Why is the Darwin Term in the Fine Structure Hamiltonian Not Hermitian?

[eoghan]

Hi there! I'm solving the dirac equation to get the fine structure hamiltonian of the hydrogen atom. In the hamiltonian there is this term:
$$\frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r}$$

This term gives rise to some difficulty because it is not hermitian. So Darwin proposed to use instead the symmetrical combination:
$$\frac{1}{2}\left[\left(\frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r}\right)+[\left(\frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r}\right)^*\right]=\frac{\hbar^2}{8m^2c^2}\nabla^2V$$

but the adjoint of $$\frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r}$$ is the adjoint of the spatial derivative which is anti-hermitian, so the symmetric combination should be 0... where am I wrong??

 

[mark726]

Hello! It's great to see that you are working on the fine structure Hamiltonian of the hydrogen atom. The term you mentioned, \frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r}, is indeed not Hermitian, which can cause some difficulty in solving the Dirac equation.

You are correct that Darwin proposed using a symmetrical combination to address this issue. However, it is important to note that the adjoint of \frac{\hbar ^2 e}{4m^2c^2}\frac{dV}{dr}\frac{\partial}{\partial r} is not just the adjoint of the spatial derivative. It also includes the adjoint of the potential term, which is Hermitian.

So, the symmetric combination is not necessarily 0, but it does result in a simpler form, as you pointed out. This is because the potential term and its derivative are both Hermitian, so their adjoints are the same as the original terms.

I hope this helps clarify your confusion. Keep up the great work on solving the Dirac equation!