Editing Riesz representation theorem (section)

== Extending the bra–ket notation to bras and kets ==

{{Main|Bra–ket notation}}

Let <math>\left(H, \langle\cdot, \cdot \rangle_H\right)</math> be a Hilbert space and as before, let <math>\langle y\, | \,x \rangle_H := \langle x, y \rangle_H.</math> 
Let 
<math display=block>\begin{alignat}{4}
\Phi :\;&& H &&\;\to    \;& H^* \\[0.3ex]
        && g &&\;\mapsto\;& \left\langle \,g\mid \cdot\, \right\rangle_H = \left\langle \,\cdot, g\, \right\rangle_H \\
\end{alignat}</math>
which is a bijective antilinear isometry that satisfies
<math display=block>(\Phi h) g = \langle h\mid g \rangle_H = \langle g, h \rangle_H \quad \text{ for all } g, h \in H.</math>

'''Bras'''

Given a vector <math>h \in H,</math> let <math>\langle h\, |</math> denote the continuous linear functional <math>\Phi h</math>; that is, 
<math display=block>\langle h\, | ~:=~ \Phi h</math> 
so that this functional <math>\langle h\, |</math> is defined by <math>g \mapsto \left\langle \,h\mid g\, \right\rangle_H.</math> This map was denoted by <math>\left\langle h \mid \cdot\, \right\rangle</math> earlier in this article. 

The assignment <math>h \mapsto \langle h |</math> is just the isometric antilinear isomorphism <math>\Phi ~:~ H \to H^*,</math> which is why <math>~\langle c g + h\, | ~=~ \overline{c} \langle g\mid ~+~ \langle h\, |~</math> holds for all <math>g, h \in H</math> and all scalars <math>c.</math> 
The result of plugging some given <math>g \in H</math> into the functional <math>\langle h\, |</math> is the scalar <math>\langle h\, | \,g \rangle_H = \langle g, h \rangle_H,</math> which may be denoted by <math>\langle h \mid g \rangle.</math><ref group=note>The usual notation for plugging an element <math>g</math> into a linear map <math>F</math> is <math>F(g)</math> and sometimes <math>Fg.</math> Replacing <math>F</math> with <math>\langle h\mid :=~ \Phi h</math> produces <math>\langle h\mid(g)</math> or <math>\langle h \mid g,</math> which is unsightly (despite being consistent with the usual notation used with functions). Consequently, the symbol <math>\,\rangle\,</math> is appended to the end, so that the notation <math>\langle h\mid g \rangle</math> is used instead to denote this value <math>(\Phi h) g.</math></ref> 

'''Bra of a linear functional'''

Given a continuous linear functional <math>\psi \in H^*,</math> let <math>\langle \psi\mid</math> denote the vector <math>\Phi^{-1} \psi \in H</math>; that is, 
<math display=block>\langle \psi\mid ~:=~ \Phi^{-1} \psi.</math> 

The assignment <math>\psi \mapsto \langle \psi\mid</math> is just the isometric antilinear isomorphism <math>\Phi^{-1}  ~:~ H^* \to H,</math> which is why <math>~\langle c \psi  + \phi\mid ~=~ \overline{c} \langle \psi\mid ~+~ \langle \phi\mid~</math> holds for all <math>\phi, \psi \in H^*</math> and all scalars <math>c.</math> 

The defining condition of the vector <math>\langle \psi | \in H</math> is the technically correct but unsightly equality
<math display=block>\left\langle \, \langle \psi\mid \, \mid g \right\rangle_H ~=~ \psi g \quad \text{ for all } g \in H,</math>
which is why the notation <math>\left\langle \psi \mid g \right\rangle</math> is used in place of <math>\left\langle \, \langle \psi\mid \, \mid g \right\rangle_H = \left\langle g, \, \langle \psi\mid \right\rangle_H.</math> With this notation, the defining condition becomes
<math display=block>\left\langle \psi\mid g \right\rangle ~=~ \psi g \quad \text{ for all } g \in H.</math>

'''Kets'''

For any given vector <math>g \in H,</math> the notation <math>| \,g \rangle</math> is used to denote <math>g</math>; that is, 
<math display=block>\mid g \rangle : = g.</math> 

The assignment <math>g \mapsto | \,g \rangle</math> is just the identity map <math>\operatorname{Id}_H : H \to H,</math> which is why <math>~\mid c g + h \rangle ~=~ c \mid g \rangle ~+~ \mid h \rangle~</math> holds for all <math>g, h \in H</math> and all scalars <math>c.</math> 

The notation <math>\langle h\mid g \rangle</math> and <math>\langle \psi\mid g \rangle</math> is used in place of <math>\left\langle h\mid \, \mid g \rangle \, \right\rangle_H ~=~ \left\langle \mid g \rangle, h \right\rangle_H</math> and <math>\left\langle \psi\mid \, \mid g \rangle \, \right\rangle_H ~=~ \left\langle g, \, \langle \psi\mid \right\rangle_H,</math> respectively. As expected, <math>~\langle \psi\mid g \rangle = \psi g~</math> and <math>~\langle h\mid g \rangle~</math> really is just the scalar <math>~\langle h\mid g \rangle_H ~=~ \langle g, h \rangle_H.</math>