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

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


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

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?

  • $\begingroup$ duplicate. $\endgroup$ – Cosmas Zachos Jan 4 '19 at 11:59
  • $\begingroup$ 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. $\endgroup$ – user2723984 Jan 4 '19 at 12:01
  • $\begingroup$ 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. $\endgroup$ – Toffomat Jan 4 '19 at 12:14
  • $\begingroup$ 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 $\endgroup$ – user2723984 Jan 4 '19 at 12:23
  • $\begingroup$ Yes, the momentum is Hermitean, so the derivative anti Hermitean, as in that proof. Follow the signs. $\endgroup$ – 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 $$ <A^\dagger \phi, \chi>= <\phi, A\chi>. $$ When $<\phi,\chi>= \int_{-\infty}^\infty \phi^*\chi\,dx$ an integration by parts gives $$ <\phi, \partial_x \chi>= <- \partial_x \phi, \chi> $$ so $\partial_x^\dagger=-\partial_x$.

  • $\begingroup$ 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? $\endgroup$ – user2723984 Jan 4 '19 at 13:38
  • $\begingroup$ @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 ${\mathbb C}^4 \otimes L^2 [\mathbb R^4]$ $\endgroup$ – mike stone Jan 4 '19 at 13:44
  • $\begingroup$ 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 $\mathbb{C^4}$? $\endgroup$ – user2723984 Jan 4 '19 at 13:46
  • $\begingroup$ @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. $\endgroup$ – mike stone Jan 4 '19 at 13:49
  • $\begingroup$ Thank you, my confusion was ultimately about which space and inner products I was dealing with $\endgroup$ – user2723984 Jan 4 '19 at 13:52

One can choose gamma-matrices in such a way that they do not satisfy the hermiticity condition.


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