11

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?

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
PPR
  • 2,004
  • 2
    you need to know that $\gamma^{\mu \dagger}=\gamma^{\mu}$ for all $\mu=0,1,2,3$, then it's straight forward. – Martin Dec 01 '13 at 12:05
  • Martin: How do you show that? – PPR Dec 01 '13 at 12:07
  • 8
    Isn't $\gamma^{1,2,3}$ anti-hermitian? – lionelbrits Dec 01 '13 at 12:24
  • It depends on the representation you choose. – PPR Dec 01 '13 at 12:25
  • 2
    $\gamma^{\mu\dagger}=\gamma^0 \gamma^{\mu} \gamma^0$ 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. – user26143 Dec 01 '13 at 12:32
  • 1
    Without the information of Hermicity of gamma matrices, how to ensure the Hermicity of Hamiltonian in Dirac equation? – user26143 Dec 01 '13 at 12:36
  • Well, an odd number of gamma matrices will switch sign under hermitian adjoint, and the $i$ will take care of the sign. The anti-commutation relation will take care of re-ordering all the terms. – lionelbrits Dec 01 '13 at 13:17
  • Related meta post: http://meta.physics.stackexchange.com/q/5258/2451 – Qmechanic Dec 03 '13 at 01:46
  • What are you willing to assume about the hermiticity of the other gamma matrices? If you don't have some such hypothesis, then what exactly do you mean by ${}^\dagger$, then? – Emilio Pisanty Dec 03 '13 at 01:57
  • 1
    @CrazyBuddy This question is not off-topic. Qmechanic also thinks the same. See his comment on this meta post. – user10001 Dec 03 '13 at 04:00
  • @user10001: Done. After Emilio's revision, I think it's worthy to reopen. Cheers :) – Waffle's Crazy Peanut Dec 03 '13 at 04:09

2 Answers2

8

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.

Neuneck
  • 9,129
  • 1
    Dear Neuneck, thanks for your response. I'm missing one thing: how do you show that $(\gamma^0)^2=\mathbb{1}$ leads to $(\gamma^0)^\dagger = \gamma^0$, and that $(\gamma^j)^2=-\mathbb{1}$ leads to $(\gamma^j)^\dagger = - \gamma^j$? – PPR Dec 01 '13 at 14:17
  • @Psycho_pr I think the easiest way to see it is that is true for the Dirac representation (as detailed on wikipedia) and use the fact that all representations of the Clifford algebra are unitarily equivalent, i.e. related by a transformation $$\gamma_\mu \to U^\dagger \gamma_\mu U$$ with $U^\dagger U = \mathbf 1$, which leaves the hemicity properties invariant. – Neuneck Dec 02 '13 at 11:20
  • Hrmm, but if we are going to look at a particular representation anyway, why did we need to know that $(\gamma^0)^2 = \mathbb{1}$ and that $(\gamma^j)^2=-\mathbb{1}$? We could have just shown directly the hermiticity properties in the Dirac (or Weyl) representations and then shown it remains invariant under unitary transformations. Am I missing something? – PPR Dec 02 '13 at 16:25
  • 1
    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. – Neuneck Dec 02 '13 at 16:41
  • 1
    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. – user26143 Dec 02 '13 at 16:42
  • You can prove the Hermicity (and anti-Hermicity) from the old-fashioned Dirac equation, $H=\mathbf{\alpha} \cdot \mathbf{p} + \beta m$ (under +--- metric), $\gamma^0:=\beta, \gamma^i:=\beta \alpha^i$, $\beta^{\dagger}=\beta$, $\alpha^{i\dagger}=\alpha^i$ and the anticommuting relations of $\alpha,\beta$ matrices. – user26143 Dec 02 '13 at 16:46
  • @user26143, that's exactly what I tried to do. At Neuneck: So if I understand correctly the scheme is like so: $(\gamma^0)^2=\mathbb{1}$ and $(\gamma^j)^2=-\mathbb{1}$ out of the commutation relations and the gamma matrices must be unitary because of the reason you mentioned, so, together these two facts imply the Hermiticity relations. So with these two facts we never used any particular representation (except, maybe, when proving that the gamma matrices are unitary). – PPR Dec 02 '13 at 16:49
  • OK, if one could use the information of a particular representation (plus the unitary equivalence) then the Hermicity is a corollary... Nevertheless I would say the Hermicity is needed to justify Dirac representation is a correct representation...... – user26143 Dec 02 '13 at 17:05
  • 3
    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? – Emilio Pisanty Dec 03 '13 at 00:57
  • @EmilioPisanty, according to Neuneck, $\gamma^0$ is both idempotent and unitary. These two facts together, supposedly, imply Hermiticity. Is this true? – PPR Dec 03 '13 at 06:36
  • 2
    Yes. For unitary matrices with $A^2 =1 $ we have $$ A = A^{-1} = A^\dagger$$. – Neuneck Dec 03 '13 at 06:54
  • 1
    @user26143 You can check that the Dirac rep. is a 'correct representation' by brute forcing them through the anticommutation relations. As I mentioned, unitarity is a corrollary for finite dimensional reps of the clifford algebra (although I still can't name the theorem, sorry) – Neuneck Dec 03 '13 at 06:58
  • 2
    @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. – Emilio Pisanty Dec 03 '13 at 12:10
  • 1
    @Neuneck I am not sure if anticommuting relation of gamma matrices is sufficient for the Dirac equation, although one could say Dirac field is just spinor representation of Poincare group. From the Hamiltonian $$H= \int d^3 x \psi^{\dagger} [ - i \gamma^0 \vec{\gamma} \cdot \vec{{\nabla}} + m \gamma^0 ] \psi = \int d^3 x \psi^{\dagger} [ \gamma^0 \vec{\gamma} \cdot \vec{p} + m \gamma^0 ] \psi$$, to ensure $H^{\dagger}=H$, I need the Hermicity of $\gamma$. Would you provide a reference that unitarity is corrollary for finite dimensional reps of the clifford algebra? Thanks – user26143 Dec 03 '13 at 13:56
5

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).

Wouter
  • 5,259