# Quantization of orbital angular momentum

Probably a very simple question, but I can't find the answer on the Internet. I know nearly to nothing about quantum mechanics, but in statistical physics I'm confronted with the idea that the orbital angular momentum is quantized as $$L = \sqrt{(n(n+1)}\hbar,$$ with $n$ being an integer.

But in the Bohr description of the hydrogen atom I was also confronted with $$L = n\hbar,$$ with $n$ being an integer, which seems to contradict the first one. (If I look for it on the Internet, it is stated that the first $L$ is the norm of the angular momentum, while the second $L$ is only the $z$-component. But I don't see how those two can be different.)

And what I'm also wondering about is, if the electron of the hydrogen atom is in ground state, which one will be its rotational momentum? And because $\mu = \frac{eL}{2m}$, with $\mu$ being the magnetic momentum, what will be the magnetic momentum? It seems that both formulas give different values.

I hope it's possible to give a simple (intuitive) explanation for this, because I have no experience at the moment with quantum-mechanical operators, ...

-
–  Qmechanic Apr 11 '13 at 20:24
I see you've already got one answer, so hopefully that helps... but honestly, I think the fact that you don't have much experience with quantum mechanical operators does severely handicap your ability to understand why the z component of the angular momentum is less than the norm (for $n > 0$). In other words, I think there's a limit on how simply it's possible to explain this concept. –  David Z Apr 11 '13 at 23:47

## 2 Answers

According to quantum mechanics, if you make a measurement of the magnitude squared of the orbital angular momentum of the electron in the hydrogen atom, then possible outcomes of the measurement are the discrete values $$L^2 = \ell(\ell+1)\hbar^2, \qquad \ell\geq 0$$ Given that this is the outcome of the measurement of the magnitude squared, there is a finite sequence of $2\ell+1$ possible outcomes for the corresponding measured value of the $z$ component, namely $$L_z = \ell_z\hbar, \qquad \ell_z = -\ell, -\ell+1, \dots, \ell-1, \ell$$ Now, you say that

it is stated that the first L is the norm of the angular momentum, while the second L is only the z-component. But I don't see how those two can be different

Even in the classical case, these are usually different since $$L^2 = L_x^2 + L_y^2 + L_z^2$$ so that $$L_z^2 = L^2 - L_x^2 - L_y^2$$ In other words, even in the classical case, the magnitude of the $z$ component is less than the magntude of the angular momentum unless the $x$ and $y$ components are zero. One of the main differences between the classical and quantum cases, however is that in the quantum case, the largest possible value of $L_z^2$ for a given value $L^2 = \ell(\ell+1)\hbar^2$ is $$(L_z^\mathrm{max})^2 = \ell^2\hbar^2$$ which is strictly less than $L^2$ unless $\ell=0$. You might find this image from this wiki page useful for getting a feel for this fact.

-
Thank you very much, that was very clear. I have a question though: in the Bohr model they state that $L = n\hbar$ with $n$ the principal quantum number. So for an electron in the ground state of the hydrogen atom: n = 1 and $L = \hbar$, but for n = 1, the angular quantum number (which I assume to be $l$) has to be 0, so $L$ has to be 0. I was also thinking that $L_z = L$ because in the Bohr model, they take the rotational motion of the electron in the $xy$-plane, so $L_x = L_y = 0$. But probably I can't make such assumptions in quantum mechanics. –  yarnamc Apr 11 '13 at 21:30
@yarnamc the Bohr model is a useful model - like the Rutherford "planetary system" atom - but it has severe limitations and it is not an accurate representation of reality. In particular, while in quantum mechanics $L_x=L_y=0$ does hold in the mean, their squares $L_x^2$ and $L_y^2$ have positive expectation values. In semiclassical cases there are orbits more-or-less confined to planes, but even then they are "smeared out" a bit; the ground state is way far from semiclassical and the Bohr model does not hold very well. –  Emilio Pisanty Apr 11 '13 at 22:21
Ah, thank you: it's clear now :) –  yarnamc Apr 12 '13 at 11:07

Back in those early days, Bohr hadn't really thought about L as a vector. He just wanted to get an integer number of wave cycles to fit into an orbit. Thus, the simple formula. It would work fine if we took $L$ to mean $L_z$.

Later, as theorists worked out the implications of angular momentum in three dimensions, using operators, eigenstates, and all that, they realized that atomic orbital wavefunctions had to fit nicely onto spherical surfaces. The $L_z$ formula was just for the part of the wave fitting around the 'equator' of the atom. That's when the more complex formula for total angular momentum came about.

-
Thank you for the response :). –  yarnamc Apr 12 '13 at 11:08