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Stylord

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- Homework Statement
- Let ##\rho , \phi , z## be the cylindrical coordinates of a spinless particle (##x=\rho \times \cos(\phi) ,y=\rho \times \sin(\phi) ; \rho \geq 0 , 0 \leq \phi \lt 2\pi##. Assume that the potential energy of this particle depends only on ##\rho##, and not on ##\phi## and z.

a. Write, in cylindrical coordinates, the differential operator associated with the Hamiltonian.Show that H commutes with Lz, and Pz.Show from this that the wave functions associated with the stationary states of the particle can be chosen in the form:

##\phi_{n,m,k}(\rho,\phi,z)=f_{n,m}(p) e^{im\phi} e^{ikz}##

- Relevant Equations
- See below

Hi, everyone.

Please check the following questions (extracted of the cohen Tanpoudji)

for the first question, here my Hamiltonian operator.

It's easy to see that it commutes with Lz and Pz.

Now we can determine a common eigenvector basis for these 3 operators.

For the angular part we need to solve

Lz

Pz

The resolution of the differential equation (1) and (2) gives us the angular part we see on the homework statement.

for the part ##e^{im*\phi}##

for the part ##e^{ikz}## I can't determine a condition and I don't know which values it can take.

So if someone have an hint thanks in advance !

Please check the following questions (extracted of the cohen Tanpoudji)

for the first question, here my Hamiltonian operator.

It's easy to see that it commutes with Lz and Pz.

Now we can determine a common eigenvector basis for these 3 operators.

For the angular part we need to solve

Lz

**g(théta,phi)**=m##\hbar##**g(théta,phi)**(1)Pz

**g(théta,phi)**=k##\hbar##**g(théta,phi)**(2)The resolution of the differential equation (1) and (2) gives us the angular part we see on the homework statement.

for the part ##e^{im*\phi}##

**, the wave function needs to be continuous in all the space**so ##e^{2im*\pi}##=1,consequently, m need to be an integerfor the part ##e^{ikz}## I can't determine a condition and I don't know which values it can take.

So if someone have an hint thanks in advance !

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