Ground state of 3 noninteracting Fermions in an infinite well

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SUMMARY

The discussion centers on the ground state of three non-interacting identical spin 1/2 fermions in a one-dimensional infinite potential well, as described in Zettili's Quantum Mechanics, page 477. A configuration presented in the text was questioned for its validity due to the presence of identical particles in the same state, which contradicts the Pauli exclusion principle. The participants concluded that the determinant representation of the wave function must be antisymmetric under particle exchange, and a correction was suggested to ensure that two fermions occupy the ground state with opposite spins while the third occupies the first excited state.

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RicardoMP
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In Zettili's Quantum Mechanics, page 477, he wants to determine the energy and wave function of the ground state of three non-interacting identical spin 1/2 particles confined in a one-dimensional infinite potential well of length a. He states that one possible configuration of the ground state wave function is the one as presented in the .PNG.
But this shows that there are particles in the same state, despite being fermions. Also, by expanding the determinant, the result isn't anti-symmetric under an exchange of a pair of particles. Is there something wrong here?
 

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RicardoMP said:
In Zettili's Quantum Mechanics, page 477, he wants to determine the energy and wave function of the ground state of three non-interacting identical spin 1/2 particles confined in a one-dimensional infinite potential well of length a. He states that one possible configuration of the ground state wave function is the one as presented in the .PNG.
But this shows that there are particles in the same state, despite being fermions. Also, by expanding the determinant, the result isn't anti-symmetric under an exchange of a pair of particles. Is there something wrong here?
You are correct. The second row of the determinant is identical to the first which will of course make the determinant zero. This looks like a typo to me. The second row wavefunctions should have subscripts "3" instead of "1" for this to make sense.
 
Are the first and second rows really identical? The spins for the first two terms of each mentioned row have different spin states. Otherwise yes, the determinant would be zero. However, despite the configuration being one that doesn't cancel the determinant, it is one that involves identical fermions in the same state. Indeed, I agree this might be a typo.
 
RicardoMP said:
Are the first and second rows really identical?
They are not. I was too hasty. Let me think for a moment about what it should be in determinant form.
 
I think it should be
$$\begin{vmatrix}
\psi_1(x_1)|+> & \psi_1(x_2)|+> & \psi_1(x_3)|+> \\
\psi_1(x_1)|-> & \psi_1(x_2)|-> & \psi_1(x_3)|-> \\
\psi_2(x_1)|+> & \psi_2(x_2)|+> & \psi_2(x_3)|+>
\end{vmatrix}$$
This is antisymmetric under particle exchange and places two fermions in the ground state with opposite spins and one in the first excited state with spin up.
 

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