What is the Hamiltonian density for a massive Dirac field?

[Dixanadu]

Hey guys,

So here's the deal. Consider the Lagrangian

$\mathcal{L}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$

where $\bar{\psi}=\psi^{\dagger}\gamma^{0}$.

I need to find the Hamiltonian density from this, using

$\mathcal{H}=\pi_{i}(\partial_{0}\psi_{i})-\mathcal{L}$

So I get the following:

$\mathcal{H}=i\bar{\psi}\gamma^{i}\nabla\psi+\bar{\psi}\psi m$

But my teacher writes

$\mathcal{H}=-i\bar{\psi}\gamma^{i}\nabla\psi+\bar{\psi}\psi m$

And I don't know how he gets that minus factor. The only part where I could be going wrong is when I expand $\gamma^{\mu}\partial_{\mu}$...I'm using $\gamma^{\mu}\partial_{\mu}=\gamma^{0}\partial_{0}-\gamma^{i}\partial_{i}$...because the metric signature is (+,---). But I guess this is wrong?

 

[AndyW]

Hi there,

Thank you for sharing your question about the Lagrangian and Hamiltonian densities. It looks like you have a good understanding of the equations and have made some progress in finding the Hamiltonian density. However, I see where the confusion may be coming from.

When expanding \gamma^{\mu}\partial_{\mu}, it is important to remember that the \gamma^{\mu} matrices are anti-commuting. This means that when you switch the order of the matrices, you will get a minus sign. So when you expand \gamma^{\mu}\partial_{\mu}=\gamma^{0}\partial_{0}-\gamma^{i}\partial_{i}, the second term should actually be -\gamma^{i}\partial_{i} due to the anti-commuting property.

This is why your teacher's Hamiltonian density has a minus sign in front of the first term, because they correctly expanded \gamma^{\mu}\partial_{\mu}=\gamma^{0}\partial_{0}+\gamma^{i}\partial_{i}. This may seem like a small detail, but it is important in order to have the correct sign for the Hamiltonian density.

I hope this helps clarify things for you. Keep up the good work and don't hesitate to ask for help when needed. Science is all about collaboration and learning from each other. Best of luck with your studies!