Angular momentum of hydrogen atom with Schrodinger Equation

[James Brady]

If we were to assume that the electron moves around the proton with radius a, the Schrodinger equation becomes:

$\frac{1}{a^2}\frac{d^2\psi}{d\phi^2} + \frac{2m}{\hbar^2}|E|\psi = 0$

The question in my textbook asks me to solve the above equation to obtain values of energy and angular momentum for the electron. Energy I could do:

$\psi = c_1cos(k\phi) + c_2sin(k\phi)$ where $k^2=\frac{2ma^2}{\hbar^2}$

By knowing that phi cannot have more than one value at any radial coordinate around the atom, the energy is quantized:

$E = \frac{n\hbar^2}{2ma^2}$, which is in good agreement with the Bohr model.

The process up to here makes sense to me. However, how can I now find the angular momentum?? The idea of a wave having angular momentum is absurd to me, but I do not even know where to start working on the mathematics of it all.

 

[drvrm]

If we were to assume that the electron moves around the proton with radius a, the Schrodinger equation becomes:

1a2d2ψdϕ2+2mℏ2|E|ψ=01a2d2ψdϕ2+2mℏ2|E|ψ=0\frac{1}{a^2}\frac{d^2\psi}{d\phi^2} + \frac{2m}{\hbar^2}|E|\psi = 0

The question in my textbook asks me to solve the above equation to obtain values of energy and angular momentum for the electron. Energy I could do:

ψ=c1cos(kϕ)+c2sin(kϕ)ψ=c1cos(kϕ)+c2sin(kϕ)\psi = c_1cos(k\phi) + c_2sin(k\phi) where k2=2ma2ℏ2k2=2ma2ℏ2k^2=\frac{2ma^2}{\hbar^2}

the equation does not look like a schrodinger wave equation for H-atom.

the wave function has all three spherical polar coordinate dependence,namely r,theta, and phi.
the interaction term is there- coulomb interaction with 1/r dependence - your equation looks like free particle equation.
the hamiltonian operator is missing- first clarify your model problem then proceed ahead take a textbook or see net-resource on H-atom.

 

[James Brady]

I am solving problem 2

 

[drvrm]

I am solving problem 2

so you write down schrodinger eq and constrain it to move it in a circle - now you don't have a quantum picture - its a classical motion as done by Bohr.
then take the path of Bohr to quantise (artificially) the circular orbits as well as quantum condition for angular momentum J =nh
i don't think it has any features going tosatisfy the basics of quantum mechanics - just a numerical exercise.
the angular momentum will be moment of the momentum of the rotating particle taken about an axis passing through the centre perpendicular to the plane.
one can hardly call it a schrodinger picture.

 

[James Brady]

Right yeah it's not real life, I understand that. The question is tho, can I use mathematics to solve for angular momentum given this non-real scenario? I was able to solve for energies by recognizing that k must be quantized in order to have consistent boundary conditions. Is it possible to solve for L in a similar manner?

 

[QuantumQuest]

Rearrange the equation $$\frac{1}{a^2}\frac{d^2 \psi}{d\phi^2} + \frac{2m}{\hbar^2}\lvert E\rvert\psi = 0$$ so:
$$\frac{d^2 \psi}{d\phi^2} = -\frac{2ma^2}{\hbar^2}\lvert E\rvert\psi$$ so $$\psi(\phi) = e^{iC\phi}$$
where $C = \frac{\sqrt{2m\lvert E\rvert}a}{\hbar}$
I leave it to you to give values as to find $\lvert E\rvert$.
Angular momentum is $L = m\upsilon a = \hbar C$. Now fill in the final detail for angular momentum.

EDIT: Used ## for inline equations.

 

[James Brady]

OK cool, I solved for energy in my original post and using the relationship $E = \frac{1}{2}mv^2, v = \sqrt{\frac{2E}{m}}$ L can be solved for $L = \hbar\sqrt{n}$ where the square root of m is just k or the angular velocity. so $L =\hbar k$

Thanks a lot dude, appreciate the help.

 

[QuantumQuest]

In order $\psi(\phi)$ to have a physical meaning, $$\psi(\phi) = \psi(\phi + 2\pi)$$ So, what does this imply? Find kinetic energy $\lvert E\rvert$ (note: $E - U = K = \lvert E\rvert$) from this and then angular momentum $L$, as a function that includes $n$, where $n$ is any integer. Finally, compare both of them with the Bohr model.

EDIT: Used ## for inline equations.

 

[DrClaude]

Note: use ## ## instead of $$ $$ for inline equations.