# Correct method for splitting path integral in two

In 3D point particle quantum mechanics we have that the propagator can be represented as a path integral $$\langle x|e^{-iHt}|y\rangle =\int_{\gamma(0)=x}^{\gamma(t)=y}\mathcal{D}[\gamma(\tau)]e^{iS[\gamma,\dot{\gamma}]}$$ I have two questions regarding how one can split up the propagator into two or more paths

1) What is the correct procedure for splitting the path integral in two? I have derived a formula specific to time independent Hamiltonians to be $$\langle x|e^{-iHt}|y\rangle =\frac{1}{t}\int_0^t dt'\int d^3z\int_{\gamma'(t')=z}^{\gamma'(t)=y}\mathcal{D}[\gamma'(\tau')]e^{iS[\gamma',\dot{\gamma}']}\int_{\gamma(0)=x}^{\gamma(t')=z}\mathcal{D}[\gamma(\tau)]e^{iS[\gamma,\dot{\gamma}]}\quad \text{Eq}.(1)$$ where now $t'$ has the interpretation as the time at which the particle passes through the intermediate point $z^i$ and we sum over all such possible times and intermediate points. Is this formula correct? Is there a more general form?

2)On page 33, Eq.97 of Balitsky's paper https://arxiv.org/pdf/hep-ph/0101042.pdf he chooses a specific component of the intermediate point to be zero (which I will take to be $z^1$ below) and states the result to be $$\langle x|e^{-iHt}|y\rangle =\int_0^t dt''\int d^3z\delta(z^1)\int_{\gamma'(t')=z}^{\gamma'(t)=y}\mathcal{D}[\gamma'(\tau')]e^{iS[\gamma',\dot{\gamma}']}\int_{\gamma(0)=x}^{\gamma(t)=z}\mathcal{D}[\gamma(\tau)]\dot{z}^1(t)e^{iS[\gamma,\dot{\gamma}]}$$ where one now acquires the velocity factor $\dot{z}^1(t')$ at the point $z^1=0$. This result appears to be in conflict with Eq.(1) above, leading me to doubt Eq.(1) even further. Is there a difference between breaking up a path integral in two and not specifying the intermediate point (integrating over the intermediate point) versus specifying that the particle was at a particular intermediate point before reaching the final point?

(Note for anyone who consults the equation referenced in Balitsky's paper: He is working in the Schwinger proper time formalism of QFT which leads to a 4+1 point particle quantum mechanics description of the propagator)

One can then always trivially multiply by $$1 = \frac{1}{t}\int_0^t dt''$$ to find $$\langle x|e^{-iHt}|y\rangle=\frac{1}{t}\int_0^t dt''\int_{-\infty}^\infty dz\,\int_{\gamma(t_1)=z}^{\gamma(t)=y}\mathcal{D}[\gamma(\tau)]e^{iS[\gamma,\dot{\gamma}]}\int_{\gamma(0)=x}^{\gamma(t_1)=z}\mathcal{D}[\gamma(\tau)]e^{iS[\gamma,\dot{\gamma}]}.$$