# Energy levels for two particles in infinite potential well

by AStaunton
Tags: energy, infinite, levels, particles, potential
 P: 105 problem is: (a)write down the spatial or orbital for two-non interacting particles, with the same mass, in a one dimensional well, where the potential energy is zero for 0
 HW Helper P: 2,322 No, they're not valid. An energy level is just that: a definite energy. It doesn't matter how that energy was achieved or what the components of the system look like; the only thing that matters is the value of the energy. In this case: $$E_{N}=\frac{N^{2}\pi^{2}\hbar^{2}}{8ma^{2}}$$ where N^2=n^2+n_bar^2, and the lowest energy levels correspond to the lowest values of N.
 P: 105 OK using this: N^2=n^2+n_bar^2 and it is not allowed that n is equal to nbar?
HW Helper
P: 2,322
Energy levels for two particles in infinite potential well

 Quote by AStaunton OK using this: N^2=n^2+n_bar^2 and it is not allowed that n is equal to nbar?
It is allowed. Every possible value of N^2 corresponds to one energy level. It doesn't matter what values n and n_bar take, or whether they're the same.
 P: 105 OK, so in that case weren't my assumptions that I stated at the bottom of my first post valid? it seems that you are saying the same thing... unless I'm not understanding you properly... ie. would you agree that the four energy levels I stated E_1system to E_4system are indeed the four lowest energies?
 HW Helper P: 2,322 I wouldn't agree because you posted: $$E_{3system}=5E_{0}$$ $$E_{4system}=5E_{0}$$ Those represent the same energy level, because they have the same energy.
P: 188
 Quote by ideasrule It is allowed. Every possible value of N^2 corresponds to one energy level. It doesn't matter what values n and n_bar take, or whether they're the same.
However, this is only true if the particles are identical bosons, right? If they're fermions we have to take the Pauli principle into account.

(Although I see now that it was stated in the OP that degenerate levels are allowed, implying that we're dealing with bosons.)

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