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Why does the $\pi$-flux state have time-reversal symmetry?

+ 3 like - 0 dislike
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It's known that the $\pi$-flux state of the antiferromagnetic Heisenberg model on the square lattice is an important concept. The $\pi$-flux state is described by the (simplified) mean-field Hamiltonian $$H=t_1f_{1\sigma}^\dagger f_{2\sigma}+t_2f_{2\sigma}^\dagger f_{3\sigma}+t_3f_{3\sigma}^\dagger f_{4\sigma}+t_4f_{4\sigma}^\dagger f_{1\sigma}+H.c.$$, where $t_i=\left | t \right |e^{i\frac{\pi}{4}}(i=1,2,3,4)$, and the spin-1/2 operator is $\mathbf{S}_i=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$.

It's obvious that the mean-field Hamiltonian $H$ is not invariant under time-reversal operation( $T$ ), say $H\neq THT^{-1}$, and $H$ is also not $SU(2)$ gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$. So due to what reason, the projected spin-state $\psi_{spin}=\hat{P}\psi_{MF}$ is time-reversal invariant? Where $\psi_{MF}$ is the ground state of the mean-field Hamiltonian $H$ and $\hat{P}=\prod (2\hat{n}_i-\hat{n}_i^2)$ is the projection to the spin subspace.

Remarks: Here the effect of translation(with one lattice spacing along the $\hat{x}$ or $\hat{y}$ direction) on the Hamiltonian $H$ is the same as the effect of time-reversal $T$. Thus, if the spin-state $\psi_{spin}$ has $T$ symmetry, it must also have the translation symmetry.

Thanks in advance.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
asked Nov 23, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]
Most voted comments show all comments
Maybe one of the tags could be traded for research-level ?

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Dimensio1n0
@DIMension10 Yes, this question is not explicitely related to neither superconductivity nor gauge theory. Also, I believe $t_i$ is ill-defined. A better definition should be $t_{n}=te^{\mathbf{i}\pi/4}$, for all the $n$, and $\mathbf{i}^{2}=-1$. Tell me if I'm wrong.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
@ Oaoa Isn't my definition of $t_i$ the same as yours?

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
@ Oaoa Here we enlarged the spin Hilbert space by introducing the spinon operators $f_{i\sigma}$, and hence introduced many unphysical states(gauge redundancy). And to get the physical spin-state, we must perform the projection on the mean-field states in the end. This high-energy gauge structure is known as SU(2).

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
@K-boy Sorry for being hard sometimes. Your notations are not exactly wrong, they are just confusing, since $i$ appears as an index for $t_{i}$ and as the imaginary number such that $i^{2}=-1$ in the exponential. I prefer to define the complex quantity with a bold faced letter $\mathbf{i}$. Thanks for your comment on SU(2) gauge structure. Nevertheless, your question is not really on the gauge aspect, nor on the superconductivity aspect of the article you cited in your question, whereas DiMension10 was looking for a tag to erase, hence my previous remark. Please feel free to re-edit the ...

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
… the Dimension10's modification(s) if you wish. Please also see the modification of my answer below about the transformation of the operators, and feel free to comment further in any case.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
@ Oaoa Dear Oaoa, my question is essentially interested in the symmetry of the projected spin-state $\psi_{spin}$ rather than the symmetry of the mean-field Hamiltonian $H$ or its ground-state $\psi_{MF}$. Just because different mean-field ansats of $t_i$(and hence different mean-field Hamiltonians) may have the same projected spin-state, the projected spin-state $\psi_{spin}$ can have more symmetries than the mean-field Hamiltonian $H$. So the nature of my question is indeed a gauge problem.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy

2 Answers

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I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with

$$H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4} & 0\\ 0 & e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4}\\ e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4} & 0 \end{array}\right)$$

and $F^{\dagger}=\left(\begin{array}{cccc} f_{1}^{\dagger} & f_{2}^{\dagger} & f_{3}^{\dagger} & f_{4}^{\dagger}\end{array}\right)$. Then, one has

$$H_{\pi}=\dfrac{t}{\sqrt{2}}\left[\left(1+\tau_{x}\right)\otimes\eta_{x}-\left(1-\tau_{x}\right)\otimes\eta_{y}\right]$$

where the $\eta$ and $\tau$ are the usual Pauli matrices.

Time reversal symmetry operator -- when it exists -- is defined as an anti-unitary operator which commutes with the Hamiltonian. Such an operator can be defined as $T=\mathscr{K}\tau_{z}\otimes\mathbf{i}\eta_{y}$ and thus $H$ is time reversal symmetric. $\mathscr{K}$ is the anti-unitary operator $\mathscr{K}\left[\mathbf{i}\right]=-\mathbf{i}$ and thus $\mathscr{K}\left[\eta_{y}\right]=-\eta_{y}$. One verifies that $\left[H_{\pi},T\right]=0$ as it must.

Please tell me if I started with the wrong Hamiltonian.

A few words about the definition (as follow from the comment below): The time-reversal operator is defined as I did, i.e. one applies it to the Hamiltonian $H_{\pi}$, (call it the Hamiltonian density if you wish, since in my way of writing $H=F^{\dagger}\cdot H_{\pi}\cdot F$, the dots should include summation(s) over phase-space-time [delete as appropriate]). You could prefer to define the action of an operator as transforming the operators (or the wave-function). But you should not use both definitions at the same time. It is clear that you can not do both, since otherwise you transform $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot \left(U \cdot H_{\pi} \cdot U^{\dagger}\right) \cdot U\cdot F = H$ trivially, whatever (anti-)unitary transformation $U$ you choose. It is clear that what your are looking for is something like $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot H_{\pi} \cdot U\cdot F \sim H$ and you see what I just said: apply the transformation to the Hamiltonian (density) or to the fields, but not both. In condensed matter we usually choose the convention I gave to you: we transform the Hamiltonian. One of the reasons is that the operators (especially the fermionic creation/annihilation ones) are seen as encoding the statistics of the fields, whereas the Hamiltonian encodes the dynamics, and it is simple imagination to change the dynamics.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
answered Nov 24, 2013 by Oaoa (50 points) [ no revision ]
@ Oaoa Thanks for your answer. Yes, you gave the correct form of the Hamiltonian. But I think the antiunitary time-reversal operator should be defined as:$f_{i\uparrow}\rightarrow f_{i\downarrow}, f_{i\downarrow}\rightarrow -f_{i\uparrow}$ and similarly $f_{i\uparrow}^\dagger \rightarrow f_{i\downarrow}^\dagger , f_{i\downarrow}^\dagger \rightarrow -f_{i\uparrow}^\dagger$.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
@ Oaoa Using your notation of the Hamiltonian and according to my understanding of time-reversal operation, the Hamiltonian under time-reversal is changed as $H=F^\dagger \cdot H_{\pi} \cdot F \rightarrow THT^{-1}=F^\dagger \cdot H_{\pi}^* \cdot F\neq H$.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
@K-boy I think you miss the point. As long as you do not want to define what a time-reversal symmetry is for you, there is no way to discuss. A symmetry makes the transformed Hamiltonian similar to the original one. That is in essence what you said in your answer, with your SU(2) redundancy. I included this redundancy in the structure of $H_{\pi}$ and I defined time-reversal symmetry in my answer. Maybe you used a local-in-space definition (something like $T=\mathscr{K}\mathbf{i}\sigma_{y}$ applied to each lattice site) but there is no need for a so-restrictive definition...

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
… In contrary, I used a non-local-in-space generalisation, and I included the sub-lattice (in your language) redundancy in the $\tau$'s matrices (in my language). I don't know whether the distinction between our two (equivalent) approaches is important or not. It should not, since it is a pure rhetoric problem, whereas the math underneath are the same.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
+ 2 like - 0 dislike

Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong.

Let's rewrite the Hamiltonian as $H(\psi_i)=\sum_{<ij>}(\psi_i^\dagger\chi_{ij}\psi_j+H.c.)$, where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$ and $\chi_{ij}=\begin{pmatrix} t_{ij} & 0\\ 0 & -t_{ij}^* \end{pmatrix}$. And divide the square lattice into two sublattices(nearest-neighbour sites belong to different sublattices) denoted as $A$ and $B$. Now it's easy to see that $$TH(\psi_i)T^{-1}=H(G_i\psi_i),G_i\in SU(2)$$, with $G_i=\begin{cases} i\sigma_y& \text{ if } i\in A \\ -i\sigma_y& \text{ if } i\in B \end{cases}$ or $G_i=\begin{cases} -i\sigma_y& \text{ if } i\in A \\ i\sigma_y& \text{ if } i\in B \end{cases} .$ Thus, the projected spin-state $\psi_{spin}$ indeed has the time-reversal symmetry as well as the translation symmetry.

Remarks: In fact, as long as the mean-field Hamiltonian $H(\psi_i)$ on the suqare lattice has the above form(containing only nearest-neighbour terms)

(1)with $\chi_{ij}=\begin{pmatrix} t_{ij} & \Delta_{ij}\\ \Delta_{ij}^* & -t_{ij}^* \end{pmatrix}$, the mean-field Hamiltonian $H(\psi_i)$ always satisfies the above identity under time-reversal transformation, and thus the projected spin-state always has the time-reversal symmetry.

(2)on the other hand, if $\chi_{ij}=\begin{pmatrix} t_{ij} & 0\\ 0 & -t_{ij}^* \end{pmatrix}$, where $t_{ij}$ are parametrized by four complex parameters $t_{1,2,3,4}$ as shown in the Fig.1. in the paper, as long as $t_{1,2,3,4}$ have equal magnitudes(no need for equal phase), then one can also show that the projected spin-state has the translation symmetry.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy
answered Nov 25, 2013 by Kai Li (975 points) [ no revision ]
Good answer. I've the feeling it is exactly the same as mine, simply rephrased. You defined the SU(2) as a sub lattice pseudo-symmetry (perhaps better to say redundancy), whereas I included it in a $4\times 4$ matrix notation in my $H_{\pi}$ matrix. Then at the end it is only a matter of rhetoric. Note in particular that your $G_{i}=\pm \mathbf{i}\sigma_{y}$ is the same as my $T\propto \tau_{z}\otimes\mathbf{i}\eta_{y}$. Which picture is the most fruitful for which purpose is an other interesting question I believe ; I've no answer to it.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user Oaoa
@ Oaoa I agree with you now, thanks.

This post imported from StackExchange Physics at 2014-03-09 08:39 (UCT), posted by SE-user K-boy

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