Momentum eigenvalues and eigenfunctions

[White_M]

Homework Statement

For the following wave functions:
ψ_{x}=xf(r)
ψ_{y}=yf(f)
ψ_{z}=zf(f)

show, by explicit calculation, that they are eigenfunctions of Lx,Ly,Lz respectively, as well as of L^2, and find their corresponding eigenvalues.

Homework Equations

I used:
L_{x}=-ih(y\partial/\partial z-z\partial/\partial y)
L_{y}=-ih(z\partial/\partial x-x\partial/\partial z)
L_{z}=-ih(x\partial/\partial y-y\partial/\partial x)

for solving:

L_{z}|ψ_{z}>=lz|ψ_{z}>
L_{x}|ψ_{x}>=lx|ψ_{x}>
L_{y}|ψ_{y}>=ly|ψ_{y}>


and
L^2|ψ>=l^2|ψ>

where:
L^2=L_{x}^2+L_{y}^2+L_{z}^2

The Attempt at a Solution

For instance for Lz:
ψ_{z}(r)=<r|z>=zf(r)

L_{z}|ψ_{z}>=-ih(x\partial/\partial y-y\partial/\partial x) zf(r)=lz|ψ_{z}>

Is that correct?

 

[lightgrav]

I think they want you to "explicitly" take partial derivatives of the function z f(r) .

 

[White_M]

-ih (x(\partialz\\partialy f(r)+z \partialf(r)/\partialy)-y(\partialz\\partialx+z\partialf(r)\\partialx))=-ih (xz\partialf(r)\\partialy-yz \partialf(r)\\partialx)=-ih (x\partial\\partialy-y\partial\\partialx)zf(r)

Which is the same as the initial expression.

What am I missing?

Thanks