# Difference between quantum numbers $j$ and $m$ and the four others

I am confused about what is the difference between the quantum numbers $$j$$ and $$m$$ and the other four quantum numbers: the principal quantum number $$n$$, angular momentum $$m$$, etc.?

From Quantum Mechanics: Concepts and Applications by N. Zettilli, it says:

The results obtained thus far can be summarized as follows: the eigenvalues of $$\hat{\vec{J}}^2$$ and $$J_z$$ corresponding to the joint eigenvectors $$\lvert j, m\rangle$$ are given, respectively, by $$\hbar^2 j(j+1)$$ and $$\hbar m$$:

$$\boxed{\hat{\vec{J}}\vphantom{J}^2 \lvert\ j, m\rangle = \hbar^2 j(j+1)\lvert\ j, m\rangle\qquad\text{and}\qquad\hat{J}_z\lvert\ j, m\rangle = \hbar m\lvert\ j, m\rangle,}\tag{5.48}$$

where $$j = 0, 1/2, 1, 3/2, \ldots$$ and $$m = -j, -(j-1), \ldots, j-1, j$$. So for each $$j$$ there are $$2j+1$$ values of $$m$$. For example, if $$j = 1$$ then $$m$$ takes the three values $$-1$$, $$0$$, $$1$$; and if $$j = 5/2$$ then $$m$$ takes the six values $$-5/2$$, $$-3/2$$, $$-1/2$$, $$1/2$$, $$3/2$$, $$5/2$$. The values of $$j$$ are either integer or half-integer. We see that the spectra of the angular momentum operators $$\hat{\vec{J}}\vphantom{J}^2$$ and $$\hat{J}_z$$ are discrete. Since the eigenstates corresponding to different angular momenta are orthogonal, and since the angular momentum spectra are discrete, the orthonormality condition is

$$\boxed{\langle j', m'\ \lvert\ j, m\rangle = \delta_{j',j}\delta_{m',m}.}\tag{5.49}$$

Where $$J$$ is the general angular momentum operator.

My questions basically are:

• How is the principal quantum number $$n$$ related to $$j$$? Meaning, what is the difference between the two? I know $$n$$ can't be half an integer, so why can $$j$$ be half-integer?

• And also, perhaps this is a bit off-topic, but what is the difference between general angular momentum operator $$J$$ vs. angular momentum operator $$L$$?

N. Zettilli says:

\boxed{\begin{aligned}\ [\hat{L}_x,\hat{L}_y] &= i\hbar \hat{L}_z, & [\hat{L}_y,\hat{L}_z] &= i\hbar \hat{L}_x, & [\hat{L}_z,\hat{L}_x] &= i\hbar \hat{L}_y\end{aligned}}\tag{5.8}

As mentioned in Chapter 3, since $$\hat{L}_x$$, $$\hat{L}_y$$, and $$\hat{L}_z$$ do not commute, we cannot measure them to arbitrary accuracy.

Note that the computations of the above commutators were derived by expressing the orbital angular momentum in the position representation, but since these are operator relations, they must be valid in any representation. In the following section, we are going to consider the general formalism of angular momentum, a formalism that is restricted to no particular representation.

Meaning that $$L$$ is in the "position representation". But isn't $$J$$ also in the position representation since it is still angular momentum? Or is it that since $$J = L + S$$ (where $$S$$ is the spin) then it is no longer in "position representation"? Then what "representation" is it in?

I may be late to reply, but I would really appreciate any discussion or explanation of these questions. I am really confused about this. A similar question was this here but it wasn't quite the same question.

• Good question, but could you please transcribe the text and equations from your images into the post using MathJax? There shouldn't be any images containing large blocks of text. (Although it's not such a big problem in this case because the images are quite readable, as opposed to some other posts :-) Apr 1, 2015 at 6:06
• General editorial comment: writing "and etc." is like saying "ATM machine". The "et" in "etc." already means "and".
– hft
Apr 1, 2015 at 6:09

If $L$ is in the "position representation" (call its eigenstates "$|lm_l\rangle$"), and it $S$ is in the "spin representation" (call its eigenstates "$|sm_s\rangle$"). Then, first thing to remember is that these spaces ("position" and "spin") are completely independent, in that it makes sense to talk about the direct product space: $$|lm_l\rangle\otimes|sm_s\rangle$$ It is in this space that $J$ acts. The operator $J$ can be written very explicitly to show it is the extension of the operators $L$ and $S$ into the direct product space: $$J\equiv L\otimes \mathcal{I_s}+\mathcal{I_l}\otimes S\;,$$ where $\mathcal{I_l}$ and $\mathcal{I_s}$ are the identity operators on the "position" and "spin" spaces, respectively. However, in practice we usually leave the identity operators implicit... mostly to confuse the non-initiated...