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?
 A: 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.
A: A determinant is alternating, or totally skew-symmetric. This is why it works well for defining the state of a fermionic system. What you need for a bosonic system is a symmetrizer, and thus the related notion of symmetric algebra.
