Timeline for Symmetry transformations of states and operators
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Oct 26 at 13:39 | history | suggested | Chandra Prakash | CC BY-SA 4.0 |
added equation for clarification
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| Oct 26 at 8:56 | review | Suggested edits | |||
| S Oct 26 at 13:39 | |||||
| Jul 24, 2021 at 7:16 | history | bounty ended | SRS | ||
| Jul 17, 2021 at 23:21 | comment | added | Lucas Baldo | @Andrea Thanks :) | |
| Jul 17, 2021 at 23:20 | comment | added | Lucas Baldo | @Andrea In other words, if you are just making a reference frame change, then $A_{F'} = x -100$ and $\vert x \rangle_{F'} = \vert x+100\rangle_{F} $ in that case, such that $\langle A \rangle_{F'} = x = \langle A \rangle_F$. | |
| Jul 17, 2021 at 23:20 | comment | added | Andrea | Right on. Great answer | |
| Jul 17, 2021 at 23:16 | comment | added | Lucas Baldo | The representation $A = x$ is tied to a reference frame (that of the measurement apparatus, for example). Unless you were to move the measurement apparatus with yourself when going from frame $F$ to frame $F'$, then you would move away from both the measurement apparatus and the particle, such that the expected value of the measurement stays the same. You could consider moving the measurement apparatus with yourself when making the reference frame change, but then you would be making a physical transformation on the system, which fits into the second category explained in my answer. | |
| Jul 17, 2021 at 22:04 | comment | added | Andrea | But way, if the origin of F is 100m away from the origin of F’, and $A=x$, wouldn’t you want $\langle A \rangle_F \neq \langle A \rangle_{F’}$? | |
| Jul 17, 2021 at 2:04 | history | answered | Lucas Baldo | CC BY-SA 4.0 |