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In J.J. Sakurai's Modern Quantum Mechanics, the same operator $X$ acts on both, elements of the ket space and the bra space to produce elements of the ket and bra space, respectively. Mathematically, an operator is simply a map between two spaces.

So, how can the same operator act on the ket as well as the bra space?

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    $\begingroup$ Sakurai's well-written and insightful book and the mathematical theory of linear operators in topological vector spaces are two parallel roads, i.e. they never meet. My advice, use the book by Galindo and Pascual. They respect the mathematical foundations of QM. $\endgroup$ – DanielC Jul 10 '18 at 15:45
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This is what computer scientists would call (ad-hoc) polymorphic or “overloaded” functions: basically, an operator $X$ on the Hilbert space $\mathcal{H}$ is not just one function $X: \mathcal{H}\to\mathcal{H}$, but a family of two functions $$ X = \{ X_{\mathcal{H}}, \quad X_{\mathcal{H}^\ast} \} $$ with $$ X_{\mathcal{H}}: \mathcal{H}\to\mathcal{H}, \quad X_{\mathcal{H}^\ast}: \mathcal{H}^\ast\to\mathcal{H}^\ast $$ Since a Hilbert space's dual space (the “space of bra-vectors”) is a different space from $\mathcal{H}$ itself, it's always unambiguous which of the two functions you mean when applying $X$ to either a ket or a bra.

And the definition of $X_{\mathcal{H}^\ast}$ follows directly from the one of $X_{\mathcal{H}}$ vice versa. This is easy to see in one direction: $$ \bigl(X_{\mathcal{H}^\ast}(\langle f|)\bigr)(|v\rangle) = \langle f| \bigl(X_\mathcal{H}(|v\rangle)\bigr) $$ In the other direction, we need to invoke the Riesz representation theorem: for any $|v\rangle \in \mathcal{H}$, let $$\begin{align} \langle X_v^\ast| & \in\mathcal{H}^\ast \\ \langle X_v^\ast| &:= X_{\mathcal{H}^\ast}\bigl(\langle . , v\rangle\bigr). \end{align}$$ where by $\langle . , v\rangle$ I mean the function $$\begin{align} v^\ast & \in \mathcal{H}^\ast \\ v^\ast(w) & := \langle w,v\rangle \end{align}$$ (This is now not an application of dual-vectors, but the actual scalar product that comes with the Hilbert space!)

Then Riesz tells us that this corresponds to one unique element $X_v\in\mathcal{H}$, so we can define $$ X_{\mathcal H}(|v\rangle) := X_v. $$ And because physicists are lazy (good trait, although some overdo it), they avoid all the “obvious” parentheses etc., and just assume the reader will know how the operator needs to be applied, which space it lives in, etc.. Swapping between a bra- and a ket-version of a state implicitly always invokes the Riesz representation theorem, but this is seldom talked about.

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  • $\begingroup$ Thanks for the response. You defined $v^*(w)$ as $<w,v>$. I know that there exists a bijective map between the ket and the bra space. When defining the function that way, has a particular map been assumed? $\endgroup$ – D12ac Jul 3 '18 at 16:09
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    $\begingroup$ @D12ac no need to assume anything there, the map follows from the scalar product (which is pertaining to the Hilbert space). $\endgroup$ – leftaroundabout Jul 3 '18 at 16:23
  • $\begingroup$ I tried searching for this on stack, but couldn't find a solution. So, the definition $ v^*(w) := <w,v>$ is for the functional operation of a bra element, defined using the inner product on the (ket(?)) Hilbert Space. What about the application of dual vectors, then? Will the quantity $<v^*|w>$ be equal to $<w,v>$? $\endgroup$ – D12ac Jul 5 '18 at 5:11
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    $\begingroup$ Well, this is a bit murky. When physicists write $\langle v|$, they actually mean already the dual-ised version, so that $\langle v|w\rangle \equiv \langle v,w\rangle$. That's where the bra-ket notation comes from in the first place. One could make an argument that it would be better to ditch those brakets and instead just use dual vectors explicitly as functions (with generous information about what variable belongs to which space). This avoids confusion, in particular when order matters as it does in complex spaces. $\endgroup$ – leftaroundabout Jul 5 '18 at 10:01
  • $\begingroup$ Speaking of which, I'm not sure if I got it wrong here... $\langle w,v\rangle$ or $\langle v,w\rangle$? You could as a good exercise check that the cross-definitions are actually consistent, i.e. show that $X_{\mathcal{H}^{\ast\ast}} = X_\mathcal{H}$. $\endgroup$ – leftaroundabout Jul 5 '18 at 10:06
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The action (to the right) on the ket space naturally induces an action (to the left) on the bra space. The bra space is the dual space of the ket space (that is the space of linear functionals over the kets). We can simply define $\left< \psi \right| X$ by its action on kets (or, since it's linear, on a basis of ket space and linear extension): $$ \big(\left< \psi \right| X\big) \left| \phi \right> := \left< \psi \right| \big( X \left| \phi \right>\big).$$ Such definitions are also common in pure mathematics. If you want to be pedantic, you can use some notation, e.g. $\iota(X)$, for die induced operator on the dual space.

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Just think of the operator as a matrix (as per your "matrix-elements" tag). Then it operates to the right on a column-vector $\left|\mbox{ket}\right>$, and to the left on a row-vector $\left<\mbox{bra}\right|$

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  • $\begingroup$ Thanks for your reply. I know that the matrix formulation would work; I was uncomfortable with the notion of an operator, as it has been named, acting in the two spaces. $\endgroup$ – D12ac Jul 3 '18 at 16:03
  • $\begingroup$ In support of this answer: If you work in a finite-dimensional Hilbert space (e.g. lattice problems with periodic boundary conditions, finitely many coupled spins), this correspondence gets exact, since then the operators can exactly be represented as matrices and the bras can exactly be represented as row-vectors. This shows that the vector-matrix-approach presents a valid intuition. $\endgroup$ – Sebastian Riese Jul 4 '18 at 8:19

protected by Qmechanic Jul 2 '18 at 17:18

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