Normalization of a symmetric wavefunction

[phosgene]

Homework Statement

I need to find the normalization constant $N_{S}$ of a symmetric wavefunction

$ψ(x_{1},x_{2}) = N_{S}[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]$

assuming that the normalization of the individual wavefunctions $ψ_{a}(x_{1})ψ_{b}(x_{2}), ψ_{a}(x_{2})ψ_{b}(x_{1})$ are both just 1 and not orthogonal.

Homework Equations

For a symmetric wavefunction $ψ_{a}(x_{1})ψ_{b}(x_{2}) = ψ_{a}(x_{2})ψ_{b}(x_{1})$

3. Attempt at solution

I do the normalization and get $N_{S}^2∫_{-∞}^{∞} 2 {|ψ_{a}(x_{1})ψ_{b}(x_{2})|}^{2} + ψ*_{a}(x_{1})ψ*_{b}(x_{2})ψ_{a}(x_{2})ψ_{b}(x_{1}) + ψ*_{a}(x_{2})ψ*_{b}(x_{1})ψ_{a}(x_{2})ψ_{b}(x_{1})dx_{1}dx_{2}=1$

Now, since the wavefunctions are symmetric, $ψ*_{a}(x_{1})ψ*_{b}(x_{2})ψ_{a}(x_{2})ψ_{b}(x_{1}) = ψ*_{a}(x_{2})ψ*_{b}(x_{1})ψ_{a}(x_{2})ψ_{b}(x_{1})$

Since the answer is supposed to be $N_{S}=1/\sqrt{2}$, I'm guessing that $ψ*_{a}(x_{1})ψ*_{b}(x_{2})ψ_{a}(x_{2})ψ_{b}(x_{1}) = 0$, but I don't know why this would be.

 

[MisterX]

Well, you might have saved yourself some work by just applying the equality at the beginning, I think.

$ψ(x_{1},x_{2}) = N_{S}[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]$
$ψ_{a}(x_{1})ψ_{b}(x_{2}) = ψ_{a}(x_{2})ψ_{b}(x_{1})$
$ψ(x_{1},x_{2}) = 2N_{S}ψ_{a}(x_{1})ψ_{b}(x_{2})$But in case you were wondering what to do with your later expression

$\left(a^* + b^*\right)\left(a + b\right) = a^*a +b^*b + a^*b + ab^*$

$= |a|^2 + |b|^2 + (a^*b) + (a^*b)^*$
Adding a number to its complex conjugate eliminates the imaginary component and doubles the real component.
$= |a|^2 + |b|^2 + 2Re\left\{a^*b \right\}$

Consider what happens when $a = b$.

Edit: perhaps I'm missing something though

 

[phosgene]

Well, you might have saved yourself some work by just applying the equality at the beginning, I think.

$ψ(x_{1},x_{2}) = N_{S}[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]$
$ψ_{a}(x_{1})ψ_{b}(x_{2}) = ψ_{a}(x_{2})ψ_{b}(x_{1})$
$ψ(x_{1},x_{2}) = 2N_{S}ψ_{a}(x_{1})ψ_{b}(x_{2})$

But if I do this and try to normalize it, won't I get

$N_{S}^2∫_{-∞}^{∞}4|ψ_{a}(x_{1})ψ_{b}(x_{2})|^2 dx_{1}dx_{2}$

which gets me $N_{S}=1/2$

But in case you were wondering what to do with your later expression

$\left(a^* + b^*\right)\left(a + b\right) = a^*a +b^*b + a^*b + ab^*$

$= |a|^2 + |b|^2 + (a^*b) + (a^*b)^*$
Adding a number to its complex conjugate eliminates the imaginary component and doubles the real component.
$= |a|^2 + |b|^2 + 2Re\left\{a^*b \right\}$

Consider what happens when $a = b$.

Edit: perhaps I'm missing something though

$2Re(a*b) = 2|a|^2, 2|b|^2$ and then overall I would get $4|a|^2$ or $4|b|^2$, which again gives me $N_{S}=1/2$, but I need $N_{S} = 1/ \sqrt{2}$ so I'm not really sure what to do

 

[MisterX]

Perhaps I replied without sufficient knowledge. It might be that the number of states has been halved (identical particles).

 

[tannerbk]

$ψ(x_{1},x_{2}) = N_{S}[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]$

This is the wavefunction for two identical particles, specifically bosons. The wave function is required to satisfy the symmetry equation you stated. To solve, we have $|N_{S}|^2∫[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]^{*}[ψ_{a}(x_{1})ψ_{b}(x_{2}) + ψ_{a}(x_{2})ψ_{b}(x_{1})]dx_{1}dx_{2}=1$. Expand this equation out carefully. It is always possible to orthonormalize using Gram-Schmidt so that $∫|ψ_{a}(x_{1})^2|=1$ and $∫ψ_{a}(x_{1})^{*}ψ_{b}(x_{1})=0$. This should give you $N_{S} = 1/√2$.