It is known that the different spatial components of the angular momentum don't commute with each other. $$ [L_x,L_y] \propto L_z \\ [L_y,L_z] \propto L_x \\ [L_z,L_x] \propto L_y $$ Also it is known that the uncertainty depends on the commutator. $$ \Delta L_x \Delta L_y \geq \frac{1}{2} \lvert \langle[L_x,L_y]\rangle\rvert $$ This makes me wonder. Knowing, say, $L_z$, how does one put a number on the uncertainty in $L_y$?

For 1D it is straightforward. $$ \Delta p = \frac{\hbar}{\Delta x} $$ I can figure out $\Delta x$, width of the slit, for example, and then know the uncertainty in $p$.

However, to determine the uncertainty in $L_y$ , I need to know $L_x$, but because I know $L_z$ there is uncertainty associated with knowing $L_x$ proportional to $L_y$, since $ [L_z,L_x] \propto L_y $. But I don't know $L_y$!

How would you figure this out?


1 Answer 1


For the angular momentum there is no lower bound for the product $(\Delta L_a)_\psi (\Delta L_b)_\psi$ differently from $x$ and $p$. Indeed there are states $\psi$ such that $L_a\psi =0$ for $a=x,y,z$ simultaneously. I am referring to the states with $L^2\psi =0$ which imply $$(\Delta L_a)_\psi (\Delta L_b)_\psi=0$$ So you cannot write something like $$ (\Delta L_x)_\psi (\Delta L_y)_\psi \geq k$$ for any constant $k>0$ independent from the state $\psi$ as it happens for $x$ and $p$. From this example you can see how, generally speaking, there exist simultaneous eigenstates of incompatible observables, even if you cannot find an orthonormal (Hilbert) basis of these vectors.

  • $\begingroup$ If $(\Delta L_x)_\psi =0$, indeed there is no positive constant $k$ such $ (\Delta L_x)_\psi (\Delta L_y)_\psi \geq k$ because the product of zero with anything is zero ! But that does certainly not mean that you can get simultaneous eigenstates of incompatible observables ! $\endgroup$
    – Alfred
    Commented Feb 2, 2022 at 23:31
  • $\begingroup$ I never asserted it. I only said that there may be some common eigenvectors of two incompatible observables. What it is impossible is a Hilbert basis of common eigenvectors. Furthermore, in the specific case, there is no analogue of the uncertainty principle though the observables are incompatible. $\endgroup$ Commented Feb 3, 2022 at 7:06

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