Why doesn't the wave function change when φ increases by 2π?

[m-i-t-o]

Hi everyone, Can anybody solve my simple problems of quantum:
Usually we say, that wave function Ψ is dependent on r,θ,Φ .But this is just a coordinate system,or more than that. Imagination to this is qite difficult.
Moreover somewhere in a book I have read that if Ψ = f(r,θ) exp(imΦ),then on increment of 2pi in Φ does not change wave function.? Why?

 

[cmos]

Usually we say, that wave function Ψ is dependent on r,θ,Φ .But this is just a coordinate system,or more than that.

Did you mean to ask the question: "Is this just a coordinate system,or more than that?"
If so, then it looks like you are just presenting spherical coordinates. Instead of your axes defining three edges of a box (as in Cartesian coordinates), your three axes define a sphere.

Moreover somewhere in a book I have read that if Ψ = f(r,θ) exp(imΦ),then on increment of 2pi in Φ does not change wave function.? Why?

This just means that the specific wavefunction you have is sinusoidally periodic in Φ.

 

[Fredrik]

You can think of points in space as something independent of coordinates, and coordinate systems as functions that assign three numbers to each point in space. If p is a point in space and f and g are coordinate systems, then

$$\psi(p)=(\psi\circ f^{-1})(f(p))=(\psi\circ g^{-1})(g(p))$$

If f assigns the cartesian coordinates and g the spherical coordinates, then we can write $(\psi\circ f^{-1})(x,y,z)=(\psi\circ g^{-1})(r,\theta,\phi)$.

So you can think of your wave function as the composition of a coordinate independent wave function and the inverse of a coordinate system.