## Quantum mechanics - a free particle

Hello everyone!

If we measure the position of a particle in a free space, and say we find that it is at x0,

what is the wavefunction right after the measurement in x representation?

shouldn't it be delta (x-x0), because delta functions are the eigenfunctions of the position operator?

Another question is how do I find the eigenfunctions of the Hamiltonian of a free particle in x representation?

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Recognitions:
Gold Member
Homework Help
Staff Emeritus
 Quote by maria clara Hello everyone!
Hello! :)

 If we measure the position of a particle in a free space, and say we find that it is at x0, what is the wavefunction right after the measurement in x representation? shouldn't it be delta (x-x0), because delta functions are the eigenfunctions of the position operator?
Yes, up to a normalization constant I think. I don't have my quantum book handy so you might want to check on this. But the wavefunction will at least be proportional to a delta function.

 Another question is how do I find the eigenfunctions of the Hamiltonian of a free particle in x representation?
You solve the Schrodinger equation with $V(x)=0$.

 thank you! I solved the Schrodinger equation and got the function: Aexp(ikx)+Bexp(-ikx). This means that any private case (like A=0 and B=1) is an eigenfunction of the Hamiltonian. Can I conclude that the Hamiltonian has an infinite number of eigenfunctions? I also notice that there are only two basic "types" here - exp(ikx) and exp(-ikx). Does it mean that these two form a basis of the Hamiltonian eigenfunctions space? is it considered as a Hilbert space?