Of course, for any problem all the paths are not going to be counted one by one, but it will be used a simple plausible way.
To understand how this works, it is conventional to start with the most general free particle problem whose Lagrangian is \[L = \frac{m\dot{x}^2}{2}\,\,\tag{2.1}\]
The technique is to split the action into two parts of which one of them is the classical action and the other is treated as the variational part. For this, it is defined the arbitrary path as \[x(t) = x_c(t)+y(t)\,\,\,\tag{2.2}\] where $x_c(t)$ represents the actual classical path. Substituting this in our Lagrangian and expanding it in terms of Taylor expansion, \[L(\dot{x}) = L(\dot{x}_c(t)+\dot{y}(t)) = L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\,\tag{2.3}\] The expansion is exact since L is quadratic in $\dot{x}$. Thus, it is possible to write the action integral as, \[S = \int_{t_1}^{t_2} \,dt \left(L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\right)\,\,\tag{2.4}\]
The first term can be denoted as,
\[ S_c = \int_{t_1}^{t_2}dt\,L(\dot{x}_c)\,\,\tag{2.5}\]
Using (2.1)
\[\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}}\vert_{\dot{x}_c} = mass = const.\,\tag{2.6}\]
and using integral by parts
\[\int_{t_1}^{t_2}dt\,\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} = \left[\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c}y(t)\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}dt\, \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right)y \]
for the lagrangian
\[ \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right) = m\dot{x}_c = 0 \]
Which results finally,
\[S = S_c + \frac{m}{2} \int_{t_1}^{t_2} dt \,\dot{y}^2\,\tag{2.7}\]
and
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}S_c} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.8}\]
where it is changed the integration variable to $dy(t)$ using (2.2)
The classical action is calculated to be
\[S_c = \frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}\,\tag{2.9}\]
which finally yields,
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.10}\]
The integral limit points out the fact that the deviation from the classical path at the end points is zero.
The integral over $y(t)$ is independent of $x_1$ and $x_2$. Its value depends only on $t_1$ and $t_2$ , since the entire problem is time translation invariant,
\[A(t_2-t_1) = \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.11}\]
and
\[K(x_2,t_2;x_1,t_1)= A(t_2-t_1)e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}}\,\tag{2.12}\]
To determine A(t) for $t_1=0$ making use of the group property (1.23) and (1.19),
\[ \delta(x_2-x_1) = K(x_2,t;x_1,t) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K(x,0;x_1,t)\] \[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K^*(x_1,t;x,0) \,\tag{2.13}\]
Substituting for K and using (2.12),
\[K(x_2,t;x,0)= A(t)e^{\frac{i}{\hbar}S_c(x_2,t;x,0)}\]
\[K^*(x_1,t;x,0)= A^*(t)e^{\frac{-i}{\hbar}S_c(x_1,t;x,0)}\]
and
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx \,\left\vert{A(t)}\right\vert^2 e^{\frac{i}{\hbar}\left(S_c(x_2,t;x,0)-S_c(x_1,t;x,0)\right)}\,\tag{2.14}\]
When $x_2 = x_1+\Delta{x}$ the argument of exponential can be modified with Taylor expansion as,
\[ S_c(x_2,t;x,0)-S_c(x_1,t;x,0) = \frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} \Delta{x_1}\]
Considering $\Delta{x_1}\rightarrow\,0$ all higher order terms are neglected.
Substituting for $S_c$,
\[\frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} = \frac{\partial(\frac{m\,(x_1-x)^2}{2\,t})}{\partial{x_1}} = \frac{m\,(x_1-x)}{t} = \gamma(x) \,\tag{2.15}\]
Since $\gamma(x)$ is linear function of $x$, its derivative $\frac{d\gamma}{dx}$ is independent of $x$. Using this information to write,
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} d\gamma\,\left\vert\frac{dx}{d\gamma}\right\vert \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)\left(x_2-x_1\right)}\,\tag{2.16}\]
From Fourier transform,
\[\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\beta(x_2-x_1)} \,d\beta = \delta(x_2-x_1)\,\tag{2.17}\]
for $\beta = \frac{\gamma}{\hbar}$,
\[\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} e^{\frac{i}{\hbar}\gamma(x_2-x_1)} \,{d\gamma} = \delta(x_2-x_1)\,\tag{2.18}\]
Thus, multiplying and dividing by the factor $2\pi\hbar$ in (2.16),
\[ \delta(x_2-x_1) = \int_{-\infty}^{\infty} \frac{d\gamma}{2\pi\hbar} \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)(x_2-x_1)}\frac{2\pi\hbar}{\left\vert\frac{d\gamma}{dx}\right\vert}\,\tag{2.19}\]
Comparing both the left and right side,
\[\delta(x_2-x_1) = \delta(x_2-x_1)\frac{2\pi\hbar|A(t)|^2}{\left\vert\frac{d\gamma}{dx}\right\vert}\]
so that,
\[|A(t)|^2 = \frac{1}{2\pi\hbar}\left\vert\frac{d\gamma}{dx}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{-m}{t}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{\partial^2S_c(x_1,t;x,0)}{\partial{x}\partial{x_1}}\right\vert\,\tag{2.20}\]
The phase can be chosen such that,
\[A(t) = \sqrt{\frac{m}{2i\pi\hbar{t}}}\]
Thus, it is determined the propagator for a free particle as,
\[K(x_2,t_2;x_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[S_c = \frac{m\,(x_2-x_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.21}\]
In general for three dimensions,
\[K(r_2,t_2;r_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(r_2-r_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.22}\]
For fixed $x_1=0\,\,,t_1=0$ using (3.5) the propagator for a free particle should reduce to the Schr\"{o}dinger wave function for a free particle. Thus, the probability amplitude of the free particle from the propagator is,
\[\psi(x,t) = K(x,t,0,0) = \sqrt{\frac{m}{2i\pi\hbar{t}}}e^{\frac{i}{\hbar}\frac{m\,x^2}{2\,t}}\,\tag{2.23}\]
If it is considered a specific point $(x_0,t_0)$, then the classical momentum at this point is $$p_0 = mv_0 = m\frac{x_0}{t_0}$$ with energy $$E = \frac{mv_0^2}{2}= \frac{mx_0^2}{2t_0^2}$$ The change in phase in the vicinity of $(x_0,t_0)$ is then using again Taylor expansion,
\[
\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} exp\frac{i\,m}{\hbar\,2}\left[\frac{x_0^2}{t_0}+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{x}}\right\vert_{(x_0,t_0)}(x-x_0)+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{t}}\right\vert_{(x_0,t_0)}(t-t_0)+...\right]
\]
The bracketed term reduces to (neglecting higher order terms),
\[\left[\frac{x_0^2}{t_0}+ \frac{2x_0}{t_0}(x-x_0) -\frac{x_0^2}{t_0^2}(t-t_0)\right]\] Simplifying with arithmetic,
\[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,exp{\frac{i}{\hbar}\left[m\frac{x_0}{t_0}x - \frac{m}{2}{\frac{x_0^2}{t_0^2}t}\right]}\]
Thus, the wave function varies in the immediate vicinity of $(x_0,t_0)$ according to \[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,e^{\frac{i}{\hbar}\left[p_0x - E_0t\right]}\,\tag{2.24}\]
which is the well-known Einstein-de Broglie relation, according to which a particle with momentum $p$ and energy $E$ is assigned a wave function with wave length and wave number respectively $\lambda = \frac{h}{p}\,\,\rightarrow\,\, k = \frac{2\pi}{\lambda}$ Similarly with frequency and angular frequency $\nu = \frac{E}{h}\,\,\rightarrow\,\,\omega = {2\pi\nu}$
So,
\[ e^{i\left(kx- \omega{t}\right)}= e^{i\left(\frac{2\pi}{\lambda}x- {2\pi\nu}t\right)} = e^{\frac{i}{\hbar}(px-Et)} \tag{2.25}\]
which is equivalent to (2.24).
Schr\"{o}dinger’s time evolution equation
So far, everything is discussed in position basis. Whereas it needs to be in momentum or energy basis for explicit derivation of Schr\"{o}dinger’s time evolution equation. This is achieved using the group property of our propagators,
\[K(x,t;p,0) = \int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,K(x',0;p,0)\,\tag{2.26}\]
For fixed initial momentum $p$ at time $t$,
\[K(x,t;p,0) = \chi_{p,0}(x,t)=\int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,\chi_{p,0}(x',0)\,\tag{2.27}\]
Let us take this as an ansatz for the transformation amplitude,
\[\chi_{p,0}(x,0) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}xp}\,\tag{2.28}\]
Substituting this in (2.27) with the corresponding substitution for $K$,
\[\chi_{p,0}(x,t) =\int_{-\infty}^{\infty}\sqrt{\frac{m}{2i\pi\hbar{(t)}}}e^{\frac{i}{\hbar}\frac{m\,(x-x')^2}{2\,t}} \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{i}{\hbar}x'p} \,dx'\tag{2.29}\]
As $x'$ is the integrating variable, the x' terms are combined together and simplified using arithmetic,
\[\chi_{p,0}(x,t) = e^{\frac{i}{\hbar}\left(xp-\frac{p^2}{2m}t\right)}\sqrt{\frac{m}{2i\pi\hbar{t}}}\sqrt{\frac{1}{2\pi\hbar}} \int_{-\infty}^{\infty}e^{\frac{i}{\hbar}\frac{m}{2\,t}\left[x'-\left(x-\frac{pt}{m}\right)\right]^2}\,\tag{2.30}\]
changing the variable to $u = \left[x'-\left(x-\frac{pt}{m}\right)\right]$ and making use of Gaussian Integral,
\[\chi_{p,0}(x,t) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[xp-\frac{p^2}{2m}t\right]}\,\tag{2.31}\]
In three dimensions,
\[\chi_{p,0}(x,t) = \left(\frac{1}{2\pi\hbar}\right)^{\frac{3}{2}} e^{\frac{i}{\hbar}\left[r.p-\frac{p^2}{2m}t\right]}\tag{2.32}\]
Again using the group property with momentum arguments,
\begin{align*}
K(p_2,t;p_1,0) &=\int_{-\infty}^{\infty} dx'\,K(p_2,t;x,t) \,K(x,t;p_1,0) \\ &= \int_{-\infty}^{\infty} dx\,K(p_2,0;x,0) \,\chi_{p_1,0}(x,t) \\ & = \int_{-\infty}^{\infty} dx\,\chi^*_{p_2,0}(x,0) \,\chi_{p_1,0}(x,t) \\ &= \int_{-\infty}^{\infty}dx\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{-i}{\hbar}\left[p_2x\right]}\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[p_1x-\frac{p_1^2}{2m}t\right]}
\end{align*}
Using Fourier transform,$$ \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} dx \,e^{\frac{-i}{\hbar}x(p_2-p_1)} = \delta(p_2-p_1)$$ which results,
\[K(p_2,t;p_1,0) = \delta(p_2-p_1) \,e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}}\,\tag{2.33}\]
From this, it can be shown that $K(p_2,t;p_1,0)$ satisfies Schr\"{o}dinger equation,
\[i\hbar\frac{\partial{K(p_2,t;p_1,0)}}{\partial{t}} = \delta(p_2-p_1)\frac{p_1^2}{2m}e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}} = \frac{p_2^2}{2m} K(p_2,t;p_1,0) \,\tag{2.34}\]
Thus, it is shown the equivalence between Schr\"{o}dinger formulation and path integral formulation.
It can also be checked that,
\[i\hbar\frac{\partial}{\partial{t}}\chi_{p,0}(x,t) = \frac{p^2}{2m}\chi_{p,0}(x,t)\,\tag{2.35}\]
and
\[i\hbar\frac{\partial}{\partial{t}}K(x,t;x',0) = \frac{-\hbar^2}{2m}K(x,t;x',0)\,\tag{2.36}\]
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter
To understand how this works, it is conventional to start with the most general free particle problem whose Lagrangian is \[L = \frac{m\dot{x}^2}{2}\,\,\tag{2.1}\]
The technique is to split the action into two parts of which one of them is the classical action and the other is treated as the variational part. For this, it is defined the arbitrary path as \[x(t) = x_c(t)+y(t)\,\,\,\tag{2.2}\] where $x_c(t)$ represents the actual classical path. Substituting this in our Lagrangian and expanding it in terms of Taylor expansion, \[L(\dot{x}) = L(\dot{x}_c(t)+\dot{y}(t)) = L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\,\tag{2.3}\] The expansion is exact since L is quadratic in $\dot{x}$. Thus, it is possible to write the action integral as, \[S = \int_{t_1}^{t_2} \,dt \left(L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\right)\,\,\tag{2.4}\]
The first term can be denoted as,
\[ S_c = \int_{t_1}^{t_2}dt\,L(\dot{x}_c)\,\,\tag{2.5}\]
Using (2.1)
\[\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}}\vert_{\dot{x}_c} = mass = const.\,\tag{2.6}\]
and using integral by parts
\[\int_{t_1}^{t_2}dt\,\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} = \left[\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c}y(t)\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}dt\, \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right)y \]
for the lagrangian
\[ \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right) = m\dot{x}_c = 0 \]
Which results finally,
\[S = S_c + \frac{m}{2} \int_{t_1}^{t_2} dt \,\dot{y}^2\,\tag{2.7}\]
and
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}S_c} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.8}\]
where it is changed the integration variable to $dy(t)$ using (2.2)
The classical action is calculated to be
\[S_c = \frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}\,\tag{2.9}\]
which finally yields,
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.10}\]
The integral limit points out the fact that the deviation from the classical path at the end points is zero.
The integral over $y(t)$ is independent of $x_1$ and $x_2$. Its value depends only on $t_1$ and $t_2$ , since the entire problem is time translation invariant,
\[A(t_2-t_1) = \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.11}\]
and
\[K(x_2,t_2;x_1,t_1)= A(t_2-t_1)e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}}\,\tag{2.12}\]
To determine A(t) for $t_1=0$ making use of the group property (1.23) and (1.19),
\[ \delta(x_2-x_1) = K(x_2,t;x_1,t) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K(x,0;x_1,t)\] \[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K^*(x_1,t;x,0) \,\tag{2.13}\]
Substituting for K and using (2.12),
\[K(x_2,t;x,0)= A(t)e^{\frac{i}{\hbar}S_c(x_2,t;x,0)}\]
\[K^*(x_1,t;x,0)= A^*(t)e^{\frac{-i}{\hbar}S_c(x_1,t;x,0)}\]
and
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx \,\left\vert{A(t)}\right\vert^2 e^{\frac{i}{\hbar}\left(S_c(x_2,t;x,0)-S_c(x_1,t;x,0)\right)}\,\tag{2.14}\]
When $x_2 = x_1+\Delta{x}$ the argument of exponential can be modified with Taylor expansion as,
\[ S_c(x_2,t;x,0)-S_c(x_1,t;x,0) = \frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} \Delta{x_1}\]
Considering $\Delta{x_1}\rightarrow\,0$ all higher order terms are neglected.
Substituting for $S_c$,
\[\frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} = \frac{\partial(\frac{m\,(x_1-x)^2}{2\,t})}{\partial{x_1}} = \frac{m\,(x_1-x)}{t} = \gamma(x) \,\tag{2.15}\]
Since $\gamma(x)$ is linear function of $x$, its derivative $\frac{d\gamma}{dx}$ is independent of $x$. Using this information to write,
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} d\gamma\,\left\vert\frac{dx}{d\gamma}\right\vert \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)\left(x_2-x_1\right)}\,\tag{2.16}\]
From Fourier transform,
\[\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\beta(x_2-x_1)} \,d\beta = \delta(x_2-x_1)\,\tag{2.17}\]
for $\beta = \frac{\gamma}{\hbar}$,
\[\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} e^{\frac{i}{\hbar}\gamma(x_2-x_1)} \,{d\gamma} = \delta(x_2-x_1)\,\tag{2.18}\]
Thus, multiplying and dividing by the factor $2\pi\hbar$ in (2.16),
\[ \delta(x_2-x_1) = \int_{-\infty}^{\infty} \frac{d\gamma}{2\pi\hbar} \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)(x_2-x_1)}\frac{2\pi\hbar}{\left\vert\frac{d\gamma}{dx}\right\vert}\,\tag{2.19}\]
Comparing both the left and right side,
\[\delta(x_2-x_1) = \delta(x_2-x_1)\frac{2\pi\hbar|A(t)|^2}{\left\vert\frac{d\gamma}{dx}\right\vert}\]
so that,
\[|A(t)|^2 = \frac{1}{2\pi\hbar}\left\vert\frac{d\gamma}{dx}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{-m}{t}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{\partial^2S_c(x_1,t;x,0)}{\partial{x}\partial{x_1}}\right\vert\,\tag{2.20}\]
The phase can be chosen such that,
\[A(t) = \sqrt{\frac{m}{2i\pi\hbar{t}}}\]
Thus, it is determined the propagator for a free particle as,
\[K(x_2,t_2;x_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[S_c = \frac{m\,(x_2-x_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.21}\]
In general for three dimensions,
\[K(r_2,t_2;r_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(r_2-r_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.22}\]
For fixed $x_1=0\,\,,t_1=0$ using (3.5) the propagator for a free particle should reduce to the Schr\"{o}dinger wave function for a free particle. Thus, the probability amplitude of the free particle from the propagator is,
\[\psi(x,t) = K(x,t,0,0) = \sqrt{\frac{m}{2i\pi\hbar{t}}}e^{\frac{i}{\hbar}\frac{m\,x^2}{2\,t}}\,\tag{2.23}\]
If it is considered a specific point $(x_0,t_0)$, then the classical momentum at this point is $$p_0 = mv_0 = m\frac{x_0}{t_0}$$ with energy $$E = \frac{mv_0^2}{2}= \frac{mx_0^2}{2t_0^2}$$ The change in phase in the vicinity of $(x_0,t_0)$ is then using again Taylor expansion,
\[
\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} exp\frac{i\,m}{\hbar\,2}\left[\frac{x_0^2}{t_0}+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{x}}\right\vert_{(x_0,t_0)}(x-x_0)+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{t}}\right\vert_{(x_0,t_0)}(t-t_0)+...\right]
\]
The bracketed term reduces to (neglecting higher order terms),
\[\left[\frac{x_0^2}{t_0}+ \frac{2x_0}{t_0}(x-x_0) -\frac{x_0^2}{t_0^2}(t-t_0)\right]\] Simplifying with arithmetic,
\[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,exp{\frac{i}{\hbar}\left[m\frac{x_0}{t_0}x - \frac{m}{2}{\frac{x_0^2}{t_0^2}t}\right]}\]
Thus, the wave function varies in the immediate vicinity of $(x_0,t_0)$ according to \[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,e^{\frac{i}{\hbar}\left[p_0x - E_0t\right]}\,\tag{2.24}\]
which is the well-known Einstein-de Broglie relation, according to which a particle with momentum $p$ and energy $E$ is assigned a wave function with wave length and wave number respectively $\lambda = \frac{h}{p}\,\,\rightarrow\,\, k = \frac{2\pi}{\lambda}$ Similarly with frequency and angular frequency $\nu = \frac{E}{h}\,\,\rightarrow\,\,\omega = {2\pi\nu}$
So,
\[ e^{i\left(kx- \omega{t}\right)}= e^{i\left(\frac{2\pi}{\lambda}x- {2\pi\nu}t\right)} = e^{\frac{i}{\hbar}(px-Et)} \tag{2.25}\]
which is equivalent to (2.24).
Schr\"{o}dinger’s time evolution equation
So far, everything is discussed in position basis. Whereas it needs to be in momentum or energy basis for explicit derivation of Schr\"{o}dinger’s time evolution equation. This is achieved using the group property of our propagators,
\[K(x,t;p,0) = \int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,K(x',0;p,0)\,\tag{2.26}\]
For fixed initial momentum $p$ at time $t$,
\[K(x,t;p,0) = \chi_{p,0}(x,t)=\int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,\chi_{p,0}(x',0)\,\tag{2.27}\]
Let us take this as an ansatz for the transformation amplitude,
\[\chi_{p,0}(x,0) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}xp}\,\tag{2.28}\]
Substituting this in (2.27) with the corresponding substitution for $K$,
\[\chi_{p,0}(x,t) =\int_{-\infty}^{\infty}\sqrt{\frac{m}{2i\pi\hbar{(t)}}}e^{\frac{i}{\hbar}\frac{m\,(x-x')^2}{2\,t}} \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{i}{\hbar}x'p} \,dx'\tag{2.29}\]
As $x'$ is the integrating variable, the x' terms are combined together and simplified using arithmetic,
\[\chi_{p,0}(x,t) = e^{\frac{i}{\hbar}\left(xp-\frac{p^2}{2m}t\right)}\sqrt{\frac{m}{2i\pi\hbar{t}}}\sqrt{\frac{1}{2\pi\hbar}} \int_{-\infty}^{\infty}e^{\frac{i}{\hbar}\frac{m}{2\,t}\left[x'-\left(x-\frac{pt}{m}\right)\right]^2}\,\tag{2.30}\]
changing the variable to $u = \left[x'-\left(x-\frac{pt}{m}\right)\right]$ and making use of Gaussian Integral,
\[\chi_{p,0}(x,t) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[xp-\frac{p^2}{2m}t\right]}\,\tag{2.31}\]
In three dimensions,
\[\chi_{p,0}(x,t) = \left(\frac{1}{2\pi\hbar}\right)^{\frac{3}{2}} e^{\frac{i}{\hbar}\left[r.p-\frac{p^2}{2m}t\right]}\tag{2.32}\]
Again using the group property with momentum arguments,
\begin{align*}
K(p_2,t;p_1,0) &=\int_{-\infty}^{\infty} dx'\,K(p_2,t;x,t) \,K(x,t;p_1,0) \\ &= \int_{-\infty}^{\infty} dx\,K(p_2,0;x,0) \,\chi_{p_1,0}(x,t) \\ & = \int_{-\infty}^{\infty} dx\,\chi^*_{p_2,0}(x,0) \,\chi_{p_1,0}(x,t) \\ &= \int_{-\infty}^{\infty}dx\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{-i}{\hbar}\left[p_2x\right]}\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[p_1x-\frac{p_1^2}{2m}t\right]}
\end{align*}
Using Fourier transform,$$ \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} dx \,e^{\frac{-i}{\hbar}x(p_2-p_1)} = \delta(p_2-p_1)$$ which results,
\[K(p_2,t;p_1,0) = \delta(p_2-p_1) \,e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}}\,\tag{2.33}\]
From this, it can be shown that $K(p_2,t;p_1,0)$ satisfies Schr\"{o}dinger equation,
\[i\hbar\frac{\partial{K(p_2,t;p_1,0)}}{\partial{t}} = \delta(p_2-p_1)\frac{p_1^2}{2m}e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}} = \frac{p_2^2}{2m} K(p_2,t;p_1,0) \,\tag{2.34}\]
Thus, it is shown the equivalence between Schr\"{o}dinger formulation and path integral formulation.
It can also be checked that,
\[i\hbar\frac{\partial}{\partial{t}}\chi_{p,0}(x,t) = \frac{p^2}{2m}\chi_{p,0}(x,t)\,\tag{2.35}\]
and
\[i\hbar\frac{\partial}{\partial{t}}K(x,t;x',0) = \frac{-\hbar^2}{2m}K(x,t;x',0)\,\tag{2.36}\]
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter
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