When dealing with separate non-interacting systems in quantum mechanics (i.e. with distinct Hamiltonians acting on different Hilbert spaces like $\hat{H}=\hat{H}_1 + \hat{H}_2$), one can often write the wavefunction as a tensor product

$\left| \Psi\right\rangle=\left| \phi_1\right\rangle\left| \phi_2\right\rangle$

However, if the two systems in question are, for example, two bunches of indistinguishable fermions, then this would appear not to be possible, as it would violate the required antisymmetry of the wavefunction. If the two states $\left|\phi_{1/2}\right\rangle$ which are eigenvalues of $\hat{H}_{1/2}$ are Slater determinants, then the tensor product $\left|\phi_2\right\rangle\left|\phi_1\right\rangle$ is no longer a Slater determinant. In this situation, one would expect that the resulting state is some kind of Slater determinant involving orbitals in both Hilbert spaces. If the states in each system are multi-determinantal, then the situation is even more complicated.

The question is then:

Is there some kind of generalisation of tensor products to the situation of separate fermionic systems which produces states of the correct symmetry?


1 Answer 1


There are two ways to proceed in this situation. The first one is to consider general states $|\psi_1\rangle\otimes|\psi_2\rangle$ and treat the permutation symmetry generated by

$$ P=\begin{bmatrix}0&I\\I&0\end{bmatrix} $$

as a "gauge symmetry", i.e. you need to restrict that the physical states are "annihilated" by this generator. More concretely you need to impose an extra condition for physical states given by:

$$ P|\psi\rangle=\pm|\psi\rangle\qquad or \qquad (P\pm 1)|\psi\rangle=0 $$

For the minus sign, this will lead to the Slater determinant. Also, you should impose that all physical observables $\mathcal{O}$ should be invariant under $P$:

$$ \mathcal{O}P = P\mathcal{O} $$

In this approach it is not clear why we are allow to consider states that are not invariant under $P$ (i.e. allowing the minus sign). It is also not clear why we can only chose those phases. One can agree that experimentally this is what we observe.

The second way to proceed will make this discussion about the phases more clear. The idea is to restrict the quantum numbers to be invariant under permutation, and than obtain the Hilbert space out of this quantum numbers. So in this approach there is no permutation operator since the Hilbert space is the physical Hilbert space from start.

Suppose the quantum numbers are the position of each particle and the sign of the $S_z$ component of the $1/2$-spin. The states can be labelled by the center of mass $(x+y)$, by the relative distance $(x-y)$ and by the following combination of sings: $++$, $+-$ or $--$.

$$ |\psi\rangle=\sum_{(x+y),(x-y)}\psi^{\pm\pm}(x+y,x-y)|(x+y),(x-y),\pm\pm\rangle + \psi^{+-}(x+y,x-y) |(x+y),(x-y),+-\rangle $$

The relative distance should be restricted to be positive in order to avoid over-counting (avoid summing over permutation). This restriction will make the topology of the space of quantum number $x-y$ non-trivial. The anti-symmetry of fermions will be more obscure in this approach and will be described by how states changes under holonomies in the space of quantum numbers. There will be non-trivial closed paths in the space of quantum numbers and states are free to acquire a phase that represent the homotopy group. For 3 dimensions turns out that the homotopy group will be $Z_2$ so the only allowed phases are $\pm 1$, up to a global phase that can be absorbed in the definition of the states. For $2$ dimensions there the homotopy group is $Z$ so you have the possibility of arbitrary phase, the well know anyons. You can see more about this second approach here.

  • $\begingroup$ I'm not sure I'm following what the conclusion is here. Is the answer then just to take the product state and shuffle around the fermions among the occupied states in the original separate states (with the appropriate sign) to get the correct final state? $\endgroup$
    – phi1123
    Commented May 18, 2019 at 16:53
  • $\begingroup$ Yes, one way of proceeding is exactly this one: "take the product state and shuffle around the fermions among the occupied states in the original separate states" This will lead to the Slater determinant. No matter how separated the fermions/bosons are, you should always do that if you want to measure both systems. You can show that in given cases this will be negligible and then you can ignore this shuffle, but in principle you should always do that. $\endgroup$
    – Nogueira
    Commented May 18, 2019 at 20:41

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