Multi-particles system wave function As we all know there is a symmetric and antisymmetric of 2 particles system states.
\begin{align}
\text{symmetric} & \ \rightarrow & |μ,ν⟩=\frac{|μ⟩|ν⟩+|ν⟩|μ⟩}{\sqrt{2}}
\\
\text{antisymmetric} & \ \rightarrow & |μ,ν⟩=\frac{|μ⟩|ν⟩-|ν⟩|μ⟩}{\sqrt{2}}
\end{align}
Are there also other types of multi system with complex coefficients?
For example:
Two particles $|μ⟩$ and $|ν⟩$  s.t
$$
|μ,ν⟩=\frac{|μ⟩|ν⟩+i|ν⟩|μ⟩}{\sqrt{2}}
$$
This wave function keeps orthonormal to work out.
$$
⟨μ,ν|μ,ν⟩ = 1/2(⟨v|⟨μ|μ⟩|v⟩ + i⋅⟨v|⟨μ|ν⟩|μ⟩ + (i^*)⋅⟨μ|⟨ν|μ⟩|v⟩ + i⋅(i^*)⟨μ|⟨v|v⟩|μ⟩)  
$$
The two middle components are equal to 0 as |μ⟩ orthogonal to |ν⟩.
So, we end up with
$$
⟨μ,ν|μ,ν⟩ = 1/2(⟨v|⟨μ|μ⟩|v⟩ + i⋅(i^*)⟨μ|⟨v|v⟩|μ⟩) = 1/2(1+i⋅(i^*)) = 1/2(1+1) = 1.
$$
The example I have shown is clearly neither a symmetrical nor anti-symmetrical case.
 A: I think that when we want to take about the symmetry of a wave function the order of the kets are important. |u,v> = -|v,u> in the antisymmetric case, however through your example you switched the two without changing the sign meaning you have considered the symmetric case. Try again but this time when you want to permute |u> and |v>, you should put a minus sign.
A: Alright I have an answer.
If the particles are identical and indistinguishable.
Then probability is preserved if we swap the particles:
$|\Psi_{12}|_{(1,2)}^2  = \Psi_{21}|_{(2,1)}^2$
$\Psi_{12(1,2)}\cdot\Psi^*_{12(1,2)}  = \Psi_{21(2,1)}\cdot\Psi^*_{21(2,1)} $
If we take my general example with complex coefficient
$\Psi_{12} = 1/\sqrt2 \cdot (|1⟩|2⟩ + c|2⟩|1⟩)$   where c∊Complex
$\Psi^*_{12(1,2)} = 1/\sqrt2 \cdot ((|1⟩|2⟩)^* +c^*(|2⟩|1⟩)^*)$
$\Psi_{12(1,2)}\cdot\Psi^*_{12(1,2)} = 1/2 \cdot ((|1⟩|2⟩)(|1⟩|2⟩)^* + (|1⟩|2⟩)c^*(|2⟩|1⟩)^* +  c|2⟩|1⟩(|1⟩|2⟩)^* + c|2⟩|1⟩ c^*(|2⟩|1⟩)^*  ) $
$\Psi_{12(1,2)}\cdot\Psi^*_{12(1,2)} = 1/2 \cdot ((|1⟩|2⟩)(|1⟩|2⟩)^* + (|1⟩|2⟩)c^*(|2⟩|1⟩)^* +  c|2⟩|1⟩(|1⟩|2⟩)^* + |c|^2|2⟩|1⟩ (|2⟩|1⟩)^*  )$
by the same calculation:
$\Psi_{21(2,1)}\cdot\Psi^*_{21(2,1)} = 1/2 \cdot ((|2⟩|1⟩)(|2⟩|1⟩)^* + (|2⟩|1⟩)c^*(|1⟩|2⟩)^* +  c|1⟩|2⟩(|2⟩|1⟩)^* + |c|^2|1⟩|2⟩ (|1⟩|2⟩)^*  )$
Because |c|^2 = 1 to keep the wave function to be normalized. These components are equals:
$|c|^2|1⟩|2⟩ (|1⟩|2⟩)^* = (|1⟩|2⟩)(|1⟩|2⟩)^*$  and
$|c|^2|2⟩|1⟩ (|2⟩|1⟩)^* = (|2⟩|1⟩)(|2⟩|1⟩)^*$
We want that these also be equal to satisfy our identical particles.
$(|1⟩|2⟩)c^*(|2⟩|1⟩)^* = c|1⟩|2⟩(|2⟩|1⟩)^*$ To satisfy this equation we need to choose coefficient c s.t $c=c^*$
To sum up, c must apply:
$$|c|^2=1$$
$$c=c^*$$
So for identical particles $c={-1,+1}$
By thus, my example where $c=i$  is not true.
