Does the sign on the exponentials matter in the solution of Dirac's equation? The general solution of the Dirac equation is a linear combination of plane waves. The Positive Frequency Solutions are of the form
\begin{equation*}
\psi(x) = u(p)e^{-ipx},\quad\text{with}\quad p^2 = m^2\quad\text{and}\quad p^0>0
\end{equation*}
whereas the Negative Frequency Solutions are of the form
\begin{equation*}
\psi(x) = v(p)e^{+ipx},\quad\text{with}\quad p^2 = m^2\quad\text{and}\quad p^0>0
\end{equation*}
On Tong's QFT notes (p.106), the sign of the exponential is reversed:

He later flipped the sign back in eqn.5.25. My question is does the sign of exponential matter in rge plane wave expansion? If we associate $u(p)$ with $e^{+ipx}$, do we still have the same anticommutation relations:
$$
\{\psi_a(x),\psi_b(y)\} = \{\psi_a^\dagger(x),\psi_b^\dagger(y)\} = 0\quad
\{\psi_a(x),\psi_b^\dagger(y)\} = \delta_{ab}\delta^3(x-y)
$$
 A: What matters here is positive/negative frequency - that is energy. The exponents in the Tong's equations contain only coordinate part, whereas $px$ in the first two equations is the product of two four-vectors.
A: The sign of the $u$-solution depends on the chosen Minkowski-metric $\eta_{ij}$. If the metric is $\eta_{ij}=diag(1,-1,-1,-1)$ called West coast convention, then the sign of the exponential of the $u$-solution written as product of 4-vectors is negative. This convention is for instance chosen in the book of Peskin & Schroeder on QFT. The 4-product of momentum and position is according Srednicki $p\cdot x = \omega t-\mathbf{p}\mathbf{x}$
However, if the Minkowski-metric is chosen to be $\eta_{ij}=diag(-1,1,1,1)$ called east coast convention, then the sign in the exponential of the $u$-solution written as product of 4-vectors is positive. This convention is for instance chosen in the book of Srednicki on QFT. The 4-product of momentum and position is according Srednicki $p\cdot x = \mathbf{p}\mathbf{x} -\omega t$
Actually in west coast convention the $u$-solution is written as $\psi(x) = u\cdot e^{-ipx}$  whereas in east coast convention it is written as $\psi(x) = u\cdot e^{ipx}$ where in both cases  the product of 4-vectors $p=(E,\mathbf{p})$ and $x=(t,x)$ is used. Given the 4-vector products above, the space-position part of the $u$-solution turns out to be always $\sim e^{i\mathbf{p}\mathbf{x}}$.
The sign of the exponential of the $v$-solution is always opposite to the corresponding sign of the $u$-solution. Of course the existence of the negative energy solutions is independent of the sign convention.
