sum_i n_{i\uparrow}n_{i\downarrow}.$$ The two parameters $t$ and $U$ for the hopping and onsite interaction are usually both assumed to be of the order of electron Volts. If one wants to introduce an external magnetic field, a Zeeman coupling term is introduced: $$H= -t \sum_i c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_{i} + U\sum_i n_{i\uparrow}n_{i\downarrow} + h_B \sum_i (n_{i\uparrow} - n_{i\downarrow})$$

If this term should have any significance, the parameter $h_B$ should be of the same order of magnitude as $t$ and $U$, so also in the eV range.

Well, the Zeeman coupling term in SI units usually reads (see e.g. here) $$H_{Z}= g_s \mu_B\; B \; \hat{s}_z = \frac{1}{2} g_s \mu_B\; B \; (\hat{n}_\uparrow - n_\downarrow)$$ With $\mu_B$ the Bohr magneton. One can then identify $h_B \hat{=} \frac{1}{2} g_s \mu_B\; B = \mu_B\; B$ (because $g_s=2$).

If now $h_B=1$eV, then $B$ would have to be of the order of $20000$T! That appears to be completely unreasonable.

Where did I make my mistake?

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You presume that the order of magnitude of a relevant magnetic field should be the same as the order of magnitude of the other parameters. That intuition is correct if there is only one parameter in play, e.g. if $U$ is very small. Indeed let's take $U=0$ such that we have a free fermion hopping problem. At half-filling, we have two (disconnected) half-filled bands (one for spin up, the other for spin down). If we now start adding a small magnetic field, it indeed won't do too much: it will only slightly favor spins along the magnetic field. To really cause a phase transition, i.e. to polarize our spins, we need to make the magnetic field of the order of the band width (which completely forces one of the two bands to be full, the other empty). So in that case if $t$ is of the order eV, then so must the magnetic field be, which corresponds to a ridiculously large magnetic field as you point out.

But interactions change everything! Indeed, it's well-known that for large $U$ and at half-filling, our effective Hamiltonian becomes the Heisenberg antiferromagnet $H = \frac{4t^2}{U} \sum \mathbf{S}_i \mathbf \cdot \mathbf{S}_{i+1}$. To fully polarize this system, we need to only apply a magnetic field of the order of $\sim \frac{t^2}{U}$ (more precisely and generally, the critical magnetic field at half-filling is $2\left( \sqrt{t^2+U^2} - U \right)$, ref: page 198 of The one-dimensional Hubbard Model).

Still, despite $\frac{t^2}{U}$ being a smaller quantity, it typically still corresponds to quite large magnetic fields. Even if we take the rather weak $t = .5$ eV and strong $ U = 10$ eV, we would need a magnetic field corresponding to $0.025$ eV, or $\sim 500$ Tesla! So part of the answer is: you're right, the magnetic fields needed to cause a phase transition are quite big (at least at half filling, it goes down quite a bit at lower filling, again see the above reference). Note though: that is to actually get a phase transition, this of course implies you can already start seeing appreciable effects with lower fields.

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