# Feynman's Propagator calculation

alex23
Homework Statement:
Prove that ##A=\sqrt{\frac{m}{i2\pi\hbar\Delta t}}## is the coeficient in Feynman's Propagator for the movement of a particle of mass ##m##, in a potential ##U(x,t)## from ##(x_1,t_1)## to ##(x,t)## with the limit ##\Delta t=t-t_1 \rightarrow 0##.
Relevant Equations:
Postulate of propagation:
##\left<x_2t_2|x_1t_1\right>=\int_{-\infty }^{+\infty }\left<x_2t_2|xt\right>\cdot\left<xt|x_1t_1\right>dx##
Feynnman's Propagator:
##K(x,t;x_1,t_1)\equiv\left<xt|x_1t_1\right>=Ae^{\frac{iS}{\hbar}}=\sqrt{\frac{m}{i2\pi\hbar\Delta T}}exp\left[\frac{i}{\hbar}\left( \frac{m(x-x_1)^2}{2\Delta t}-U\left(\frac{x+x_1}{2},t\right)\Delta t \right) \right]##
##\int_{-\infty }^{+\infty } e^{ax^2+bx}=\sqrt{\frac{\pi}{-a}}e^{\frac{-b^2}{4a}}##
##A(\Delta t /2)=\sqrt 2 A(\Delta t)##
The first thing I have to consider is that, since ##\Delta t \rightarrow 0##, the potential ##U## is not going to contribute and we can consider it to be ##0##.
Next thing I did was calculate ##\left<xt|x_1t_1\right>## directly with the definition considering what I said before, and I got:
$$\left<xt|x_1t_1\right>=Aexp\left[\frac{i}{\hbar}\left( \frac{m(x-x_1)^2}{2\Delta t}\right) \right]$$
Now I calculated the same thing using the postulate and calculating the integral with the help of the solved integral, and considering a position ##(x',t')## at ##\Delta t/2## from both ##(x,t)## and ##(x_1,t_1)##:
$$\left<xt|x_1t_1\right>=\int_{-\infty }^{+\infty }\left<xt|x't'\right>\cdot\left<x't'|x_1t_1\right>\ dx' \ =$$
I introduced the definition and made the next calculations:
$$= \ \int_{-\infty }^{+\infty }A(\Delta t/2)exp\left[\frac{i}{\hbar}\left( \frac{m(x-x')^2}{2\frac{\Delta t}{2}}\right) \right]\cdot A(\Delta t/2)exp\left[\frac{i}{\hbar}\left( \frac{m(x'-x_1)^2}{2\frac{\Delta t}{2}}\right) \right] \ dx' \ =$$
$$= \ \int_{-\infty }^{+\infty }A^2(\Delta t/2)exp\left[\frac{im}{\hbar\Delta t}\left( (x-x')^2+(x'-x_1)^2\right) \right] \ dx' \ =$$
$$= \ A^2(\Delta t/2)\int_{-\infty }^{+\infty }exp\left[\frac{im}{\hbar\Delta t}\left( x^2+{x'}^2-2xx'+{x'}^2+{x_1}^2-2x'x_1 \right) \right] \ dx' \ =$$
$$= \ A^2(\Delta t/2)exp\left[\frac{im}{\hbar\Delta t}\left( x^2+{x_1}^2 \right) \right] \int_{-\infty }^{+\infty }exp\left[\frac{im}{\hbar\Delta t}2{x'}^2-2(x+x_1)\frac{im}{\hbar\Delta t}x' \right] \ dx' \ =$$
Where I can use the integral solution given before to get:
$$=\ \ A^2(\Delta t/2)exp\left[\frac{im}{\hbar\Delta t}\left( x^2+{x_1}^2 \right) \right]\sqrt{\frac{i\pi\hbar\Delta t}{2m}}exp\left[ 4\left( x+x_1\right) ^2(\frac{im}{\hbar\Delta t})^2\frac{\hbar\Delta t}{8im} \right] \ =$$

$$=\ A^2(\Delta t/2)\sqrt{\frac{i\pi\hbar\Delta t}{2m}}exp\left[\frac{im}{\hbar 2\Delta t}\left( 2x^2+2{x_1}^2 +(x+x_1)^2 \right) \right]$$

Now, comparing the two expressions that I got for ##\left<xt|x_1t_1\right>## :

$$\left<xt|x_1t_1\right>=A(\Delta t)exp\left[\frac{i}{\hbar}\left( \frac{m(x-x_1)^2}{2\Delta t}\right) \right]=A^2(\Delta t/2)\sqrt{\frac{i\pi\hbar\Delta t}{2m}}exp\left[\frac{im}{\hbar 2\Delta t}\left( 2x^2+2{x_1}^2 +(x+x_1)^2 \right) \right]$$

It is very evident that I had to make a mistake with my calculations since the exponentials don't cancel each other.
In case they did, we would arrive to the desired conclusion using the given relation for ##A(\Delta t /2)##:

$$A(\Delta t)=A^2(\Delta t/2)\sqrt{\frac{i\pi\hbar\Delta t}{2m}} \Rightarrow A(\Delta t)=2A^2(\Delta t)\sqrt{\frac{i\pi\hbar\Delta t}{2m}}\Rightarrow 1=A(\Delta t)\sqrt{\frac{2i\pi\hbar\Delta t}{m}}$$
And finally arriving to the desired result of:
$$A=\sqrt{\frac{m}{i2\pi\hbar\Delta t}}$$

So I know I'm either applying something wrong or I made a mistake on my calculations. I went over my math several times and I've done it multiple times from scratch without checking my old math and I'm still arriving to the same result. Is there something I am not taking into account? Or is there something I am missing to get to where I want?
Any help pointing me towards my mistake or what I'm missing would be greatly appreciated since I'm starting to lose my mind on this one.