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A professor told me that most physicists assume that all identical fermions are in completely orthogonal states. If that is true, then does that mean that that the total wave function is highly localized for a closed box of a huge number of electrons? I get confused between when your supposed multiply wavefunctions and when your supposed to add wavefunctions. I know that for one electron, adding all orthogonal states of that one electron in a closed box would yield a highly localized spatial wavefunction.

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Orthogonal states in what space? "All fermions" in the sense of "all electrons" or in the sense of "the electron, the muon, the up quark..."? Why would orthogonality of states imply high localization (momentum eigenstates are also orthogonal and highly non-localized, for example)? More information, please! – ACuriousMind Jul 17 '14 at 18:21
I think your professor might have said or meant completely antisymmetric N-particle states. As to the second question, the answer is no, you do not simply add the single-particle wavefunctions. Your N-particle wavefunction is a function of the form $\Psi(x_1,x_2,...,x_N)$ and it can't be generally written as a product of single-particle wavefunctions. – Void Jul 17 '14 at 18:55
Is there a way to add observables? – linuxfreebird Jul 17 '14 at 18:57

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What you refer to is probably to ground state of a infinite potential wall. If this is the case, then there, in the ground state, the particles are not localized. You can find the solution of the orthogonal ground-states here:

We can reduce the Problem to a one dimensional case, that doesn't change the Problem or solution to much.

With orthogonal states is meant that the integral over the wave function vanishes. So in the case of a one dimensional box this is fulfilled for the given sinus functions. That doesn't imply that the two particles may not be found at the some position in space. If that would be the case, the electrons would have to be localized. But this is not the requirement for orthogonal states.

I hope this helps to clarify this.

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I assumed that if the electrons were closed in a box, the wavefunction was forced to be zero at the boundary. I also assumed that if the wavefunction must be zero at the closed impenetrable boundary, then the only eigenstates that are allowed are discretely spaced moment states. Sorry I think this is more of a quantum dot problem. – linuxfreebird Jul 17 '14 at 19:11
The function is for the infinite quantum well zero. If you look at a finite height, like in a quantum dot, this is not the case. In the case of energy eigenvalues, localization is not possibly, as this would require an superposition of different sinus-functions. – user3384414 Jul 17 '14 at 19:19
I apologize. I was confusing adding and multiplying wavefunctions. I was thinking of adding wave spatial observables for all of the different fermion states in a impenetrable box. If there was such a thing as a spatial wave observable of a fermion, then I think the sum of all of whatever this spatial wave observable would represent something that is highly localized. – linuxfreebird Jul 17 '14 at 19:26

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