# Wave function of a free particle

1. Jun 17, 2005

### Feynmanfan

A short question:

I've learned that the wave function corresponding to a free particle has this form:

Psi(x,0)=1/sqrt(2*Pi)*Integral[g(k)*E^(ikx)dx] (i can't write it in Latex, sorry)

Is it just for the free particle, or any quantum state of a system can be represented in this form, where g(k) is the fourier transf. of Psi?

Can any function of x, represent the state of the free particle?

2. Jun 17, 2005

### dextercioby

It's a wave packet.It's an element of the $\tilde{M}$ which is the space of the continuous linear functionals on the nuclear subspace $M$.

That is $g(k)\in \tilde{M}$ and $\Psi (x,0)\in\tilde{M}$.

The free particle's (nonrelativistic) hamiltonian is a densly defined essentially self-adjoint linear operator with a completely continuous spectrum.It is unbounded,therefore the correct mathematical description of the system (free particle)is achived through the construct of rigged Hilbert space.The dual space of the nuclear subspace can be realized as a Hilbert space itself,by defining the scalar product of functionals and "moving" the Hamiltonian from $M$ to $\tilde{M}$.

As for quantum states,they are built from the fundamental theorems of Wigner & Bargmann.The quantum states are elements of the Rigged Hilbert space of the unitary/antiunitary ray/projective representations of the symmetry groups of the Hamiltonian.

Daniel.

Last edited: Jun 17, 2005