Spin operator and spin quantum number give different values, why?

[43arcsec]

TL;DR Summary

Spin operator measures hbar/2

Spin quantum number, s(s+1)hbar (for spin 1/2) measures sqrt(3)/2hbar

Assume spin 1/2 particle

So the spin operator gives +/- hbar/2

eg. S |n+> = +/- hbar/2 |n+>

But S= s(s+1) hbar = sqrt(3)/2 hbar

So I'm off by a factor of sqrt(3).

I suspect I am missing something fundamental about my understanding of spin.

My apologies and thanks in advance.

 

[PeterDonis]

the spin operator gives +/- hbar/2

What spin operator?

But S= s(s+1) hbar = sqrt(3)/2 hbar

Where does s(s+1) come from? (Also, shouldn't there be a square root in there somewhere?)

 

[43arcsec]

Sorry for the lack of clarity.

Let's take S_z |z;+> = hbar/2 |z;+> but I think all spin operators in any direction have the same eigenvalue, hbar/2

And yes, mea culpa, I left off the sqrt. As you said, it's

S = sqrt(s(s+1))hbar I thought this was a common formuler for spin, it's on the wiki spin page.

Thanks for the quick response, I appreciate it.

 

[PeterDonis]

Let's take S_z |z;+> = hbar/2 |z;+> but I think all spin operators in any direction have the same eigenvalue, hbar/2

Yes.

S = sqrt(s(s+1))hbar I thought this was a common formuler for spin, it's on the wiki spin page.

But what does "S" mean, physically? Obviously it doesn't mean "the result of a spin measurement in a particular direction", since that will always be ##\pm \hbar / 2##. So what does it mean?

 

[43arcsec]

Thanks again for the response, I appreciate it.

I think you hit the essence of my problem, although I can't give you an anwers.

A spin operator measures the value of a particles spin.

The spin formula by quantum number, S, is:

"The expression √(s(s+1))ħ represents the magnitude of the spin angular momentum for a particle with a given spin quantum number, denoted as s." (chatGPT)

The magnitude of of S, the spin angular momentum. I am not sure what I am missing.

If the magnitude of S is sqrt(s(s+1))hbar, than when you measure it with the spin operator, why wouldn't you get this magnitude.

 

[PeterDonis]

A spin operator measures the value of a particles spin in a particular direction.

See the bolded addition. It's crucial.

The spin formula

What "spin formula"? What does this formula represent, physically?

chatGPT

This is not a valid reference. ChatGPT is not reliable.

The magnitude of of S

By itself, as you seem to have already grasped, this does not tell you anything useful.

Have you looked in any QM textbooks to see what, if anything, this "magnitude" ##S = \sqrt{s(s + 1)}## might actually represent?

Note, btw, that in addition to big-S, you also have this thing, small-s, which for the case under discussion is ##\hbar / 2##, and which does exactly match the magnitude of the result you get when you measure spin in a particular direction. So small-s seems to have a straightforward physical meaning as a "magnitude of spin", just in a particular direction. So big-S, which has a different magnitude, must be "magnitude of spin" in some other sense, or as a result of measuring some other operator. What other sense/operator might that be?

 

[PeterDonis]

the spin operator

You keep saying "the" spin operator. But I have already pointed out to you that the "spin operator" that gives ##\hbar / 2## as a result is a spin operator in a particular direction. And it should be obvious that there is more than one such operator. (In fact there are an infinite number of them, since there are an infinite number of possible directions in which you could measure the spin.) So saying "the" spin operator is obviously wrong. You need to specify which operator.

And since any spin operator in a particular direction gives the value ##\hbar / 2## when you measure it, which is different from ##\sqrt{s(s + 1)}##, then the latter value, if it's a measurement result at all, must be a measurement result of some different spin operator, one that isn't even in the set of "spin operators in all possible particular directions".

 

[43arcsec]

ok, so I think I almost have it, thanks for the help.

S^2=Sx^2+Sy^2+Sz^2
S^2 |z+>= Sx Sx |z+>+SySy|z+>+SzSz|z+>=(hbar/2)^2 * 3 |z+>
S=hbar/2*sqrt(3)

If I got this right, then shouldn't
S=Sx+Sy+Sz ?

And if you go through the same calculation, you get
S=3 hbar/ 2

Thanks for hanging in there with me

 

[PeterDonis]

S^2=Sx^2+Sy^2+Sz^2

Yes, that's right. And this ##S## operator is the one whose eigenvalues are ##\sqrt{s(s+1)}##.

shouldn't
S=Sx+Sy+Sz ?

No. Take the square root of the expression you wrote for ##S^2##. Does it give ##S_x + S_y + S_z##?

 

[DrClaude]

If I got this right, then shouldn't
S=Sx+Sy+Sz ?

Additional hint: ##\hat{\mathbf{S}}## is a vector operator.

 

[43arcsec]

Thanks Peter and DrClaude, I think you may have led me to my mistake.

When I looked at S=Sx+Sy+Sz and squared both sides, I dropped the cross terms, ie, SxSy, SxSz, SySz thinking they are orthogonal. But these are 2x2 matrices, so they are not 0. In fact, they are all (hbar/2)^2.

This appears to be what accounts for my discrepancy S=sqrt(3)/2 hbar (correct) and S=3/2 hbar (incorrect).

Thanks again for your patience, but please let me know if I'm still missing something. I still may be missing some subtleties, although I may be being kind to myself with that description.

 

[vanhees71]

The point is that ##\hat{\vec{S}}## is an operator-valued (axial) vector. The commutation relations between its components (with respect to a Cartesian basis) are
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{s}_l,$$
i.e., no two components commute. But you can show by using the algebra of commutators that
$$[\hat{\vec{s}}^2,\hat{s}_j]=0$$
for ##j \in \{1,2,3 \}##. That implies that you can get simultaneous eigenvectors of ##\hat{\vec{s}}^2## and ##\hat{s}_3## (the choice of the the 3rd component is conventional).

Further it's clear that ##\hat{s}_j^2## is a positive semidefinite operator, i.e., for any ket ##|\psi \rangle##
$$\langle \psi|\hat{s}_j^2|\psi \rangle \geq 0.$$
Think about, how to prove this!

This implies that if ##|s,\sigma \rangle## is a simultaneous normalized eigenvector for ##\hat{\vec{s}}^2## and ##\hat{s}_3## with
$$\hat{\vec{s}}^2|s,\sigma \rangle =s(s+1) \hbar^2 |s,\sigma \rangle, \quad \hat{s}_3 |s,\sigma \rangle=\sigma \hbar |s,\sigma \rangle$$
you have
$$\hbar^2 \sigma^2 = \langle s,\sigma|\hat{s}_3^2|s,\sigma \rangle \leq \langle s,\sigma|\hat{\vec{s}}^2|s,\sigma \rangle = \hbar^2 s(s+1).$$
This inequality and the fact that
$$\hat{s}_{\pm} = \hat{s}_1 \pm \mathrm{i} \hat{s}_2$$
raises and lowers the eigenvalue ##\sigma## by one unit ##\hbar##, which you can prove using the commutation relations of these operators with ##\hat{s}_3##, leads (with pure algebra!) to the conclusion that the eigenvalues are given by ##s \in \{0,1/2,1,\ldots \}## and for a given ##s## that ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}##.

For ##s=1/2## you thus have ##\sigma \in \{-1/2,1/2 \}##.

 

[DrClaude]

When I looked at S=Sx+Sy+Sz and squared both sides, I dropped the cross terms, ie, SxSy, SxSz, SySz thinking they are orthogonal. But these are 2x2 matrices, so they are not 0. In fact, they are all (hbar/2)^2.

Pedantic point: the ##\hat{S}_\xi## operators are not 2x2 matrices, but they can be represented as 2x2 matrices (for a spin-1/2 system). But this representation is not unique, as it will depend on the choice of basis.

As for ##\hat{S}^2##, the thing is that ##\hat{\mathbf{S}}## it is a vector operator,
$$
\hat{\mathbf{S}} = \hat{S}_x \mathbf{u}_x +\hat{S}_y \mathbf{u}_y +\hat{S}_z \mathbf{u}_z
$$
where the ##\mathbf{u}_\xi## are unit vectors along the Cartesian coordinates, and what is meant by the square of a vector operator is the dot product:
$$
\hat{S}^2 = \hat{\mathbf{S}} \cdot \hat{\mathbf{S}},
$$
so
$$
\hat{S}^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2
$$