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In Witten's review paper "Fermion path integrals and topological phases", the Dirac equation (Eq(2.2)) is $$(\gamma^{\mu}D_{\mu}-m)\psi=0$$ which appears very strange to me. Initially I thought this is the equation in Euclidean metric, but this is not true. My argument is the following: Assuming we Wick rotate from the Minkowski metric $(-,+,+)$, then both $\gamma^0$ and $\partial_0$ will contribute an $i$, which will cancel each other out. Then overall we do not get an extra $i$ to cancel the $i$ in $i\gamma^{\mu}\partial_{\mu}$.

Another attempt: maybe we can extract a factor of $i$ out of the gamma matrices in a particular basis. This however, doesn't work. The reason is that after factoring out an $i$ from each of the gamma matrices, the new set of "gamma matrices" will no longer satisfy the Clifford algebra because $\{\gamma^{\mu}/i,\gamma^{\nu}/i\}=-2\eta^{\mu\nu}$.

Any help would be greatly appreciated.

========= Edit with detailed answer=============

Based on Madmax's answer, I elaborate on what is happening here. In short, whether there is an $i$ or not depends on whether the Minkowski metric used is $(+,-,-,-,...)$ or $(-,+,+,+,...)$. Apparently, people call these west coast convention and east coast convention\footnote{\url{https://www.math.columbia.edu/~woit/wordpress/?p=7773&cpage=1}}.

The basic idea of Dirac equation is to factorize the Klein Gordon (KG)equation, and the crucial thing here is that the KG equation takes slightly different forms in the two different conventions. In the many-plus metric, the KG eq takes the form $(-\partial^2+m^2)\psi=0$, whereas in the many-minus metric, it takes the form $(\partial^2+m^2)\psi=0$.

The we can easily see that in the many-minus metric, the Dirac equation should be $(i\gamma^{\mu}\partial_{\mu}-m)\psi=0$. Simply act $(-i\gamma^{\mu}\partial_{\mu}-m)$ on the left to obtain the KG eq in the many-minus metric. Very similarly, we can check that the Dirac equation has to take the form $(\gamma^{\mu}\partial_{\mu}-m)\psi=0$ in order to recover the KG equation in the many-plus convention.

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    $\begingroup$ It might be silly, but have you considered the possibility that it might be a simple typo? In fact in equation (2.3) for two fermions Dirac field, the $i$ is there. $\endgroup$ – mollusca96 Nov 13 '19 at 22:39
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    $\begingroup$ It's not a typo, you can refer to my edits above $\endgroup$ – M. Zeng Nov 14 '19 at 0:16
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Whether there is an $i$ $$ (i\gamma^{\mu}D_{\mu}-m)\psi=0 $$ or no $i$ $$ (\gamma^{\mu}D_{\mu}-m)\psi=0 $$ in the Dirac equation is determined by the metric:

  • For $(\gamma^0)^2 = I$, there is $i$
  • For $(\gamma^0)^2 = -I$, there is no $i$

In Witten's paper, the metric is (-, +, +), therefore no $i$.


Added note:

Since the Hermitian of $\gamma^\mu$ is defined as: $$ (\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0, $$ $\gamma^0$ is non-Hermitian in the $(\gamma^0)^2 = -I$ case: $$ (\gamma^0)^\dagger = \gamma^0\gamma^0\gamma^0 = -\gamma^0. $$


One more note:

The choice (in 4D case) between west coast metric (+, -, -, -) and east coast metric (-, +, +, +) is usually regarded as a matter of convention or personal taste. However, one should note that the respective Clifford algebra Cl(1,3) and Cl(3, 1) are not isomorphic to each other. Instead:

  • Cl(1,3) is isomorphic to $M_2(H)$, 2*2 matrices of quaternions.
  • Cl(3,1) is isomorphic to $M_4(R)$, 4*4 matrices of real numbers.

Does the different isomorphism have bearing on the physics? John Baez says yes (see the link in the OP). But I would like to know your take.

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