9
$\begingroup$

Is it possible to show that ${\gamma^5}^\dagger = \gamma^5$, where $$ \gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3,$$ using only the anticommutation relations between the $\gamma$ matrices, $$ \left\{\gamma^\mu,\,\gamma^\nu\right\}=2\eta^{\mu\nu}\,\mathbb{1},$$ and without using any specific representation of this algebra and a unitary invariance argument, as is usually done?

$\endgroup$
  • 2
    $\begingroup$ you need to know that $\gamma^{\mu \dagger}=\gamma^{\mu}$ for all $\mu=0,1,2,3$, then it's straight forward. $\endgroup$ – Martin Dec 1 '13 at 12:05
  • 6
    $\begingroup$ Isn't $\gamma^{1,2,3}$ anti-hermitian? $\endgroup$ – lionelbrits Dec 1 '13 at 12:24
  • 2
    $\begingroup$ $\gamma^{\mu\dagger}=\gamma^0 \gamma^{\mu} \gamma^0$ and $[\gamma^{\mu},\gamma^{\nu}]_+=2\eta^{\mu\nu}$. For +--- metric, $\gamma^0$ is Hermitian and $\gamma^i,i=1,2,3$ is anti-Hermitian. For -+++ metric, $\gamma^0$ is anti-Hermitian and $\gamma^i$ is Hermitian. Perhaps it is not due to representation. $\endgroup$ – user26143 Dec 1 '13 at 12:32
  • 1
    $\begingroup$ Without the information of Hermicity of gamma matrices, how to ensure the Hermicity of Hamiltonian in Dirac equation? $\endgroup$ – user26143 Dec 1 '13 at 12:36
  • 1
    $\begingroup$ @CrazyBuddy This question is not off-topic. Qmechanic also thinks the same. See his comment on this meta post. $\endgroup$ – user10001 Dec 3 '13 at 4:00
7
$\begingroup$

As the comments explained, you need to know a few properties of the $\gamma$ matrices. First of all, from $$ \{\gamma_\mu, \gamma_\nu\} = 2 \eta_{\mu \nu} \mathbf{1}_4$$ you can infer that (depending on the metric but not on the representation of the dirac algebra!) in (+---) metric $\gamma_0$ is hermitian (hint: look at the $\mu = 0, \nu = 0$ component of the above equation), while the $\gamma_i$ ($i = 1, 2, 3$) are anti-hermitian. (In -+++ metric, this would be interchanged). And this should allow you to solve the problem.

The hermicity properties can be condensed into $$ \gamma_\mu^\dagger = \gamma_0 \gamma_\mu \gamma_0$$ which just reproduces the above if you take the commutation properties into account.

$\endgroup$
  • 1
    $\begingroup$ The full proof involves showing that the gamma matrices are the basis of a (finite dimensional representation of an) infinite group. Therefore they must be unitary and using this in addition to the relations I gave implies the conjugation properties. -- Sorry that I didn't mention this before. $\endgroup$ – Neuneck Dec 2 '13 at 16:41
  • 1
    $\begingroup$ I guess you tried to follow Peskin's QFT, p50 "The matrix $\gamma^5$ has the folllowing properties, all of which can be vertified using (3.68) and the anticommuting relation (3.22)". I don't think that's enough. One has to use the Hermicity properties of gamma matrices. The Hermicity properties of gamma matrices is a requirement for the Hermicity of Dirac Hamiltonian. It's natural to use them. Once we adopt the Hermicity of gamma matrices, the proof of $\gamma^{5\dagger}=\gamma^5$ is straightforward. $\endgroup$ – user26143 Dec 2 '13 at 16:42
  • 2
    $\begingroup$ Just a quick note, squaring to the identity does not imply a matrix is hermitian. A counterexample is $$M=\begin{pmatrix}-1&-2\\ \phantom{-}0&\phantom{-}1\end{pmatrix}.$$ Does this answer still hold in light of that fact? $\endgroup$ – Emilio Pisanty Dec 3 '13 at 0:57
  • 1
    $\begingroup$ Yes. For unitary matrices with $A^2 =1 $ we have $$ A = A^{-1} = A^\dagger$$. $\endgroup$ – Neuneck Dec 3 '13 at 6:54
  • 2
    $\begingroup$ @Psycho_pr That much is correct. However, you should note that if you assume $\beta$ is hermitian then you're already there. Unless you admit a hypothesis relating the gamma matrices to the inner product structure you will never prove your result. $\endgroup$ – Emilio Pisanty Dec 3 '13 at 12:10
5
$\begingroup$

Try using the definition of $\gamma^5$ and just apply the conjugation. Remember that conjugation flips the order of the matrices, which means you want to change their order before applying the conjugation.

Then realize that the anticommutation relations give you a way of interchanging two gamma matrices, giving only a minus sign (provided the indices are different).

Lastly, note that, independent of your convention, you have an uneven number of anti-hermitian matrices in this expression (while the rest is hermitian).

$\endgroup$

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.