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If I have two identical bosons, one in the single-particle state [itex]\phi_{a}(x)[/itex] and the other in the single-particle state [itex]\phi_{b}(x)[/itex], the two-particle state of the system would then be :

[itex]\psi(x_{1},x_{2}) = \frac{1}{\sqrt{2}}\big(\phi_{a}(x_{1})\phi_{b}(x_{2}) + \phi_{b}(x_{1})\phi_{a}(x_{2})\big)[/itex]

right ?

Then if both bosons are in the same single-particle state, say [itex]\phi_{a}(x)[/itex], then the two-particle state according the equation above, would be :

[itex]\psi(x_{1},x_{2}) = \sqrt{2}\phi_{a}(x_{1})\phi_{a}(x_{2})[/itex]

But this two-particle state isn't normalized, due to factor of [itex]\sqrt{2}[/itex].

I can't see were I go wrong, someone who can help ?

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# The state of two identical bosons

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