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

• I
• 43arcsec
In summary, the spin operator gives +/- hbar/2. But S=s(s+1) hbar = sqrt(3)/2 hbar, so when you measure spin with the spin operator, you get a different magnitude than what is indicated by the spin operator.
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.

43arcsec said:
the spin operator gives +/- hbar/2
What spin operator?

43arcsec said:
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?)

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.

43arcsec said:
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.

43arcsec said:
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?

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.

43arcsec said:
A spin operator measures the value of a particles spin in a particular direction.
See the bolded addition. It's crucial.

43arcsec said:
The spin formula
What "spin formula"? What does this formula represent, physically?

43arcsec said:
chatGPT
This is not a valid reference. ChatGPT is not reliable.

43arcsec said:
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?

43arcsec said:
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".

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

43arcsec said:
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)}##.

43arcsec said:
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##?

43arcsec said:
If I got this right, then shouldn't
S=Sx+Sy+Sz ?
Additional hint: ##\hat{\mathbf{S}}## is a vector operator.

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.

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 \}##.

dextercioby
43arcsec said:
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
$$\hat{S}^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2$$

vanhees71

## 1. What is a spin operator and how does it relate to spin quantum number?

A spin operator is a mathematical operator used to describe the spin of a particle. It is related to the spin quantum number, which is a quantum number that describes the intrinsic angular momentum of a particle. The spin operator and spin quantum number are related through the equation S^2 = ħ^2s(s+1), where S is the spin operator and s is the spin quantum number. This equation shows that the spin operator and spin quantum number give different values because they are describing different aspects of the particle's spin.

## 2. Why do the spin operator and spin quantum number give different values?

The spin operator and spin quantum number give different values because they are measuring different properties of the particle's spin. The spin quantum number describes the intrinsic angular momentum of the particle, while the spin operator describes the direction of the spin. This means that the spin quantum number is a scalar quantity, while the spin operator is a vector quantity. As a result, they give different values but are both necessary to fully describe the spin of a particle.

## 3. How do the spin operator and spin quantum number affect the behavior of a particle?

The spin operator and spin quantum number affect the behavior of a particle by determining its spin state. The spin quantum number determines the possible values of the spin, while the spin operator determines the direction of the spin. Together, they determine the spin state of the particle, which can affect its interactions with other particles, such as in quantum entanglement experiments.

## 4. Can the spin operator and spin quantum number be used interchangeably?

No, the spin operator and spin quantum number cannot be used interchangeably. While they are related, they are measuring different aspects of the particle's spin and have different units. The spin quantum number is a dimensionless quantity, while the spin operator has units of angular momentum. They are both necessary to fully describe the spin of a particle.

## 5. How do the spin operator and spin quantum number relate to the Pauli exclusion principle?

The spin operator and spin quantum number are related to the Pauli exclusion principle through the fact that the spin quantum number determines the possible values of the spin, and the Pauli exclusion principle states that no two particles can have the same set of quantum numbers. This means that particles with different spin quantum numbers can occupy the same energy state, but particles with the same spin quantum number cannot. The spin operator is used to describe the spin of particles, which is one of the quantum numbers used in the Pauli exclusion principle.

• Quantum Physics
Replies
2
Views
651
• Quantum Physics
Replies
26
Views
2K
• Quantum Physics
Replies
3
Views
348
• Quantum Physics
Replies
1
Views
759
• Quantum Physics
Replies
12
Views
1K
• Quantum Physics
Replies
1
Views
705
• Quantum Physics
Replies
3
Views
938
• Quantum Physics
Replies
1
Views
744
• Quantum Physics
Replies
32
Views
2K
• Quantum Physics
Replies
5
Views
465