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 Exercise [22.25] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [22.25]
The 'bra' (i.e. dual, or Hermitian conjugare) version of a state:


\{a,b\} \;\; = \;\; a\left|\boldsymbol{\uparrow}\right> \;\; + \;\; b\left|\boldsymbol{\downarrow}\right>

is

\{a,b\}^* \;\; = \;\; \overline{a}\left<\boldsymbol{\uparrow}\right| \;\; + \;\; \overline{b}\left<\boldsymbol{\downarrow}\right|



So the Hermitian scalar product is (by its (anti-)linearity):


\left<\{a,b\}|\{w,z\}\right> \;\; = \;\;<br />\overline{a}w\left<\boldsymbol{\uparrow}|\boldsymbol{\uparrow}\right> \;\; + \;\;<br />\overline{a}z\left<\boldsymbol{\uparrow}|\boldsymbol{\downarrow}\right> \;\; + \;\;<br />\overline{b}w\left<\boldsymbol{\downarrow}|\boldsymbol{\uparrow}\right> \;\; + \;\;<br />\overline{b}z\left<\boldsymbol{\downarrow}|\boldsymbol{\downarrow}\right>




...Then using the given orthogonality and normalization identities


\left<\boldsymbol{\uparrow}|\boldsymbol{\downarrow}\right> = \left<\boldsymbol{\downarrow}|\boldsymbol{\uparrow}\right> = 0,<br />\;\;\;\;\;\;\;\;\;\;\;\;<br />\left<\boldsymbol{\uparrow}|\boldsymbol{\uparrow}\right> = \left<\boldsymbol{\downarrow}|\boldsymbol{\downarrow}\right> = 1



we have:


\left<\{a,b\}|\{w,z\}\right> \;\; = \;\; \overline{a}w \;\; + \;\; \overline{b}z



as required.



\blacksquare


02 Dec 2012, 06:42
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