When we say electron has spin of $\frac{1}{2}$, is that the value of the total spin of electron, or the projection on z axis, or the spin quantum number?

When we say "electron has spin of $\frac{1}{2}\hbar$", is that the value of the total spin or projection? Also, sometimes people say just "spin 1/2" without $\hbar$.

Is spin quantum number $s$ analogous to the l (total orbital angular momentum) or to the $m_s$ (projection of l).

I am confused because when I am trying to learn addition of angular momeneta (eg. in j-j coupling) where we use formula: $$\vec{j}=\vec{l}+\vec{s}$$ to get total angular momentum for the particle and then we sum all into:

$$\vec{J}=\sum \vec{j}$$

What is the $s$ in this context? I mean in the equation: $\vec{j}=\vec{l}+\vec{s}$ since we are summing it with $\vec{l}$ then it must be spin projection on z axis right?


4 Answers 4


When we say that the electron has "spin half," we mean half of the quantum of angular momentum, $\hbar$. A good quantum mechanics text or other reference will help you derive that the Laplacian operator transforms into spherical coordinates like \begin{align} \nabla^2 &= \left(\frac\partial{\partial x}\right)^2 + \left(\frac\partial{\partial y}\right)^2 + \left(\frac\partial{\partial z}\right)^2 \\ &= \frac 1{r^2}\frac\partial{\partial r}\left(r^2\frac\partial{\partial r}\right) + \frac 1{r^2 \sin^2\theta}\left(\frac\partial{\partial\theta}\right)^2 + \frac 1{r^2 \sin\phi}\frac\partial{\partial\phi}\left(\sin\phi\frac\partial{\partial\theta}\right) \end{align} The angular parts of this operator act on the spherical harmonics to give eigenvalue $\ell(\ell+1)$ for integer $\ell$. This means that the effective form of the kinetic energy operator is \begin{align} \frac{\hbar^2}{2m}\nabla^2 &= \frac{\hbar^2}{2mr^2}\frac\partial{\partial r}\left(r^2\frac\partial{\partial r}\right) + \boxed{\frac{\hbar^2}{2mr^2}{\ell(\ell+1)}} \end{align} In the limit of large $\ell$, the term in the box is the same as the orbital kinetic energy for a point mass $m$ rotating some $r$ from the center of motion with angular momentum $L\sim\ell\hbar$.

This argument is what lets us say things like "$\hbar$ is the quantum of angular momentum," or "angular momentum comes in lumps, and the size of each lump is $\hbar$." Since $\hbar$ is the only quantum of angular momentum, sometimes we only count quanta and leave the unit off. Same as when someone quotes you a price and gives the value but not the currency ("I'll take your car off this tow truck for fifty-five").

Spin angular momentum falls naturally out of the Dirac question in a surprisingly elegant way. You get the same quantum, $\hbar$. However the Dirac equation describes objects whose intrinsic angular momentum is $\hbar/2$. Therefore the projection $m_s$ of the electron spin along any axis can be $\pm\frac12\hbar$, but never zero.

I think this might clarify your search for guidance on the rules for summing vector angular momenta.

  • $\begingroup$ Thanks for the thorough answer. So to sum it up, the "1/2 spin" for electron means that electron has $\pm\frac{1}{2}\hbar$ spin projection along z axis while the "total spin" for electron is then $\frac{\sqrt{3}}{4}\hbar$. Is this correct? $\endgroup$
    – matori82
    Commented Jul 5, 2016 at 17:43
  • $\begingroup$ It's correct to say that the eigenvalue of the total spin operator $\hat s^2$ is $\frac 34\hbar^2$. There are people who observe that $\frac{\sqrt 3}2 > \frac12$ and use this relationship to "explain" why electron spin can't be entirely directed along any one axis, but like many such inviting arguments there are some subtleties to beware of. The best way to say it is the way that most people say it: the electron has total spin $\hbar/2$, and always has projection $\pm\hbar/2$ onto any axis, and angular momentum in quantum mechanics is subtle. $\endgroup$
    – rob
    Commented Jul 5, 2016 at 19:22
  • $\begingroup$ For instance: twice $\sqrt3/2$ is more than 3/2, but there's no way to combine two electrons to get a spin greater than $\hbar$. $\endgroup$
    – rob
    Commented Jul 5, 2016 at 19:30
  • 2
    $\begingroup$ Dear @rob, two electrons admit states with the total $S=1$ (a triplet) whose $S^2=j(j+1)=2\hbar^2$, so $|\vec S|$ is equal to $\sqrt{2}\hbar$, contradicting your assertion. Otherwise your quotes around "explain" are inappropriate, too - the assertion may be proven rigorously. The larger eigenvalue of $S^2$ than $S_z^2$ implies that the state can't be an eigenstate of $S_x,S_z$ with a vanishing eigenvalue - and because the expectation value of at least either $S_x^2$ or $S_y^2$ is non-negative, the other is non-negative as well due to the uncertainty principle from $[S_x,S_z]=-i\hbar S_y$ etc. $\endgroup$ Commented Jul 8, 2016 at 16:09
  • 1
    $\begingroup$ @LubošMotl Your statements are true and correct and exactly the explanation I had in mind. I meant that there is no way to combine two spins $hbar/2$ to end up with a spin larger than $3\hbar/2$, which was intended to be uncontroversial (and correct, since $\sqrt2<3/2$). The subtlety is too complex for me to put in this comment box. $\endgroup$
    – rob
    Commented Jul 9, 2016 at 13:03

Given an angular momentum operator with components $S_1, S_2, S_3$ and commutation relations $[S_i, S_j] = \sum_k \epsilon_{ijk}S_k$, where $\epsilon_{ijk}$ are structure constant of the $\mathfrak{su}(2)$ algebra, the Casimir operator $S^2 = S_1^2+S_2^2+S_3^2$ can be diagonalised simultaneously with any of the original components $S_j$ onto their eigenstates $|\psi\rangle$. Furthermore, the below holds: $$ S^2|\psi\rangle = \hbar^2 s(s+1)|\psi\rangle \qquad S_j|\psi\rangle = \hbar m_j|\psi\rangle. $$ The value $s$ is said to be the spin of the state, $m_j$ being its projection onto the $j$-direction. According to the structure constants and the Lie-algebras the angular momentum operators close, different values of $s$ are allowed. In the case of electrons we have $s=1/2$.


To explain simply, without getting into the details of rotation symmetry groups, etc: When one says $s=1/2$, or $m_s=1/2$ or $m_s=-1/2$, we are specifying a quantum number which describes how the eigenvalues of spin operators behave.

If we specify an eigenket of the spin angular momentum operators $|s,m_s\rangle = |1/2, \pm 1/2\rangle$, with operators $\hat{s^2}$ and $\hat{s_z}$ then $$\hat{s^2} |1/2, \pm 1/2\rangle = \frac{3}{4} \hbar^2~ |1/2, \pm 1/2\rangle \\ \hat{s_z} |1/2, \pm 1/2\rangle = \pm\frac{1}{2} \hbar~ |1/2, \pm 1/2\rangle $$

When one is combining quantized angular momenta, there are rules from the symmetry groups that help us determine the allowed quantum numbers. The allowed quantum numbers follow a triangle rule. Suppose we want to find the allowed quantum numbers for a state resulting from the combination of two angular momenta $|\ell_1, m_{\ell 1}\rangle$ and $|\ell_2, m_{\ell 2}\rangle$: $$ \ell_1 +\ell_2 \ge \ell_{new} \ge |\ell_1 - \ell_2|$$ with integer steps between the minimum and maximum, and z-component quantum number merely sum: $$ m_{\ell \mathrm{new}}=m_{\ell 1}+m_{\ell 2}$$.

Where $\ell$ represents any type of angular momentum quantum number (spin, orbital, spin-orbit combo, etc).


The explanation is very simple. Based on the Stern- Gerlach experiment spin of 1/2 simply means that if you fire electrons through his apparatus... 1/2 of the electrons will spin up and the other 1/2 will spin down

  • $\begingroup$ Welcome to stack exchange. Your answer is of course correct. However, taking into account that the question is rather old, I'd like you to post only answers which contribute a substantial idea. However, I don't understand why others are downvoting the answer without leaving a comment $\endgroup$
    – Semoi
    Commented Jan 21, 2020 at 23:00
  • 1
    $\begingroup$ Actually this is ambiguous. If the particle were spin-1, would 1/1 of the particles go up, and 1/1 go down? Of course not! $\endgroup$ Commented Jan 21, 2020 at 23:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.