# Spin up/down issue

1. Dec 1, 2009

### mccoy1

Surely the spin can only either be up or down?
I've been thinking about it lately and I think this iview s incorrect.
It's incorrect partly because an atom/molecule/electron actually doesn't "reside" in any specific energy level/wavefunction, therefore it doesn't make sense to view it as having specific up or down spin.

2. Dec 1, 2009

### ansgar

These are unrelated questions, first you must specify WHAT spin? S,J,L ? And for what kind of particle? An elementary fermion has S = 1/2 and thus it can have as values +1/2 or -1/2 (i.e. up and down)

Secondly, the question what a certain atom/electron etc has as spin value is not even a question in QM, the question we can ask is what the outcome of measurements can become. E.g. a measurment of the spin S of an electron can only give you values "up" & "down", but asking what the electron "has" (prior to measurment) is not a question that can be asked in QM (at least not in the Copenhagen interpretation of QM)

SO I think your confusion lies here.

3. Dec 2, 2009

### mccoy1

Thanks Ansgar for the reply but I think you didn’t get what I’m talking about or else I haven’t done a good job in explaining myself.
First these questions are related. Think about it. H(hat)ψ+1/2 = -γВ0[Іz(hat)ψ+1/2].
Secondly, it’s clear that I’m talking about electrons which we all know has up or down spin (or so we think) represented by ±1/2ћ.
Third, I didn’t ask what certain atoms or electrons have spin values. All that I was saying was that it’s incorrect to think of electron as having definite spin up or spin down because there’s nothing like that in real world. When doing NMR experiment, not all electrons point along the applied magnetic field....some point along the x, y-axis. Also if electron has absolute up or down spin, how come we can prepare it to point at some direction (angle) and still emits photon? Believe it or not, there’s no pure down or pure up.
Too I didn’t ask what electron ‘has’ prior to the measurement...oh God, I don’t think you read my post before replying to it!
In conclusion, In QM books, you always have nice diagrams showing the energy levels of the associated wavefunctions and we are fooled into believing that the molecule/atom(or whatever you like) belongs to one of those associated wavefunction. This representation is just wrong. Anyone with experience with NMR spectroscopy will understand what I’m talking about.

4. Dec 2, 2009

### DrChinese

Who says it has a definite spin value prior to its measurement? The x, y and z axes are arbitrary anyway.

5. Dec 2, 2009

### peteratcam

The spin-1/2 degree of freedom has a 2-dimensional Hilbert space.
Eigenstates of hermitian operators often prove to be a useful basis for the space of states.

It is entirely conventional that the eigenstates of Sz are taken as the basis. The eigenstates of Sz are often called up and down. (and what a stupid set of names too)

As always in quantum mechanics, the state can be any element of the vector space of states.
So in general, the spin state is a|up> + b|down>, when expressed in the Sz basis. You could choose any other basis.

"Surely the spin can only either be up or down?" makes such little sense as a question I don't know what to say to it.

6. Dec 2, 2009

### ytuab

Can the quantum mechanics (QM) predict all the correct energy levels and wavefunctions (of multi-electron atoms, and molecules) theoretically?

That of the simple one-electron atom hydrogen can be theoretically predicted, I think
But the values of the multi-electron atoms are determined empirically, (or theoretically + empirically).

For example, the spin precession, the spin-orbital interaction, and the electron-electron interaction ...etc.
To be in accord with the experimental value, QM changes the kind and combination ratio of Lande-g-factor coupling (L-S, L-J, J-J) cleverly.

So I think we can not compare the QM theory with the experimental results purely.
I also don't know why the angle of the spin precession is fixed. ( S / S(S+1) )

Last edited: Dec 2, 2009
7. Dec 2, 2009

### mccoy1

And who asked whether the electron has definite spin value prior to measurement?

The "arbitrariness" of the choice of axis(say x, it could be y or z, or k) just means that we arbitarily label x as the axis along which the magnetic field is pointing(just as a reference axis).But the fact that we can prepare the electron at any configuration by rotating the magnetic poles at some angle and still releases a photon has nothing to do with x, z or y being arbitray axes.

8. Dec 2, 2009

### mccoy1

Don't worry..it's correct.

9. Dec 3, 2009

### alxm

Yes, absolutely.

Nonsense. Just because the one-electron (clamped-nucleus, nonrelativistic) atom is the only one that's analytically soluble doesn't mean the rest are 'empirical'. A numerical solution to an equation is in no way 'empirical'.

The non-relativistic ground-state energy of Helium is -2.903724377034119598311159245194404446696925310. Accurate enough for you?
(I can't be bothered to look up a relativistic one.. it exists. Even ones corrected for QFT effects, which are chemically insignificant, exist.)

None of the methods used for calculating these things are empirical, and the only semi-empirical method in wide use is the B3LYP density functional, which is only 'empirical' to the extent that two theoretically-derived methods whose errors cancel were mixed to an empirically-chosen extent. And even that's getting less 'empirical' as some of those 8 or so parameters have been retroactively justified from theory.

Methods like full-CI are completely exact within numerical limits. The fact that quantum mechanics is the 'final theory' as far as the properties of atoms and molecules are concerned, is not at dispute any more than the idea that classical mechanics is the 'final theory' of billiard balls.

And for the record, one of my co-workers got her PhD on ab-initio calculations of g-tensors.

10. Dec 3, 2009

### DrChinese

Uh, you did. You asked whether the view that "Surely the spin can only either be up or down?" is correct. The answer is that it is not. An electron lacks a definite up or down spin until observed in some arbitrary direction.

11. Dec 3, 2009

### ytuab

OK. alxm. May I go back to "spin" ? (Because I have noticed one thing.)
The splitting pattern under the magnetic field is important and useful for examining the each orbital property. Right?
If you say that the quantum mechanics(QM) is the final theory,
The Lande g-factors (spin-orbital interaction) of all the atoms are theoretically derived ? (not as I said above?)

And even in the simplist case as follows.
The Sodium D-line dublet (3p3/2, 3p1/2 >>>> 3s ) is the best and simplist example of the Lande g-factor and is often explained in the anomalous Zeeman effect section of the QM textbooks.

But if you look at the way of deriving the Lande g-factor formula carefully, you will probably notice something strange.

The magnetic moment parallel to the external magnetic field is, ($$\beta_{B}$$ is the Bohr magneton, $$j, l, s$$ are the total, orbital and spin angular momentum.)

$$\beta_{B} ( l \cos(l,j) + 2s \cos(s,j) ) = \beta_{B} j g = \beta_{B} j ( \frac{l}{j} \cos(l,j) + \frac{2s}{j} \cos (s,j) )$$

So, Lande g factor is
$$g = \frac{l}{j} \cos(l,j) + \frac{2s}{j} \cos(s,j) = 1 + \frac{j^2 + s^2 - l^2}{2 j^2}$$

Here, we change as follows, $$j^2, l^2, s^2 \rightarrow j(j+1), l(l+1), s(s+1)$$. So final Lande g-factor is,
$$g= 1 + \frac{j(j+1) + s(s+1) -l(l+1)}{2j(j+1)}$$

But only the remaining j (of the upper $$\beta_{B} j g$$ )is not changed as $$j \rightarrow \sqrt{j(j+1)}$$.

I think this is strange and inconsistent. And if it is strange, even the simplist case of the anomalous Zeeman effect can not be explained by QM?

How do you think about it? If I'm mistaken, please tell me.

12. Dec 3, 2009

### alxm

Actually, I'd like you to provide references to explain what these 'empirical' methods are, and why you think quantum mechanics doesn't work for many-electron systems, despite having an 80-year track record of doing exactly that.

First: The inconsistency is yours: You've confused the operator $$\hat{J}$$ with the total angular momentum $$J$$.

Second, as a parenthesis: This is not the usual derivation. This is a fairly hand-waving derivation I've only seen in one textbook (An appendix to Littlefield's "Atomic and nuclear physics: An introduction" - which typographically does properly distinguish operators and values). If you want a more rigorous derivation, see for instance Chapter XVI of Messiah book II, or Chapter XV of Landau-Lifgarbagez.

I think it's arrogant in the extreme to assume that quantum mechanics fails to describe atoms just because you didn't understand a derivation.

13. Dec 3, 2009

### kote

^ What he said. Although I prefer "measured" to "observed" .

14. Dec 3, 2009

### ytuab

For example, in page 705 of the Historical Development of Quantum Theory by Jagdish Mehra
------------------------------
Goudsmit and Back later analyzed the Zeeman effects of several multiplet spectra.
They found, for example that only in the case of silicon and the ground states of neon and argon, did pure (K,J)-coupling exist; in the case of lead terms they found either (J,J) coupling or more complex, in which J1 was first coupled to K2, and then the resultant J1+K2 was coupled to R2 (1926).
-------------------------------
Sorry. It's a little old. (But as you know, the form of the Lande-g factor has not changed since 1920's.)

As I said, The magnetic moment parallel to the external magnetic field is,
$$\beta_{B} ( l \cos(l,j) + 2s \cos(s,j) ) = \beta_{B} ( l \frac{j^2 + l^2 -s^2}{2jl} + 2s \cos(s,j)) = \beta_{B}(\frac{3j^2 + s^2 - l^2}{2j}) \rightarrow \beta_{B}(\frac{3j(j+1)+s(s+1)-l(l+1)}{2j}$$

And this is equal to
$$\beta_{B} ( l \cos(l,j) + 2s \cos(s,j) ) = \beta_{B} j ( \frac{l}{j} \cos(l,j) + \frac{2s}{j} \cos (s,j) ) = \beta_{B} j (\frac{3j^2 +s^2-l^2}{2j^2}) \rightarrow \beta_{B} (\frac{3j(j+1)+s(s+1)-l(l+1)}{2(j+1)} )$$

But those results are different. The correct answer is the latter.
I think whether operator or not is not important here. (The Lande-g factor is caused by the precession.)

Last edited: Dec 3, 2009
15. Dec 4, 2009

### alxm

That quote does not support anything that you claimed. Again: Where are these 'empirical' methods you're talking about?

LS-coupling is an approximation. That is not news. Just further evidence you haven't actually understood the derivation in the textbooks, which are invariably explained at the same time as the LS-coupling scenario.

Then why should anyone waste their time explaining what you've gotten wrong? You clearly don't understand the basics of the math involved, and you don't appear willing to learn, either.

16. Dec 4, 2009

### ytuab

Then you can predict when you shoud use J-J coupling insted of L-S coupling in the anomalous Zeeman effect theoretically using QM? (not empirically.)
It's impossinle.

First, Do you undestand the meaning of "J-J coupling" ?

Probably you don't understand the "real" meaning of the Lande-g-factor.
In Lande-g-factor, "precession" is important.
The more important issue is in "another place" you said above.

Sorry. I'm now in a hurry. So I replied later.