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The state of two identical bosons

  1. Aug 4, 2014 #1
    Hi PF.
    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 ?
     
  2. jcsd
  3. Aug 4, 2014 #2
    That's not an issue. You can always renormalize the wave function. All the wave functions must have unitary norm
     
  4. Aug 4, 2014 #3
    Thanks Einj, that helped a lot.
    I just found it strange that my textbook would write the normalization factor as [itex]\frac{1}{\sqrt{N!}}[/itex], for N identical bosons.. But this only seems to be the case where the bosons isn't occupying the same single-particle state.

    It would be more accurate/general to write the two-particle state as :[itex]\psi(x_{1},x_{2}) = A\big(\phi_{a}(x_{1})\phi_{b}(x_{2}) + \phi_{b}(x_{1})\phi_{a}(x_{2})\big)[/itex].

    And again thanks :-)
     
  5. Aug 4, 2014 #4
    Yes indeed. If I remember correctly the right normalization for a bosonic wave function is:
    $$
    \Psi(x_1,...,x_N;\alpha_1,...,\alpha_N)=\frac{1}{\sqrt{N!\prod_{k=1}^N n_k!}}\sum_{\sigma}\sigma\left(\phi_{\alpha_1}(x_1)\phi_{\alpha_2}(x_2)\cdots \phi_{\alpha_N}(x_N)\right),
    $$
    where [itex]\sigma[/itex] are all the possible permutations and [itex]n_k[/itex] is the number of bosons in the same state [itex]\alpha_k[/itex].

    I hope I'm not missing anything, but that was the main idea.
     
  6. Aug 4, 2014 #5
    Yes, exactly... This gives the correct normalized wavefunction I was looking for. The lecture notees I'm reading has just dropped the factor of [itex]\frac{1}{\sqrt{\Pi_{k=1}^{N}n_{k}!}}[/itex].

    Enjoy the rest of your day :-)
     
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