Timeline for Ensemble interpretation and density matrices - why is it impossible to distinguish two equivalent density matrices?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22, 2023 at 16:59 | vote | accept | Kobamschitzo | ||
| Apr 21, 2023 at 20:28 | answer | added | Bohan Xu | timeline score: 1 | |
| Apr 21, 2023 at 20:05 | answer | added | hft | timeline score: 0 | |
| Apr 21, 2023 at 19:57 | answer | added | benrg | timeline score: 1 | |
| Apr 21, 2023 at 15:46 | comment | added | FlatterMann | @Kobamschitzo I can construct a Dirac delta function out of a sum of cosines or white noise, depending on how I choose the phases. What is the difference? The difference is the sample that I take in the interval around zero. In reality no matter what phases I choose, there will be a strong peak in this function somewhere, I just choose not to look for it. | |
| Apr 21, 2023 at 15:34 | comment | added | Kobamschitzo | @march No, it depends on the choice of directions of measurement. If measuring in {0,1} direction, then we will get only 0 and 1 for the system 0+1 | |
| Apr 21, 2023 at 15:30 | comment | added | march | Forget about density matrices for the moment. What about pure state? Do you think that if the system is in state $|0\rangle + |1\rangle$, then you can only get 0 and 1 as the result of a measurement of any observable? | |
| Apr 21, 2023 at 15:30 | comment | added | march | @Kobamschitzo. You are misinterpreting the meaning of the density matrix. The first one does not mean that you can only get 0 and 1 as the result of a measurement, because what if you were measuring $\sigma_x$ instead of $\sigma_z$? Then you should be able to get "+" and "-" (so to speak) as the result of the measurement, and again, both of those density matrices will yield the same results. | |
| Apr 21, 2023 at 15:26 | answer | added | march | timeline score: 1 | |
| Apr 21, 2023 at 15:26 | comment | added | Kobamschitzo | @FlatterMann if I construct an end symbol of 1000 pure states st 500 each are states 0 and 1; this ensemble must be different from another ensemble where 500 each states are +x and -x. This is because the first ensemble should give me values 0 and 1 while the latter something else; even though the statistical average remains same | |
| Apr 21, 2023 at 15:24 | comment | added | Kobamschitzo | @FlatterMann Good point. But it seems there is some distinction here. The properties or the eigenvalues of these states must be objective, unlike potential (mathematical construct) and velocities (observer dependent or subjective upto a point). | |
| Apr 21, 2023 at 14:59 | comment | added | FlatterMann | A classical potential is only unique up to a constant, all inertial systems are equivalent etc.. We have many situations in physics where the same physical reality is open to multiple equivalent mathematical descriptions. This does not look any different from that to me. | |
| Apr 21, 2023 at 14:58 | comment | added | LPZ | In both cases, $\rho=\frac{1}{2}Id$. It's just a question of basis choice of decomposition. If you fix the basis, the matrix is uniquely determined by its coefficients. It's just a question of mathematical convenience. Sometimes, it is even better not to use a orthonormal basis (or even a basis). | |
| Apr 21, 2023 at 14:48 | history | asked | Kobamschitzo | CC BY-SA 4.0 |