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Consider the hamiltonian $$ H = - \frac{1}{2} \nabla^2 + V. $$ The potential $V : (\mathbb{R^3})^N \to \mathbb{R}$ is symmetric, so for each eigenvalue, there is an antisymmetric eigenvector. There is not any spin operator inside $V$. It can be assumed that the hamiltonian describes a fermionic many-body problem.

Suppose I give you $2^N$ functions $f_i : (\mathbb{R}^{3})^N \to \mathbb{C}$ for $i = 1,2,3,\cdots,2^N$ that are square-integrable and eigenvectors of $H$ with eigenvalue $E_i$. If $\sigma_j \in \{1/2,-1/2\}$ for each $j \in \{1,2,3,\cdots, 2^N\}$; how would you find out which $f_i$ corresponds to the spin configuration $\sigma_1\sigma_2\cdots\sigma_N$?

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  • $\begingroup$ "for each eigenvalue, there is an antisymmetric eigenvector" I'm not sure I understand that, can you please elaborate? What does it mean for a vector to be antisymmetric? $\endgroup$
    – Prof. Legolasov
    Jul 29, 2020 at 9:20
  • $\begingroup$ @Prof.Legolasov Suppose $S_k$ is the symmetric group on $k$-letters and $H^1$ the first sobolov space. If $\tau \in S_k$, define a group action of $S_k$ on $H^1(\mathbb{R}^k)$ by $\tau f(x_1,\cdots,x_k) = f(x_{\tau(1)},\cdots,x_{\tau(k)})$. Definition 1: An $f \in H^1(\mathbb{R}^k)$ is said to be antisymmetric if $\tau f = \mathrm{sign}(\tau)f$. Definition 2: Consider am inner product space $W$ over a field $F$. If $H : W \to W$ is an unabounded operator and $f \in D(H)$, then $f$ is said to be an eigenvector if there is an $\lambda \in F$ such that $Hf = \lambda f$. $\endgroup$
    – Mikkel Rev
    Jul 29, 2020 at 9:35
  • $\begingroup$ Basically: If you interchange the arguments of $f$ $k$ times, the sign of $f$ changes sign $k$ times. Such a function is an eigenvector of an operator $H$ if there is a number $\lambda$ such that $H - \lambda$ fails really bad to be invertible. $\endgroup$
    – Mikkel Rev
    Jul 29, 2020 at 9:43
  • $\begingroup$ so totally antisymmetric over the interchange of $N$ coordinates? $\endgroup$
    – Prof. Legolasov
    Jul 29, 2020 at 9:48
  • $\begingroup$ @Prof.Legolasov Short answer: $\tau$ does not act on the index of $f_i$. Long answer: My hope, in the end, is to work out how $\tau$ must act on $i$ and $x_i$ jointly if I wish to combine the $2^N$ functions into a single function. But that is more complicated, because then we need to be really careful how we define non-degeneracy etc. $\endgroup$
    – Mikkel Rev
    Jul 29, 2020 at 9:52

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(I'm improvising here -- please treat what follows with a degree of healthy skepticism)

So lets say your wavefunction is a multiplet of $2^N$ components -- functions over $\mathbb{R}^{3N}$, with suitable integration condition imposed.

Your Hamiltonian $H$ acts on each component individually, and its action on differnt components is the same, by construction.

The spin operators (read, the operators that correspond to projections of the physical spin on some fixed axis) for any of the $N$ particles only shuffle the components and don't change any of the functions.

Therefore, the spin operator commutes with the Hamiltonian.

The two operators must therefore have joint eigenvalues.

Which means that your problem is essentially unsolvable, unless you allow the spin operator to enter $H$ as e.g. in the Pauli equation.

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