Off diagonal element of density matrix

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In a two-level quantum system, the off-diagonal elements of the density matrix are represented as ρ_{12} = |2⟩⟨1| and ρ_{21} = |1⟩⟨2|. For a pure state f = a|0⟩ + b|1⟩, the density operator includes an off-diagonal component ab*|0⟩⟨1|, where ab* = ⟨f|0⟩⟨1|f⟩, assuming |0⟩ and |1⟩ are orthogonal. In the case of a mixture of pure states f1 and f2 with probabilities p1 and p2, the off-diagonal contribution is given by p1⟨f1|0⟩⟨1|f1⟩ + p2⟨f2|0⟩⟨1|f2⟩. The discussion seeks to confirm the accuracy of these expressions related to the density matrix's off-diagonal elements.
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For two level system , let denotes the ground state as 1 and exctied state as 2, for writing the office off-diagonal matrix element for the density operator, shall it be

\rho_{12} = |2\rangle\langle 1|

and

\rho_{21} = |1\rangle\langle 2|

?
 
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From a pure state f = a|0> + b|1> one gets the density opérator
(a|0> + b|1>)(a*<0| + b*<1|)
with a off diagonal part ab*|0><1|
the off diagonal matrix element is ab* = <f |0><1|f> (with |0> and |1> orthogonal)
With a mixture of pure state f1,f2 with probabilities p1, p2
we get p1<f1 |0><1|f1> + p2<f2 |0><1|f2>
I hope there is no error in my answer!
 
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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