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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?

Thanks in advance:blushing:
 

Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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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 [itex]V(x)=0[/itex].
 
thank you!:smile:

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?
 

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