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Explaining quantized angular momentum?

  1. May 1, 2014 #1
    Hey there,

    I'm having trouble understanding where two of the formulas for angular momentum of an electron in the hydrogen atom tie together. The first relation L = nh/(2∏) which comes from the de Broglie momentum-wavelength relation and the requirement for the electron wavelength to equal an integer number of the path circumference. The other I have come into contact with is: L = (√(l(l+1))h/(2∏) from the Schrodinger equation.

    I understand the derivation of both, I just don't see how they are the same. I thought this n was the principle quantum number?

    Thanks for your help!
     
  2. jcsd
  3. May 1, 2014 #2

    WannabeNewton

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    The eigenvalues of the total angular momentum ##L^2## are ##\hbar^2 l(l + 1)## where ##l = 0,1,2,3,...## and the eigenvalues of the ##z## component of angular momentum ##L_z## are ##\hbar m## where ##m = -l,...,l## for a given ##l##. You're confusing the total angular momentum and the ##z## component of the angular momentum-they aren't the same; ##L^2 = L_x^2 + L_y^2 + L_z^2## involves all three components of the angular momentum, not just ##L_z##.
     
  4. May 1, 2014 #3
    Thanks for the reply!

    The z component angular momentum is actually what got me confused in the first place leading me to reevaluate my understanding of the concept and then my question.

    But Im reffering to the equation L= n[itex]\hat{}h[/itex] which comes from linking L = pr where p = h/λ and 2∏r = nλ. where n is the principle quantum number (or so I thought) .
     
  5. May 1, 2014 #4
    [itex]\hat{h}[/itex] rather, sorry.
     
  6. May 1, 2014 #5

    jtbell

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    Staff: Mentor

    Also note that your first formula comes from assuming that the electron travels in a circular orbit, and imposing de Broglie's standing-wave condition. In reality, the electron doesn't travel in a circular orbit, or indeed in a classical orbit at all.

    Quantization of Lz comes from solving the Schrödinger equation in spherical coordinates, and requiring that the ψ function be continuous as you go around the nucleus in an azimuthal direction. This happens to give superficially the same result as the de Broglie derivation.
     
  7. May 1, 2014 #6
    Okay that makes sense. thank you. I understand the solution to the Schrodinger equation in spherical co-ordinates for L = √l(l+1) [itex]\hat{h}[/itex] yet my textbook ignores the derivation for the z component and that's where I've been stuck most of the day. Do you know where I can find its derivation?

    Thank you so much for the replies by the way, this forum is amazing.
     
  8. May 1, 2014 #7

    jtbell

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    Staff: Mentor

    I don't know how much of this stuff you've seen already, but here's an outline. It might help you find your way around a QM or "modern physics" textbook that covers the hydrogen atom, or give you some things to Google for:

    1. Solve the Schrödinger equation in spherical coordinates to find the stationary-state wavefunctions ##\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)\Theta_{lm}(\theta)\Phi_m(\phi)## where n, l, m are the quantum numbers which obey certain rules that emerge from the details of the solution.

    2. Starting from the operators for position and momentum:
    $$\hat x = x\\
    \hat p_x = -i\hbar\frac{\partial}{\partial x}$$
    (and similarly for y and z components) construct operators for the (orbital) angular momentum from the operators for position and momentum by using ##\hat{\vec L} = \hat{\vec r} \times \hat{\vec p}## (which corresponds to the classical formula ##\vec L = \vec r \times \vec p##). In particular we work with the operators ##\hat L_z## for the z-component and ##\hat L^2## for the square of the magnitude because the solutions of the SE in step 1 turn out to be eigenfunctions of these operators, that is:
    $$\hat L_z \psi_{nlm} = L_z \psi_{nlm}\\
    \hat L^2 \psi_{nlm} = L^2 \psi_{nlm}$$

    3. Apply the operators from step 2 to the ##\psi_{nlm}## from step 1, to find that the ##\psi_{nlm}## are indeed eigenfunctions, with
    $$L_z = m \hbar\\
    L^2 = l(l+1)\hbar$$
     
  9. May 1, 2014 #8

    jtbell

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    Staff: Mentor

    Actually, ##\hbar##. (\hbar in LaTeX) :smile:

    \hat is often used to denote operators, as I did in my preceding post. It's also often used for unit vectors, so beware of the context!
     
  10. May 2, 2014 #9
    Great post to all. Appreciate the resource.
     
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