# Homework Help: Showing functions are eigenfunctions of angular momentum.

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1. Jan 25, 2015

### Robsta

1. The problem statement, all variables and given/known data
Verify by brute force that the three functions cos(θ), sin(θ)e and sin(θ)e−iφ are all eigenfunctions of L2 and Lz.

2. Relevant equations

I know that Lz = -iћ(∂/∂φ)
I also know that an eigenfunction of an operator if, when the operator acts, it leaves the function unchanged apart from a multiplicative factor (the eigenvalue)

3. The attempt at a solution

So, Lzcos(θ) = -iћ(∂/∂φ)cos(θ)
But I think that equals zero. There's no component of cos(θ) in the φ direction. There different variables. So I think that when the differential operator acts on it, it makes 0. But then the function isn't left unchanged. Can somebody help me resolve this in my head?

2. Jan 25, 2015

### Simon Bridge

You have $L_z\psi = 0\psi$ so $\psi$ is an eigenfunction of $L_z$ with eigenvalue ...

3. Jan 25, 2015

### Robsta

Eigenvalue 0. Does that mean that any function orthogonal to the phi direction is an eigenfunction then? Because its differential Thanks for your help!

4. Jan 25, 2015

### Simon Bridge

What does the eigenvalue of 0 mean - physically? What does $L_z$ measure?

5. Jan 25, 2015

### Robsta

Lz measures the angular momentum about the vertical (z) axis. The eigenvalue is the actual amount of ang. momentum. I've managed to crack this question now, thanks a lot for your help :)

6. Jan 25, 2015

### Simon Bridge

Put more precisely: since phi is the angle the total angular momentum makes with the z axis, then any wavefunction with no phi dependence will also have no component of angular momentum along the z axis. Hence, zero eigenvalue.

Notice the emphasis on physics (well, geometry) rather than mathematics (calculus) - yes it comes out zero because of the maths, but the Universe does not care about what our calculations yield. The maths is just a model - it is trying to describe Nature. Look for the truth in Nature.

Similarly ...
The "z axis" does not have to be vertical - you are not dealing with gravity here - it is determined by the orientation of the apparatus doing the measuring. It can as easily (and more usually) be along the direction of motion or the direction of an applied field. It's just a common label for an "axis of interest". You should abandon ideas about the orientation of the Cartesian axes.

The phi direction is not mentioned in the problem ... it is not actually a direction since knowing phi does not tell you where to look, but describes a set of infinitely many directions from the z axis. It is important to distinguish between a direction (component of a vector) and the dependence that the magnitude of the component has.

i.e. $\vec B = (0, kx^2, 0)$ has a y direction that depends on the x component of position.