I want to derive the Klein-Gordon Green function equation $$(\Box_b + m^2) D_F(x_b - x_a) = - i \delta^4(x_b - x_a)$$ by using the same steps taken when fixing the 'exact' Green function of the non-relativistic Schrodinger equation. I just keep finding a problematic extra factor of 2 that will not go away.
In the Klein-Gordon case the derivation is as follows:
Given $\Psi(x) = \Psi^+(x) + \Psi^-(x)$ satisfying the Klein-Gordon equation $(\Box + m^2) \Psi(x) = 0$, let us just focus on $\Psi^+(x)$. The Feynman propagator $D_F(x_b - x_a)$ can be used to evolve $\Psi^+(x_a)$ into (Bjorken and Drell, Eq. (9.13)): $$\theta(t_b - t_a) \Psi^+(x_b) = i \int d^3 \mathbf{x}_a D_F(x_b - x_a) \overleftrightarrow{\partial}_{t_a} \Psi^+(x_a).\tag{9.13}$$ where $t_b \geq t_a$, and $D_F(x_b - x_a)$ is technically not yet defined/specified.
We now apply $(\Box_b + m^2)$ to both sides.
Using $$\partial_{t_b}[\theta(t_b - t_a) \Psi^+(x_b)] = \delta(t_b - t_a) \Psi^+(x_b) + \theta(t_b - t_a) \partial_{t_b} \Psi^+(x_b)$$ $$\partial_{t_b}^2[\theta(t_b - t_a) \Psi^+(x_b)] = \partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b) + 2 \delta(t_b - t_a) \partial_{t_b} \Psi^+(x_b) + \theta(t_b - t_a) \partial_{t_b}^2 \Psi^+(x_b)$$ we see the left-hand side satisfies $$(\Box_b + m^2)[\theta(t_b - t_a) \Psi^+(x_b)] = \partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b) + 2 \delta(t_b - t_a) \partial_{t_b} \Psi^+(x_b).$$
This tells me to try $$(\Box_b + m^2) D_F(x_b - x_a) = - i \delta^4(x_b - x_a)$$ on the right-hand side, however when I try this I get $$(\Box_b + m^2) i \int d^3 \mathbf{x}_a D_F(x_b - x_a) \overleftrightarrow{\partial}_{t_a} \Psi^+(x_a) = i(-i) \int d^3 \mathbf{x}_a \delta^4(x_b - x_a) \overleftrightarrow{\partial}_{t_a} \Psi^+(x_a) = \delta(t_b - t_a) \overleftrightarrow{\partial}_{t_a} \Psi^+(t_a,\mathbf{x}_b) = - \partial_{t_a} \delta(t_b - t_a) \Psi^+(t_a,\mathbf{x}_b) + \delta(t_b - t_a) \partial_{t_a} \Psi^+(x_a,\mathbf{x}_b) = \partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b) + \delta(t_b - t_a) \partial_{t_b} \Psi^+(x_b) .$$ Clearly I am missing a factor of $2$ in the second computation. Any ideas?
Edit: One disgraceful idea is to leave the answer on the RHS as $$(\Box_b + m^2) i \int d^3 \mathbf{x}_a D_F(x_b - x_a) \overleftrightarrow{\partial}_{t_a} \Psi^+(x_a) = \partial_{t_b} \delta(t_b - t_a) \Psi^+(t_a,\mathbf{x}_b) + \delta(t_b - t_a) \partial_{t_a} \Psi^+(t_a,\mathbf{x}_b) $$ and on the LHS to take the computation as $$\partial_{t_b}[\theta(t_b - t_a) \Psi^+(x_b)] = \delta(t_b - t_a) \Psi^+(x_b) + \theta(t_b - t_a) \partial_{t_b} \Psi^+(x_b) = \delta(t_b - t_a) \Psi^+(t_a,\mathbf{x}_b) + \theta(t_b - t_a) \partial_{t_b} \Psi^+(x_b)$$ which leads to $$\partial_{t_b}^2[\theta(t_b - t_a) \Psi^+(x_b)] = \partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b) + \delta(t_b - t_a) \partial_{t_b} \Psi^+(t_a,\mathbf{x}_b) + \theta(t_b - t_a) \partial_{t_b}^2 \Psi^+(x_b)$$ which gives the right answer on both sides now (assuming we can set $\partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b) = \partial_{t_b} \delta(t_b - t_a) \Psi^+(t_a,\mathbf{x}_b)$, this step is another rabbit hole, compare Peskin (2.56) for example where they deal with this by simply doing IBP despite no time integral appearing anywhere, and it seems unavoidable... Note in Peskin's (2.56) they also get this factor of 2 which is important in what they do, so the above idea is pretty disgraceful, however it also appears to give the right answer here...). Any objection/agreement to this leave it as a comment.
Here I am mimicking the following non-relativistic derivation:
Given a wave function $\Psi(x)$, $x = (t,\mathbf{x})$, satisfying the non-relativistic Schrodinger equation $$[i \frac{\partial}{\partial t} - \hat{H}(x)]\Psi(x) = 0,$$ with $\hat{H} = \hat{H}_0 + \hat{V}$, the non-relativistic propagator $G(x_b;x_a)$ evolves a wave function $\Psi(x)$ from $x_a$ to $x_b$ with $t_b \geq t_a$ via $$\theta(t_b - t_a) \Psi(x_b) = i \int d^3 \mathbf{x}_a G(x_b;x_a) \Psi(x_a)$$ where $i G(x_b;x_a)$ is not yet specified/defined. From \begin{align*} \begin{split} [i \frac{\partial}{\partial t_b} - \hat{H}(x_b)][\theta(t_b - t_a) \Psi(x_b)] &= i \delta(t_b - t_a) \Psi(x_b) + \theta(t_b - t_a) [i \frac{\partial}{\partial t_b} - \hat{H}(x_b)]\Psi(x_b) \\ &= i \delta(t_b - t_a) \Psi(x_b) \\ &= i \int d^3 \mathbf{x}_a [i \frac{\partial}{\partial t_b} - \hat{H}(x_b)] G(x_b;x_a) \Psi(x_a) \end{split} \end{align*} we see we should set $$[i \frac{\partial}{\partial t_b} - \hat{H}(x_b)] G(x_b;x_a) = \delta^4 (x_b - x_a).$$
The Klein-Gordon case obviously has to be true for $t_b \geq t_a$, not just $t_b > t_a$ (if I say $t_b > t_a$, then technically the $\delta(t_b - t_a)$ vanish, and so I'm left with the $\partial_{t_b} \delta(t_b - t_a) \Psi^+(x_b)$ terms, but this is not good enough). This post discusses resolving a factor of 2 problem arising because $\theta(0) = 1/2$, I don't see that resolving things and in fact it causes problems in the non-relativistic case.
References:
- Bjorken and Drell, Relativistic Quantum Mechanics, Eq. (9.13).
