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Normalization of a symmetric wavefunction

  1. Jun 2, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to find the normalization constant [itex]N_{S}[/itex] of a symmetric wavefunction

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

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

    2. Relevant equations

    For a symmetric wavefunction [itex]ψ_{a}(x_{1})ψ_{b}(x_{2}) = ψ_{a}(x_{2})ψ_{b}(x_{1})[/itex]

    3. Attempt at solution

    I do the normalization and get [itex]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[/itex]

    Now, since the wavefunctions are symmetric, [itex]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})[/itex]

    Since the answer is supposed to be [itex]N_{S}=1/\sqrt{2}[/itex], I'm guessing that [itex]ψ*_{a}(x_{1})ψ*_{b}(x_{2})ψ_{a}(x_{2})ψ_{b}(x_{1}) = 0[/itex], but I don't know why this would be.
  2. jcsd
  3. Jun 2, 2013 #2
    Well, you might have saved yourself some work by just applying the equality at the beginning, I think.

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

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

    [itex]\left(a^* + b^*\right)\left(a + b\right) = a^*a +b^*b + a^*b + ab^*[/itex]

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

    Consider what happens when [itex]a = b[/itex].

    Edit: perhaps I'm missing something though
    Last edited: Jun 2, 2013
  4. Jun 2, 2013 #3
    But if I do this and try to normalize it, won't I get

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

    which gets me [itex]N_{S}=1/2[/itex]

    [itex]2Re(a*b) = 2|a|^2, 2|b|^2[/itex] and then overall I would get [itex]4|a|^2[/itex] or [itex]4|b|^2[/itex], which again gives me [itex]N_{S}=1/2[/itex], but I need [itex]N_{S} = 1/ \sqrt{2}[/itex] so I'm not really sure what to do
  5. Jun 2, 2013 #4
    Perhaps I replied without sufficient knowledge. It might be that the number of states has been halved (identical particles).
  6. Jun 2, 2013 #5
    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 [itex] |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 [/itex]. Expand this equation out carefully. It is always possible to orthonormalize using Gram-Schmidt so that [itex]∫|ψ_{a}(x_{1})^2|=1 [/itex] and [itex] ∫ψ_{a}(x_{1})^{*}ψ_{b}(x_{1})=0 [/itex]. This should give you [itex] N_{S} = 1/√2 [/itex].
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