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I need to find the wave function for two non-interacting particles of mass $m_1$ and $m_2$ in 1D infinite box (well) of length $L$, where the positions of the particle is given by $x_i$ ($i$ being $1,2$).

I have found the wave function using the method of separation of variables:

$$\Psi(x_1,x_2) = C \sin\left({\frac{n_1 \pi x_1}{L}}\right) \sin\left({\frac{n_2 \pi x_2}{L}}\right)$$

How should I normalise this to obtain $C$? Should I integrate with respect to to $\text{d}x_1 \text{d}x_2$ as if this were a 2D problem or is there some other way?

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    $\begingroup$ I've edited your question to include your equation in MathJax, do try to do this from now on? It makes your questions easier to search (and read!) for other users. $\endgroup$
    – Philip
    Aug 25, 2020 at 7:24
  • $\begingroup$ Yes, you are right. In addition, you can consider in detail a case with even $n_1$ and odd $n_2$, for instance. $\endgroup$ Aug 25, 2020 at 7:53
  • $\begingroup$ Right in the sense that it is identical to a 2D problem with Integral done wrt to $dx_1dx_2$ ? $\endgroup$
    – Lost
    Aug 25, 2020 at 8:21

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Positions of individual particles are just orthogonal degrees of freedom of the system. In general, to obtain a scalar product of two states from the appropriate product of wavefunctions you should just integrate over all of the degrees of freedom (and sum over discrete DoF's)—not forgetting about Jacobian determinant in curvilinear coordinates.

Thus, whether $\Psi(x_1,x_2)$ represents a state of two 1D particles or of a single 2D particle, doesn't matter for the purpose of normalization or scalar multiplication. The square of norm is simply

$$\langle\Psi|\Psi\rangle=\int|\Psi(x_1,x_2)|^2\,\mathrm{d}x_1\mathrm{d}x_2.$$

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  • $\begingroup$ The above expression ,as I interpret it, represents probability of finding 1st particle in $dx_1$ *and (reperesenting multiolication law of probabilites for independent events) the other in $dx_2$. So from this reasoning , lets say if I wanted to find the probability of finding one particle in $dx_1$ then that would just be identical to a 1-D scenario since they are non-interacting , right? $\endgroup$
    – Lost
    Aug 25, 2020 at 9:56
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    $\begingroup$ @Lost if you need probability distribution for particle 1 (regardless of position of particle 2), you need to integrate out the other degrees of freedom, i.e. your function will be $\rho(x_1)=\int |\Psi(x_1,x_2)|^2\,\mathrm{d}x_2.$ $\endgroup$
    – Ruslan
    Aug 25, 2020 at 9:58

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