# Particle in a cylindrically symmetric potential (Quantum mechanics)

• Stylord
In summary, the Hamiltonian operator for the three-state system is easy to determine and it has a common eigenvector basis. The wavefunction for the energy eigenstate is a product of two other wavefunctions and the wavefunction for free movement in z is independent of the well in ##\rho,\phi##.
Stylord
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
Lzg(théta,phi)=m##\hbar##g(théta,phi)(1)
Pzg(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 integer
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 !

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In z direction the state is free wave which has momentum eigenvalue and energy eigenvalue.
$$E=\frac{k^2\hbar^2}{2m}$$

If the wave function was ##e^{ikz}## and if I injected this in the schrödinger equation, yes I will obtain this result. But the wavefunction is ##e^{ikz}e^{im\phi}f(\rho)## and if I write the eigenvalue equation of the Hamiltonian I will not obtain this, so I don't understand why we can say that. Could you enlighten me please ?

Stylord said:
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!
There is no boundary condition that restricts the value of ##k##.

Stylord said:
If the wave function was eikz and if I injected this in the schrödinger equation, yes I will obtain this result. But the wavefunction is eikzeimϕf(ρ) and if I write the eigenvalue equation of the Hamiltonian I will not obtain this, so I don't understand why we can say that. Could you enlighten me please ?
Hamiltonian is written as sum of
$$H=H(\rho,\phi)+H(z)$$
Wavefunction of energy eigenstate is written as product of
$$\psi=\eta(\rho,\phi)\xi(z)$$
$$H\psi =\xi H(\rho,\phi)\eta +\eta H(z)\xi=(E(\rho,\phi)+E(z))\xi \eta=(E(\rho,\phi)+E(z))\psi$$
Round quantum well in ##\rho,\phi## and free movement in z are independent. Sum of their energy is the system energy, e.g. ground state of the well and kinetic energy of ##\frac{p_z^2}{2m}##.

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Ohhh it's a very good idea ! Now I Can finish the exercice, Thank you !

## What is a cylindrically symmetric potential in quantum mechanics?

A cylindrically symmetric potential in quantum mechanics is a potential energy function that depends only on the radial distance from a central axis and is independent of the angular coordinate around that axis. This symmetry simplifies the problem, often allowing it to be expressed in cylindrical coordinates (r, θ, z), where the potential V(r, z) depends only on the radial distance r and the axial coordinate z.

## How is the Schrödinger equation modified for a cylindrically symmetric potential?

For a cylindrically symmetric potential, the Schrödinger equation is typically expressed in cylindrical coordinates. The time-independent Schrödinger equation becomes:

(-ħ²/2m) [ (1/r) ∂/∂r (r ∂ψ/∂r) + (1/r²) ∂²ψ/∂θ² + ∂²ψ/∂z² ] + V(r, z)ψ = Eψ

Here, ψ is the wave function, ħ is the reduced Planck's constant, m is the mass of the particle, V(r, z) is the potential energy, and E is the energy eigenvalue. The equation reflects the cylindrical symmetry by separating the radial, angular, and axial components.

## What are common applications of cylindrically symmetric potentials?

Cylindrically symmetric potentials are commonly found in various physical systems, including quantum wires, nanowires, and certain types of quantum dots. They are also used in modeling the behavior of particles in magnetic fields (such as in the Aharonov-Bohm effect) and in studying the electronic properties of cylindrical nanostructures like carbon nanotubes.

## How do you solve the Schrödinger equation for a cylindrically symmetric potential?

To solve the Schrödinger equation for a cylindrically symmetric potential, one typically employs the method of separation of variables. The wave function ψ(r, θ, z) can be separated into radial, angular, and axial components: ψ(r, θ, z) = R(r)Θ(θ)Z(z). Each component satisfies its own differential equation. The angular part often results in solutions involving exponential functions (e.g., e^(imθ)), while the radial and axial parts may require special functions like Bessel functions or numerical methods for more complex potentials.

## What are the boundary conditions for a particle in a cylindrically symmetric potential?

The boundary conditions for a particle in a cylindrically symmetric potential depend on the physical setup of the problem. Common boundary conditions include the wave function vanishing at infinity (for bound states) and continuity conditions at interfaces or boundaries of the potential. For example, in a

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