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The following diagram is taken from this Wikipedia page:

enter image description here

It illustrates how we may only know the total orbital angular momentum $L$ (so the radius on our sphere in $L$ space) and the z-component of the orbital angular momentum $L_z$ which has to be $m\hbar$ for integer $m$, so our orbital angular momentum vector is restricted to the circles shown.

However, I know we cannot hope to know $L_x$ and $L_y$ as well, but surely they too must be of the form $m\hbar$ and so the circles shown should really just be limited to discrete points that satisfy this on that circle.

Why is this not correct?

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The reason the $x$ and $y$ components lie on circles is because the expectation values for the components are zero (here you should check the math in the book you're learning from):

$$\langle L_x\rangle=\langle L_y\rangle=0$$

When one talks about expectation values then one should always consider an experiment. In your case an experiment would be that we first prepare your state as shown in the picture, where $L_z$ can take on any value $m\hbar$. Next we measure the $L_x$ and $L_y$ components. If we repeat this - preparation, measuring - a gazillion times then we get as an average value for the $x$ and $y$ component the value $0$ for both.

This means that after you prepared your state in a state of definite $L_z$ and then measure the $x$ and $y$ components that these can point in any direction if you measure them. This is why the $x$ and $y$ components lie on circles: then for each possibility that one component points the one way there is an equal probability that this component points the other way. So for each $L_z$ both $L_x$ and $L_y$ lie on a circle. If you had only discrete points then the expectation values would not be zero.

When you prepare your state such that $L_x$ has definite values then you get the same picture as above, only with $x$ instead of $z$. But then my argument is the same, because if you want the $x$ component to be precise then

$$\langle L_z\rangle=\langle L_y\rangle=0$$

and both the $z$ and $y$ components lie on circles. So, everytime you know that one component has the form $m\hbar$, where $-l\leq m \leq l$, you know that the others are not of this form.

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There are states that correspond to eigenvalues of $\hat{L}_x$ or $\hat{L}_y$. But they are (almost always) of no interest: By the convention of what we mean by x,y, and z, at best only $\hat{L}$ and $\hat{L}_z$ are simultaneous eigenstates of the sort of systems usually considered (for example eigenstates of an atomic Hamiltonian). Hence for such eigenstates, $L_x$ and $L_y$ are smeared out; in the corresponding basis, the states of interest are in a superposition defined by being simultaneous eigenstates to $\hat{L}$ and $\hat{L}_z$.

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  • $\begingroup$ But my point is we don't know $L_x$ and $L_y$ but we still know they must be quantised so why are they permitted to take on a whole circle of combinations, and not just a few points, i.e $(i,j)\hbar$ for integer $i$ and $j$? $\endgroup$ – Watw Apr 5 '15 at 21:07
  • $\begingroup$ They are quantized, but not in discrete points: Their orientation is not (in the states usually considered). Hence you have all the superpositions that actually are relatively well-represented by the circle-like representation you linked to. The point is that you are free to rotate the points you think of about the $z$-axis, which corresponds to forming superpositions of them (think adding/interpolating between spherical harmonics). $\endgroup$ – pyramids Apr 5 '15 at 21:11
  • $\begingroup$ So is it not true that $L_x=m\hbar$ if we know that this holds for $L_z$? Does the fact we can only know $L$ and $L_z$ mean we can't even know if $L_x$ and $L_y$ are quantised? Can I think about it this way? $\endgroup$ – Watw Apr 5 '15 at 21:12
  • $\begingroup$ I think you will need to go through an actual calculation to see the issue. Sorry, but I can't be bothered (at least not now). You may want to read up about Pauli matrices (the core of the angular momentum operators) and apply them to a state such as a spin-1/2 up or down state and you'll see. $\endgroup$ – pyramids Apr 5 '15 at 21:18
  • $\begingroup$ I have been struggling with this whole thing with spin angular momentum too, so hopefully It will become clear soon, thanks anyway. $\endgroup$ – Watw Apr 5 '15 at 21:20

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