The state of two identical bosons

[Wuberdall]

Hi PF.
If I have two identical bosons, one in the single-particle state $\phi_{a}(x)$ and the other in the single-particle state $\phi_{b}(x)$, the two-particle state of the system would then be :

$\psi(x_{1},x_{2}) = \frac{1}{\sqrt{2}}\big(\phi_{a}(x_{1})\phi_{b}(x_{2}) + \phi_{b}(x_{1})\phi_{a}(x_{2})\big)$

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

$\psi(x_{1},x_{2}) = \sqrt{2}\phi_{a}(x_{1})\phi_{a}(x_{2})$

But this two-particle state isn't normalized, due to factor of $\sqrt{2}$.
I can't see were I go wrong, someone who can help ?

 

[Einj]

That's not an issue. You can always renormalize the wave function. All the wave functions must have unitary norm

 

[Wuberdall]

Thanks Einj, that helped a lot.
I just found it strange that my textbook would write the normalization factor as $\frac{1}{\sqrt{N!}}$, 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 :$\psi(x_{1},x_{2}) = A\big(\phi_{a}(x_{1})\phi_{b}(x_{2}) + \phi_{b}(x_{1})\phi_{a}(x_{2})\big)$.

And again thanks :-)

 

[Einj]

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 $\sigma$ are all the possible permutations and $n_k$ is the number of bosons in the same state $\alpha_k$.

I hope I'm not missing anything, but that was the main idea.

 

[Wuberdall]

Yes, exactly... This gives the correct normalized wavefunction I was looking for. The lecture notees I'm reading has just dropped the factor of $\frac{1}{\sqrt{\Pi_{k=1}^{N}n_{k}!}}$.

Enjoy the rest of your day :-)