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  Why $ U_\theta \mid 0 \rangle_M $ is boost invariant in $ER=EPR$ ?

+ 3 like - 0 dislike

The question is from $ER=EPR$  http://lanl.arxiv.org/abs/1306.0533, page 37, the authors claim that $ U_\theta \mid 0 \rangle_M $ is boost invariant. I did not see the reasons. Sorry for my naive question.

asked Jul 19, 2014 in Theoretical Physics by C Thone (110 points) [ revision history ]

1 Answer

+ 2 like - 0 dislike

The answer is in the same document, § 2.1 AdS Black Holes , page 4

The thermofield hamiltonian (2.2) generates boosts which are translations of the usual hyperbolic angle $\omega$. One can think of the boost as propagating upward on the right side of the Penrose diagram, and downward on the left. The state (2.1) is an eigenvector of $H_{tf}$ with eigenvalue zero, and is therefore boost invariant.

and then page 37

For any non-trivial $\theta$ these are states that are boost invariant and differ from the Minkowski vacuum.

those above.

answered Oct 21, 2017 by igael (360 points) [ no revision ]

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