Is any Hamiltonian time reversal symmetric?

Question

Of course it shouldn't be. Please shoot me down.

Let $H$ be a single-particle Hamiltonian. It is time reversal symmetric if there exists some $U_T$ satisfying:

$$ \begin{split} U_T^{\dagger}\, U_T &= 1, \, U_T^{*} \, U_T = \pm 1; \\ U_T^{\dagger} \, H^{*} \, U_T &= H. \end{split} $$

My understanding based on Schnyder et. al. is that I don't have to discuss specifically how spin, momentum or whatnot transform under time reversal. I can keep the state labels all abstract, and just focus on the above condition.

Being a Hamiltonian, $H$ should be Hermitian. It is therefore diagonalizable by some unitary $J$, and it has only real eigenvalues. Let $\tilde{H}$ be the diagonal form:

$$ J \, H \, J^{\dagger} = \tilde{H} = \tilde{H}^{*} = J^{*} \, H^{*} \, J^{t}. $$

Now if I just identify $U_T = J^{t} J$. This guy is obviously unitary, turns $H^{*}$ into $H$, and satisfies $U_T^{*} \, U_T = 1$. Thus one can define a time-reversal symmetry for any Hamiltonian as long as it is Hermitian.

I am very confused at this point. It will be greatly appreciated if anyone can tell me what goes wrong here.

--edit--

@hft suggests in the comment that my question is similar to this one.

Well, in that question, the answer reached was "if a Hamiltonian is unitarily equivalent to its complex conjugate, it is time reversal symmetric." Here I just showed that any Hamiltonian is unitarily equivalent to its complex conjugate.

Not only is that previous question not helpful, I have serious issue with that answer.

--edit 2--

My confusion is largely cleared thanks to @Andrew and @mikestone. Let me summarize my new-found understanding here.

As far as the mathematical classification goes, one must first define "time reversal" with respect to a particular basis and stick with it throughout. $U_T$ comes pre-determined, and all we can do is to test whether a Hamiltonian is invariant under this particular $U_T$. Physically speaking, we know what time reversal is before we have the Hamiltonian.

(Otherwise, as demonstrated above, I can always cook up a $U_T$ with all the correct properties for any Hamiltonian. The whole thing just becomes meaningless.)

Consider the physical example of a 4-site spinless fermion hopping problem with magnetic flux. The Hilbert space is spanned by $| n \rangle$, with periodic identification $| n + 4 \rangle = | n \rangle$. Given the physical meaning, we would want to take simply $U_T = 1$.

The Hamiltonian is $$ H = \sum_{n=1}^{4} t | n \rangle \langle n+1 | + h.c. $$ In the presence of a magnetic flux, $t$ is complex. Then this Hamiltonian is not invariant with $U_T = 1$.

While I can cook up some $\tilde{U_T}$ that seemingly keeps $H$ invariant, that "time reversal" operation mixes lattice sites. Is it a "hidden symmetry" of this particular $H$? Yes. Can we find an entire class of Hamiltonians that share the same symmetry? You bet. But is it a physically useful thing to do? Not quite. For one thing, it is easily broken by a site-dependent chemical potential, a rather mundane perturbation.

More importantly, if I decide that time reversal is instead given by that $\tilde{U_T}$, I have to stick with it. With this definition, the $t \in \mathbb{R}$ model becomes T-non-invariant instead. As @mikestone remarked, mathematically you will come up with a similar classification scheme anyway, but you may not want it that way for physical reasons.

Answer 1

The algebra here is, as far as I can see correct, but there is a conceptual issue that has confused me in the past. In @Vokaylop's construction we find a $U_T$ that is constructed for this particular hamiltonian. In other words $U_T$ depends on $H$. The Dyson-Altland-Zirnbauer classification is set up to descibe classes of Hamilonians, and so given a $U_T$ we want to find all matrices that satisfy $U_TH^*U_T^{-1} =H$ for that fixed $U_T$. For example the Gaussian Orthogonal ensemble class is that which given a fixed basis of the Hilbert space consists of all matrices that have real entries in that fixed basis. This class is the above construction for the choice $U_T={\mathbb I}$.

The choice $U_T={\mathbb I}$ may seem special, but we can write the condition $U_TH^*U_T^{-1} =H$ as $U_TH^tU_T^{-1} =H$ and under change of basis for $H$ we find that $U\to V^tUV$ and a complex symmetric matrix that squares to ubity can always be diagonalized to ${\mathbb I}$ in this way, so $U_T= {\mathbb I}$ is not a special case.

In other words the set of matrices that are real (and so symmetric) in the basis that gives @Vokaylop's diagonal $\tilde H$ are counted as "time reversal" invariant for his $J^tJ$, but this may not be a physical time reversal operation.

Answer 2

There’s no problem with your time-reversal symmetry definition or your observation about the Hermiticity of $H$ and its diagonalizability.

Which brings us to your choice of $U_T = J^t J$:

a) it does satisfy $U_T^† H^* U_T = H$, but not necessarily all the conditions for a time-reversal operator.

b) to be specific - the condition $U_T^* U_T = \pm 1$ isn’t guaranteed by this construction.

Time-reversal symmetry imposes additional constraints on the structure of the Hamiltonian beyond just Hermiticity.

Not all Hermitian Hamiltonians are time-reversal symmetric. A system with a magnetic field breaks time-reversal symmetry for example.

The existence of a time-reversal symmetry often implies degeneracies in the energy spectrum (Kramers degeneracy for half-integer spin systems), which not all Hamiltonians possess. So, while all time-reversal symmetric Hamiltonians are Hermitian, not all Hermitian Hamiltonians are time-reversal symmetric.

Answer 3

Your condition $U_T^\dagger H^* U_T = H$ is inconsistent with the notion that $U_T$ is a symmetry at all. A symmetry of $H$ means the operator representing that symmetry must commute with $H$. Therefore: $$ U_T^\dagger H U_T=U_T^\dagger U_T H = H \neq H^* $$ Perhaps you could clarify where exactly in the paper you retrieved that expression from, maybe you misread it or accidentally modified it.