# 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 \begin{equation} \langle x|e^{-iHt}|y\rangle =\int_{\gamma(0)=x}^{\gamma(t)=y}\mathcal{D}[\gamma(\tau)]e^{iS[\gamma,\dot{\gamma}]} \end{equation} 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 \begin{equation} \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) \end{equation} 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 \begin{equation} \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}]} \end{equation} 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)

## 1 Answer

Yes, your formula in 1) looks right to me.

Working in 1D for simplicity, \begin{align} \langle x|e^{-iHt}|y\rangle & = \langle x|e^{-iH(t-t_1)}e^{-iHt_1}|y\rangle \\ & = \langle x|e^{-iH(t-t_1)}\Big(\int_{-\infty}^\infty dz\,|z\rangle\langle z|\Big)e^{-iHt_1}|y\rangle \\ & = \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}]}. \end{align}

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

Yes, there's a big difference between fixing the intermediate point to something specific compared to integrating over the intermediate point. What Balitsky is doing seems to be particular for his specific shock problem at hand.