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 4×4 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.

## Best 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).

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$.