Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

Born Scattering Amplitude from Path Integral

+ 3 like - 0 dislike
251 views

I am stuck on an exercise from "Condensed Matter Field Theory" by Altland and Simons on path integral. The exercise asks to obtain a perturbative expansion for the scattering amplitude

\(\langle \mathbf{p'}| U(t \rightarrow \infty,t' \rightarrow -\infty)| \mathbf{p} \rangle\)

of a free particle from a short-range central potential \(V(r)\) and show that the first-order term recovers the Born scattering amplitude

\(- i \hbar e^{-i(t-t')E(p)/\hbar} \delta(E(p)-E(p')) \langle \mathbf{p'}|V|\mathbf{p}\rangle\)

Here is my attempt

\(\begin{align} & \langle \mathbf{p'}|U|\mathbf{p}\rangle \\ =& \int d \mathbf{x} \int d \mathbf{x'} \langle \mathbf{p'}| \mathbf{x'} \rangle \langle \mathbf{x'} |U| \mathbf{x} \rangle \langle \mathbf{x} |\mathbf{p}\rangle \\ =& \int d \mathbf{x} \int d \mathbf{x'} \langle \mathbf{p'}| \mathbf{x'} \rangle G(\mathbf{x'},\mathbf{x};t',t) \langle \mathbf{x} |\mathbf{p}\rangle \\ \end{align}\)

Now we can expand the propagator

\(\begin{align} & G(\mathbf{x'},\mathbf{x};t',t) \\ =& \int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_t^{t'} dt' ~ (\frac{1}{2} m \mathbf{\dot x^2}-V) ] \\ =& \int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_t^{t'} dt' ~ \frac{1}{2} m \mathbf{\dot x^2} ] \mathrm{exp}[-\frac{i}{\hbar} \int_t^{t'} dt' ~V ] \\ \approx& \int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_t^{t'} dt' ~ \frac{1}{2} m \mathbf{\dot x^2} ] (1 - \frac{i}{\hbar}\int_t^{t'} dt' ~V) \end{align}\)

I am currently stuck at this step. How should I proceed?

asked Oct 15, 2014 in Theoretical Physics by chichane (15 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

You have to interchange the integration over paths and over time to obtain

\(G(\mathbf{x}_1,\mathbf{x}_0;t_1,t_0)= \int_{t_0}^{t_1} dt\int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_{t_0}^{t_1} ds ~ \frac{1}{2} m \mathbf{\dot x^2} ] (- \frac{i}{\hbar} ~V(\mathbf{x}(t)))\)

The inner integral can be interpreted as a propagator again, except that the particle is now scattered at the potential V at the place x(t).

The inner path integral corresponds to the propagator of a free particle, and we can write

\(\int \mathcal{D} \mathbf{x}~\mathrm{exp}[\frac{i}{\hbar} \int_{t_0}^{t_1} ds ~ \frac{1}{2} m \mathbf{\dot x^2} ] = G_0(\mathbf{x_1},\mathbf{x_0};t_1,t_0) \)\(= \int dxdt' G_0(\mathbf{x_1},\mathbf{x};t_1,t') \cdot G_0(\mathbf{x},\mathbf{x_0};t',t_0) \delta(\mathbf{x}-\mathbf{x}(t))\delta(t-t')\)

This is the amplitude for a particle traveling from the point x0 to some point x and then to x1.

Another way of writing this is

\(\langle x_1 |U| x_0\rangle = \int dx \langle x_1 |U_0| x\rangle \langle x |U_0| x_0\rangle (-\frac{i}{\hbar}V(x))\)

Now, the Born scattering amplitude is just the Fourier transform of this. The free propagator \(U_0\) will give the terms involving energy, whereas the addition multiplication with \(V(x)\)corresponds to the additional matrix element. 

answered Oct 15, 2014 by Greg Graviton (665 points) [ revision history ]
edited Oct 16, 2014 by Greg Graviton
Thanks a lot for your answer. When you say switch to momentum eigenbasis, does that mean I need to take the fourier transform? Could you give me a little more detail? Thanks.

I have expanded my answer, though I was too lazy to expand the Fourier transform. Does this help?

Yes! I get it now. Thank you!

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...