# Showing Dirac Hamiltonian is hermitian

I'm trying to show that $H_D = -i\boldsymbol{\alpha}.\nabla+\beta m$ is hermitian.

Its given that $$\gamma^{0\dagger}=\gamma^0$$ $$\boldsymbol\gamma^\dagger=-\boldsymbol\gamma$$

What i've done is: $$H_D^\dagger = +i\boldsymbol\alpha^\dagger.\nabla+\beta^\dagger m$$ $$=+i(\gamma^0\boldsymbol\gamma)^\dagger.\nabla+\beta m$$ $$=+i(\boldsymbol\gamma^\dagger\gamma^{0\dagger}).\nabla+\beta m$$ $$=-i(\boldsymbol\gamma\gamma^0).\nabla+\beta m$$

Well clearly i'm doing somthing wrong because $\boldsymbol\gamma\gamma^0\neq\gamma^0\boldsymbol\gamma$

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$\nabla^\dagger = - \nabla$ and $\bf{\gamma} \gamma^0 = - \gamma^0 \bf{\gamma}$ – Prahar Jun 12 '13 at 19:03
why $\nabla^\dagger=-\nabla$ @Prahar – Aftnix Jun 12 '13 at 19:12
Please give in a answer form so that i can accept it. – Aftnix Jun 12 '13 at 19:13
The operator $p = i \nabla$ is hermitian (this is the observable momentum), so $p = p^\dagger$, so $(i \nabla)^\dagger = (i \nabla)$. So there is an sign error in the first line. At the end, you have to add a step, that is $\gamma \gamma_0 = -\gamma_0 \gamma = \alpha$ – Trimok Jun 12 '13 at 19:34

## 1 Answer

So the key is to understand that $\nabla^\dagger = - \nabla$. To see why this should be true, we go back to the definition of the adjoint of an operator, namely $$\left< \phi \right|\left. A \psi \right> = \left< A^\dagger\phi \right|\left. \psi \right> \implies \int d^dx \phi(x)^* {\hat A} \psi(x) = \int d^d x \left( {\hat A}^\dagger \phi(x) \right)^* \psi(x)$$

In particular, for the differential operator

$$\int \phi^* \nabla \psi = - \int \nabla \phi^* \psi = \int \left( - \nabla \phi \right)^* \psi \implies \nabla^\dagger = - \nabla$$

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Have you done integration by parts in $\int\phi^*\nabla\psi=-\int\nabla\phi^*\psi$? – Aftnix Jun 12 '13 at 19:50
Yes. I have assumed that the boundary term drops out. – Prahar Jun 12 '13 at 19:59