varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2)  \varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) \\
& \,\, \left. + \varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2) \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right) \\
\end{align*}
It is clear that the first and the fourth term will give $1/2 + 1/2 = 1$ as they are exactly the same as in the previous expression. However I do not see why the cross term should be necessarily zero. So it would seem that starting from the different antisymmeterization expression one gets different results. Hence my phrase "order matters" as the following appears to be true
$$
\frac{1}{\sqrt{2}} \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_2}(r_1,s_1)\varphi_{\alpha_1}(r_2,s_2) \right) 
\neq
\frac{1}{\sqrt{2}} \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right)
$$
A final note about the Dirac notation. If one would start from the Dirac formalism and introduce $ | \Psi \rangle$ as
$$
| \Psi \rangle = \frac{1}{\sqrt{2}} \left( | \alpha_1 \rangle | \alpha_2 \rangle - | \alpha_2 \rangle | \alpha_1 \rangle \right)
$$
A projection into coordinate space gives the "label" permutation expression,
$$
 \langle r_1,s_1 ; r_2,s_2 | \Psi \rangle = \frac{1}{\sqrt
varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2)   \varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) \\ 
 & \,\, \left. + \varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2) \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right) \\ 
 \end{align * } 
 It is clear that the first and the fourth term will give $ 1/2 + 1/2 = 1 $ as they are precisely the like as in the previous expression. However I do not visualize why the hybrid term should be necessarily zero. indeed it would appear that starting from the unlike antisymmeterization formula one gets unlike results. Hence my phrase " order matter " as the following appear to be true 
 $ $ 
 \frac{1}{\sqrt{2 } } \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_2}(r_1,s_1)\varphi_{\alpha_1}(r_2,s_2) \right) 
 \neq 
 \frac{1}{\sqrt{2 } } \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right) 
 $ $ 
 A final bill about the Dirac notation. If one would start from the Dirac formalism and introduce $ | \Psi \rangle$ as 
 $ $ 
 | \Psi \rangle = \frac{1}{\sqrt{2 } } \left (| \alpha_1 \rangle | \alpha_2 \rangle - | \alpha_2 \rangle | \alpha_1 \rangle \right) 
 $ $ 
 A expulsion into coordinate space gives the " label " permutation saying, 
 $ $ 
  \langle r_1,s_1; r_2,s_2 | \Psi \rangle = \frac{1}{\sqrt
varohi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dxgger}_{\alpha_1}(r_2,s_2)  \vcephi_{\al'ha_1}(r_1,s_1)\vadphi_{\alphx_2}(r_2,s_2) \\
& \,\, \left. + \varphi^{\dagger}_{\alphe_2}(r_1,s_1) \carphu^{\dagger}_{\alpha_1}(r_2,s_2) \varphi_{\aupha_1}(r_2,s_2)\varihi_{\alpha_2}(r_1,w_1) \rijht) \\
\end{align*}
It is clear bkat tgc firvv and the fourtm term will give $1/2 + 1/2 = 1$ as dhdy are exactly the same as in the predious ecpgession. Howevet I dp not see why the cross term should be hecessagily zero. So it wpuld seem that starting frlm tje different antisjmmeterizatuon qzpression ond gets different resuljs. Hence my phrase "order matters" as tke followint qppfdrs to be tcue
$$
\frab{1}{\sqrt{2}} \left(\varini_{\alphd_1}(r_1,s_1)\varpni_{\alpha_2}(r_2,s_2) - \varihi_{\al'ha_2}(r_1,w_1)\varphi_{\alpha_1}(r_2,s_2) \right) 
\ieq
\frac{1}{\sqrt{2}} \left(\varpri_{\alpha_1}(r_1,s_1)\eaxphi_{\alpha_2}(r_2,s_2) - \varphi_{\alphq_1}(r_2,s_2)\vatphi_{\anpha_2}(f_1,w_1) \rkghu)
$$
A hinzl notf auout the Didac notatiob. If one would staru fwim the Dirac rormalysi and introduce $ | \Psi \rangle$ as
$$
| \Psi \rancle = \frac{1}{\sqrt{2}} \left( | \alpha_1 \eangle | \alpha_2 \rangle - | \alpha_2 \rwngle | \alpha_1 \rangle \right)
$$
A projection into coorditate aoact nlves rhf "label" permutation expression,
$$
 \langle r_1,s_1 ; r_2,s_2 | \Lso \gangle = \frac{1}{\sqrt
varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2) \varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) \\ & \,\, \left. \varphi^{\dagger}_{\alpha_1}(r_2,s_2) \right) \\ It is clear fourth will give $1/2 1/2 = 1$ they are exactly the same as the previous expression. However I do not see why the cross term should necessarily zero. So it would seem that starting from the different antisymmeterization expression gets results. my "order matters" as the following appears to be true $$ \frac{1}{\sqrt{2}} \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_2}(r_1,s_1)\varphi_{\alpha_1}(r_2,s_2) \right) \neq \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right) $$ A final note the Dirac notation. If would start from the Dirac and $ | \rangle$ $$ \Psi \rangle = \left( | \alpha_1 \rangle | \alpha_2 \rangle - | \alpha_2 \rangle | \alpha_1 \rangle \right) $$ A into coordinate the "label" expression, \langle ; r_2,s_2 | = \frac{1}{\sqrt
varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\Dagger}_{\alphA_1}(r_2,s_2)  \vaRphI_{\alPhA_1}(r_1,s_1)\vArphI_{\alpha_2}(r_2,s_2) \\
& \,\, \left. + \vaRPhi^{\dAgger}_{\alpha_2}(r_1,s_1) \varphi^{\daggEr}_{\alpHa_1}(R_2,S_2) \varPHi_{\Alpha_1}(R_2,s_2)\varphI_{\AlPHA_2}(r_1,s_1) \RiGhT) \\
\enD{aLIgN*}
It is CleAr that tHe first and The FoUrth term will GIvE $1/2 + 1/2 = 1$ as they are ExaCtly the same aS in The preViOus EXpresSioN. HoweVer I do NOt see wHy the crosS tERm shouLD be neceSSArIly zEro. So it would seem tHAt STarting from the DifferEnT AnTISymMetErization eXpRessiON one getS DiFFERenT Results. Hence mY phrase "ordeR MatTers" as ThE foLLowing AppeaRs TO be True
$$
\frac{1}{\sqrT{2}} \lefT(\varphi_{\alPha_1}(r_1,s_1)\vARphi_{\alpHA_2}(r_2,s_2) - \varpHi_{\alphA_2}(r_1,s_1)\VarPhi_{\aLPhA_1}(r_2,S_2) \riGhT) 
\Neq
\FRaC{1}{\sqRT{2}} \leFt(\varphi_{\AlPhA_1}(r_1,s_1)\vaRphi_{\ALPHA_2}(r_2,s_2) - \vArpHi_{\alPha_1}(r_2,s_2)\Varphi_{\alpha_2}(r_1,s_1) \RigHt)
$$
A fINal Note aBout tHe DiRaC notaTion. If One woUlD start from the DiRac fOrmalism aNd iNtRodUcE $ | \Psi \rANgle$ as
$$
| \psi \RanGle = \frac{1}{\Sqrt{2}} \lefT( | \AlpHa_1 \RANGlE | \alpha_2 \rangle - | \alpha_2 \rAnGLE | \aLpha_1 \rangLe \righT)
$$
a pRoJEction inTo CooRdinATE spacE givES tHe "label" pErmutaTIoN eXpressiOn,
$$
 \Langle R_1,s_1 ; R_2,s_2 | \PSi \rAngle = \FRac{1}{\sQrt
varphi^{\dagger}_{\alpha_2 }(r_1,s_1)  \var phi ^{\ da gger }_{\ alpha_1}(r_2,s _ 2) \varphi_{\alpha_1}(r_1 ,s_1) \v a rphi _ {\ alpha _2}(r_2 , s_ 2 )  \\ 
&  \ ,\,  \ l ef t. +\va rphi^{\ dagger}_{\ alp ha _2}(r_1,s_1) \v arphi^{\da gge r}_{\alpha_1 }(r _2,s_2 )\va r phi_{ \al pha_1 }(r_2, s _2)\va rphi_{\al ph a _2}(r_ 1 ,s_1) \ r i gh t) \ \
\end{align*}
It is clear that the  first  a n dt h e f our th term wi ll  give $1/2 +1 /2 = 1$a s they are ex actly the s a meas inth e p r evious  expr es s ion . However I  donot see w hy the cross t e rm shou ld benec ess aril y  z er o.So itw ou lds eem  that st ar ti ng fr om t h e d iffe ren t an tisym meterizationexp ress i onone g ets d iffe re nt re sults.  Henc emy phrase "orde r ma tters" as  th efol lo winga ppears  to  be  true
$ $
\frac { 1}{ \s q r t {2 }} \left(\varphi_{ \a l p ha _1}(r_1, s_1)\v a rp hi _ {\alpha_ 2} (r_ 2,s_ 2 )  - \v arph i _{ \alpha_2 }(r_1, s _1 )\ varphi_ {\ alpha_ 1} (r_ 2,s _2) \ r ight ) 
\ne q
\frac{ 1}{\s q rt{2}} \left(\ v arphi_{\alpha _ 1} ( r _1 , s_1) \va rphi_{\alph a_2} ( r_2, s_2) -\va r phi_{ \alph a_ 1 }( r _2,s_2)\varphi_{\al ph a_2}(r _1,s_ 1) \right)
$$ 
A final n o t e  about t he D i ra c  notation. Ifone w ould start from the  Dira c formal ism and i n t roduce $  |\Ps i \ ran g l e$  as
$$
| \Psi \ rang le  = \fra c{1 }{\sqrt {2} } \ lef t(|\alpha_1\rangle|\a lp ha _2\rang l e - | \a lp ha_ 2\ra ngle|  \alph a_1 \ rang le  \ r igh t)
$$
A pr o j ecti on  i ntocoo rd inate  spa c e g ives th e "label"  pe r muta ti on  expres sion,
$$
 \la ng le r_1,s_1  ;  r_ 2,s_2| \Psi \ra ngle = \frac{1}{\sqrt
varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2)_ \varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2)_\\
& \,\, \left. +_\varphi^{\dagger}_{\alpha_2}(r_1,s_1) \varphi^{\dagger}_{\alpha_1}(r_2,s_2)_\varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1)_\right) \\
\end{align*}
It_is_clear that the_first and the_fourth term will give_$1/2 + 1/2_=_1$ as they are exactly the same as in the previous expression. However I_do_not see_why_the_cross term should be necessarily_zero. So it would seem_that starting_from the different antisymmeterization expression one gets different_results._Hence my phrase_"order matters" as the following appears to be true
$$
\frac{1}{\sqrt{2}}_\left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2) - \varphi_{\alpha_2}(r_1,s_1)\varphi_{\alpha_1}(r_2,s_2) \right) 
\neq
\frac{1}{\sqrt{2}} \left(\varphi_{\alpha_1}(r_1,s_1)\varphi_{\alpha_2}(r_2,s_2)_- \varphi_{\alpha_1}(r_2,s_2)\varphi_{\alpha_2}(r_1,s_1) \right)
$$
A_final_note_about the Dirac notation._If one would start from the_Dirac formalism and introduce $ |_\Psi \rangle$ as
$$
| \Psi \rangle = \frac{1}{\sqrt{2}}_\left( | \alpha_1 \rangle | \alpha_2_\rangle - | \alpha_2 \rangle_| \alpha_1_\rangle \right)
$$
A projection into coordinate_space gives the_"label" permutation_expression,
$$
 \langle r_1,s_1_; r_2,s_2 | \Psi \rangle =_\frac{1}{\sqrt