Some notes on scattering 01

— 1. Back! —

After a very long hiatus I’m definitely going back to doing research in Physics and blogging. Yes, I know that I promised this a lot of times in the past but this time it is for real.

In December I went to Portugal and talked with my future advisor and we decided this time it’d be for real. We talked about the work he has done lately, namely with his latest PhD. student, and we tried to figure out what would be my research program.

We talked about a few possibilities and narrowed it down to basically two or three options. To tell you the truth I still don’t know which one of the options I’ll end up choosing and this is so because all of the time I was away from Physics.

Thus from the time being I’ll mainly do some catching up and I really hope that in one or two months time I’ll be able to look into an article and figure out what’s going on.

In order to do the catching up I mentioned previously I’ll need to read texts mainly on Quantum Mechanics and Field Theory. The texts I’ll read are:

— 3. Scattering Theory —

This set of class notes aims to be a concise introduction to the theory of quantum scattering in the non-relativistic and relativistic formalism.

The second section of the class notes is about the mathematical description of particles in quantum mechanics under the guise of waves. First we describe one dimensional waves and later on the formalism is expanded to include three dimensional waves.

— 3.1. One dimensional wave packet —

In the instant ${t=0}$, a particle can be described in quantum mechanics by the use of a wave packet: $\displaystyle \psi(x,t=0)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\varphi(k)e^{ikx} \ \ \ \ \ (1)$

Where ${\varphi (k)}$ is the Fourier transform. As an example in the lectures the Fourier transform is taken to be $\displaystyle \varphi (k) = \begin{cases} \dfrac{1}{\sqrt{2\Delta k}} \quad |k-\bar{k}|\leq 0 \Delta k \\ 0 \quad\quad\quad |k-\bar{k}|>0 \end{cases} \ \ \ \ \ (2)$

Now for that choice of the Fourier transform the wave function becomes {\begin{aligned} \phi(x) &= \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\varphi(k)e^{ikx} \\ &= \dfrac{1}{\sqrt{2\pi}}\dfrac{1}{\sqrt{2\Delta k}}\int_{\bar{k}-\Delta k}^{\bar{k}+\Delta k}dke^{ikx} \\ &= \dfrac{1}{2\sqrt{\pi\Delta k}ix}\left( e^{i(\bar{k}+\Delta k)x}-e^{i(\bar{k}-\Delta k)x} \right) \\ &=\dfrac{1}{2\sqrt{\pi\Delta k}ix}e^{ikx}\left( e^{i\Delta kx}-e^{-i\Delta kx} \right) \\ &= \dfrac{1}{2\sqrt{\pi\Delta k}ix}e^{ikx}\left(\cos(\Delta kx)+i\sin(\Delta kx)-\cos(\Delta kx)+i\sin(\Delta kx) \right)\\ &= \dfrac{e^{ikx}}{\sqrt{\pi\Delta k}}\dfrac{\sin(\Delta kx)}{x} \end{aligned}}

— 3.2. Momentum —

The average momentum of a particle is defined in quantum mechanics by the following relationship $\displaystyle =\int_{-\infty}^{+\infty}dx\psi^*(x)\left\lbrace-i\dfrac{\partial }{\partial x}\right\rbrace\psi(x) \ \ \ \ \ (3)$

For the wave packet expression 1 and the Fourier transform example 2 that was chosen it follows {\begin{aligned} &=\int_{-\infty}^{+\infty}dx\psi^*(x)\left\lbrace-i\dfrac{\partial }{\partial x}\right\rbrace\psi(x) \\ &= \int_{-\infty}^{+\infty}dx \left\lbrace \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\varphi^*(k)e^{-ikx} \right\rbrace\left\lbrace-i\dfrac{\partial }{\partial x}\right\rbrace \left\lbrace\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk'\varphi(k')e^{ik'x}\right\rbrace \\ &=\int_{-\infty}^{+\infty}dx \left\lbrace \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\varphi^*(k)e^{-ikx} \right\rbrace \left\lbrace\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk'\varphi(k')k'e^{ik'x}\right\rbrace\\ &= \int_{-\infty}^{+\infty}dk\int_{-\infty}^{+\infty}dk'\varphi^*(k)\varphi(k')k'\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}dx e^{i(k'-k)x}\\ &=\int_{-\infty}^{+\infty}dk\int_{-\infty}^{+\infty}dk'\varphi^*(k)\varphi(k')k'\delta(k'-k)\\ &= \int_{-\infty}^{+\infty}dk|\varphi(k)|^2k\\ &= \int_{\bar{k}-\Delta k}^{\bar{k}+\Delta k}dk\dfrac{k}{2\Delta k}\\ &= \dfrac{1}{2\Delta k}\left( (\bar{k}+\Delta k)^2-(\bar{k}-\Delta k)^2 \right)\\ &= \dfrac{1}{4\Delta k}(\bar{k}^2+\Delta k^2+2\bar{k}\Delta k-\bar{k}^2-\Delta k^2+2\bar{k}\Delta k)\\ &= \bar{k} \end{aligned}}

The previous derivation shows that the average value of the momentum of a particle is ${\bar{k}}$ which is the central value of the ${k}$ distribution.

As I was studying the lecture notes a part about the previous derivation bothered me. I really didn’t like that we had to use the function given as an example of the Fourier transform to achieve the result. This result is patently valid for a more general class of Fourier transforms and a more general proof must exist.

The facts that make this result stand are:

1. The Fourier transform must be normalizable ${\left(\displaystyle\int_{-\infty}^{+\infty}|\varphi(k)|^2=1\right)}$.
2. The Fourier transform must be symmetric around ${\bar{k}}$.

Introducing the change of variable ${u=k-\bar{k}}$ it is ${du=dk}$ and ${\varphi (u)=\varphi (-u)}$ which means that ${\varphi (u)}$ is an even function.

Therefore our derivation can be: {\begin{aligned} &=\int_{-\infty}^{+\infty}dk|\varphi(k)|^2k\\ &= \int_{-\infty}^{+\infty}du|\varphi(u+\bar{k})|^2(u+\bar{k})\\ &= \bar{k}\int_{-\infty}^{+\infty}|\varphi(u+\bar{k})|^2+\int_{-\infty}^{+\infty}du |\varphi(u+\bar{k})|^2u\\ &= \bar{k}+0\\ &= \bar{k} \end{aligned}}

Where ${\displaystyle\bar{k}\int_{-\infty}^{+\infty}|\varphi(u+\bar{k})|^2=\bar{k}}$ follows from the fact that ${\varphi (k)}$ is normalized (and a shift doesn’t alter the nature of an integral) and ${\int_{-\infty}^{+\infty}du |\varphi(u+\bar{k})|^2u=0}$ follows from the fact that ${|\varphi(u+\bar{k})|^2u}$ is an odd function. Physics, Math, Life and Poetry. Lately it has been more about Physics and Math though...
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2 Responses to Some notes on scattering 01

1. António Morais says:

I do like the way you simplified a lot such derivation! Indeed, expression (2) and the calculation that follows (3) are somehow long and not satisfactorily generic. Arguments 1 and 2, and the comments below, provide the necessary and sufficient conditions that all text books should have.

Keep on the good work and share with us what you learn about scattering 😉

• ateixeira says: