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In every book or pdf I read about the path integral, I see:

$$ \begin{array}{ccl} \langle q_2,t_2|\hat{q}(t_1)|q_0,t_0\rangle & = & \displaystyle{\iint} dq_3dq_4\langle q_2,t_2|q_3,t_1\rangle\langle q_3,t_1|\hat{q}(t_1)|q_4,t_1\rangle\langle q_4,t_1|q_0,t_0\rangle\\ & = &\displaystyle{\iint}dq_3dq_4\langle q_2,t_2|q_3,t_1\rangle q_3\delta(q_3-q_4)\langle q_4,t_1|q_0,t_0\rangle\\ &=&\displaystyle{\int}dq_4\langle q_2,t_2|q_4,t_1\rangle q_4\langle q_4,t_1|q_0,t_0\rangle\\ &=&\displaystyle{\int dq_4\left(q_4\int_{q(t_0)=q_0}^{q(t_1)=q_4}\mathcal{D}[q]e^{iS[q]} \int_{q(t_1)=q_4}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}\right)} \end{array} \tag{1}$$

and then, they directly write:

$$\langle q_2,t_2|\hat{q}(t_1)|q_0,t_0\rangle = \int_{q(t_0)=q_0}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}q(t_1)\tag 2$$

I guess they use:

$${\displaystyle \langle q_{2},t_{2}|q_{0},t_{0}\rangle = \int d q_4 \langle q_{2},t_{2}|q_{4},t_{1}\rangle \langle q_{4},t_{1}|q_{0},t_{0}\rangle}\tag 3$$

Since it's not explained where I looked, I suppose this is trivial, but I can't manage to find (2) from (1)... I tried integration by part, but it doesn't seem to work.

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You have to use the following property of path integrals, $$ \int\limits_{q(t_0)=q_0}^{q(t_1)=q_1} [dq] = \int\limits_{-\infty}^\infty dx \int\limits_{f(t_0)=q_0}^{f(T)=x} [df] \int\limits_{g(T)=x}^{g(t_1)=q_1} [dg] \qquad \qquad \forall ~ T \in [t_0,t_1]. \tag{1} $$

There are several ways to prove this, but I will simply explain this to you intuitively. The LHS is the integral over all possible functions $q:[t_0,t_1]\to{\mathbb R}$ which satisfy the boundary conditions $q(t_0) = q_0$ and $q(t_1) = q_1$. Let us denote this set by ${\cal F}(t_0,q_0 ; t_1 ,q_1)$.

The set ${\cal F}$ can be equivalently described as follows. We first choose a random point $T \in [ t_0 , t_1 ]$. The set ${\cal F}(t_0,q_0 ; t_1 ,q_1)$ can be broken up into infinitely many disjoint subsets ${\cal F}_x(t_0,q_0 ; t_1 ,q_1)$ $$ {\cal F}(t_0,q_0 ;t_1 ,q_1) = \sum_{x\in{\mathbb R}} {\cal F}_{T,x} (t_0,q_0 | t_1 ,q_1) , \qquad {\cal F}_{T,x}(t_0,q_0 ; t_1 ,q_1) = \{ q \in {\cal F}(t_0,q_0 ; t_1 ,q_1) ~| ~ q(T) = x\} . $$ In other words, ${\cal F}_{T,x} (t_0,q_0 | t_1 ,q_1)$ is the set of all functions $[t_0,t_1]\to {\mathbb R}$ such that $q(t_0)=q_0$, $q(t_1)=q_1$ and $q(T) = x$. Now, it is clear that $$ {\cal F}_{T,x} (t_0,q_0 | t_1 ,q_1) = {\cal F} (t_0,q_0 ; T , x ) \cup {\cal F} (T,x; t_1 , q_1 ) $$ If this is NOT clear, take a pause here and think about it. Drawing some graphs for the possible functions on both sides might help.

It then follows $$ {\cal F}(t_0,q_0 ;t_1 ,q_1) = \sum_{x\in{\mathbb R}} {\cal F} (t_0,q_0 ; T , x ) \cup {\cal F} (T,x; t_1 , q_1 ) $$ The path integral formula (1) represents precisely this property. Again, pause here and think a bit more about why this is the case.

To answer the question in the comments, we have \begin{align} &\int\limits_{-\infty}^\infty dx x \int\limits_{q(t_0)=q_0}^{q(T)=x} [dq] \int\limits_{q(T)=x}^{q(t_1)=q_1} [dq] \\ &\qquad =\int\limits_{-\infty}^\infty dx \int\limits_{q(t_0)=q_0}^{q(T)=x} [dq] \int\limits_{q(T)=x}^{q(t_1)=q_1} [dq] x \\ &\qquad =\int\limits_{-\infty}^\infty dx \int\limits_{q(t_0)=q_0}^{q(T)=x} [dq] \int\limits_{q(T)=x}^{q(t_1)=q_1} [dq] q(T) \\ &\qquad = \left( \int\limits_{-\infty}^\infty dx \int\limits_{q(t_0)=q_0}^{q(T)=x} [dq] \int\limits_{q(T)=x}^{q(t_1)=q_1} [dq] \right) q(T) \\ &\qquad = \int\limits_{q(t_0)=q_0}^{q(t_1)=q_1} [dq] q(T) \end{align}

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  • $\begingroup$ Thanks for your answer! If i get it correctly, your last equation says the same thing that the equation (3) on my post? A path from a to b can be thought as a path from a to c and then from c to b. But to get all the path that go from a to b, one has to consider every c possible and this is why you do a sum on $x$ in your last equation right? But, something I don't understand is that, at the last line of my equation (1), the integration over $dq_4$ contains a $q_4$. And so, we can't use the equation (1) of your post? I am missing something... $\endgroup$
    – Syrocco
    Jun 2 '21 at 18:40
  • $\begingroup$ @Syrocco Yes! it is. Right, so you replace $q_4$ with $q(t_1)$ which you can do because that's the boundary condition. Then, you do use your equation (3). The leftover $q(t_1)$ then appears in your path integral (2). $\endgroup$
    – Prahar
    Jun 2 '21 at 18:42
  • $\begingroup$ So you get: $\displaystyle{\int dq_4\left(q(t_1)\int_{q(t_0)=q_0}^{q(t_1)=q_4}\mathcal{D}[q]e^{iS[q]} \int_{q(t_1)=q_4}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}\right)} $ but to use (3), don't you have to exclude $q(t_1)$ from the integral (so you have to suppose it's constant when $q_4$ varies, which isn't true?)? In all case, I would get: $A = q(t_1)\displaystyle{\int dq_4\int_{q(t_0)=q_0}^{q(t_1)=q_4}\mathcal{D}[q]e^{iS[q]} \int_{q(t_1)=q_4}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]}}=q(t_1)\int_{q(t_0)=q_0}^{q(t_2)=t_2}\mathcal{D}[q]e^{iS[q]} $ which doesn't even makes sense? I'm confused... $\endgroup$
    – Syrocco
    Jun 2 '21 at 18:53
  • $\begingroup$ $q(t_1)$ is not a constant. It is the value of the function at $t_1$ and you're integrating over all functions so you cannot take it out of the integral. Let me edit my answer to discuss this. $\endgroup$
    – Prahar
    Jun 2 '21 at 18:54
  • $\begingroup$ You're a fast writer! Alright, I think I get it, thanks! $\endgroup$
    – Syrocco
    Jun 2 '21 at 19:01
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To see this most clearly, it's necessary to go back to the discretized form of the path integral. There, the measure $\mathcal{D}[q]$ really means $\prod_{i = 1}^{N-1} dq_i$ (up to some normalization). So, discretizing the total time into increments of length $\varepsilon$, with $t_1 = t_0 + N \varepsilon$ and $t_2 = t_1 + M \varepsilon$, and letting $q_i = q(t_0 + \varepsilon i)$, the integrals you wrote take the form $$ \int dq_4 \int_{q(t_0) = q_0}^{q(t_1) = q_4} \mathcal{D}[q] \int_{q(t_1) = q_4}^{q(t_2) = q_2} \mathcal{D}[q] = \int dq_N \int \prod_{i = 1}^{N-1} dq_i \prod_{i = 1}^{M-1} dq_{N+i} = \int \prod_{i = 1}^{N+M-1} dq_i = \int_{q(t_0) = q_0}^{q(t_2) = q_2} \mathcal{D}[q] $$ More intuitively, your integral (1) says "sum over all paths $q(t)$ subject to the constraint $q(t_4) = q_4$, then sum over all possible $q_4$." Upon summing over $q_4$, the total set of paths summed over will no longer have any constraint on $q_4$.

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