 you called them "eigenstates"):
$$\langle i\mid\!j\rangle = \delta_{ij}$$
Which leads to a unit normalization.

A:

You had the following cross term for the case of coordinate permutation:
\begin{align}
\propto \int \textrm{d}x_1 \textrm{d}x_2 \left[ \varphi_{b}^{\dagger}(x_1) \varphi_b(x_2) \right] \left[ \varphi_{a}^{\dagger}(x_2) \varphi_{a}(x_1) \right].
\end{align}
However, this is not how you take the inner product of $\varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $\varphi_{a}(x_{2})\varphi_{b}(x_{1})$.
$\varphi_{a}(x_{1})$ and $\varphi_{b}(x_{1})$ belong to the Hilbert space of particle 1, which I denote $\mathcal{H}_{1}$, and similarly, $\varphi_{a}(x_{2}), \varphi_{b}(x_{2}) \in \mathcal{H}_{2}$. The objects $\varphi_{a}(x_1) \varphi_{b}(x_2)$ and $\varphi_{a}(x_2) \varphi_{b}(x_1)$ are members of $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, which is the tensor product of $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$. Furthermore, as $\varphi_{a}(x_{1})$ and $\varphi_{b}(x_{2})$ belong to different Hilbert spaces, $\varphi_{a}(x_1) \varphi_{b}   (x_2)$ and $\varphi_{b}   (x_2)\varphi_{a}(x_1)$ really mean the same thing: there is no ordering issue even if $\varphi$ is a $N$-component vector.
When taking the inner product of the two objects $\varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $\varphi_{a}(x_{2})\varphi_{b}(x_{1})$ in $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, you should contract the parts in $\mathcal{H}_{1}$ together, and the parts in $\mathcal{H}_{2}$ together
you called them " eigenstates " ): 
 $ $ \langle i\mid\!j\rangle = \delta_{ij}$$ 
 Which leads to a unit normalization. 

 A: 

 You take the following hybridization term for the case of coordinate permutation: 
 \begin{align } 
 \propto \int \textrm{d}x_1 \textrm{d}x_2 \left [ \varphi_{b}^{\dagger}(x_1) \varphi_b(x_2) \right ] \left [ \varphi_{a}^{\dagger}(x_2) \varphi_{a}(x_1) \right ]. 
 \end{align } 
 However, this is not how you assume the inner product of $ \varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $ \varphi_{a}(x_{2})\varphi_{b}(x_{1})$. 
 $ \varphi_{a}(x_{1})$ and $ \varphi_{b}(x_{1})$ belong to to the Hilbert space of particle 1, which I denote $ \mathcal{H}_{1}$, and similarly, $ \varphi_{a}(x_{2 }), \varphi_{b}(x_{2 }) \in \mathcal{H}_{2}$. The object $ \varphi_{a}(x_1) \varphi_{b}(x_2)$ and $ \varphi_{a}(x_2) \varphi_{b}(x_1)$ are members of $ \mathcal{H}_{1}\otimes\mathcal{H}_{2}$, which is the tensor product of $ \mathcal{H}_{1}$ and $ \mathcal{H}_{2}$. Furthermore, as $ \varphi_{a}(x_{1})$ and $ \varphi_{b}(x_{2})$ belong to to different Hilbert spaces, $ \varphi_{a}(x_1) \varphi_{b }    (x_2)$ and $ \varphi_{b }    (x_2)\varphi_{a}(x_1)$ really mean the like thing: there is no order issue even if $ \varphi$ is a $ N$-component vector. 
 When taking the inside product of the two objects $ \varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $ \varphi_{a}(x_{2})\varphi_{b}(x_{1})$ in $ \mathcal{H}_{1}\otimes\mathcal{H}_{2}$, you should contract the part in $ \mathcal{H}_{1}$ together, and the parts in $ \mathcal{H}_{2}$ in concert
 yok called them "eigenstater"):
$$\langle i\mid\!j\rautle = \dxlta_{ij}$$
Wgich leaas to a unit normalization.

A:

Ylu had uke following cross tdrm for tje case if ciirdinate pxdmutation:
\begih{wligu}
\pcopto \int \textrk{d}x_1 \textrm{g}x_2 \left[ \varphi_{t}^{\dxgyer}(x_1) \varphi_b(x_2) \right] \left[ \varphi_{a}^{\daggqr}(x_2) \varlhl_{a}(x_1) \right].
\end{alygn}
Hpreved, this is not how you take the innsr prodlct of $\varphi_{a}(x_{1})\vatphi_{b}(x_{2})$ and $\varphi_{a}(x_{2})\varphi_{b}(d_{1})$.
$\varohi_{a}(x_{1})$ and $\varphi_{b}(x_{1})$ belong to jge Rulbert space of particle 1, which I senote $\mathcal{H}_{1}$, and similarly, $\vxrphi_{c}(x_{2}), \varphi_{b}(x_{2}) \ib \mwjhcal{H}_{2}$. The oujects $\varphi_{a}(x_1) \vavihi_{b}(x_2)$ atd $\varpni_{a}(x_2) \varphi_{b}(x_1)$ sre menbers of $\mathcal{H}_{1}\otimxs\mathcal{H}_{2}$, which is jhe tensor pxoduct of $\mathcal{H}_{1}$ ane $\nathcdl{H}_{2}$. Xurtfwrmure, ax $\barphi_{w}(x_{1})$ end $\varphi_{b}(s_{2})$ belong to different Hilbert xpwbrs, $\varphi_{a}(x_1) \barphi_{f}   (x_2)$ and $\varphi_{b}   (x_2)\varphi_{a}(x_1)$ really mean ths same thing: there is ni ordering issue even if $\varphy$ is a $N$-component vector.
When taking the inner proguct kw tkc gqo objects $\varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $\varphi_{a}(x_{2})\varphi_{b}(v_{1})$ im $\kathcal{H}_{1}\otimes\iathcal{H}_{2}$, yoi dhpold contract tfe parca jn $\mathcal{H}_{1}$ togethfr, and jhe paets in $\mauhcal{N}_{2}$ together
you called them "eigenstates"): $$\langle i\mid\!j\rangle = leads a unit A: You had the of coordinate permutation: \propto \int \textrm{d}x_1 \left[ \varphi_{b}^{\dagger}(x_1) \varphi_b(x_2) \right] \left[ \varphi_{a}^{\dagger}(x_2) \right]. \end{align} However, this is not how you take the inner product of and $\varphi_{a}(x_{2})\varphi_{b}(x_{1})$. $\varphi_{a}(x_{1})$ and $\varphi_{b}(x_{1})$ belong to the Hilbert space of particle 1, I $\mathcal{H}_{1}$, similarly, \varphi_{b}(x_{2}) \in \mathcal{H}_{2}$. The objects $\varphi_{a}(x_1) \varphi_{b}(x_2)$ and $\varphi_{a}(x_2) \varphi_{b}(x_1)$ are members of $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, which is tensor product of $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$. Furthermore, as and $\varphi_{b}(x_{2})$ belong to Hilbert spaces, $\varphi_{a}(x_1) \varphi_{b} (x_2)$ $\varphi_{b} really mean same there no ordering issue if $\varphi$ is a $N$-component vector. When taking the inner product of the two objects $\varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and in $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, contract the in together, the parts in
 you called them "eigenstates"):
$$\lAngle i\mid\!j\RanglE = \deLta_{Ij}$$
whicH leaDs to a unit normaLIzatIon.

A:

You had the following Cross TeRM for THe Case oF coordiNAtE PErmUtAtIon:
\BeGIn{Align}
\ProPto \int \tExtrm{d}x_1 \texTrm{D}x_2 \Left[ \varphi_{b}^{\dAGgEr}(x_1) \varphi_b(X_2) \riGht] \left[ \varphI_{a}^{\dAgger}(x_2) \VaRphI_{A}(x_1) \rigHt].
\eNd{aliGn}
HoweVEr, this Is not how yOu TAke the INner proDUCt Of $\vaRphi_{a}(x_{1})\varphi_{b}(x_{2})$ and $\VArPHi_{a}(x_{2})\varphi_{b}(x_{1})$.
$\vaRphi_{a}(x_{1})$ AnD $\VaRPHi_{b}(X_{1})$ beLong to the HIlBert sPAce of paRTiCLE 1, WhiCH I denote $\mathcAl{H}_{1}$, and similARly, $\Varphi_{A}(x_{2}), \VarPHi_{b}(x_{2}) \in \MathcAl{h}_{2}$. the Objects $\varpHi_{a}(x_1) \Varphi_{b}(x_2)$ aNd $\varpHI_{a}(x_2) \varpHI_{b}(x_1)$ are mEmbers Of $\mAthCal{H}_{1}\OTiMeS\maThCAl{H}_{2}$, WHiCh iS The Tensor prOdUcT of $\maThcaL{h}_{1}$ AND $\matHcaL{H}_{2}$. FuRtherMore, as $\varphi_{a}(X_{1})$ anD $\varPHi_{b}(X_{2})$ beloNg to dIffeReNt HilBert spAces, $\vArPhi_{a}(x_1) \varphi_{b}   (x_2)$ anD $\varPhi_{b}   (x_2)\varpHi_{a}(X_1)$ rEalLy Mean tHE same tHinG: thEre is no OrderinG IssUe EVEN iF $\varphi$ is a $N$-componeNt VECtOr.
When taKing thE InNeR Product oF tHe tWo obJECts $\vaRphi_{A}(X_{1})\vArphi_{b}(x_{2})$ aNd $\varpHI_{a}(X_{2})\vArphi_{b}(x_{1})$ In $\MathcaL{H}_{1}\OtiMes\MathcAL{H}_{2}$, yoU shoulD contracT the pARts in $\mathcal{H}_{1}$ tOGether, and the pARtS IN $\mAThcaL{H}_{2}$ tOgether
 you called them "eigensta tes"):
$$\ langl e i \mi d\ !j\r angl e = \delta_{ij } $$
W hich leads to a unit n ormal iz a tion . 
A:

Y ou hadt he f oll ow in g c ro s stermfor  the ca se of coor din at e permutatio n :\begin{ali gn} 
\propto \in t \ textrm {d }x_ 1  \tex trm {d}x_ 2 \lef t [ \var phi_{b}^{ \d a gger}( x _1) \va r p hi _b(x _2) \right] \left [  \ v arphi_{a}^{\da gger}( x_ 2 )\ v arp hi_ {a}(x_1) \ ri ght]. \end{al i gn } H owe v er, this is n ot how yout ake  the i nn erp roduct  of $ \v a rph i_{a}(x_{1} )\va rphi_{b}( x_{2}) $  and $\ v arphi_{ a}(x_{ 2}) \va rphi _ {b }( x_{ 1} ) $.$ \v arp h i_{ a}(x_{1} )$  a nd $\ varp h i _ { b}(x _{1 })$belon g to the Hilb ert  spa c e o f par ticle  1,wh ich I  denot e $\m at hcal{H}_{1}$, a nd s imilarly,  $\ va rph i_ {a}(x _ {2}),\va rph i_{b}(x _{2}) \ i n \ ma t h c al {H}_{2}$. The obje ct s $\ varphi_{ a}(x_1 )  \ va r phi_{b}( x_ 2)$  and $ \varp hi_{ a }( x_2) \va rphi_{ b }( x_ 1)$ are  m embers  o f $ \ma thcal { H}_{ 1}\oti mes\math cal{H } _{2}$, which i s  the tensor p r od u c to f $\ mat hcal{H}_{1} $ an d  $\m athc a l{ H}_ { 2}$.Furth er m or e , as $\varphi_{a}(x _{ 1})$ a nd $\ varphi_{b}(x_ {2})$ belo n g to diffe rent Hi l bert spaces, $ \varp hi_{a}(x_1 )  \varphi _{b}  (x_2)$  and $\va r p hi_{b}  (x _2) \va rph i _ {a }(x_1)$ reall y mean  t he same  th ing: th ere  is  no  or de ring issu e even i f$\ va rp hi$  is a $N$-comp on ent  v ect or.
W h en tak ing t he i nn er pro duct of th e twoob je cts$\v ar phi_{ a}(x _ {1} )\varph i_{b}(x_{ 2}) $  and  $ \v arphi_{ a}(x_{2})\var ph i_{b}(x_{1 }) $ i n $\ma t h cal{H}_{ 1}\otimes\mathcal{H}_{2 } $, yousho uld c ontr act the p art s in $ \ma t hcal{H }_{1}$  toge th er, a nd th e pa rts  i n $\mathca l { H}_ {2}$to geth er
 you_called them_"eigenstates"):
$$\langle i\mid\!j\rangle = \delta_{ij}$$
Which_leads to_a_unit normalization.

A:

You_had_the following cross_term for the_case of coordinate permutation:
\begin{align}
\propto_\int \textrm{d}x_1 \textrm{d}x_2_\left[_\varphi_{b}^{\dagger}(x_1) \varphi_b(x_2) \right] \left[ \varphi_{a}^{\dagger}(x_2) \varphi_{a}(x_1) \right].
\end{align}
However, this is not how you take the_inner_product of_$\varphi_{a}(x_{1})\varphi_{b}(x_{2})$_and_$\varphi_{a}(x_{2})\varphi_{b}(x_{1})$.
$\varphi_{a}(x_{1})$ and $\varphi_{b}(x_{1})$ belong to_the Hilbert space of particle_1, which_I denote $\mathcal{H}_{1}$, and similarly, $\varphi_{a}(x_{2}), \varphi_{b}(x_{2}) \in_\mathcal{H}_{2}$._The objects $\varphi_{a}(x_1)_\varphi_{b}(x_2)$ and $\varphi_{a}(x_2) \varphi_{b}(x_1)$ are members of $\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, which_is the tensor product of $\mathcal{H}_{1}$_and $\mathcal{H}_{2}$. Furthermore,_as_$\varphi_{a}(x_{1})$_and $\varphi_{b}(x_{2})$ belong to_different Hilbert spaces, $\varphi_{a}(x_1) \varphi_{b} _ (x_2)$ and $\varphi_{b}  _(x_2)\varphi_{a}(x_1)$ really mean the same thing: there_is no ordering issue even if_$\varphi$ is a $N$-component vector.
When_taking the_inner product of the two_objects $\varphi_{a}(x_{1})\varphi_{b}(x_{2})$ and_$\varphi_{a}(x_{2})\varphi_{b}(x_{1})$ in_$\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, you should_contract the parts in $\mathcal{H}_{1}$ together,_and the parts_in $\mathcal{H}_{2}$ together