# Division by an operator in non-relativistic limit of Dirac equation?

Suppose we are dealing with the Dirac equation and we are solving it for a system with only electric field. The equations are: $$\left(i \hbar \frac{\partial}{\partial{t}}\right)\phi = c \left(\sigma \cdot P\right) \chi + V \phi + mc^2\phi$$ $$\left(i \hbar \frac{\partial}{\partial{t}}\right)\chi = c \left(\sigma \cdot P\right) \phi + V \chi - mc^2\chi$$

The energy $$E$$ is $$i \hbar \frac{\partial}{\partial{t}}$$ and we rewrite these equation in this way:

$$\left( E - V - mc^2 \right)\phi = c \left(\sigma \cdot P\right) \chi$$ $$\left( E - V + mc^2 \right)\chi = c \left(\sigma \cdot P\right) \phi$$

In order to get the NR limit ($$E-V << mc^2$$) we "divide" the second equation in this way: $$\chi = c \frac{\left(\sigma \cdot P\right)}{\left( E - V + mc^2 \right)}\phi.$$

But what's the mathematical meaning of this step? $$E$$ is still the derivative with respect to time, why can we "divide" in this way?

Suppose we are dealing with the Dirac equation...

The energy E is $$i \hbar \frac{\partial}{\partial{t}}$$ and we rewrite these equation in this way: $$\left( E - V - mc^2 \right)\phi = c \left(\sigma \cdot P\right) \chi$$ $$\left( E - V + mc^2 \right)\chi = c \left(\sigma \cdot P\right) \phi$$

In order to get the NR- limit ($$E-V << mc^2$$) we "divide" the second equation in this way: $$\chi = c \frac{\left(\sigma \cdot P\right)}{\left( E - V + mc^2 \right)}\phi$$

But what's the mathematical meaning of this step? E is still the derivative with respect to time, why can we "divide" in this way?

It's called a "Green's function."

As a simple pedagogical example, suppose that there exists a function G(t,t') such that: $$\left(\frac{d}{dt}+m^2\right)G(t,t')=\delta(t-t')\;,$$ where $$\delta(t-t')$$ is a dirac delta function.

Then the particular solution to $$\left(\frac{d}{dt}+m^2\right)a(t) = b(t)$$ is $$a(t) = \int dt' G(t,t')b(t')\;,$$ which you might write, rather formally, as: $$a = \frac{1}{\frac{d}{dt}+m^2}b$$

And you might formally write: $$G = \frac{1}{\frac{d}{dt}+m^2}\;,$$ where the "inverse" is supposed to be understood in the sense of a formal "continuous matrix inverse": $$\mathbf{a}=\mathbf{G}\mathbf{b}$$ $$\sim a_i = \sum_{j}G_{ij}b_j$$ $$\sim a(t)=\int dt' G(t,t')b(t')$$

• 2 questions: 1) Can I analytically compute the Green function for the Dirac equation? 2) Why do I get the correct physical result after the the taylor expansion (using E<<<m)? For example we get at first order the Pauli Equation, at the second order we get the fine structure hamiltonian. It's strange to me that we get such beautiful results starting from a formal scripture of the real green function. Commented Oct 13, 2023 at 23:49
• 1) Yes. It is well known. It is also called the electron "propagtor" in QED. See: en.wikipedia.org/wiki/Propagator#Spin_1%E2%81%842.
– hft
Commented Oct 14, 2023 at 0:04
• 2) You had better get the correct non-relativistic result in the non-relativistic limit or else the whole thing is wrong.
– hft
Commented Oct 14, 2023 at 0:05
• I guess this is why Dirac was so highly regarded
– hft
Commented Oct 14, 2023 at 0:06
• energy is not the time derivative (which is not an operator on the Hilbert space): Also, if energy is a derivative, what should the non-relativistic limit, as written in the question, mean?? Commented Oct 14, 2023 at 6:57

$$E=\hbar \omega$$ is the effect of an ansatz with a single frequency mode

$$(\phi,\chi) = e^{i E t/\hbar} (a(x),b(x))$$

So $$E$$ is a real number. Your equation allows the calculate the small lower component, if the dominating upper component as a function of x is known.

In order to get it, you have to calculate the second order time derivatives, eliminate the lower component and its derivatives and solve the resulting Pauli-Klein-Gordon-eigenvalue equation for the upper component $$H a = E^2$$ a. Very hard algebra and ODE stuff.