Graduate Two equivalent statements of time reversal symmetric Hamiltonian

[hokhani]

TL;DR

How $[H,\Theta] =0$ is equivalent to $H=H^*$?

Time reversal invariant Hamiltonians must satisfy $[H,\Theta]=0$ where $\Theta$ is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as $H=H^*$. How these two conditions are identical?

 

[pines-demon]

TL;DR: How $[H,\Theta] =0$ is equivalent to $H=H^*$?

Time reversal invariant Hamiltonians must satisfy $[H,\Theta]=0$ where $\Theta$ is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as $H=H^*$. How these two conditions are identical?

Be careful, what is the argument? If you write $H(t)$ or $H(\mathbf r)$ it is not the same as $H(\mathbf k)$ ($\mathbf k$ being the momentum/wavenumber). In Fourier space $H(-\mathbf k)=H^*(\mathbf k)$ is an equivalent condition to time-reversal invariance.

 

[hokhani]

$H(t)$ or $H(\mathbf r)$ it is not the same as $H(\mathbf k)$ ($\mathbf k$ being the momentum/wavenumber).

From the Eq. (7.18) in that text, it seems that by $H$ the authors mean the Hamiltonian in the position space. Also, as I remember, it was discussed previously in one of my last threads that $H$ is an operator and writing in the functional form $H(variable)$ doesn't make sense.

 

[pines-demon]

From the Eq. (7.18) in that text, it seems that by $H$ the authors mean the Hamiltonian in the position space. Also, as I remember, it was discussed previously in one of my last threads that $H$ is an operator and writing in the functional form $H(variable)$ doesn't make sense.

Sorry this gets confusing because we mix various stuff in second quantization, your $H$ is in some basis, let me use hats for operators. When I write $H(\mathbf k)$ I mean that you can write the usual Hamiltonian as $\hat H=\sum_k \hat{c}^\dagger_k H(k) \hat{c}_k$ with some creation operators $\hat c^\dagger_k$. Note that $H^*$ makes no sense if $H$ is an operator.