Explaining quantized angular momentum?

[RKlintworth]

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!

 

[WannabeNewton]

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 $L_z$.

 

[RKlintworth]

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 I am reffering to the equation L= n$\hat{}h$ 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) .

 

[RKlintworth]

$\hat{h}$ rather, sorry.

 

[jtbell]

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 L_z 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.

 

[RKlintworth]

Okay that makes sense. thank you. I understand the solution to the Schrodinger equation in spherical co-ordinates for L = √l(l+1) $\hat{h}$ 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.

 

[jtbell]

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
$$L_z = m \hbar\\
L^2 = l(l+1)\hbar$$

 

[jtbell]

$\hat{h}$ rather, sorry.

Actually, $\hbar$. (\hbar in LaTeX)

\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!

 

[richardhale]

Great post to all. Appreciate the resource.