Timeline for Could you help me understand this paper (PRL 106:136806)?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2014 at 12:43 | comment | added | hyd | Could you please demonstrate this using the expressions I gave in my question post ? Thanks. | |
| Oct 11, 2014 at 9:32 | comment | added | Trimok | The function $f_−(\vec r)$ is odd under the reflection relatively to the line defect. So $f_−(\vec r)=0$ for $\vec r$ being on the line defect. Maybe in my answer, I should have use $ \vec r_\perp$ instead of $\vec r$. I have updated the end of the answer. | |
| Oct 11, 2014 at 9:32 | history | edited | Trimok | CC BY-SA 3.0 |
Clarification
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| Oct 9, 2014 at 11:52 | comment | added | hyd | Not for all but only $\vec{r}$ on the line defect (see the original question part). | |
| Oct 9, 2014 at 11:21 | comment | added | Trimok | $\langle\vec{r}|-\rangle$ is not equals to $0$. Replace the notation $|-\rangle$ by $|f_-\rangle$. The complex number $\langle\vec{r}|f_-\rangle$ means simply $f_-(\vec{r})$. It is not possible that $f_-(\vec{r})=0$ (for all $\vec r$) because it would mean $|f_-\rangle=0$. As explained in my answer, it is only $f_-(\vec 0)$ which is equals to zero, because the function $f_-(\vec{r})$ is odd under reflection. | |
| Oct 5, 2014 at 2:22 | comment | added | hyd | I also understand the meaning of $\Gamma G \Gamma G^{\dagger}$. But why should be acted on the states of the electrodes ? | |
| Oct 5, 2014 at 2:20 | comment | added | hyd | I understand $|-\rangle$ is odd under reflection. But I could not thence deduce $\langle\vec{r}|-\rangle = 0$ explicitly using the expressions given above. | |
| Oct 4, 2014 at 14:46 | comment | added | Trimok | @Hai-YaoDeng : For the last part of your question, I think that one may see that like a kind of closed path. Your are starting from one side of the line defect, you get a $\Gamma$ to go to the other side, a propagator $G$ to move along the line defect, an other $\Gamma$ to come back to the initial side of the line defect, then an other propagator $G^\dagger$ coming back to the initial point. | |
| Oct 4, 2014 at 14:41 | comment | added | Trimok | It is said in the text : "As the reflection oper- ator maps A sites onto B sites, and vice versa", so $|-\rangle \sim (|A\rangle - |B\rangle)$ is odd under reflection. | |
| Oct 3, 2014 at 13:22 | comment | added | hyd | Thanks for your post. I understand your point. But how do you show that from the expression (given in the original post) of $|-\rangle$? | |
| Oct 3, 2014 at 13:06 | history | answered | Trimok | CC BY-SA 3.0 |