I was wondering if someone could give a simple explanation of what is meant by time reversal invariance. Is it analogous to spatial translational symmetry? If so, how? By spatial translational symmetry I mean the following. Suppose, for example, one has a solid consisting of an array of ions and electrons. If we pick a coordinate system we can write the Hamiltonian of the solid in terms of the coordinates of the ions and electrons. If we translate the origin of our coordinate system our Hamiltonian will be expressed using new coordinates, however, the Hamiltonian will have the same form. Is there a similar understanding for time reversal invariance?
[Physics] Meaning of Time Reversal Symmetry
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Starting with some background information from Wikipedia, we have that under time reversal the position is unchanged while the momentum changes sign.
In quantum mechanics we can express the action of time reversal on these operators as $\Theta\,\mathbf{x}\,\Theta^\dagger = \mathbf{x}$ and $\Theta\,\mathbf{p}\,\Theta^\dagger = -\mathbf{p}$. It is worth mentioning here that the time reversal operator, $\Theta$, is anti-unitary, which allows it to be expressed as $\Theta = UK$ where $U$ is unitary and $K$ is the complex conjugation operator.
As for the creation/annihilation operators used in second quantization the sign changes under $\Theta$ would suggest a transformation of $a_r \rightarrow a_r$ and $a_k \rightarrow a_{-k}$. If you are worried about the fact that $k$ represents a crystal momentum and not a true momentum you can just take the position transformation, which is perhaps more trustworthy, and use $a_k = \sum_r \, a_r \,\mathrm{exp}[ -\mathrm{i} k\cdot r]$ to verify $a_k \rightarrow a_{-k}$ directly.
Using these transformations you should be able to verify that the tight-binding Hamiltonian is invariant under time reversal in position and momentum space for a lattice with or without a basis. Keep in mind that you would generally take the complex conjugate of the coefficients in $H$, however in your case $t$ and $\epsilon_k$ are both real. Its important to remember though, mostly to make sure $H$ stays hermitian.
As far as your comment about $\sigma_y$, this is only necessary if you include spin. Spin changes sign under time reversal so $\Theta\,\mathbf{S}\,\Theta^\dagger = -\mathbf{S}$. In this case, we can formally write $\Theta = \mathrm{exp}[-i \pi J_y]\,K$, which is probably the relation you are alluding to.
According to J.J. Sakurai's Modern Quantum Mechanics one possible convention for the time-reversed angular momentum states is $\Theta | j,m\rangle = (-1)^m |j,-m\rangle$. This suggests that with spin indices the creation/annihilation operators transform like $a_{r,m} \rightarrow (-1)^m\,a_{r,-m}$ and $a_{k,m} \rightarrow (-1)^m \, a_{-k,-m}$ under time reversal. From what I understand, most spin Hamiltonians will be invariant under this transformation. An example when this is not the case would be in the presence of an external magnetic field which couples to the spins through a $\mathbf{S}\cdot \mathrm{B}-$like term.
It is interesting how even in the absence of an external field the groundstate of spin Hamiltonians can still sponanteously break the time reversal symmetry present in $H$, but rather than discuss this myself I will direct you to this very well written answer.
Indeed, one of the definitions of spontaneous symmetry breaking is in terms of its susceptibility:
Suppose we add a symmetry breaking perturbation $h \; \delta H$ to our Hamiltonian (as you do), if $$ \lim_{h \to 0} \lim_{N \to \infty} \langle m \rangle \neq 0 $$ then we say our system has spontaneous symmetry breaking.
(Note: $N$ is the number of spins in our system. Indeed, on a mathematical level, non-analyticities can only arise in the thermodynamic limit.)
What is special is that any arbitrarily small perturbation will do. Imagine you have a million spins. If the state is originally in a symmetric state (i.e. not symmetry broken yet), then even if I just apply an arbitrarily small magnetic field on a single spin, the whole system will choose that orientation.
You suggest that the fact one in principle needs the environment to 'make the choice' that this is not really spontaneous. It is true that in that philosophical sense of the word, the direction of magnetization is not 'spontaneous'. But what can be called spontaneous in the universe? If I perfectly balance an egg, then the direction it will eventually roll when it loses its balance is spontaneous (or not spontaneous) in exactly the same sense. And note that once the egg has rolled down (and stopped), the tiny perturbations in the air which influenced its original direction are now no longer sufficient to change its position. I.e.: after the `spontaneous' process, the system is now stable.
The same thing happens in the above magnet: once it has chosen a direction of magnetization, then changing the applied magnetic field on that single spin I mentioned before will not change the total magnetization. So in that sense it is not true that it is so susceptible! One needs to apply an extensive magnetic field (i.e. a field that acts on most of the spins) to change the direction of the magnetization.
That is what is so funny about these systems:
An arbitrarily small perturbation can create a magnetization, but it cannot change it!
On a more quantum-mechanical note, if one has a Hamiltonian whose ground state should display spontaneous symmetry breaking, then if one takes the ground state to be in a symmetric superposition (which one can always do), then this state has ridiculously long entanglement. These are called cat states (in reference to Schrodinger's cat). This is a natural consequence of the above: an interaction with a single spin has to influence all spins at once, which is only possible if every single spin is entangled with every other spin. An example is the state $|\uparrow \uparrow \uparrow \cdots \rangle + |\downarrow \downarrow \downarrow \cdots \rangle$. (Indeed: an interaction with a single spin will collapse this 'cat state' to a product state, and then it is clear that any subsequent single-spin interaction cannot flip the state to the other product state.) Indeed, the way symmetry breaking phases are classified in one spatial dimension is in terms of these entanglement properties [Schuch et al., 2010].
Best Answer
Time reversal essentially means a system looks the same if you reverse the flow of time. The only difference beeing that things like velocity go in the opposite direction. In condensed matter systems it is represented as a unitary matrix times complex conjugation $\mathcal{T} = U\mathcal{K}$. A simple system that follows T-symmetry would be a system described by a real Hamiltonian (if we're ignoring spin, anyway.) If you include spin, for spin 1/2 particles it is represented as $\mathcal{T}=i\sigma_y \mathcal{K}$ and then we can have certain complex Hamiltonians as well.
As examples, a quantum spin Hall insulator is a system that preserves T-symmetry (because of the spins and their chirality when we reverse time we see the same system as before). A system which breaks T-symmetry is a ferromagnet. In this case the spin reversal is again the culprit, but because we do not see the same system as before, it has broken the symmetry. Hope this makes it a bit more clear!