Griffiths says that you can have only 1 well defined component of the angular momentum because of the uncertainty principle. From the uncertainty principle, we get that

$$ \sigma_{L_x}\sigma_{L_y} \geq \frac{\hbar}{2}|\langle L_z\rangle | $$

If we know $L_z$, can't we make $\sigma_{L_x}$ as small as possible while letting the deviation in the $y$ component get bigger, essentially making it so that we know 2 components?

Also from the uncertainty principle,

$$ \sigma_{L_y}\sigma_{L_z} \geq \frac{\hbar}{2}|\langle L_x\rangle | $$

So if we know the $z$ component, that means its deviation is zero, making the right hand side $\langle L_x \rangle$ zero as well?


2 Answers 2


That you can only ever know one of the three components of angular momentum is best seen not through the uncertainty principle, but on the states themselves.

Since $[L_i,L_j] = \epsilon_{ijk} L_k$, the three momentum operators are pairwise not simultaneously diagonalizable (since simultaneous diagonalizability implies that the operators commute), meaning there is no eigenstate of an $L_i$ that would also be an eigenstate of any of the others. Hence you can only ever know one one angular momentum component for certain, since "knowing for certain" means "having an eigenstate".

Well, actually, the non-commutativity only says that there's no eigenbasis, but there could be some shared eigenvectors. However, one can check for the angular momentum operators that no such state - except the one with zero total angular momentum - exists.

To see this from the uncertainty principle, you have to look at what, actually, the relations say. We have the three relations

$$ \sigma_x \sigma_y \ge \lvert \langle L_z \rangle \rvert $$

$$ \sigma_z \sigma_x \ge \lvert \langle L_y \rangle \rvert $$

$$ \sigma_y \sigma_z \ge \lvert \langle L_x \rangle \rvert $$

where we have discarded the annoying $\frac{\hbar}{2}$ and set $\sigma_i \equiv \sigma_{L_i}$. Now, if we know the values of $L_z$, then $\sigma_z = 0$, and $\lvert \langle L_x \rangle \rvert = \lvert \langle L_y \rangle \rvert = 0$ follows immediately.

This implies that $\sigma_x$ and $\sigma_y$ are non-zero if the actual angular momenta do not identically vanish.

  • 1
    $\begingroup$ (Given the confusion caused on the other side, might I suggest using explicit $\sigma_{L_j}$s here? They're only slightly more annoying and they do help avoid ambiguity when skim-read.) $\endgroup$ Commented Jul 25, 2016 at 18:25
  • $\begingroup$ But if the expectation values are zero then can't we say that the components on the other two axis are just zero? $\endgroup$ Commented Oct 31, 2020 at 13:51
  • $\begingroup$ I agree with @Emilio Pisanty that the notation is better with $_{L_j}$. More importantly, you hand waived over the meat of the question with, "This implies that..." Please expand your statement to round out your argument. $\endgroup$ Commented Nov 30, 2021 at 23:39

The angular momentum operators obey the commutation relations $$[L_i,L_j]=i\sum_k\epsilon_{ijk}L_k$$ A nonzero commutation relation means we can't have a state vector which is an eigenstate of more than one angular momentum operator at the same time. This also leads to an uncertainty principle.


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