I post this hoping for clarification, and particularly in a context of linear algebra without too much mention of matrices.

(1) Let $V$ be a (perhaps infinite dimensional) Hilbert space and $V'$ its continuous dual. Let $(u', v)$ be the composition of a linear functional $u' \in V'$ with a vector $v \in V$ and let $\langle u, v \rangle$ be the inner product $u, v \in V$ Then by the Riesz representation theorem there is a canonical isomorphism $J:V \to V'$ such that for $u' = J(u)$, $(u', v) = \langle u, v \rangle$. So, I think that Dirac's $\langle u|v \rangle = (u', v) = \langle u, v \rangle$, I.e. it can mean either the inner product or the composition of the corresponding linear functional and the vector (since they are equal) ?

(2) Additionally, let $T:V \to V$ be a (continuous ?) linear transformation. Then there is a dual transformation defined by $T': V' \to V'$ where $(T'(u'), v) = (u', T(v)) $ for all $u' \in V'$ and $v \in V.$ Also, there is an adjoint transformation $T^*:V \to V$ defined by $T^*(v) = J^{-1}(T'(J(v)))$ with the property that $\langle u, T(v) \rangle = \langle T^*(u), v \rangle$. It follows then that $\langle u, T(v) \rangle = \langle T^*(u), v \rangle = (T'(u'), v) = (u', T(v)) $ and I assume that this is what is meant by Dirac's $\langle u|T|v \rangle$ ?

(3) Other than to confuse mathematicians, what is the idea behind conflating these expressions like this (i.e. $\langle u|v \rangle $ means $(u', v) $ or $\langle u, v \rangle$, and $\langle u|T|v \rangle$ means any of $\langle u, T(v) \rangle $, $\langle T^*(u), v \rangle $, $ (T'(u'), v) $, $ (u', T(v)) $)

  • $\begingroup$ (1) and (2) seem OK to me (haven't checked in too much detail). On (3): Which expressions are being conflated? $\endgroup$ – joriki Mar 20 '16 at 12:03
  • $\begingroup$ See Bra–ket notation. $\endgroup$ – Mauro ALLEGRANZA Mar 20 '16 at 12:06
  • $\begingroup$ @joriki. Thanks; expanded conflation in edit to post. $\endgroup$ – Tom Collinge Mar 20 '16 at 12:08

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