3
$\begingroup$

My textbook on QFT says that the Dirac equation can be used to show the following relation:

$$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}$$

I have searched around and unable to find how to prove this as it seems like it has to be assumed at some point by definition. My understanding was that this relation is a fundamental one and that it is assumed in order that the gamma matrices generate a matrix representation of the Clifford algebra, so it is a mathematical assumption rather than something which you derive from a physical equation. One approach I started is to take the Dirac equation and then multiply as follows:

$$(i\gamma^{\nu}\partial_{\nu}-m)\psi=0$$

$$(i\gamma^{\mu}\partial_{\mu}+m)(i\gamma^{\nu}\partial_{\nu}-m)\psi=0$$

$$-(\gamma^{\nu}\gamma^{\mu}\partial_{\nu}\partial_{\mu}+m^2)\psi=0$$

Is there some way to use this to show the given identity?

$\endgroup$
  • $\begingroup$ What textbook are you using? I'd agree that one usually looks at the problem the other way around, but if you (for example) demand that applying the Dirac operator twice yields the Klein-Gordon equation, then the necessary anti-commutation relations follow. $\endgroup$ – J. Murray Mar 19 '18 at 6:03
  • $\begingroup$ Thanks, I think I've got it now: the metric is in the definition of the d'Alembertian so you just operate twice and then compare with KG. $\endgroup$ – Tom Mar 19 '18 at 6:22
  • $\begingroup$ Yes. Roughly speaking, Dirac wanted to "factor" the Klein-Gordon equation to yield an equation which was first-order in time, so he posited the above form for some unknown $\gamma^\mu$, and then deduced the necessary constraints on them - which are precisely the defining characteristics of what we now call the Dirac algebra. $\endgroup$ – J. Murray Mar 19 '18 at 6:39
  • 1
    $\begingroup$ Check out the second chapter of Freemans Dysons Lectures on Advanced QM where he derives the anti-commutation rules following Diracs original reasoning. Its available online on the arxiv $\endgroup$ – Mozibur Ullah Sep 23 '18 at 18:24
3
$\begingroup$

Even if this is similar, but this answer should be clearer, as it was to me.

We are here. \begin{eqnarray*} (\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu + m^2)\psi &=& 0\\ (\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu + m^2)\psi &=& 0 \end{eqnarray*} Adding both the equations, \begin{eqnarray*} [(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu) + 2m^2]\psi &=& 0\\ \end{eqnarray*} Dividing by 2, \begin{eqnarray*} \left[\frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu) + m^2\right]\psi &=& 0\\ \end{eqnarray*} and comparing with the Klein Gordon equation, \begin{eqnarray*} (\partial^\mu \partial_\mu+ m^2)\psi &=& 0\\ \Rightarrow (g^{\mu\nu} \partial_\nu \partial_\mu+ m^2)\psi &=& 0\\ \end{eqnarray*} we get, \begin{eqnarray*} g^{\mu\nu} \partial_\nu \partial_\mu &=& \frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu)\\ &=& \frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\nu \partial_\mu) {\text{ :as $\partial_\nu \partial_\mu =\partial_\mu \partial_\nu$},}\\ &=& \frac{1}{2}(\gamma^\nu \gamma^\mu +\gamma^\mu \gamma^\nu )\partial_\nu \partial_\mu\\ \end{eqnarray*} So, we have \begin{eqnarray*} (\gamma^\nu \gamma^\mu +\gamma^\mu \gamma^\nu ) &=& 2 g^{\mu\nu}\\ \Rightarrow \{\gamma^\nu, \gamma^\mu \} &=& 2 g^{\mu\nu}\\ \end{eqnarray*}

$\endgroup$
2
$\begingroup$

To be honest I think in this case the best proof is by direct computation. The gamma matrices are

$$ \begin{equation} \gamma^{0}=\begin{pmatrix} 1 & 0 & 0 & 0\newline 0 & 1 & 0 & 0\newline 0 & 0 & -1 & 0\newline 0 & 0 & 0 & -1 \end{pmatrix}\, \quad \gamma^{1}=\begin{pmatrix} 0 & 0 & 0 & 1\newline 0 & 0 & 1 & 0\newline 0 & -1 & 0 & 0\newline -1 & 0 & 0 & 0 \end{pmatrix}\, \end{equation} $$ and $$ \begin{equation} \gamma^{2}=\begin{pmatrix} 0 & 0 & 0 & -i\newline 0 & 0 & i & 0\newline 0 & i & 0 & 0\newline -i & 0 & 0 & 0 \end{pmatrix}\, \quad \gamma^{3}=\begin{pmatrix} 0 & 0 & 1 & 0\newline 0 & 0 & 0 & -1\newline -1 & 0 & 0 & 0\newline 0 & 1 & 0 & 0 \end{pmatrix}. \end{equation} $$

Direct calculation shows that $$ \{\gamma^{0},\gamma^{0}\} = \gamma^{0}\gamma^{0} + \gamma^{0}\gamma^{0} = 2\eta^{00}\mathbb{I}_{4}\,$$

where $\eta^{00}=1$ and $\mathbb{I}_{4}$ is the $4\times 4$ identity matrix. Furthermore, direct calculation shows that

$$ \{\gamma^{0},\gamma^{i}\} = \gamma^{0}\gamma^{i} + \gamma^{i}\gamma^{0} = 2\eta^{0i}\mathbb{I}_{4}\, = 0_{4,4}\,$$

where $\eta^{0i}=0$ for $i=1, 2, 3$ and $0_{4,4}$ is the $4\times 4$ matrix with all zero entries. Additional calculations show that

$$ \{\gamma^{i}, \gamma^{i}\} = 2\eta^{ii}\mathbb{I}_{4} $$

and that

$$ \{\gamma^{i}, \gamma^{j}\} = 2\eta^{ij}\mathbb{I}_{4}=0_{4,4}\,, $$ where $\eta^{ii}=-1$ and $\eta^{ij}=0$ for $i\ne j$ with both $i$ and $j$ taking values from $1, 2, 3\,.$

The results $$ \{\gamma^{0},\gamma^{0}\} = 2\eta^{00}\mathbb{I}_{4} $$ $$ \{\gamma^{0},\gamma^{i}\} = 2\eta^{0i}\mathbb{I}_{4}=0_{4,4} $$ $$ \{\gamma^{i},\gamma^{i}\} = 2\eta^{ii}\mathbb{I}_{4}, $$ $$ \{\gamma^{i},\gamma^{j}\} = 2\eta^{ij}\mathbb{I}_{4}=0_{4,4} $$ can be summarised into the single formula

$$ \{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4} $$

where $\eta^{\mu\nu}$ satisfies

$$ \begin{equation} \eta^{\mu\nu}=\begin{pmatrix} 1 & 0 & 0 & 0\newline 0 & -1 & 0 & 0\newline 0 & 0 & -1 & 0\newline 0 & 0 & 0 & -1 \end{pmatrix}\,. \end{equation} $$ This means $\eta^{\mu\nu}$ is the metric tensor of the Minkowski space-time of special relativity.

I prefer the expression $\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4}$ instead of $\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}$ (or $\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$ as the poster wrote) which gives the false impression that $\{\gamma^{\mu}, \gamma^{\nu}\}$ is just a number since for any chosen values of the pair ($\mu,\nu)$ the entry in $g^{\mu\nu}=\eta^{\mu\nu}$ is equal to 0 or $\pm 1$. Clearly this is not the case since $\{\gamma^{\mu}, \gamma^{\nu}\}$ involves the sum of products of $4\times 4$ matrices.

Declaration: I did not come up with the notation $\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4}$ myself. I saw it on the Wikipedia entry for gamma matrices (https://en.wikipedia.org/wiki/Gamma_matrices) earlier today. I do note though that the two QFT books I have to hand use the notation $\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$ (Itzykson & Zuber) and $\{\gamma^{\mu},\gamma^{\nu}\} = -2g^{\mu\nu}$ (Srednicki, where $g^{\mu\nu} = \mbox{diag}(-1,1,1,1)\,$) but again I think this notation is confusing.

$\endgroup$
1
$\begingroup$

The second-order derivative is $g^{\mu\nu}\partial_\nu\partial_\mu$, but since $\partial_\nu\partial_\mu$ is symmetric the symmetrised coefficients match, viz. $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=g^{\mu\nu}+g^{\nu\mu}=2g^{\mu\nu}$.

$\endgroup$
0
$\begingroup$

Just write $$ \gamma^\mu\gamma^\nu=\frac{1}{2}\{\gamma^\mu,\gamma^\nu\}+\frac{1}{2}[\gamma^\mu,\gamma^\nu] $$ and note that the last term is antisymmetric.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.