# Rationale of the position operator?

1. Sep 6, 2012

### PerpStudent

Why is the position operator of a particle on the x axis defined by x multiplied by the wave function? Is there an intuitive basis for this or is it merely something that simply works in QM?

2. Sep 6, 2012

### Jolb

Multiplication by x is not in general the position operator. "x multiplied by the wave function" would be the correct recipe if by wave function you specifically mean the particle's state ket projected onto the position basis, ψ(x)=<x|ψ>. But often people speak about other "wavefunctions" such as the particle's state representation in momentum space, ψ'(p)=<p|ψ>. In the momentum basis, the (one-dimensonal) position operator is $$\hat{x}=i\hbar\frac{\partial }{\partial p}$$

Multiplication by x is the position operator only in the position basis. Measuring the position of a particle causes the particle's state to collapse into an eigenstate of the position basis. This means it has a definite position. If you know a particle's position exactly, let's say it's at q along the x-axis, then the probability of finding the particle at q is 100% and the probability of finding it anywhere else is 0%. So if a particle has the definite position q, then its wavefunction in the position basis must look like a dirac delta function centered at q, which would be written δ(x-q).

Multiplying the dirac delta function by x,
xδ(x-q) = qδ(x-q)
since δ(x-q) is zero everywhere except at x=q. You should be able to see that in the position basis, the delta functions are the eigenfunctions of multiplication by x and their eigenvalues are equal to their position.

Last edited: Sep 6, 2012
3. Sep 8, 2012

### PerpStudent

Thanks, that's very helpful.

4. Sep 8, 2012

### Dickfore

Look at the 2 formulas for the expectation of the position operator:
$$\langle \hat{x} \rangle = \int_{-\infty}^{\infty}{x \, \psi^{\ast}(x) \, \psi(x) \, dx} = \int_{-\infty}^{\infty}{\psi^{\ast}(x) \, \hat{x} \, \psi(x) \, dx}$$
This has to be true for all possible wave functions. Comparing the 2, we conclude that:
$$\hat{x} \, \psi(x) = x \, \psi(x)$$
This is generally true for every operator in its own eigenbasis representation.