I'm trying to find the ground state and first excited state for 2 identical bosons in an infinite square well. I know that both states are degenerate and the spatial and spin parts of the wave functions must both be either symmetric or antisymmetric. The spatial part of the wave function is easy to construct, but I'm having trouble with the spin part $\,\chi(m_1,m_2).$

I used a Clebsch-Gordon table to find the 9 $\,\chi(s,m)$ states in terms of $\,\chi(m_1,m_2)$ for $s=0,1,2.$ In addition, I found the $s=0,2$ states to be symmetric, and the $s=1$ state to be antisymmetric, so I figure the rest shouldn't be hard. But I'm confused because I am asked to "choose the spin part of the wave function, $\,\chi(m_1,m_2)$, to be an eigenvector of total spin, i.e. a simultaneous eigenvector of $\mathbf{S}^2$ and of $\mathbf{S}_z$, where $\mathbf{S}\equiv\mathbf{S}_1+\mathbf{S}_2$."

Does that mean I should find $\,\chi(m_1,m_2)$ states in terms of $\,\chi(s,m)$ states instead? How do I tell which are symmetric and antisymmetric if the uncoupled states are composed of coupled states with differing $s$ values?


1 Answer 1


What I get from the question is you want to find the wave functions of the ground and first excited states of a system of two non-interacting spin-1 bosons in an infinite square well potential.

Because the particles are non-interacting, you get the usual solutions for the infinite square well with quantized eigenvalues $E_n^{\sigma} = \frac{\hbar^2\pi^2}{2ma^2}n^2, \ n \in \{1,2,...\}, \ \sigma \in\{ \uparrow, 0, \downarrow\}$, degenerate in the spin index so $\psi_{n_i}(\vec{x}, \sigma) = \phi_{n_i}(\vec{x})\chi(\sigma)$ and the total wave function can be written as $$ \Psi_{n_1,n_2}(\vec{x}_1, \sigma_1; \vec{x}_2, \sigma_2) \equiv \Phi_{n_1,n_2}(\vec{x}_1,\vec{x}_2) X(\sigma_1, \sigma_2). $$ Since we have bosons, $\Psi$ must be symmetric under permutation of $(\vec{x}_1,\sigma_1)$ and $(\vec{x}_2, \sigma_2)$, so each $\Phi$ and $X$ must be either symmetric or anti-symmetric, as long as their product is symmetric.

For the ground state, the lower energy is clearly achieved with $n_1 = n_2 = 1$ and $\Phi_{1,1}(\vec{x}_1,\vec{x}_2) = \prod_{i\in\{1,2\}} \phi_1(\vec{x}_i)$ is symmetric, so $X$ must be as well. For the first excited state, you need $n_1 (n_2) = 1, \ n_2 (n_1) = 2$. You can construct $\Phi$ to be either symmetric or anti-symmetric, like $$ \Phi^{\pm}_{1,2}(\vec{x}_1,\vec{x}_2) = \frac{1}{\sqrt{2}}\left(\phi_1(\vec{x}_1)\phi_2(\vec{x}_2) \pm \phi_1(\vec{x}_2)\phi_2(\vec{x}_1) \right), $$ so the corresponding spin part has to be symmetric or anti-symmetric.

Now, the spin states can be expressed in any basis because the energy is independent of the spins. The basis of eigenvectors of the total spin are already in described in either symmetric or anti-symmetric, so it's a natural basis to take.

Taking a Clebsch-Gordan table, you can see that, as you said, total spin $s \in \{0,2\}$ are symmetric while $s = 1$ is anti-symmetric. The reason for this is that when combining two identical particles of spin $j$, the highest spin you can create is both particles maximally aligned $X(s=2j, m_s=2j) = X(m_1=j, m_2 = j)$, which is symmetric. To create the other vector of the subset $X(s = 2j, m_s)$, you apply $J_-$, a symmetric operator. on both sides, so all the states generated are symmetric.

To construct the next lower subset (here $X(s=2j-1, m_s)$), you take $$ X(s=2j, m_s=2j-1) = \frac{1}{\sqrt{2}}\left( X(m_1=j, m_2=j-1) + X(m_1=j-1, m_2=j) \right) $$ and $X(s=2j-1, m_s=2j-1)$ must be made of the same two vectors, but be orthogonal, so the only way is to take the anti-symmetric version with a $-$ instead of a $+$, so the subset $X(s=2j-1, m_s)$ is anti-symmetric and so on.

So for your ground state, you need $X$ symmetric so any $X(s \in \{0,2\}, m_s)$ and for your first excited state, either $X$ symmetric if $\Phi$ is symmetric, or any anti-symmetric $X(s=1,m_s)$ if $\Phi$ is anti-symmetric.


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