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  Born Scattering Amplitude from Path Integral

+ 3 like - 0 dislike

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 (775 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!

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