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Consider two non-interacting distinguished particles in one-dimensional infinity square potential. Suppose the particles have the same mass $m$, and the potential is zero in the region $\frac{a}{2}>x>-\frac{a}{2}$. In the coordinate of $x_1,x_2$, the Schrodinger eq. $$-\frac{\hbar^2}{2m}\partial^2_{x_1}\psi(x_1,x_2)-\frac{\hbar^2}{2m}\partial^2_{x_2}\psi(x_1,x_2)=E\psi(x_1,x_2)$$

can be seperated and the solution is

$$\psi(x_1,x_2)=\frac{2}{a} \sin [\frac{n_1\pi}{a}(x_1+\frac{a}{2})] \sin [\frac{n_2\pi}{a}(x_1+\frac{a}{2})]$$ $$E_{n_1n_2} =\frac{\pi^2\hbar^2}{2ma^2}(n_1^2+n^2_2)$$

For $n_1=n_2=1$, we get the ground statewave function and energies.

My question is how to sovle the above problem in center of mass frame?

In center of mass frame the schrodinger equation becomes: $$-\frac{\hbar^2}{2M}\partial^2_{R}\psi(R,r)-\frac{\hbar^2}{2\mu}\partial^2_{r}\psi(R,r)=E\psi(R,r)$$ Naively if we require wave function is nonvanshing in the region $$-\frac{a}{2} < R<\frac{a}{2}\\ -a <r<a $$ Then we will find the ground energy is half of $E_{11} =\frac{\pi^2\hbar^2}{2ma^2}(1+1)$. What's wrong wth the center of mass frame calculation?

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    $\begingroup$ I don't have the solution yet. But clearly, what is wrong with your approach is that the limits of the box for $r$ depend on the value of $R$, e.g. when $R = 0$, $-a < r < a$, but when $R = \pm a/2$, $r = 0$ (otherwise one particle leaves the box). The boundary condition on $r$ depend on $R$. The problem is not separable. If you regard $V(|x_j|\ge a/2) = \infty$, then you can see that for the internal motion the potential in $r$ changes with the values of $R$. $\endgroup$ – perplexity Dec 4 '13 at 13:53
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    $\begingroup$ note that center of mass coordinates are most useful when their is a interaction potential between two particles which depends on their separation. Here you have objects in an external potential so as perplexity says, center of mass coordinates don't work so well here. $\endgroup$ – Brian Moths Dec 5 '13 at 15:10
  • $\begingroup$ Thanks for comments. Maybe it is not a good question. $\endgroup$ – thone Dec 5 '13 at 15:37
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    $\begingroup$ I think the question is quite fine, and @perplexity should turn his comment into an answer. $\endgroup$ – Ruslan Dec 5 '13 at 16:00

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