I Spin quantum numbers: correct names and formulae?

1. Mar 25, 2017

N88

I'm having trouble finding the correct names for the 4 standard labels: n, l, m, s.
Where might I find, free online: their correct names (ie, no colloquialisms or short cuts) and the accepted formula for each (to be sure I use them correctly)?
Thanks.

2. Mar 25, 2017

blue_leaf77

You can denote any quantum numbers in any system by any character - you can for instance denote the first four quantum numbers out of 4N quantum numbers for a N-electron atom by n, l, m, s. So, without specifying which system is being considered it's actually impossible to give thoughts on it.
If by those 4 numbers, you mean the usual notation for hydrogen-like system, they are usually called "principal", "angular momentum", "magnetic", and "spin" quantum numbers respectively. If you had taken a chemistry class in highschool, I think it's hard to miss this.
There is no explicit formula of the form $n=\ldots$, $l=\ldots$, $m=\ldots$, and so on for them since they are actually eigenvalues of some operators. To find them, you need to solve the corresponding eigenvalue equation.

3. Mar 26, 2017

Staff: Mentor

The Hyperphysics web site is usually reliable for basic stuff like this:

http://hyperphysics.phy-astr.gsu.edu/hbase/qunoh.html

[added] Note that Hyperphysics uses a different set of labels ($n,\ell,m_\ell,m_s$) from yours ($n,\ell,m,s$). I think perhaps physics books (at least quantum-physics books) tend to use the first set, and chemistry books the second. In the first set, one could also include $s$ but it's always the same for electrons (1/2). $\ell$ and $m_\ell$ have the same relationship to each other, with respect to orbital angular momentum, that $s$ and $m_s$ have with respect to spin angular momentum, as described in one of my posts that you've seen elsewhere.

Last edited: Mar 26, 2017
4. Mar 26, 2017

N88

Again, many thanks; that's exactly what I wanted! As a QM beginner, I need to be very sure about these sort of details, especially re my new favourite formula:

My thanks again; N88

5. Mar 26, 2017

blue_leaf77

I don't think the square root of the operator $S^2$ has a useful meaning in quantum mechanics, physicists rarely address this quantity. For one thing, given the matrix of $S^2$ there are multiple possibility to construct an operator $A$ defined to follow $A^2 = S^2$ due to the multiple choice of the sign of each diagonal element of $A$. Second, if a ket $|v\rangle$ is an eigenvector of $S^2$, it doesn't necessarily lead to $|v\rangle$ being an eigenvector of $A$.

6. Mar 26, 2017

N88

I'd be pleased if you and jtbell would discuss your point in detail.
Reason: In my private studies (allowing the spin of a polarised spin-half particle to be $\pm\hbar/2$ in relation to a given orientation) I find the expectation of

very useful as the average total angular momentum.

In QM (without averaging) it is taken to be the total angular momentum http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c1

It looks to me that this is related to the point that you are making? Your critical comments would be very welcome.

7. Mar 26, 2017

blue_leaf77

Looking at the hyperphysics link, it looks to me that the quantity $L$ is just defined to be the positive square root of the eigenvalue of the operator $L^2$. That is, $L$ is just a number while $L^2$ is an operator. Since $L$ is aroused as a defined quantity, it does not necessarily have any physical meaning.
I think you have been misled on that part. If you had calculated an expectation value, you must have the operator whose averaged measured value you calculated. I want to know which operator whose expectation value you calculated resulting in $\sqrt{0.5(0.5+1)}\hbar$?

8. Mar 26, 2017

N88

Thanks for this. I'm an engineer (familiar with vectors and vector-products). So I'm privately studying QM from that perspective (to see how far it takes me). That means that the operators that I work with are the detectors/polarisers (as in EPRB), operating on pristine particles and polarising them.

So all my variables have physical meaning and I derive the RHS of the relation we are discussing; the LHS follows from s = 1/2.

A similar idea (I have some different ideas) is to be found in the peer-reviewed http://zfn.mpdl.mpg.de/data/Reihe_A/53/ZNA-1998-53a-0637.pdf -- page 647 and eqns (74)-(75).

EDIT: In case this helps: I arrive at $\tfrac{\left\langle \lambda^{2}\right\rangle }{3}=1.$ Since $\lambda$ is a non-negative magnitude, I write: $\left\langle \lambda\right\rangle= \sqrt{3}$.

Then, since $\boldsymbol\lambda= \tfrac{\hbar}{2}\lambda$: $\left\langle\boldsymbol\lambda\right\rangle= \sqrt{3}\tfrac{\hbar}{2}$ equals the average of the total angular momentum under QM via the relation for total angular momentum $\sqrt{0.5(0.5+1)}\hbar$: which is what is under discussion.

Thanks.

Last edited: Mar 27, 2017