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.