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.

 

[hokhani]

Note that $H^*$ makes no sense if $H$ is an operator.

Thanks, it was the key point here. If we work in the basis $|q\rangle$ where $\Theta |q\rangle =|q\rangle$, then we have:
$\Theta H |\rangle=\sum_q H_q^* \langle q|\rangle^* |q\rangle$ and
$H\Theta|\rangle=\sum_q H_q \langle q|\rangle^* |q\rangle$
so, $[H,\Theta]=0$ corresponds to $H_q=H_q^*$ and vice versa.

 

[pines-demon]

Thanks, it was the key point here. If we work in the basis $|q\rangle$ where $\Theta |q\rangle =|q\rangle$, then we have:
$\Theta H |\rangle=\sum_q H_q^* \langle q|\rangle^* |q\rangle$ and
$H\Theta|\rangle=\sum_q H_q \langle q|\rangle^* |q\rangle$
so, $[H,\Theta]=0$ corresponds to $H_q=H_q^*$ and vice versa.

For that your Hamiltonian has to be diagonal in $q$. Also you are assuming properties of $\Theta$ if not I don't get how are you getting the complex conjugate, maybe I am missing a step.

 

[hokhani]

For that your Hamiltonian has to be diagonal in $q$. Also you are assuming properties of $\Theta$ if not I don't get how are you getting the complex conjugate, maybe I am missing a step.

Right, since in that text the Hamiltonian were represented in position space where $H_r$ is diagonal, I implicitlly assumed diagonal q-representation for $\hat{H}$ as well as properties of $\Theta$, $\Theta(c|q\rangle)=c^*|q\rangle$.

 

[pines-demon]

Right, since in that text the Hamiltonian were represented in position space where $H_r$ is diagonal, I implicitlly assumed diagonal q-representation for $\hat{H}$ as well as properties of $\Theta$, $\Theta(c|q\rangle)=c^*|q\rangle$.

Please copy the Hamiltonian here to understand your question better.

 

[hokhani]

Please copy the Hamiltonian here to understand your question better.

Eq. (7.18) in the text, introduces the time reversal dependent Hamiltonian as: $H\psi(r)=[-\frac{1}{2m}(\nabla_r+ieA)^2+V(r)]\psi(r)$. So, here $H$ is in the positon space as $H=H_r=\langle r|\hat{H}| r \rangle.$

 

[pines-demon]

Eq. (7.18) in the text, introduces the time reversal dependent Hamiltonian as: $H\psi(r)=[-\frac{1}{2m}(\nabla_r+ieA)^2+V(r)]\psi(r)$. So, here $H$ is in the positon space as $H=H_r=\langle r|\hat{H}| r \rangle.$

But this $H$ still seems like an operator what does $\nabla_r^*$ mean?

 

[pines-demon]

Look at this notes from Berkeley:

• https://bohr.physics.berkeley.edu/classes/221/1112/notes/timerev.pdf

They do the usual time-reversal operator properties in first quantization and why it acts like a complex conjugation. Note they are careful to never write $H^*$ or for any operator. I think Bruus is abusing with the notation.

 

[hokhani]

But this $H$ still seems like an operator what does $\nabla_r^*$ mean?

I think if we accept the complex conjugation only acts on the numbers there is no ambiguity in this term, because at least the explicit form of the operator is clear. In contrary, the form of $H^*$ is representation-dependent and while the representation is not specified, there is no sense of complex conjugation.

 

[hokhani]

Look at this notes from Berkeley:

• https://bohr.physics.berkeley.edu/classes/221/1112/notes/timerev.pdf

I think Bruus is abusing with the notation.

The notes you sent formalizes the time reversal in a nice approach in accord with Sakurai.
The many body book of Bruus is a good reference which covers condense matter topics. However, in spite of several editions it contains a few such problems. Do you know any other alternative text at this level?

 

[pines-demon]

The notes you sent formalizes the time reversal in a nice approach in accord with Sakurai.
The many body book of Bruus is a good reference which covers condense matter topics. However, in spite of several editions it contains a few such problems. Do you know any other alternative text at this level?

Try Fetter&Walecka or Coleman.