Simple question concerning unitary transformation

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The discussion centers on the correct form of transforming an operator under infinitesimal unitary transformations, specifically whether it should be U^-1AU or UAU^-1. It highlights that both forms are valid depending on the definition of the unitary transformation U, which can act on kets or bras. The distinction is noted to be superficial, as unitary matrices form a group and both U and U^-1 are unitary. However, the choice of U affects how the operator is represented in the new basis, leading to different expressions for the matrix elements. Ultimately, the transformation's definition is crucial for correctly applying it to quantum states and operators.
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Is the transformation of an operator under INFINITESIMAL unitary transformation, U^-1AU or UAU^-1?? I saw that two books defined it differently?
 
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Remember that the unitary matrices form a group. So if U is a unitary matrix, then U^-1 is also a unitary matrix. Therefore, the distinction is superficial depending on which transformation you want to define as your U.
 
So it doesn't matter?
 
Well, it does matter a little bit. If you define U to be the unitary transformation that transforms kets |a>, then U^-1 will be the unitary transformation that makes the same transformation on the bras <b|. Using this, one should see that the matrix element <b|S|a> under a transformation goes to <b|U^-1SU|a> which means that the matrix S'=U^-1SU is the matrix that represents the operator in this new basis. If I defined U the opposite way, as U is the transformation on bras, then I get the other definition.
 
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Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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