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Is the transverse field Ising model time-reversal invariant? Specifically consider a non-integrable variant:

\begin{equation} H = -J \sum_i^{L-1} \sigma_i^z \sigma_{i+1}^z + g \sum_i^L \sigma_i^x + h \sum_i^L \sigma_i^z, \end{equation}

so that it has both a transverse and longitudinal field. $\sigma$s are the usual Pauli-matrices.

So, is it TRI?

Arguments for no:

1) The time-reversal operator is $T = K \prod_{j=1}^L (-i \sigma^y_j)$ where $K$ is complex conjugation in the $\uparrow,\downarrow$ basis. One can check that $[H,T] \neq 0$.

2) Simply recall the action of $T$ on spins. It flips $\vec{S} \to -\vec{S}$. So the transverse fields $\sigma^z$ and $\sigma^x$ flip, and the model is not invariant.

Arguments for yes:

1) One must recall that $g, h$ are external magnetic fields which are pseudovectors. So we have to flip them at the same time as flipping $\vec{S}$. So overall, yes, TRI.

2) The level statistics of the model obey well GOE (Gaussian orthogonal ensemble) statistics (see arxiv:1306.4306 for example), a class of random matrices that are supposed to describe time-reversal invariant Hamiltonians.

Puzzling issues:

1) If no, how to reconcile with GOE statistics?

2) If yes because of reason 1, that feels like it is cheating; we have to put in the way the constants $g,h$ transform by hand.

3) If yes, Kramers theorem says the spectrum should be doubly-degenerate for a chain of total spin half-integral. Quickly check in Mathematica for $L = 3,5,7\cdots$, this is untrue - the spectrum is not degenerate. Kramer's theorem doesn't hold.

So, yes or no? Or yes and no?

EDIT Also, I've heard this phrase alot: oh, the Hamiltonian is real so it is time-reversal symmetric.

Does that make sense??

I mean, if I had applied my transverse field in the $y$-direction, I don't expect any physical difference. It's just a magnetic field pointing a different way. But in the particular basis I've chosen, the model is now complex. So it has become non-time-reversal symmetric?

Oh is the statement "a real Hamiltonian is TRI" too cavalier?

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  • $\begingroup$ While it's true that a magnetic field is a pseudovector, that isn't the word you want to use in this context. Pseudovectors are defined as being even under parity inversion, but you want to say that magnetic fields are odd under time reversal. Similar idea though. $\endgroup$ – tparker Jan 12 '16 at 2:11
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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.

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  • $\begingroup$ Thank you. Could you elaborate why $T = K$ only and not $T = K \prod_i i \sigma_i^y$? This seems to go against my understanding of what the time reversal operator should do, namely, flip $\vec{S} \to - \vec{S}$. Furthermore, the answer given in this question physics.stackexchange.com/questions/78367/… also agrees with the TR operator I wrote down. $\endgroup$ – nervxxx Jan 11 '16 at 19:08
  • $\begingroup$ As I said, the physical spins indeed have $T=K\prod_i i\sigma_i^y$, and the Hamiltonian is not symmetric under this definition of $T$. If you insist on that this is the only acceptable definition of $T$, then end of story. However, to explain the properties of the spectrum (like the level statistics), one should allow generalizations of $T$. The reality condition only requires $T$ being anti-unitary, the $i\sigma_i^y$ part is important for $T^2=-1$ but otherwise not really essential. So if you choose $T=K$, then the Hamiltonian is invariant under this $T$. $\endgroup$ – Meng Cheng Jan 11 '16 at 19:37
  • $\begingroup$ I am sorry, I do not understand the logic behind defining $T = K$. Where does Reality $\implies$ $T = K$ come from? Maybe the resolution is that the so-called 'TRS' associated with the GOE ensemble is not actually the real, physical TRS; but rather, the fact that the Hamiltonian can be written completely real in an appropriate basis. In other words, if I can find an appropriate unitary transformation that brings a matrix into a completely real form, then it should (generically) be described by GOE stats. That's why rotating the transverse field from x to y is OK. Conversely, if I cannot $\endgroup$ – nervxxx Jan 11 '16 at 20:05
  • $\begingroup$ (continued) find a unitary that brings it real, then the matrix should generically be described by GUE stats (ignoring GSE or other ensembles). All these have nothing to do with whether the Hamiltonian is physically TRS or not, defined by $T = K \prod_i i \sigma_i^y$. Would that be reasonable? $\endgroup$ – nervxxx Jan 11 '16 at 20:06
  • $\begingroup$ This is what I meant. Reality means there is some anti-unitary symmetry and you are right that this means the Hamiltonian can become real in an appropriate basis, which are related to the original one by local unitary transformations. This is equivalent to say that there is some generalized $T$ in the original basis, which is $K$ times some unitary transformation, under which the Hamiltonian is invariant. $\endgroup$ – Meng Cheng Jan 11 '16 at 21:10
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I would argue that there are two different antiunitary operators that are both commonly called the "time-reversal operator," and you need to specify which one you mean. Under one definition ($T = K \prod_i i\, \sigma_i^y$), all three Pauli matrices change sign under $T$. This definition is more "physical" because spin is odd under time-reversal. (E.g. if you model spin semiclassically as a tiny loop of electric current, then time reversal reverses the direction of the current and so flips the direction of the magnetic dipole moment). Under this definition, the TQIM is not $T$-symmetric because the field term changes sign.

The second definition $(T = K)$ is a little less physical and messier conceptually, because the complex conjugation operator $K$ is basis-dependent, so $T = K$ is only defined in a particular basis (although it turns out that the resulting operator is antiunitary in any basis). Under this definition, $\sigma^y$ changes sign under $T$ but $\sigma^x$ and $\sigma^z$ do not, so the operation consists of a reflection across the $x-z$ plane in spin space. Now, we say that a Hamiltonian is $T$-symmetric if there exists a basis in spin space such that it is symmetric under this operator. (Or, as @MengChen pointed out, you can build the unitary $U$ that rotates the appropriate spin axis into the definition of $T$.) Under this definition, a time-reversal-invariant magnetic state corresponds to one that has coplanar order in spin space (in the $x-z$ plane in the appropriate basis). States with this property are easier to model numerically, because the wavefunction is purely real (in the right basis) and so arithmetic operation can be performed twice as fast. Under this definition, the QTIM is $T$-symmetric, because all the terms in the Hamiltonian lie in the $x-z$ plane in spin space.

In either case, as @MengChen pointed out, you do not change the sign of $h$, because the current generating $h$ is assumed to lie outside of the system and the time-reversal operator does not act on it. If you were to incorporate the source of the magnetic field into your Hamiltonian, then $T$-symmetry would be restored, but of course that would be a much more complicated problem. If you were to define time reversal to flip all external fields, then virtually any system would be $T$-symmetric (with the exception of a few exotic processes involving the weak nuclear force), so the concept would not useful.

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