I'm having son trouble when obtaining the Dirac equation. I am working in (1+1)-dimensional Minkowski spacetime with signature $(-, +)$ in coordinates $(t, x)\equiv(1, 2)$. I can think of two ways to obtain the Dirac Hamiltonian, but I'm having some trouble since they differ in an $i$ factor.

First way

I begin from the Dirac equation in this convention, which is: $$ (\gamma^\mu\partial_\mu-m)\psi=0\Rightarrow(\gamma^0\partial_0+\gamma^1\partial_1-m)\psi=0. $$

I follow now a single-particle approach: I want to obtain an Schrödinger-like equation, with $\psi(x)$ being the wavefunction that one gets when applying the fermionic operator to the vacuum state. The equation I want to reach is (in natural units): $$i\dot{\psi}=H_d\psi,$$ where $H_d$ is the single-particle Hamiltonian. The calculation is very simple and one arrives at: $$i\dot{\psi}=(i\gamma^0\gamma^1\partial_1-im\gamma^0)\psi\Rightarrow\boxed{H_d=i\gamma^0\gamma^1\partial_1-im\gamma^0}.$$

Second way

Now, I want a Lagrangian from which the Dirac equation can be derived. This can be achieved trivially with: $$\mathcal{L}=\bar{\psi}(\gamma^\mu\partial_\mu-m)\psi.$$ From this Lagrangian one can calculate the conjugate momentum: $$\Pi=\frac{\partial\mathcal{L}}{\partial(\partial_{0}\psi)}=\bar{\psi}\gamma^0,$$ and then one can get the field Hamiltonian density as: $$\mathcal{H}=\Pi\partial_0\psi-\mathcal{L}=\bar{\psi}\gamma^0\partial_0\psi-\bar{\psi}\gamma^0\partial_0\psi-\bar{\psi}\gamma^1\partial_1\psi+m\bar{\psi}\psi=\\=-\bar{\psi}\gamma^1\partial_1\psi+m\bar{\psi}\psi.$$

Then, using $\bar{\psi}=\psi^\dagger\gamma^0$ and integrating I arrive at: $$\boxed{H=\int dx\; \psi^\dagger(-\gamma^0\gamma^1\partial_1+m\gamma^0)\psi}.$$

My question

I was expecting to obtain: $$H=\int dx\;\psi^\dagger H_d\psi,$$ where $H_d$ is the single-particle Hamiltonian that I obtained in the first way and $H$ is the Hamiltonian "with operators" that I obtained in the second way. Nevertheless, there is an extra $i$ factor in $H_d$ that I don't recover in the second approach. Most of the resources that I find uses the sign convention $(+,\,-)$ for the metric and they don't get this problem. So I guess the problem must be in some wrong definition that I'm using and that is not valid when using the $(-,\,+)$ signature, or maybe equation $i\dot{\psi}=H_d\psi$ should have no $i$ in the LHS?


1 Answer 1


Your Dirac equation lacks a factor i. Your lagrangian lacks a factor i, as well.

  • $\begingroup$ @TopoLynch I am not referring to a sign. $\endgroup$
    – my2cts
    Commented May 2, 2022 at 20:02
  • $\begingroup$ I mean, I know that the usual Dirac equation reads $(i\gamma^\mu\partial_\mu-m)\psi=0$, but that only stands for the convention $(+,-,-,-,..)$. For example arxiv.org/abs/1609.07904 eqn. 2 is a reference where the $(-,+,+,...)$ convention is assumed and the Dirac equation lacks the $i$ factor. $\endgroup$
    – TopoLynch
    Commented May 2, 2022 at 20:08
  • $\begingroup$ @TopoLynch The two conventions differ by a factor $-1$. $\endgroup$
    – my2cts
    Commented May 2, 2022 at 20:10
  • $\begingroup$ In order to satisfy the Clifford algebra, the gamma matrices absorb the $i$ factor. Weinberg, for example, uses this Dirac equation. This issue is discussed here: physics.stackexchange.com/q/507075. $\endgroup$
    – TopoLynch
    Commented May 2, 2022 at 20:14
  • $\begingroup$ Correct. I overlooked the i factor. $\endgroup$
    – my2cts
    Commented May 2, 2022 at 22:05

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