Mathematical significance of the "Dirac conjugate" Let $\psi$ be a Dirac spinor.  The so-called "Dirac conjugate" of $\psi$ is defined to be $\widetilde{\psi}:=\psi ^*\gamma ^0$, where $^*$ denotes the adjoint and the gamma matrices $\gamma ^\mu$ comprise the essentially unique irreducible representation of $\mathcal{C}\ell (1,3)$.  Physicists introduce this, in a relatively ad hoc manner, so that the quantity
$$
\widetilde{\psi}\psi
$$
is Lorentz invariant.  This does the trick, but I have a feeling like there is something deeper going on here.
The quantity $\psi ^*\gamma ^0$ makes sense for an arbitrary Clifford algebra $\mathcal{C}\ell (1,2m-1)$, whereas the notion of Lorentz invariance is specific to the case $m=2$, so the significance of $\widetilde{\psi}$ in other dimensions is not obvious to me.  It might be the case that the significance of $\psi ^*\gamma ^0$ is unique to the case $m=2$, but I would be surprised if that were the case.
So then, what is the general mathematical significance of the Dirac conjugate $\psi ^*\gamma ^0$.
(Please let me know if explanation of any physics jargon is needed.)
 A: The Lorentz group, $SO(1,3)$, is non-compact, thus their representations are not unitary (in general).
Therefore, if you have a spinor, $\psi\in \mathcal{S}$, transforming as $\psi\mapsto S\psi$, it follows that the construction $$ \psi^\dagger \psi \mapsto \psi^\dagger S^\dagger S \psi \neq\psi^\dagger \psi,$$
since $S^\dagger\neq S^{-1}$.
This tells us that $\psi^\dagger$ does not belongs to the dual space of the spinors, $\psi^\dagger\not\in \mathcal{S}^*$.
It this point, you can realize that $$S^\dagger\gamma^0 = \gamma^0 S^{-1},$$
and this allows to define a dual spinor to $\psi$ through the construction $$\mathcal{S}^*\ni\bar{\psi}\equiv \psi^\dagger\gamma^0.$$
Hope it would be helpful.
Summary
The Dirac conjugate serves to define a dual spinor, by giving a spinor in the direct space.
A: Let $S$ denote the space of Dirac spinors with the fundamental representation of $Cl(1,3)$, so elements of $S$ are complex column 4-vectors, and $Cl(1,3)$ is acting by matrix multiplication on the left. The general significance of $\tilde\psi$ is that it is the image of $\psi$ under a linear transformation which is an equivalence of representations. 
To be more specific, suppose $e_i \mapsto \gamma(e_i)=\gamma_i$ denotes the fundamental representation above. We can define a different representation of $Cl(1,3)$ by sending $a\mapsto {\overline{\gamma(\beta(a))}}^t$ where $\beta$ is the fundamental anti-automorphism of the Clifford algebra which exchanges the order of the generators. This defines a representation of the Clifford algebra on the conjugate dual space $\bar{S}^*$. In particular, we see that $e_i\mapsto {\bar\gamma_i}^t$ in this representation.
We can seek a linear transformation $A: S \rightarrow \bar{S}^*$ which commutes with the representations (that is, we want $A$ to be an equivalence) by solving 
$A\gamma_i={\bar{\gamma_i}}^tA$
which has a unique solution up to phase (i.e. a complex scale factor). This solution turns out to agree with the matrix $\gamma_0$. 
Notice that $\bar{A}:\bar{S}\rightarrow S^*$, so we can define a pairing of elements of $S$: 
$(\phi, \psi)=<\bar{A}(\bar{\phi}),\psi>$
where $<,>$ denotes the natural pairing of a vector space with its dual space. In the typical chiral (Weyl) choice for the $\gamma$-matrices, $\gamma_0$ is symmetric and real, so the matrix $\gamma_0$ is also the matrix of this bilinear form.
So, the significance of $\tilde{\psi}$ is that it is the contraction of $\bar{\psi}$ with this bilinear form.
A useful principle I keep in mind when I think about this stuff is that whenever physicists do things in an $\textit{ad hoc}$ manner, guided by Lorentz invariance, mathematically we can achieve the same goal by thinking about equivalence of representations.
