3 added 889 characters in body edited May 4 '18 at 16:06 ohneVal 1,06466 silver badges1515 bronze badges 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 hope you getunderstand correctly this should be 1 of the pointpossible 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} 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. I hope you get the point 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} 2 added 633 characters in body edited May 4 '18 at 15:25 ohneVal 1,06466 silver badges1515 bronze badges 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. I hope you get the point 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. 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. I hope you get the point 1 answered May 4 '18 at 14:01 ohneVal 1,06466 silver badges1515 bronze badges 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.