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:
$K(x,t;x_1,t_1)\equiv\left<xt|x_1t_1\right>=Ae^{\frac{iS}{\hbar}}=\sqrt{\frac{m}{i2\pi\hbar\Delta T}}exp\left[\frac{i}{\hbar}\left( \frac{m(x-x_1)^2}{2\Delta t}-U\left(\frac{x+x_1}{2},t\right)\Delta t \right) \right]$
Helpful integral:
$\int_{-\infty }^{+\infty } e^{ax^2+bx}=\sqrt{\frac{\pi}{-a}}e^{\frac{-b^2}{4a}}$
Helpful relation for $A$:
$A(\Delta t /2)=\sqrt 2 A(\Delta t)$

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.
Thanks in advance!

 

[NateD]

A:Your derivation of the amplitude is correct, and you should get what you got. In the end, you have to plug it into your expression for the overlap and see if that works with the postulate. The postulate states that $\langle x_2t_2 | x_1 t_1 \rangle$ is the same as $\langle x_3 t_3 | x_2 t_2 \rangle \langle x_2 t_2 | x_1 t_1 \rangle$. In the limit of small time step, this means that $$\langle x_2 t_2 | x_1 t_1 \rangle = \langle x_2 t_2 | x_2 t_2 \rangle \langle x_2 t_2 | x_1 t_1 \rangle$$Now, use the result you got for $\langle x_2 t_2 | x_2 t_2 \rangle$ and the one you want to get for $\langle x_2 t_2 | x_1 t_1 \rangle$ to see if they are consistent. You should be able to find a relation between the two amplitudes that makes them equal, and gets you the desired result.(Note: I assumed $x_2$ is the same as $x'$ in your calculation).