2 added 157 characters in body edited Jan 11 '16 at 22:00 tparker 25.7k11 gold badge5555 silver badges135135 bronze badges 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-reveralreversal-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 iswould be $$T$$-symmetric, so the concept is not useful (with the exception of a few exotic processes involving the weak nuclear force), so the concept would not useful. 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 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. 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. Now, we say that a Hamiltonian is $$T$$-symmetric if there exists a basis in spin space 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-reveral-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 define time reversal to flip all external fields, then virtually any system is $$T$$-symmetric, so the concept is not useful (with the exception of a few exotic processes involving the weak nuclear force). 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. 1 answered Jan 11 '16 at 21:53 tparker 25.7k11 gold badge5555 silver badges135135 bronze badges 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 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. 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. Now, we say that a Hamiltonian is $$T$$-symmetric if there exists a basis in spin space 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-reveral-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 define time reversal to flip all external fields, then virtually any system is $$T$$-symmetric, so the concept is not useful (with the exception of a few exotic processes involving the weak nuclear force).