# Momentum and position eigenstates ; free and bound

1. Nov 8, 2015

### dyn

Hi. Hoping for some help with the following questions -

1 - Are there any momentum eigenstates in a box ? I think the answer is no because if i solve the momentum eigenvalue equation in 1-D i get Aikx but it seems impossible to get this to meet the boundary conditions
2 - As far as I know a free particle has the following wavefunction ψ = Aikx but if i solve Scrodingers time independent equation i get ψ = Aikx + B-ikx which is a standing wave and the wavefunction for a particle in a box. What am I doing wrong ?
3 - Are the position eigenstates for a free particle and a particle in a box the same ? Given by δ(x-x0 ) ?

2. Nov 8, 2015

### vanhees71

Ad 1) If you mean the box with rigid boundary conditions $\psi(0)=\psi(L)=0$, then you are right. There's no sensible momentum operator you can define in that case. The reason is quite subtle, because $-\mathrm{i} \partial_x$ is still Hermitean, but that's not sufficient, because an observable must be represented by a self-adjoint operator.

Ad 2) $\psi(x)=\exp(\mathrm{i} k x)$ cannot be a wave function describing the pure state of a particle, because it's not square integrable. It's a "generalized eigenfunction" of the momentum operator, i.e., it's a distribution in the sense of functional analysis.

Concerning the energy eigenvalue problem for the free particle you don't make any mistake, because it's degenerate, i.e., to any energy eigenvalue in the case for motion in one dimension you have two linearly independent generalized eigenstates, namely $\exp(+\mathrm{i} x \sqrt{2mE})$ and $\exp(-\mathrm{i} x \sqrt{2mE}$. So the most general eigenstate is the general superposition of these two eigenstates.

Ad 3) Again, there are no true eigenstates but only generalized ones, and indeed $u_{x_0}(x)=\delta(x-x_0)$ is this generalized position eigenstate in the position representation. That's happening for any value in the continuous spectrum of a self-adjoint operator, and both position and momentum have the purely continuous spectrum $\mathbb{R}$.

A very good first introduction to these subtleties of self-adjoint operators is

http://arxiv.org/abs/quant-ph/9907069

For a textbook on the modern formulation in terms of "rigged Hilbert spaces", see

L. Ballentine, Quantum Mechanics - A modern development

3. Nov 8, 2015

### dyn

Thanks. So solving Schrodingers equation for a free particle gives Aeikx + Be-ikx which is the same solution for a particle in a 1-D infinite well ? But in the infinite well the wave reflects giving standing wave ie wave moving to the right and one moving to the left. But with a free particle the wave cannot reflect so it must either move the left or the right but not both ?

4. Nov 9, 2015

### vanhees71

No! For a free particle the domain is $\mathbb{R}$ not $[0,L]$. It's another function space for the free than for the particle confined to a rigid box (although both are equivalent in the sense that both are separable infinite-dimensional Hilbert spaces). For the free particle all energy eigenvalues (except $E=0$) are degenerate, i.e., there are two eigenstates for each eigenvalue. It's more convenient to work with momentum eigenstates, which are energy eigenstates too, because for a free particle momentum is conserved. Momentum eigenstates are non-degenerate.