Timeline for Does the Unruh effect need the accelerating observer's Hilbert space to be smaller than the inertial one?
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| Sep 10, 2021 at 23:52 | answer | added | Gold | timeline score: 2 | |
| Sep 10, 2021 at 23:29 | comment | added | Gold | The trace is, in fact, implied. The number operator that Carroll is computing the mean value of is that associated to one of the Rindler wedges. In that sense such an operator acts as the identity in the complementary wedge. Whenever you have a composite Hilbert space ${\cal H} = {\cal H}_A\otimes{\cal H}_B$ and an observable of the form $O_A\otimes 1_B$ its mean value with respect to a state in ${\cal H}$ is the mean value of $O_A$ with respect to the state obtained by tracing out ${\cal H}_B$. | |
| Sep 10, 2021 at 22:50 | history | edited | P. C. Spaniel | CC BY-SA 4.0 |
added note
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| Sep 10, 2021 at 22:33 | comment | added | P. C. Spaniel | Also, my question points at the fact that I didn't have to specify that the Rindler basis is smaller, in my first approach. And I still got a thermal distribution. So that also confuses me. | |
| Sep 10, 2021 at 22:32 | comment | added | P. C. Spaniel | Hi! Thanks for your reply. I was actually following Carroll's calculation for the first method and I do understand that you need two wedges of Rindler to fully describe Minkowski. However, I fail to see at which point he is tracing over one wedge. Specifically, I'm looking at equation 9.163 where he calculates the expectation value. | |
| Sep 10, 2021 at 22:24 | comment | added | Gold | In the first approach you actually need modes in the two Rindler wedges to expand the inertial observer's plane waves. Then you need to trace out the left Rindler wedge as well. For a full explanation of the first approach see the chapter on quantum field theory in curved spacetimes in Carroll's GR textbook. | |
| Sep 10, 2021 at 22:07 | history | asked | P. C. Spaniel | CC BY-SA 4.0 |