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It is stated in "Modern Physics" by Tipler and Llewellyn (pp. 284-285) when discussing the quantum numbers of the Hydrogen atom:

Note the perhaps unexpected result that the angular momentum vector never points in the $z$ direction since the maximum $z$ component $m\hbar$ is always less than the magnitude $\sqrt{l(l+1)}\hbar$. This is a consequence of the uncertainty principle for angular momentum, which implies that no two components of angular momentum can be precisely known simultaneously, except in the case of zero angular momentum.

If this is the case, why isn't the uncertainty principle violated when we measure the spin of an electron to be along a certain axis?

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  • $\begingroup$ The argument by the authors is not related to spin, $\ell$ is the quantum number for the orbital angular momentum. It's allowed to be 0, while the spin $s$ is 0 only for a scalar particle, such as a pion (en.wikipedia.org/wiki/Pion) for which the internal structure is neglected. $\endgroup$ – DanielC Oct 17 '17 at 8:11
  • $\begingroup$ I do understand this - but if spin is considered a form of angular momentum, then to know its precise direction would be to violate the uncertainty principle, and that was where my confusion originated from. $\endgroup$ – Billy Kalfus Oct 17 '17 at 8:45
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We don't measure the direction of the "full spin" of the electron - we measure its component along a chosen axis.

The allowed values for the z-component of the electron spin are $\pm \frac{\hbar}{2}$, but the magnitude of the total spin is $\sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\hbar = \frac{\sqrt{3}}{2}\hbar$

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The angular momentum along a certain axis is not the full momentum, which has $3$ components. Thus, $\hat L_z$ along $\boldsymbol{\hat z}$ is not the "full angular momentum" which would also include $\hat L_y$ and $\hat L_x$.

Suppose for instance the electron has a definite value of $\hat L_z$, i.e. the state $\psi(r,\theta,\phi)$ is such that $$ \hat L_z\psi(r,\theta,\phi)=\hbar m \psi(r,\theta,\phi)\, . $$ Then $\langle L_z\rangle= \hbar m$ and $\Delta L_z=0$: the last two equalities validate the statement that the electron has definite value of $\hat L_z$ as there is no fluctuation of that value from the average value $m\hbar$.

The uncertainty relation for angular momentum then states that $$ \Delta L_x\Delta L_y\ge \frac{1}{2}\vert \langle L_z\rangle\vert =\frac{1}{2}\vert m\vert \hbar \tag{1} $$ so that neither $\Delta L_x$ nor $\Delta L_y$ can be $0$, and consequently that the electron in the state $\psi(r,\theta,\phi)$ does not have a well-defined value of $\hat L_x$ or $\hat L_y$.

If the state $\psi(r,\theta,\phi)$ has well defined angular momentum in the sense that $$ \hat L^2\psi(r,\theta,\phi)=\ell(\ell+1)\hbar^2 \psi(r,\theta,\phi)\, , $$ then the magnitude squared of $\vec L$ is well defined, and its $\boldsymbol{\hat z}$-component is well-defined, leaving the tip of $\hat yL_y+ \hat x L_x$ to lie somewhere on a circle in the $xy$ plane at height $m\hbar$, as illustrated below for $\ell=2$.

enter image description here

The tip of the angular momentum must be found equally likely anywhere on the circle. To see this, you can also use the cyclicity of (1) to find \begin{align} \Delta L_x\Delta L_z&\ge \frac{1}{2}\vert \langle L_y\rangle\vert =0\, \\ \Delta L_y\Delta L_z&\ge \frac{1}{2}\vert \langle L_x\rangle\vert =0\, \end{align} since $\Delta L_z=0$ by assumption on the state $\psi(r,\theta,\phi)$, showing that $\langle L_x\rangle = \langle L_y\rangle =0$ and justifying the random location of the tip on the circle.

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