# A Asymptotic momentum eigenstates in scattering experiments

1. May 5, 2016

### spaghetti3451

In a typical collider experiment, two particles, generally in approximate momentum eigenstates at $t=-\infty$, are collided with each other and we measure the probability of finding particular outgoing
momentum eigenstates at $t=\infty$
.

Firstly, what does it mean for the particles to be in approximate momentum eigenstates? Does it mean that there is a small spread in the distribution of the momentum of the particles, due to the particles being in a superposition of momentum eigenstates, so that the particles are rather in approximate momentum eigenstates, so to speak?

Secondly, how can we be certain that the experiment produces only momentum eigenstates at $t=\infty$ and not some superposition of momentum eigenstatesat $t=\infty$?

2. May 6, 2016

### vanhees71

Momentum eigenstates are not representing true states, because they are not square integrable. So you cannot prepare particles in an exact momentum eigenstate but only in an approximate one, i.e., as a "wave packet" which provides a square integrable wave function. With a careful analysis using such wave packets you'll have no trouble with squaring energy-momentum conserving $\delta$ distributions in the $S$-matrix elements, etc. For a concise discussion, see Peskin, Schroeder, Introduction to Quantum Field Theory. For the non-relativistic case a very good discussion is given in Messiah, Quantum Mechanics.

3. May 6, 2016

### spaghetti3451

Does this mean that there are technical difficulties associated with with squaring energy-momentum conserving $\delta$ distributions in the $S$-matrix elements for pure momentum eigenstates?

4. May 6, 2016

### vanhees71

Yes, you simply cannot square a $\delta$ distribution. It's an undefined expression! There are different ways to overcome this issue. The physical one is to use wave packets as said in my previous posting. You can as well use "box quantization", i.e., you take a finite volume, e.g., a cube and look at wave functions defined on this cube with periodic boundary conditions. Then you have a momentum operator (which you don't have if using "rigid" boundary conditions!) with discrete eigenvalues $\vec{p} \in \frac{2\pi \hbar}{L} \mathbb{Z}^3$, and the energy-momentum conserving $\delta$ functions become usual Kronecker $\delta$'s. At the very end of the calculation, i.e., after squaring the matrix element you can take the infinite-volume limit again, leading to the rule that you take the invariant-matrix element squared times one energy-momentum conserving $\delta$ distribution to get a transition rate per volume, which makes a lot of sense, because it is precisely what you need to calculate (invariant) cross sections!

5. May 6, 2016

### spaghetti3451

I presume that, by periodic boundary conditions, you mean that you consider an infinite collection of boxes filling up the universe such that the dynamics of the particle(s) in each box is the same so that the boundary conditions ought to be periodic. Do we really need to consider an infinite collection of boxes filling up the universe? Can't we only consider the dynamics of the particle(s) in a single box of finite volume and after all the necessary mathematical analysis, we take the limit of the volume of the box to infinity, in which case the boundary conditions do not need to be periodic?

Why do we not have a momentum operator with "rigid" boundary conditions?

6. May 6, 2016

### vanhees71

Of course, at the end you take the limit $L \rightarrow \infty$. It's just to regularize the $\delta$ distribution. As I said, you can stay in $\mathbb{R}^3$ and use the correct wave packets and then take the limit of wave packets that are very sharply peaked in momentum space.

The short answer to the second question is that if you define the momentum operator as generating translations, you get as its operator in the position representation $-\mathrm{i} \vec{\nabla}$ (setting $\hbar=1$ for convenience). Then the eigenstates are given by
$$-\mathrm{i} \vec{\nabla} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x}) \; \Rightarrow \; u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x}),$$
but none of these eigenfunctions satisfies the rigid boundary conditions, i.e., $\psi(\vec{x})=0$ at all edges of the cube. Thus the operator, although being Hermitean (check!) is not self-adjoint.

Another argument is that if you act with $-\mathrm{i} \vec{\nabla}$ on wave functions fulfilling the boundary conditions you usually do not get a new wave function also fulfilling the boundary conditions, i.e., the image of the operator is not in the Hilbert space of wave functions you consider, and thus the operator is not self-adjoint, but for a proper observable you need self-adjoint operators to make sense as given by the QT postulates.

Note that the free-particle Hamiltonian $\hat{H}=-\Delta/(2m)$ is a self-adjoint operator. Particularly you can find a complete set of momentum eigenfunctions (cosines and sines with the appropriate energy eigenvalues fulfilling the boundary conditions). You can expand any wave function in these eigenfunctions (leading to the Fourier series of wave functions fulfilling the boundary conditions).

7. May 6, 2016

### A. Neumaier

In practice it means that one considers one box only in which you identify opposite faces, so that in fact you work with a 3-torus. Thus what goes out at one side of the box enters again at the opposite end.