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Angular momentum of hydrogen atom with Schrodinger Equation

  1. Mar 30, 2016 #1
    • Moved from a technical forum, so homework template missing
    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.
     
  2. jcsd
  3. Mar 30, 2016 #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 text book or see net-resource on H-atom.
     
  4. Mar 30, 2016 #3
    upload_2016-3-30_11-23-24.png I am solving problem 2
     
  5. Mar 30, 2016 #4
    so you write down schrodinger eq and constrain it to move it in a circle - now you dont 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 dont 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.
     
  6. Mar 30, 2016 #5
    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?
     
  7. Mar 30, 2016 #6

    QuantumQuest

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    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.
     
    Last edited: Mar 30, 2016
  8. Mar 30, 2016 #7
    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 alot dude, appreciate the help.
     
  9. Mar 30, 2016 #8

    QuantumQuest

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    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.
     
    Last edited: Mar 30, 2016
  10. Mar 30, 2016 #9

    DrClaude

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    Staff: Mentor

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