# Angular momentum and spin unit

• I
I know that spin is a type of intrinsic angular momentum.

For electron spin is (1/2)ħ . But unit of (1/2)ħ is J.s, which is not the unit of angular momentum. Can you please explain this discrepancy?

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hilbert2
Gold Member
I know that spin is a type of intrinsic angular momentum.

For electron spin is (1/2)ħ . But unit of (1/2)ħ is J.s, which is not the unit of angular momentum. Can you please explain this discrepancy?
Actually, that is the unit of angular momentum, ##J\cdot s =\frac{kg\cdot m^2}{s}##. It's also the unit of action in classical mechanics.

Edge5
A. Neumaier
2019 Award
Actually, that is the unit of angular momentum, ##J\cdot s =\frac{kg\cdot m^2}{s}##. It's also the unit of action in classical mechanics.
... and in quantum mechanics.

Edge5
DrClaude
Mentor
For electron spin is (1/2)ħ
That is not correct. The spin quantum number is ##s=1/2##, which means that the spin angular momentum is ##\sqrt{s(s+1)} \hbar = \frac{\sqrt{3}}{2} \hbar##.

Edge5
jtbell
Mentor
More explicitly, angular momentum is a vector quantity. An electron's spin angular momentum has magnitude ##\frac {\sqrt 3} 2 \hbar## as DrClaude noted. Its component in any direction is ##\pm \frac 1 2 \hbar##.

Edge5
More explicitly, angular momentum is a vector quantity. An electron's spin angular momentum has magnitude ##\frac {\sqrt 3} 2 \hbar## as DrClaude noted. Its component in any direction is ##\pm \frac 1 2 \hbar##.
Do you mean the net magnitude of total spin vector is ##\frac {\sqrt 3} 2 \hbar## ? Do we consider spin as a 3-D vector?

I mean if spin is a vector like A(x hat) + B(y hat) + C(z hat).
Is ##\frac {\sqrt 3} 2 \hbar## = sqrt( A^2 +B^2 +C^2)

jtbell
Mentor
Do we consider spin as a 3-D vector?
Classically, spin angular momentum is a vector whose direction is along the object's axis of rotation:

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am

In QM, there's no definite axis of rotation, but we can still say that spin angular momentum has a definite magnitude (##\frac {\sqrt 3} 2 \hbar## for an electron), and certain allowable values for its component along any measurement axis (##\pm \frac 1 2 \hbar## for an electron) which we traditionally call the "z-direction" even though it could just as well be the x-direction or the y-direction or any other direction.

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

Classically, spin angular momentum is a vector whose direction is along the object's axis of rotation:

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am

In QM, there's no definite axis of rotation, but we can still say that spin angular momentum has a definite magnitude (##\frac {\sqrt 3} 2 \hbar## for an electron), and certain allowable values for its component along any measurement axis (##\pm \frac 1 2 \hbar## for an electron) which we traditionally call the "z-direction" even though it could just as well be the x-direction or the y-direction or any other direction.

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

So does that mean we can't measure ##\frac {\sqrt 3} 2 \hbar## (since it involves all direction of vectors) but we can measure it only in one direction which gives us a result of ##\pm \frac 1 2 \hbar## ?

Nugatory
Mentor
So does that mean we can't measure ##\frac {\sqrt 3} 2 \hbar## (since it involves all direction of vectors) but we can measure it only in one direction which gives us a result of ##\pm \frac 1 2 \hbar## ?
We cannot measure ##\vec{S}##, but we can measure ##|\vec{S}|^2##, the square of its magnitude, and then take the square root to get the magnitude. There's an easy visualization: imagine the set of vectors with their base at the center of the earth and their tip on the Arctic circle. All of these vectors have the same squared magnitude and the same ##z## component, but the other two components and the overall direction are not determined.