5.5.4.2 So­lu­tion com­plexsb-b

Ques­tion:

A more con­cise way of un­der­stand­ing the or­tho­nor­mal­ity of the two-par­ti­cle spin states is to note that an in­ner prod­uct like $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ equals $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$, where the first in­ner prod­uct refers to the spin states of par­ti­cle 1 and the sec­ond to those of par­ti­cle 2. The first in­ner prod­uct is zero be­cause of the or­thog­o­nal­ity of ${\uparrow}$ and ${\downarrow}$, mak­ing $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ zero too.

To check this ar­gu­ment, write out the sums over $S_{z1}$ and $S_{z2}$ for $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$ and ver­ify that it is in­deed the same as the writ­ten out sum for $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ given in the an­swer for the pre­vi­ous ques­tion.

The un­der­ly­ing math­e­mat­i­cal prin­ci­ple is that sums of prod­ucts can be fac­tored into sep­a­rate sums as in:

$$\sum_{{\rm all} S_{z1}} \sum_{{\rm all} S_{z2}} f(S_{z1}) ... ...S_{z1})\right] \left[\sum_{{\rm all} S_{z2}} g(S_{z2})\right]$$

This is sim­i­lar to the ob­ser­va­tion in cal­cu­lus that in­te­grals of prod­ucts can be fac­tored into sep­a­rate in­te­grals:

$$\begin{eqnarray*}\lefteqn{\int_{{\rm all} {\skew0\vec r}_1} \int_{{\rm all} {\... ...ec r}_2} g({\skew0\vec r}_2) { \rm d}^3 {\skew0\vec r}_2\right] \end{eqnarray*}$$

An­swer:

$$\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\v... ...= \pm\frac 12\hbar}{\uparrow}(S_{z2}){\uparrow}(S_{z2})\right]$$

and writ­ten out

$$\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\v... ...{1}{2}}\hbar){\uparrow}(-{\textstyle\frac{1}{2}}\hbar) \right]$$

and mul­ti­ply­ing out, and re­order­ing the sec­ond and third fac­tor in each term, you see it is the same as the ex­pres­sion ob­tained in the an­swer to the pre­vi­ous ques­tion,

$$\begin{eqnarray*}\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle &... ...tyle\frac{1}{2}}\hbar){\uparrow}(-{\textstyle\frac{1}{2}}\hbar). \end{eqnarray*}$$