Say you have two particles a and b with respective positions ##x_a## and ##x_b##. Particle a is in the state ##\psi_a##, and particle b is in the state ##\psi_b##. If they are distinguishable, the wavefunction is(adsbygoogle = window.adsbygoogle || []).push({});

$$\psi=\psi_a(x_a)\psi_b(x_b)$$

However, if they are identical fermions, the wavefunction is

$$\psi(a,b)=A\left(\psi_a(x_a)\psi_b(x_b)-\psi_a(x_b)\psi_b(x_a)\right)$$

If ##\psi_a=\psi_b## then ##\psi=0##. But, if ##\psi_a## is arbitrarily close to ##\psi_b## (i.e. ##\psi_a=\psi_1## and ##\psi_b=\sqrt{.99}\psi_1+.1\psi_2## where 1 and 2 denote the 1st and 2nd principal quantum numbers of some potential) then the wavefunction is not zero, and can be normalized if A is arbitrarily large. So, it would seem as though you could have an infinite number of identical fermions in just two energy levels. I know, however, that is not the case. So, I came up with a thought on why this is, but it had a problem. Could tell me if I'm on the right track, and where I am going wrong?

The constant A in the identical fermion equation must be ##\frac{1}{\sqrt{2}}## (for N particles it would be ##\frac{1}{\sqrt{N!}}##). This would require that ##\psi_a## and ##\psi_b## (and any other wavefunctions corresponding to more identical fermions) be orthogonal in order to keep ##\psi## normalized. But, this has two issues. First, as a and b get farther apart, the second identical fermions start to act distinguishable. So, ##\psi_a(x_b)## and ##\psi_b(x_a)## go to zero, and ##\psi## goes to ##\frac{1}{\sqrt{2}}\psi_a(x_a)\psi_b(x_b)##, which is not normalized (as well as not the same as the distinguishable particle equation).

I also had a thought on that; my thinking was that even as a and b get arbitrarily far apart, in theory they are still identical, and particle a being in in state ##\psi_a## and particle b being in ##\psi_a## are the same things, as the fermions are indistinguishable. Thus, ##\psi_a(x_b)\psi_b(x_a)## does not go to zero. But, since ##\psi_a## and ##\psi_b## do not overlap, they are essentially orthogonal. It appears as though this would make the identical fermion equation yield about the same practical result as the distinguishable equation.

Again, this is just my thinking, I have no idea how far off I am.

Thanks in advance!

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# I 2-particle Wavefunction and the Pauli Exclusion Principle

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