Basically, the answer is yes: $H$ is TRI because it is real. Reality condition really means that the Hamiltonian obeys a certain anti-unitary symmetry. In this case, the time-reversal operation is simply $T=K$ where $K$ is the complex conjugation. It is not the usual one($T=K\prod_i i\sigma^y_i$), and in particular $T^2=1$, so there is no Kramers' theorem and the spectrum is not doubly degenerate. The fact that level statistics follows GOE of course is a consequence of the reality condition. In fact, I think if there was a $T^2=-1$ time-reversal symmetry, the statistics would follow a different ensemble (the symplectic one, I believe).
You asked what if one changes the transverse field $g\sum_i \sigma^x_i$ to the $y$ direction. In that case, the two Hamiltonians are unitarily related (i.e. a $\pi/2$ spin rotation $U_z$ around $z$ would bring it back). Let me call the Hamiltonian with $x$ transverse field $H_x(g)$ where $g$ is the transverse field, and with $y$ transverse field $H_y(g)$. Define $U_z=e^{i\pi \sum_i\sigma^z_i/4}$, then it is easy to check that $U_z H_y(g) U_z^{-1}= H_x(g)$. Since we know $H_x^*(g)=H_x(g)$, we can easily find $H_y^*=U_z^2 H_y U_z^{-2}$. Therefore one just has to redefine the time-reversal symmetry to be $T=K U_z^2$. If you really want to break the reality condition, in a way that can not be fixed by additional unitary transformations, then one needs to turn on transverse fields along all three dimensions.
Last comment on your "Arguments for yes": the first argument you gave, namely one also flips the external parameters, does not work. In this way, there would be no time-reversal symmetry breaking, except the CP violation in the fundamental processes! When we talk about the symmetry of a Hamiltonian, we should just treat the system on its own, not with all the external devices that generate the various terms -- unless you want to consider the dynamics of these devices, but then it is a different Hamiltonian.
You are right!
If you consider whatever is producing the $\vec{B}$ field as part of the system then the whole thing is symmetric under time reversal. It is only when the $\vec{B}$ field is considered as something external (on which the time-reversal operator does not act) that the symmetry is broken.
It doesn't seem physical for K to act on one part of the system.
I don't think a 'universal' time-reversal operation is any more (or less) physical than one which acts on only a certain subsystem (can you make time run backwards?).
In practice, what time-reversal symmetry means is something like this. (I will give a classical description for ease of intuition, a quantum version can be formulated along similar lines.) Suppose I have an interacting system of $N$ particles which I prepare in state $(\vec{q}_1(0),\vec{p}_1(0))$, and after time $t$ is found to be in state $(\vec{q}_1(t),\vec{p}_1(t))$. I now prepare a second system with identical particles in state $(\vec{q}_2(0),\vec{p}_2(0))=(\vec{q}_1(t),-\vec{p}_1(t))$ (i.e. with the same positions and opposite velocities of the final state of system 1). If the interaction is time-reversal symmetric then after running the second experiment for time $t$ it is found that $(\vec{q}_2(t),\vec{p}_2(t))=(\vec{q}_1(0),-\vec{p}_1(0))$, so that the dynamics of the second experiment are exactly the reverse of those of the first experiment.
Now, if I know that the interaction is electromagnetic, and there are no external forces, then I know the results must obey time-reversal symmetry (as the laws of electrodynamics have this symmetry). However, if I choose to impose an external magnetic field on the first experiment then when I perform the second experiment I have the choice of whether to reverse this field or not.
- If I choose to reverse the field then the results show time-reversal symmetry.
- If I choose not to reverse the field then the results do not show time-reversal symmetry.
Statements 1 and 2 seem equally meaningful (and equally 'physical') to me.
Best Answer
I think the answer should be 'no'.
Because when we introduce the antiunitary time-reversal(TR) opeartor $T$ for spin-system, it should satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$ since angular-momentum should be sign reversed under TR(due to the classical correspondence). Thus, spin-spin interactions like $\mathbf{S}_i\cdot\mathbf{S}_j$ are invariant under TR.
The TR operator $T$ for the $N$-spin-$1/2$ system has a form $T=(-i)^N\sigma_1^y\sigma_2^y...\sigma_N^yK$, where $K$ is the conjugation operator. You can easily check that $T$ is antiunitary and satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$. Furthermore, $T^2=(-1)^N$, so for odd-number spin system(including single spin case), if the Hamiltonian has TR symmetry, we will arrive at the well-known Kramers theorem.