Spin quantum numbers: correct names and formulae?

[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.

 

[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 high school, I think it's hard to miss this.

the accepted formula for each

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.

 

[jtbell]

The Hyperphysics website 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.

 

[N88]

The Hyperphysics website 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.

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

 

[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 $A$.

 

[N88]

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 $A$.

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.

 

[blue_leaf77]

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.

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 find the expectation of

very useful as the average total angular momentum.

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 $\sqrt{0.5(0.5+1)}\hbar$?

 

[N88]

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 $\sqrt{0.5(0.5+1)}\hbar$?

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.