# Showing hermiticity properties of Dirac matrices using hamiltonians

I want to show $${\gamma^0}^\dagger=\gamma^0\\ {\gamma^i}^\dagger=-\gamma^i.$$

To do this I consider the Dirac equation $$(i\gamma^\mu\partial_\mu-m)\psi=0$$

and I write it as

$$i\partial_t \psi=(-i\gamma^0\gamma^i\partial_i+m\gamma^0)\psi:=H\psi$$

where I defined the Hamiltonian $$H$$. We require $$H=H^\dagger$$ i.e. $$-i\gamma^0\gamma^i\partial_i +m\gamma^0=i(\gamma^0\gamma^i)^\dagger\partial_i+m{\gamma^0}^\dagger.$$

The second term makes it clear that $${\gamma^0}^\dagger=\gamma^0$$, the first term becomes

$$-i\gamma^0\gamma^i=i{\gamma^i}^\dagger{\gamma^0}$$

or

$$\gamma^i\gamma^0={\gamma^i}^\dagger\gamma^0$$

which means $${\gamma^i}^\dagger=\gamma^i$$

Why doesn't this work? In this I'm ignoring the operator $$\partial_i$$ when taking the adjoint because the adjoint is taken in spinor space, not in $$L^2$$. Is this incorrect? Should I take $$\partial_i^\dagger=-\partial_i$$? Why?

• – Cosmas Zachos Jan 4 '19 at 11:59
• Hi, thanks for the reference. I don't see how the question you linked answers mine. The answer starts with "knowing ${\gamma^\mu}^\dagger=\gamma^0\gamma^\mu\gamma^0$", which is what I want to show. – user2723984 Jan 4 '19 at 12:01
• In your first step you seem to use the Dirca algebra, i.e. $\gamma^0 \cdot\gamma^0=\mathbb{1}$. Then the result is more or less obvious. The original drivation assumes some coefficients $\gamma^i$ and derives the algebra from the requirement that the $H^2$ is the Klein-Gordon operator. – Toffomat Jan 4 '19 at 12:14
• I derive the Dirac algebra by requiring that $(i\gamma^\mu\partial_\mu +m)(i\gamma^\mu\partial_\mu -m)=\partial^2+m^2$ but this doesn't involve the hermitian conjugate of the gamma matrices anywhere – user2723984 Jan 4 '19 at 12:23
• Yes, the momentum is Hermitean, so the derivative anti Hermitean, as in that proof. Follow the signs. – Cosmas Zachos Jan 4 '19 at 12:31

Yes. You sould use $$\partial_i^\dagger= -\partial_i$$ because that is the correct adjoint of the derivative in $$L^2[\mathbb R]$$.
Recall that the adjoint $$A^\dagger$$ with respect to an inner product $$<\phantom x,\phantom y>$$ of an operator $$A$$ is defined so that $$= <\phi, A\chi>.$$ When $$<\phi,\chi>= \int_{-\infty}^\infty \phi^*\chi\,dx$$ an integration by parts gives $$<\phi, \partial_x \chi>= <- \partial_x \phi, \chi>$$ so $$\partial_x^\dagger=-\partial_x$$.
• Then I am confused about what we're talking about. The adjoint of $H$ is the operator $H^\dagger$ such that $\langle\psi, H\psi\rangle=\langle H^\dagger \psi, \psi\rangle$, but is this the inner product on $\mathbb{C}^4$, on $L^2$ or something else? – user2723984 Jan 4 '19 at 13:38
• @user2723984 I'm not sure what you are confused about, but I added a bit to my answer about how the adjoint is defined. Ah I see your response. The inner product is on ${\mathbb C}^4 \otimes L^2 [\mathbb R^4]$ – mike stone Jan 4 '19 at 13:44
• mmh I think I understand, since in this case $H$ contains matrices, aren't we actually dealing with $L^2[\mathbb{C^4}]$ and the inner product $\langle \phi, \chi \rangle = \int dx \phi^\dagger \chi$, with the adjoint in the integral wrt the standard inner product of $\mathbb{C^4}$? – user2723984 Jan 4 '19 at 13:46
• @user2723984 Yes. I think we are both being imprcise about the apce, but it is the space of square integrable, complex, four component vector-valued function in ${\mathbb R}^3$, with the inner product you indicate. – mike stone Jan 4 '19 at 13:49