1
$\begingroup$

I attended a lecture given by Professor Wen Xiaogang. In the lecture, Prof.Wen gave an example of topological excitation:

For a state $$(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)\tag{1}$$, if it's excited into such a state:

$$(\uparrow\downarrow)\uparrow(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)\uparrow(\uparrow\downarrow)(\uparrow\downarrow)\tag{2}$$. Due to this excitation cannot be completed by a local opeator, it's called topological excitaion.

Further more, Prof.Wen says,"There's spin-$\frac{1}{2}$ excitation". From the formulas of (2), it seems to have two spin-$\frac{1}{2}$ excitations at different locations.

The question is:

However, cannot I write (2) as:

$$(\uparrow\downarrow)\uparrow\uparrow(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\downarrow\uparrow)(\uparrow\downarrow)(\uparrow\downarrow)\tag{3}$$.

In (3), it seems that there's only one spin-1 excitaion.

From experimental perspective, (2) and (3) seems to be contradictory, since (2) has half spin on two sites and (3) has one spin on one site

$\endgroup$
2
  • 1
    $\begingroup$ What is the question? $\endgroup$ – Norbert Schuch Sep 26 '20 at 9:26
  • 1
    $\begingroup$ I agree with Norbert. There isn't a clear question here (despite the edit). $\endgroup$ – Emilio Pisanty Sep 26 '20 at 11:58
4
$\begingroup$

You are abusing the brackets.

You should think about the whole thing without the brackets, which are merely a guide to the eye.

If you have a "ground state" configuration of alternating spin 1/2, $$ \uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\ , $$ you can create a spin-1 excitation by flipping one spin. The resulting configuration will have three adjacent up spins, $$ \uparrow\downarrow\uparrow\downarrow\color{red}{\uparrow\uparrow\uparrow}\downarrow\uparrow\downarrow\uparrow\downarrow\ . $$ Conversely, observe that the configuration which you describe has two places with only two adjacent spin ups, $$ \uparrow\downarrow\color{blue}{\uparrow\uparrow}\downarrow\uparrow\downarrow\uparrow\downarrow\color{blue}{\uparrow\uparrow}\downarrow\uparrow\downarrow\ . $$

Now if you look at this pattern, you can easily see that it is impossible to create it by flipping spins locally around where the two defects are located: To see this, just write the "ground state" configuration and the other one on otp of each other, $$ \uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\\ \uparrow\downarrow\color{blue}{\uparrow\uparrow}\downarrow\uparrow\downarrow\uparrow\downarrow\color{blue}{\uparrow\uparrow}\downarrow\uparrow\downarrow $$ and you immediately see that all the spins in between the two defects differ (you can shift the vacuum by one site, then the same will be true for all spins outside the region between the defects).


To put it differently, the key to the "ground state" configuration is that the spins alternate up/down, while in your bracketing interpretation, only the pairs in the bracket have to differ, whereas in the second bracketing you write, adjacent spins in adjacent brackets can be rhe same. This is not a valid state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.