# Ground state of three non-interacting fermions at an infinite well

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: $$\psi^{(0)}(x_1,x_2,x_3) = \frac{1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_1) |{+}\rangle & \psi_1(x_2) |{-}\rangle & \psi_1(x_3) |{+}\rangle \\ \psi_1(x_1) |{-}\rangle & \psi_1(x_2) |{+}\rangle & \psi_1(x_3) |{+}\rangle \\ \psi_2(x_1) |{+}\rangle & \psi_2(x_2) |{+}\rangle & \psi_2(x_3) |{-}\rangle \\ \end{vmatrix}$$

But this shows that there are particles in the same state, despite being fermions. Is there something wrong here?

• I think you're right, the determinant is not constructed correctly. I expanded it to write down all the terms, and the result is symmetric with respect to exchange of particles 1 and 2. May 4, 2018 at 14:46

As mentioned in the comments, the wavefunction as posed is symmetric in the exchange between particles 1 and 2. This can be seen directly from the wavefunction, where to avoid confusion it is easier to take explicit spin components: \begin{align} \psi^{(0)}_{i_2,i_1,i_3}(x_2,x_1,x_3) &= \frac{1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = \frac{-1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = \frac{+1}{\sqrt{3!}} \begin{vmatrix} \psi_1(x_1) \langle i_1|{+}\rangle & \psi_1(x_2) \langle i_2|{-}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_1(x_1) \langle i_1|{-}\rangle & \psi_1(x_2) \langle i_2|{+}\rangle & \psi_1(x_3) \langle i_3|{+}\rangle \\ \psi_2(x_1) \langle i_1|{+}\rangle & \psi_2(x_2) \langle i_2|{+}\rangle & \psi_2(x_3) \langle i_3|{-}\rangle \\ \end{vmatrix} \\& \\ & = + \psi^{(0)}_{i_1,i_2,i_3}(x_1,x_2,x_3) \end{align}

Because of this symmetry under exchange, the wavefunction as written is not a suitable state for identical fermions.

No, there is nothing wrong. The wavefunction is just a mixture of one particle solutions which must be anti-symmetrized because of the particles being fermions. So as it is explained in the book, two of the particles can have a groundstate wave function (the label 1, one with spin up and one with spin down) while the other has to be in the 1st excited state. (labeled $\psi_2$ in your question). If you expand the determinant you wrote, you will see that you will get all the combinations of what I just described for the 3 particles of the problem.

EDIT: You are probably expanding the determinant incorrectly. Expanding by minors through the first column for example:

$\psi^{(0)} = \frac{1}{\sqrt{3!}}\left( \psi_1(x_1)|+\rangle \otimes \psi_1(x_2)|+\rangle \otimes \psi_2(x_3)|-\rangle - \psi_1(x_1)|+\rangle \otimes \psi_1(x_3)|+\rangle \otimes \psi_2(x_2)|+\rangle + \cdots \right)$

you see there are no two states that are the same. Two states are the same if ALL particles from the first have the same wave function AND spin when compared with the second. As you said the first term here contains both particle 1 and 2 with spin + in the ground state, but it will be cancelled hopefully...

Remember tensor product is not commutative, that might also be a mistake.

EDIT2:

Yeah it is not cancelling for me so ostrichCamel might be right. The determinant can have an error. If I understand correctly this should be 1 of the possible ground states from a total of 4 that the book claims, however it appears to me that there is only 2 dictated by the spin of the particle in the excited level since one cannot know which of the 3 it is. A corrected determinant would give (for the case of $\psi_2|+\rangle$) \begin{align}\psi^{0}(x_1,x_2,x_3) \sim\; &\psi_1(x_1)|+\rangle \psi_1(x_2)|-\rangle \psi_2(x_3)|+\rangle \\ & -\psi_1(x_1)|-\rangle \psi_1(x_2)|+\rangle \psi_2(x_3)|+\rangle \\ &+\psi_1(x_3)|+\rangle \psi_1(x_1)|-\rangle \psi_2(x_2)|+\rangle \\ &-\psi_1(x_3)|-\rangle \psi_1(x_1)|+\rangle \psi_2(x_2)|+\rangle \\ &+\psi_1(x_2)|+\rangle \psi_1(x_3)|-\rangle \psi_2(x_1)|+\rangle \\ &-\psi_1(x_2)|-\rangle \psi_1(x_3)|+\rangle \psi_2(x_1)|+\rangle \end{align}

• The first term of the determinant (the way I expanded it) is the amplitude $\lvert\psi_1\psi_1\psi_2\rangle\lvert++-\rangle$. Since none of the terms cancel out, this means there's an amplitude for finding particles 1 and 2 both in the state $\lvert\psi_1,+\rangle$... May 4, 2018 at 15:03
• As ostrichCamel mentioned: by expanding the determinant I have several solutions which put two identical fermions at the ground state with the same spin. Plus, the expansion result is symmetric under the exchange of pairs of particles. May 4, 2018 at 15:09
• Answer updated @RicardoP , sorry I missed your point on the first go May 4, 2018 at 16:21