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.

  • 2
    $\begingroup$ symmetric and antisymmetric wave functions are required for indistinguishable/idental particles. But if you have a system with non-identifical particles you can describe this system with a non-symmetric and non-antisymmetric wave function. $\endgroup$
    – Davius
    Oct 17, 2022 at 9:40
  • 3
    $\begingroup$ Please use MathJax. $\endgroup$ Oct 17, 2022 at 9:43

2 Answers 2


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.

  • $\begingroup$ Please use MathJax. $\endgroup$ Oct 17, 2022 at 10:47
  • $\begingroup$ In the case of an antisymmetric wave function, we have : $$ |μ,ν⟩ = |μ⟩ |ν⟩ = - |νμ⟩ = - |ν⟩ |μ⟩ $$ The one does not equal the other, so during your example you've exchanged the two kets without changing the sign and thus presupposing that the wave function is symmetrical. If you would to try with the antisymmetric case, you would obtain : $$ ⟨μ,ν|μ,ν⟩ = 1 & ⟨ν,μ|μ,ν⟩ = -⟨μ,ν|μ,ν⟩ = -1 $$ $\endgroup$
    – HitMan01
    Oct 17, 2022 at 12:11

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.

  • $\begingroup$ As was stated below the question, please use Mathjax. Those equations are difficult to parse otherwise. $\endgroup$
    – Miyase
    Oct 17, 2022 at 13:30
  • $\begingroup$ @Miyase Actually the equations are typed with MathJax, but not in the "cleanest" style, which however is just a matter of practice. $\endgroup$ Oct 17, 2022 at 13:57
  • $\begingroup$ @Miyase - what do you mean the cleanest style? where am I wrong? Moreover, is the my proof correct? (I think that's the important issue here :) ) $\endgroup$
    – DDonkey
    Oct 17, 2022 at 14:28

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