Is the definition of the adjoint spinor $\bar{\psi}=\psi^\dagger \gamma^0$ forcing a particular choice of representation of the Dirac matrices (or a subset of the possible choices)?
More precisely, I assume (possibly incorrectly) that the adjoint of $\psi$ can always be written as $\psi^\dagger$ times some linear combination of the gamma matrices $$\bar{\psi} = \psi^\dagger c_\mu \gamma^\mu.$$ Where the usual choice of gamma matrices leads to $c_\mu = (1,0,0,0)$.
So broken into smaller steps the confusion is:
- Given a concrete choice of matrices to represent $\gamma^\mu$, how can we determine what is the adjoint spinor in terms of the spinor components?
- Is it always linear, $\bar{\psi} = \psi^\dagger c_\mu \gamma^\mu$?
- Is it always given by $c_\mu = (1,0,0,0)$ independent of the choice of gamma matrices?
This question is a follow up of comments on: spinor vs vector indices of Dirac gamma matrices
To keep this self-contained, here is a summary of thoughts for why $\bar{\psi}=\psi^\dagger \gamma^0$ may not be true for all choices of the gamma matrices.
The Dirac matrices (gamma matrices), are defined by the anticommutation relation $$\{ \gamma^\mu, \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu} I_4 $$ where $\eta^{\mu \nu}$ is the Minkowski metric (using (+ - - -) convention) and $I_4$ is the 4x4 identity matrix.
Given one set of matrices $\gamma^\mu$ that satisfy this anticommutation relation, the matrices $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ will also satisfy the anticommutation relation if $\Lambda^\mu{}_\nu$ is a Lorentz transformation :
$$\{ \gamma'^\mu, \gamma'^\nu \} = \{ \Lambda^\mu{}_\alpha \gamma^\alpha, \Lambda^\nu{}_\beta \gamma^\beta \} = \Lambda^\mu{}_\alpha \gamma^\alpha \Lambda^\nu{}_\beta \gamma^\beta + \Lambda^\nu{}_\beta \gamma^\beta \Lambda^\mu{}_\alpha \gamma^\alpha \\ = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta (\gamma^\alpha \gamma^\beta+\gamma^\beta \gamma^\alpha) = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta 2 \eta^{\alpha \beta} I_4 = 2 \eta^{\mu \nu} I_4 $$
If we assume that $\bar{\psi}\psi=\psi^\dagger \gamma^0\psi$ is a scalar for any choice of $\gamma^0$, the above appears to lead to the result that $\psi^\dagger \gamma^1\psi$ is also a scalar (or similarly with any gamma matrix). This is because we can use the above to make a new choice of gamma matrices with $\gamma'^0 = a \gamma^0 + b \gamma^1$ (where $a^2-b^2=1$). And thus would require $\psi^\dagger \gamma'^0\psi = a \psi^\dagger \gamma^0\psi + b \psi^\dagger \gamma^1\psi$ is also a scalar. Similarly then, $\psi^\dagger \gamma^\mu \psi$ would have to be a scalar for any choice of $\mu$. It seems more likely that the starting assumption, that $\bar{\psi}=\psi^\dagger \gamma^0$ independent of representation choice, is wrong.
Answer
Quick answer:
We can always define the Dirac adjoint of a spinor to be $\bar \psi := \psi^\dagger \gamma^0$. After that definition is set up, we have to additionally tell how it transforms under a Lorentz transformation and that depends on convention. For example, we could choose $\gamma'^\mu = \Lambda^\mu{}_\nu \gamma^\nu$ (note $\gamma'^0$ is no longer Hermitian) and define $\bar \psi := \psi^\dagger \gamma'^0 $, then we will see that $\bar\psi\psi$ no longer transforms as a scalar.
On the other hand, if we insist that $\bar\psi \psi$ should transform as a scalar, we should find a function of the gamma matrices $F(\gamma'^\mu)$ such that for any Lorentz transformation $\Lambda,\ \color{red}{S(\Lambda)^\dagger FS(\Lambda) \equiv F(\gamma'^\mu)}$. Then, defining $\bar \psi := \psi^\dagger F$ ensures that $\bar\psi\psi$ transforms as a scalar. The question is whether such a function always exists. With your example, you can convince yourself that the ansatz $F= c_\mu \gamma'^\mu$ will work if and only if $c_\mu \Lambda^\mu{}_j=0$ for all spatial indices $j$. This is a system of three equations $(j=1,2,3)$ with four unknowns $(c_\mu)$, and therefore possibly has many solutions. For example, if $\Lambda$ describes a boost in the $x$-direction with speed $\beta$, defining $\bar \psi := \psi^\dagger (\gamma'^0 + \beta \gamma'^1)$ makes $\psi^\dagger\psi$ a scalar. Note that $\gamma'^0 + \beta \gamma'^1 \propto \gamma^0$, so there is no contradiction.
Detailed answer:
A Clifford algebra over a $D$-dimensional spacetime equipped with the metric $g^{\mu\nu}$ is generated by $D$ hypercomplex numbers $\{\gamma^\mu\}, \mu\in \{0,\cdots,D-1\}$ defined by the following algebraic product:
$$ \{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu} \mathbb I_n\,. \tag1 $$
The algebra is $2^D$-dimensional, meaning there is a list of $2^D$ linearly independent elements, closed under multiplication, formed by various products of the $D$ hypercomplex numbers. Moreover, there is always a representation of the algebra in terms of real $n \times n$ matrices where $n=2^{[D/2]}$, the operation $[\cdot]$ taking the integer part of the given number. If the dimension $D$ is even, then this representation is the one and only irreducible representation of the algebra (up to equivalences). Furthermore, if the representation is unitary, the $2^D$ basis elements can be chosen to be Hermitian.
Without loss of generality, take the metric to be of mostly negative signature, so that $(\gamma^0)^2 = +1$ and $(\gamma^i)^2=-1$. You already begin to notice why $\gamma^0$ is special when contrasted with the rest of $\gamma$-matrices. It is exactly the same way as time is special when contrasted with space, given that the metric has a Lorentzian signature. With this in mind, for D=4 a basis for $4\times4$ matrices is given by $\{\Gamma_j\}, j\in\{1,\cdots, 16\}$ where
\begin{align*} \Gamma_j \in \{&\mathbb I_4, \gamma^0,i\gamma^1,i\gamma^2,i\gamma^3,\\ &\gamma^0\gamma^1,\ \gamma^0\gamma^2,\ \gamma^0\gamma^3,\quad i\gamma^1\gamma^2,\ i\gamma^2\gamma^3,\ i\gamma^3\gamma^1,\ \\ &i\gamma^0\gamma^1\gamma^2,\quad i\gamma^0\gamma^1\gamma^3,\quad i\gamma^0\gamma^2\gamma^3,\quad i\gamma^1\gamma^2\gamma^3,\\ &i\gamma^0\gamma^1\gamma^2\gamma^3\}\,. \tag2 \\ \end{align*}
The imaginary number $i$ has been inserted in the above array so that for all $j, (\Gamma_j)^2=+1.$ Notice that this list is closed under multiplication. This allows you to write any $4\times4$ matrix $X$ as a sum over these matrices: $X = \sum_{i=1}^{16} x^i \Gamma_i$ where $x_i = \frac14 \text{Tr}(X\Gamma_i)$. Moreover, for all $j\ne1, \text{Tr}(\Gamma_j)= 0$. Using this information, you can prove the following lemma (Cf. Schwartz Ch. 4).
Lemma:
Given two sets of gamma matrices $\{\gamma^\mu\}$ and $\{\gamma'^\mu\}$, correspondingly basis vectors $\{\Gamma_j\}$ and $\{\Gamma'_j\}$, there exists a non-singular matrix $S$ such that
$$ \boxed{\gamma'^\mu = S\gamma^\mu S^{-1}}\,, \qquad \text{ where } S=\sum_{i=1}^{16} \Gamma'_i F \Gamma_i\,, $$
and $F$ is so chosen that $S$ is non-singular.
Furthermore, $S$ is uniquely fixed (up to a numerical factor).
You have correctly identified that $\gamma'^\mu := \Lambda^\mu{}_\nu \gamma^\nu$ obeys the same commutation relation as $\gamma^\mu$, and it is precisely because of this that the above lemma tells us of the existence of a non-singular $S$ so that
$$ S^{-1} \gamma^\mu S = \Lambda^\mu{}_\nu \gamma^\nu\,, \tag3$$
which is required for the Lorentz invariance of the Dirac equation.
At this point, let us make a conventional choice (which @akhmeteli has been pointing out all along). Let us choose $\gamma^0$ such that it is Hermitian and the $\gamma^i$s such that they are anti-Hermitian. In other words, this means that $\gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$. Notice that we did not need any such assumption in the previous discussion. But this convention will simplify greatly what is about to follow.
Observe that because $\Lambda^\mu{}_\nu$ is real, and because we chose this convention,
\begin{align*} (S^\dagger\gamma^0) \gamma^\mu (S^\dagger \gamma^0)^{-1} = (S^{-1} \gamma^\mu S)^\dagger &\stackrel{(3)}{=} \Lambda^\mu{}_\nu (\gamma^\nu)^\dagger\\ \text{(Multiplying by $\gamma^0$ from both sides)} \Rightarrow (\gamma^0 S^\dagger\gamma^0) \gamma^\mu (\gamma^0 S^\dagger \gamma^0)^{-1} &= \Lambda^\mu{}_\nu \gamma^\nu \stackrel{(3)}{=}S^{-1} \gamma^\mu S\,. \\ \end{align*}
After rearranging we find that,
\begin{align*} &\Rightarrow (S\gamma^0 S^\dagger\gamma^0) \gamma^\mu (S\gamma^0 S^\dagger \gamma^0)^{-1} = \gamma^\mu \\ &\Rightarrow S\gamma^0 S^\dagger\gamma^0 = c\mathbb I_4 \\ &\Rightarrow S^\dagger\gamma^0 = c \gamma^0 S^{-1}\,, \tag4 \\ \end{align*}
where $c$ is some constant which you can convince yourself is real.
Now, if we normalize $S$ such that $\det(S)=1$, then $c^4 = 1$ or $c= \pm 1$. Let us see which situations correspond to $c=+1$ and $c=-1$. Observe the following identity.
\begin{align*} S^\dagger S = (S^\dagger\gamma^0)\gamma^0S &\stackrel{(4)}= c\gamma^0(S^{-1} \gamma^0 S)\\ &\stackrel{(3)}= c\gamma^0 \Lambda^0{}_\nu \gamma^\nu \\ &= c(\Lambda^0{}_0 \mathbb I_4 - \sum_{k=1}^3 \Lambda^0{}_k\gamma^0\gamma^k)\\ \Rightarrow \text{Tr}(S^\dagger S) &= 4c\Lambda^0{}_0 \,. \tag5 \end{align*}
Since $S^\dagger S$ has real eigenvalues (it is Hermitian) and positive-definite (since S is non-singular), its trace must be positive. This means that when $ \Lambda^0{}_0 \le -1, c=-1$ and when $ \Lambda^0{}_0 \ge +1, c=+1$.
Conclusion:
Defining $\bar \psi := \psi^\dagger \gamma^0$, we see that under a Lorentz transformation (taking $\psi \to S\psi$) that does not involve time reversal ($\Lambda^0{}_0 \ge +1$), $\bar \psi \to +\bar\psi S^{-1}$, whereas for Lorentz transformations that do reverse time ($\Lambda^0{}_0 \le -1$), $\bar \psi \to -\bar\psi S^{-1}$.
If we had not normalized $S$, we would write $\bar\psi \to c\bar\psi S^{-1}$. Then this extra factor of $c$ will need taking care of in the Lagrangian through perhaps a field redefinition.
More generally, if we did not assume the hermiticity properties of the gamma matrices, things would be much more complicated, but a definition is a definition. After defining $\bar \psi := \psi^\dagger \gamma^0$, our task would be to find its transformation rules and then implement it accordingly in the Lagrangian. Alternatively, if you wish to preserve the transformation rules, you will have to change the definition accordingly.
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