I was just reading about the free particle and how it can be represented as a superposition of momentum eigenstates and so on. I came to think about this:(adsbygoogle = window.adsbygoogle || []).push({});

Let's say we have free particle whose initial state is a Dirac delta function type of thing (localized position). The momentum distribution will then be a uniform distribution over all values. As time goes on the wave function will spread out according to the Schrödinger equation. But what happens to the momentum distribution; does it become more localized? It's just the Fourier Transform of the wave function, right?

This leads me to another question. The wave function, [tex]\Psi[/tex], seems to have some sort of preference for the "position" observable (I mean, when you take the norm-square, you get the position-distribution, you have to Fourier Transform it to get the momentum distribution and so on). Couldn't you have an equivalent "momentum-based" wave function, so that when you Fourier Transform this wave function you get the position distribution. Indeed, couldn't you have such equivalent wave functions for all observables (maybe there is, I've never met them)?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Position and momentum (basic stuff)

Loading...

Similar Threads for Position momentum basic |
---|

A Inverse momentum operator |

I Does the Schrödinger equation link position and momentum? |

I Why is position-space favored in QM? |

A Transformation of position operator under rotations |

B Measuring momentum and position in particle colliders |

**Physics Forums | Science Articles, Homework Help, Discussion**