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From Random-Matrix Theory, Hamiltonians are classified in three different ensembles depending on the spectrum statistics (Gaussian Orthogonal (GOE) , Gaussian Unitary (GUE), Gaussian Simplectic (GSE)). It is argued that the difference between Hamiltonians following GOE and those with GUE is that the later do not preserve Time Reversal Symmetry (TRS).

https://en.wikipedia.org/wiki/Random_matrix

If we consider a simple free electron model in the tight-binding approximation, we write its Hamiltonian in second quantization: \begin{eqnarray} H=-\sum_{\sigma=\uparrow,\downarrow,\langle i,j\rangle}c_{i\sigma}^{\dagger}c_{j \sigma}+\text{H.c.}+\sum_{\sigma=\uparrow,\downarrow,i}\varepsilon_{i\sigma}c_{i\sigma}^{ \dagger}c_{i\sigma}=H_0 + \sum_{\sigma=\uparrow,\downarrow,i}\varepsilon_{i\sigma}c_{i\sigma}^{ \dagger}c_{i\sigma} \end{eqnarray} where we will assume the on-site energies to depend on the spin as $\varepsilon_{i\sigma}=\sigma B_{i}$, i.e. for $\sigma=\uparrow$ we have $\varepsilon_{i\uparrow}=+B_{i}$ and for $\sigma=\downarrow$ we have $\varepsilon_{i\downarrow}=-B_{i}$, with $B_{i}$ a external field parameter that breaks translational symmetry (depends on the lattice site $i$).

Under TRS (anti-unitary transformation), if we call such transformation $T$, we must have $Tc_{j\sigma}T^{-1}=c_{j,-\sigma}^{\dagger}$, so that when we invert the arrow of time, we create a particle with opposite spin. The first term in $H_0$ is invariant under this transformation, but the second term is not, since the TRS will generally add an overall constant of energy and the minus sign in $\varepsilon_{i\sigma}$ gets swapped.

The question is, since TRS is in principle broken in such system, should its spectrum LSS follow a GUE rather than the GOE? If not, why is this the case? How would the situation change if $H$ is Hermitian, but with complex entries in $H_0$ instead? If I got this right, in principle any TRS operation can be completely defined by its action on the operators and written as $T=UK$, with $U$ a unitary matrix and $K$ complex conjugation (no transpose) $KOK^{-1}=O^{*}$ for an arbitrary operator $O$.

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2 Answers 2

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To discuss level statistics, first we consider whether the ensemble of Hamiltonians has any anti-unitary symmetry. Your Hamiltonian is invariant under the anti-unitary symmetry $T=K$ where $K$ is the complex conjugation. If there is no such symmetry, then the ensemble is GUE. GOE and GSE are distinguished by whether $T^2=1$ (GOE) or $-1$ (GSE). Since $K^2=1$, and there exists no time-reversal operation that squares to $-1$, the Hamiltonian should belong to GOE. If you replace $H_0$ with a more generic Hamiltonian with complex entries, then it goes to GUE.

Also notice that if we look at the actual spectrum of the second-quantized Hamiltonian $H$, since it is just free fermion the level statistics is very different from that of a random matrix (with the same Hamiltonian as the many-body Hilbert space).

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  • $\begingroup$ Thanks for the reply; it is in fact simple since $T=UK$ with $U=1$ is the simplest anti-unitary transformation, so you need to exhaust all possibilities. But normally, one encounters different definitions for $T$ as, for instance, in the Ising case where $U=e^{i\pi S_{y}}$ is chosen, if I am not mistaken. $\endgroup$ Jun 24, 2022 at 23:32
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Just a small note on your time-reversal transformation in a spin 1/2 system. In the second quantized picture, we have

\begin{equation} \begin{split} \hat{\mathcal{T}}\hat{c}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}_{i, \downarrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}^{\dagger}_{i, \downarrow} \\ \hat{\mathcal{T}}\hat{c}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}_{i, \uparrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}^{\dagger}_{i, \uparrow} \end{split} \end{equation}

However, as you point out, the second term will generally break TRS, i.e $\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} \neq \hat{H}$. As you are modelling a spin 1/2 Hamiltonian, $\hat{\mathcal{T}}^{2} = -1$ and $\hat{H}_{0}$ actually falls into the symplectic class (in the Altland-Zirnbauer classification, this is class AII). The onsite term (which looks equivalent to Zeeman term for a B-field in the z direction) ensures that this Hamiltonian no longer respects any anti-unitary symmetries and is indeed unitary (Altland-Zirnbauer class A).

Finally, you are correct that it is generally true that in the first quantized picture you can always write $T = \hat{\mathcal{T}}_{first quantized} = U_{T}K$.

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  • $\begingroup$ Thanks for the reply, but what can we say about its level statistics then? Shouldn't they obey GUE rather than GOE due to the break of TRS? This is where I am mostly confused. $\endgroup$ Jun 24, 2022 at 15:38

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