Exercise [22.25]

deant · 02 Dec 2012, 06:42

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):

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

...Then using the given orthogonality and normalization identities

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

we have:

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

as required.

$\blacksquare$