# Slater's determinant for Bosons/Symmetric Particles?

For Slater's determinant it is obvious how this describes two or multiple fermions/anti-symmetric particles. By definition the determinant introduces a negative sign in front of the second product.

\begin{align} \Psi(\vec{x}_1, \vec{x}_2) &= \frac{1}{\sqrt{2}}[\chi_1(\vec{x}_1)\chi_2(\vec{x}_2) - \chi_1(\vec{x}_2)\chi_2(\vec{x}_1)] \\ &= \frac{1}{\sqrt{2}}\begin{vmatrix}\chi_1(\vec{x}_1) & \chi_2(\vec{x}_1) \\ \chi_1(\vec{x}_2) & \chi_2(\vec{x}_2)\end{vmatrix} \end{align}

If we were to construct a Slater's determinant for Boson, we would have to introduce a negative sign for $\chi_2(x_1)$ or $\chi_1(x_2)$

Does Slater determinant take care of this or is there another determinant that governs how Bosonsic wave function come together?

• Put a plus where there is a minus and you obtain a symmetric, i.e. bosonic, wavefunction. However this is not a determinant (and in fact it is not called like that). This is natural since determinants have an antisymmetric structure by construction, and so are not suitable to describe bosons. Dec 27, 2014 at 9:56

A Slater determinant is by definition an antisymmetric object used to describe many-body fermionic systems. It can be written used the antisymmetrizer as $$\Psi(q_1,...,q_N) = \underbrace{\frac{1}{\sqrt{N!}} \det \{ \psi_{\nu_i}(q_j)\}}_{\text{Slater determinant}} = \sqrt{N!} \mathcal{A} \,\, \psi_{\nu_1}(q_1) ... \psi_{\nu_N}(q_N),$$ where the antisymmetrizer operator $\mathcal{A}$ is defined by $$\tag{A} \mathcal{A} \equiv \frac{1}{N!} \sum_\sigma (-1)^\sigma \hat{\sigma},$$ with the sum extending over all permutations $\sigma$ of $N$ objects: $$\hat{\sigma} \,\, \psi_1(q_1) \cdots \psi_N(q_N) \equiv \psi_{\sigma(1)}(q_1) \cdots \psi_{\sigma(N)}(q_N),$$ and we denote with $(-1)^\sigma$ the parity of the permutation $\sigma$.
In ket notation this reads $$\mathcal{A} | \nu_1, \nu_2,...,\nu_N \rangle = \frac{1}{\sqrt{N!}} \sum_\sigma (-1)^\sigma | \nu_{\sigma(1)},\nu_{\sigma(2)},...,\nu_{\sigma(N)} \rangle,$$ where $$\langle q_1,...,q_n | \nu_1,...,\nu_N \rangle = \psi_{\nu_1}(q_1) \cdots \psi_{\nu_N}(q_N).$$
So, by definition, Slater determinants are only used for fermionic systems. For bosonic systems the argument is readily extended though: we want to convert the antisymmetrizer $\mathcal{A}$ to a symmetrizer $\mathcal{S}$ which does not have the minus signs, that is, something like $$\mathcal{S} \approx \frac{1}{\sqrt{N!}} \sum_\sigma \hat{\sigma}.$$ The problem with this is that it does not preserve the normalization of the wavefunction, i.e. it is not a unitary operator. To make it one, we must add normalization factors according to the following (see also this wikipedia article): $$\mathcal{S} | n_{\nu_1}, n_{\nu_2},...,n_{\nu_N} \rangle = \sqrt{\frac{\prod_{i=1}^N n_{\nu_i}!}{N!}} \sum_\sigma | n_{\nu_{\sigma(1)}}, ... , n_{\nu_{\sigma(N)}} \rangle.$$ Note the change in notation here: $|n_{\nu_i}\rangle$ denotes a state with $n_{\nu_i}$ particles on the single-particle state $\nu_i$. This was not needed for fermionic systems because due the Pauli exclusion principle every state is occupied by at most 1 particle. As a foot note: the operation that for bosonic systems substitutes the determinant is called permanent.
• So, could a fermionic system compute a determinant of a $n$X$n$ matrix? Since, the sign in computing the determinant of a square matrix is alternating. Mar 22, 2016 at 6:53