I came across this problem in N. Zettili's Quantum Mechanics book (Chapter 9, Problem 16):

Two identical particles of spin 1/2 are enclosed in a one-dimensional box potential of length L with walls at x=0 and x=L. Find the energies of the lowest three states.

This is a solved exercise. And the solution says that

Since the two particles have the same spin, the spin part of wave function of the system must be symmetric, that is, any one of the triplet states. The overall wavefunction of two identical fermions is anti-symmetric, so the space part will be anti-symmetric.

I am confused as to why this is so? Can anyone explain why the spin part of wavefunction MUST be symmetric and can't be anti-symmetric in this case?


2 Answers 2


The ground state has the two particles in the same orbital with opposite spin. The next level has one particle promoted to the next lowest orbital. The two particles can now form a singlet or a triplet. The four states are degenerate as there is no interaction (specified). Thus the question has 6 equally valid answers. It is questionable that this is what the author had in mind.


I'm using the standard notation throughout the whole answer. The problem of adding angular momenta is essentially a change of basis, from one that diagonalizes $(S_1^2,S_2^2,S_{1z},S_{2z})$ to one that diagonalizes $(S^2,S_z,S_1^2,S_2^2).$ If you work out the problem which is given in many texts you will find the following transformation.

$$|s=1m=1,s_1=1/2 s_2=1/2\rangle =|++\rangle$$ $$|s=1m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle+|-+\rangle]$$ $$|s=1m=-1,s_1=1/2 s_2=1/2\rangle =|--\rangle$$ $$|s=0m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle-|-+\rangle]$$

The allowed values for total spin are $s=1$ and $0$,while the allowed values of $s_z$ are $\hbar,0$ and $-\hbar$.

For a system of two spins 1/2 particles the wavefunction have the following possible forms

$$\Psi = \left\{ \begin{array}{l} \psi_a\chi_s \\ \psi_s\chi_a \end{array} \right.$$ subscript $s$ and is to denote symmetric and anti-symmetric.

The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions $$ \begin{align} \psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\ k_n&=\frac{n\pi}{L}\;,\\ E_n&=\hbar \omega_n\;,\\ \omega_n&=\frac{\pi h n^2}{4L^2m}\;. \end{align} $$

Here, $|\alpha \rangle $ represents the spin state.

$$\Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)$$

The energy of state $\Psi_{n\alpha m\beta}(x_1,x_2,t)$ can be calculated as $$(H_1+H_2)\Psi_{n\alpha m\beta}(x_1,x_2,t)=(E_n+E_m)\Psi_{n\alpha m\beta}(x_1,x_2,t)$$

since each of the one-particle Hamiltonians acts on the respective one-particle wavefunction $\psi_{n\alpha}(x_1,t)$, which yields its eigenenergy $E_n$.

First consider state for which $\alpha=\uparrow$ and $\beta=\uparrow$. The ground state is the lowest-lying energy state of the system. In this case, it would correspond to $\Psi_{1\uparrow 1\uparrow}$, but this function is identically zero. Then next two lowest-lying states are $\Psi_{1\uparrow 2\uparrow}$ and $\Psi_{2\uparrow 1\uparrow}$.Thanks to the antisymmetrization, $\Psi_{1\uparrow 2\uparrow} = -\Psi_{2\uparrow 1\uparrow}$ and it represents the ground state of the system with energy $E_1+E_2$. So the first two lowest energies are $$E^0=E^1=5E_0$$

For opposite spins, we choose $\alpha=\uparrow$ and $\beta=\downarrow$. Here, the lowest lying energy state is $\Psi_{1\uparrow 1\downarrow}$ and it has energy $2E_0$.

You may wonder because this doesn't match with the answer in the textbook, So the only thing I can conclude that there is a mistake in the problem or in solution. I hope this will help you. Best wishes!

  • 2
    $\begingroup$ Yes. But the problem is, when we have two spin 1/2 particles, adding their spin can give you total spin either 1 or 0, which you have mentioned. In this problem, why aren’t we considering the singlet state, where total spin is 0? It is one of the possibilities. $\endgroup$ Oct 27, 2020 at 16:01
  • $\begingroup$ @HarshdeepSingh See the edit, Please. $\endgroup$ Oct 27, 2020 at 17:56

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