$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ A (p, q) - Tensor can be expressed in a arbitrary basis as $$T=\sum_{(i_1,\dotsb ,i_n)}T^{i_1\dotsb i_{p}}_{\quad\quad i_{p+1}\dotsb i_n}(e_{i_1}\otimes\dotsb \otimes e_{i_p}\otimes e^{i_{p+1}}\otimes\dotsb \otimes e^{i_n})$$ In Dirac-Notation that can be equivalently expressed as $$T=\sum_{(i_1,\dotsb ,i_n)}T^{i_1\dotsb i_{p}}_{\quad\quad i_{p+1}\dotsb i_n}(\ket{e_{i_1}}\otimes\dotsb \otimes \ket{e_{i_p}}\otimes \bra{e_{i_{p+1}}}\otimes\dotsb \otimes \bra{e_{i_n}})$$ However I have often seen a (1, 1) - Tensor be expressed as $$T=T^i_{\;j}\ket{e_i}\bra{e^j}$$ where Einstein summation is being used. My source of confusion is that - according to Riesz's Lemma - $\bra{e_j}$ already constitutes a dual - vector. What exactly is $\bra{e^j}$ then? Both behave differently when a vector is passed as an argument $$\bk{e_i}{e_j}=g_{ij}$$ $$\bk{e^i}{e_j}=\delta^i_j$$ so $\bra{e_i}$ and $\bra{e^i}$ can not be the same object and in calculations they also yield different results. Furthermore, in QM, where Dirac - Notation is most commonly being used, the distinction between co- and contravariant indices is not used at all (as far as I am aware; presumabely because eiganbases of hermitians are orthogonal and hence the metric is just $\delta_{ij}$; please correct me if I'm mistaken). I would be very glad if you could resolve my confusion here. My gut feeling is that some of the definitions I provided might not be entirely correct.
EDIT: I have looked at the related posts to my question and there seems to be a contradiction between the answer provided in Basis and dual basis in the Dirac notations and Ambiguity with Dirac Notation and I am still unsure whether Dirac - Notation is limited to the special case of $g_{ij} = \delta_{ij}$, since I've seen it being used in non orthonormal bases and still yield the correct result. An example would be $$g_{ij}g^{ji}=\bk{b_i}{b_j}\bk{b^j}{b^i}=\bra{b_i}\Bbb{I}\ket{b^i}=\delta^i_i=\mathrm{dim}(\mathcal{V})$$ which is the expected result. That is not the only case where a more general application of dirac-notation works.