The first thing I remember hearing about in QM was the time-independent 1-D schrodinger equation, [itex] Hψ = (\frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + V)ψ(x) = Eψ(x) [/itex]. This is an eigenvalue equation, the Hamiltonian operator H operating on the energy eigenstate ψ to produce the product of the energy eigenvalue, E, and ψ.(adsbygoogle = window.adsbygoogle || []).push({});

However, we also come to know this state ψ by another name, the "wavefunction", and we find that if we take |ψ(x)|^2 we find the probability of finding our particle at at position x.

My question is, what is it about the eigenstates of the energy operator in particular that should mean we can find out this information about the likelihood of a particle occupying a certain position x upon measurement? I don't see the connection - especially seeing as we could take a free particle (V=0) so that the energy of the particle has no dependence on position, only momentum?

In other words, why don't we take any other eigenstate for any other observable quantity, square that and use that for our position probability?

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

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

# What is the connection between energy eigenstates and position?

Loading...

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