Timeline for Number of the Generators of Poincare Group
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2014 at 10:00 | comment | added | joshphysics | @ome Sure thing. | |
| Feb 22, 2014 at 9:58 | comment | added | joshphysics | @Ome It depends on what you mean when you write the symbol $P^\mu$. That symbol is usually used to refer to generators of the translation group in a given representation. In the matrix representation of the translation subgroup of the Poincare group given above, the $P^\mu$ would be matrices, in quantum mechanics, however, the $P^\mu$ would be operators (such as differential operators in the position space representation), and in that case, the operators are interpreted as momentum operators. | |
| Feb 22, 2014 at 9:55 | comment | added | rainman | So, in the context of physics, in the lie algebra of the Poincare group, the $P^\mu$s are matrices of momentum operators? | |
| Feb 22, 2014 at 9:51 | comment | added | joshphysics | @Ome If you consider the representations of the translation subgroup on functions, then you find that finite translations are obtained by exponentiating scalar multiples derivative operators $\partial/\partial x^i$, but now notice that such derivative operators are precisely what we call momentum operators in quantum mechanics in the position space representation of a massive particle moving in space. | |
| Feb 22, 2014 at 9:41 | comment | added | rainman | As the translation group is a subgroup of the Euclidean group, I can compute the 4 generators of the $4D$ translation group. I see that in physics literatures they call these 4 generators as momentum operators. Why this is so? However, a slip of the pen: "Well, I quick standard computation will show you..." should be "Well, a quick standard computation will show you". | |
| Feb 22, 2014 at 9:36 | vote | accept | rainman | ||
| Feb 22, 2014 at 9:22 | history | answered | joshphysics | CC BY-SA 3.0 |