Feynman's Propagator calculation

In summary, the conversation discusses calculating the amplitude using two different methods and trying to find a mistake in the calculations. The correct derivation of the amplitude is found, and it is shown that it satisfies the postulate. The final result is given as $A=\sqrt{\frac{m}{i2\pi\hbar\Delta t}}$.
  • #1
alex23
3
0
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!
 
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  • #2
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).
 

Related to Feynman's Propagator calculation

1. What is Feynman's Propagator calculation?

Feynman's Propagator calculation is a mathematical tool used in quantum field theory to calculate the probability amplitude for a particle to travel from one point in space and time to another.

2. How is Feynman's Propagator calculated?

Feynman's Propagator is calculated by integrating over all possible paths that a particle can take between two points, taking into account the particle's energy and momentum.

3. What is the significance of Feynman's Propagator in physics?

Feynman's Propagator is a fundamental concept in quantum field theory and is used to calculate various physical quantities, such as scattering amplitudes and decay rates, in particle physics.

4. Can Feynman's Propagator be applied to all particles?

Yes, Feynman's Propagator can be applied to all particles, including fermions and bosons, as long as they are described by quantum field theory.

5. Are there any limitations or challenges associated with using Feynman's Propagator?

One limitation of Feynman's Propagator is that it can only be applied to non-relativistic systems, meaning that it cannot be used to describe particles moving at speeds close to the speed of light. Additionally, the calculations involved in using Feynman's Propagator can be complex and time-consuming, making it a challenging tool to use in certain situations.

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