Timeline for Norm of a jump operator
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7 at 15:27 | comment | added | Frederik vom Ende | You are of course right, thank you for pointing that out! | |
| May 7 at 15:26 | comment | added | Quantum Mechanic | @FrederikvomEnde that's a useful update - just a quick typo because I think you probably mean $\tau\to 0$ | |
| May 7 at 9:42 | comment | added | Frederik vom Ende | To complete this answer: this $\rho(t)$ comes about, e.g., via Lindblad dynamics induced by the single jump operator $$ L_1(\tau)=\begin{pmatrix}0&0\\\frac1{\sqrt\tau}&0\end{pmatrix}\,. $$ So for a single $\tau$, $L_1(\tau)$ is of course bounded (with operator norm $\frac1{\sqrt\tau}$) because every matrix is bounded, but $\|L_1(\tau)\|\to\infty$ as $\tau\to\infty$ so it is not uniformly bounded. In other words the faster the exponential decay the larger the norm of the jump operators has to be. | |
| Apr 13, 2023 at 19:50 | vote | accept | Jon Megan | ||
| Apr 13, 2023 at 13:10 | history | edited | Quantum Mechanic | CC BY-SA 4.0 |
added condition on time
|
| Apr 13, 2023 at 1:01 | history | answered | Quantum Mechanic | CC BY-SA 4.0 |