# Electron momentum distribution and wavefunction in momentum space

Does there exist any relationship between the electron momentum distribution used in above threshold ionization and the wave function in momentum space? In other words, starting with the wavefunction in momentum space $\phi(\mathrm{p})$ how can I derive an expression for $\partial^2P/\partial E\partial\theta$, where $E=p^2/2$ is the kinetic energy of the detached electron and $\theta$ is angular coordinate?

## 1 Answer

If you have the photoelectron momentum-space wavefunction $\psi(\mathbf p)$ and you have projected out the contribution from the bound states, then the momentum-space distribution $|\psi(\mathbf p)|^2$ gives you the three-dimensional distribution of velocities as measured e.g. by a velocity-map imager. To get beyond that, it depends on what it is you want do do - for example, if your distribution is not axially symmetric then the doubly-differential distribution you ask for will not be very meaningful.

To get $\partial ^2 P/\partial E \partial \theta$, the cleanest way (in my opinion) is to work inside the integral: the probability for the observed momentum $\mathbf p$ to be in some set $S$ is

\begin{align} P(\mathbf p\in S) & = \int_S |\psi(\mathbf p)|^2\mathrm d^3\mathbf p =\iiint_S|\psi(\mathbf p)|^2 p^2 \sin(\theta)\mathrm dp\mathrm d\theta\mathrm d\phi. \end{align} If $S$ is a thin slice at angle $\theta$ and energy $E$, with angular and energy widths $\Delta\theta$ and $\Delta E$, covering $\phi\in[0,2\pi]$, then \begin{align} P(\mathbf p\in S) \approx \Delta\theta\Delta E \int|\psi(\mathbf p)|^2 E \sin(\theta) \mathrm d\phi = \Delta\theta\Delta E \frac{ \partial ^2 P}{\partial E \partial \theta} \end{align} by definition of the latter, and from which you can read out its value.