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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?

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    $\begingroup$ 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. $\endgroup$ – yuggib Dec 27 '14 at 9:56
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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – user3483902 Mar 22 '16 at 6:53

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