Timeline for Measuring the quantum state of light
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 7, 2013 at 14:49 | comment | added | daaxix | I misread your second comment above and thought that you were claiming to be able to measure $C_L, C_R$, which you are not. | |
| Jan 7, 2013 at 8:49 | comment | added | Ondřej Černotík | @daaxix Note that this is not the measurement of the imaginary part of the electric field of the light (which would be completely impossible with our technology due to rapid oscillations) but the measurement of the imaginary parts of the amplitudes of the horizontally and vertically polarized components of the electric field. | |
| Jan 6, 2013 at 20:06 | comment | added | Ondřej Černotík | If you measure in both H/V and D/A bases, you measure $|c_H|^2$, $|c_V|^2$, $|c_D|^2=|c_H|^2+|c_V|^2+(c_Hc_V^\ast+c_H^\ast c_V)$, $|c_A|^2=|c_H|^2+|c_V|^2-(c_Hc_V^\ast+c_H^\ast c_V)$. (Note that the coefficients are complex.) Thus, you have four equations for four real numbers (real and imaginary parts of $c_H$, $c_V$) and can determine them completely. | |
| Jan 6, 2013 at 16:05 | comment | added | daaxix | Note that in traditional polarimetry this is called a "division of amplitude" linear Stokes polarimeter. The "linear" term is because you are not measuring any of the circular component. As you said the SNR will be worse... | |
| Jan 6, 2013 at 12:57 | history | answered | Ondřej Černotík | CC BY-SA 3.0 |