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The canonical ladder operators for, say, orbital angular momentum are something like

$$ \hat L_+ = \hat L_x + i \hat L_y $$

and it can be shown that, if $ \left| \phi \right> $ is an eigenstate of $ \hat L_z $ with eigenvalue $m \hbar$ , then $ \hat L_+ \left| \phi \right> $ will also be an eigenstate with a new eigenvalue $ (1 + m)\hbar $ .

My question is how can we be sure that this is indeed the 'next' eigenstate, and we haven't missed one with an eigenvalue $ \lambda $ where $m < \lambda < 1+m $ ? How do we know that angular momentum is quantised in units of $ \hbar $ ?

In my limited experience, ladder operators are used to demonstrate the quantisation of angular momentum. I'm sure this is an oversimplification, and that these operators have been constructed and defined for this specific purpose, but is there an obvious way to see that they generate every eigenstate just from the form of the operators?

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There is no way to see that they yield all states, because it isn't true. Whether or not they give you all the states depends on the system. For example, if you were to start with the $1s$ state of a hydrogen atom and apply these operators, you would get nowhere; you wouldn't get the $2s$, $2p$, etc. states.

The set of states you can get to, starting from one and applying these operators however you want, is called an "irreducible representation", or irrep for short. All finite-dimensional irreps have $m$ either integer, or an integer plus $1/2$, because otherwise the ladder operators keep giving you new states forever. Systems can be described by one irrep, or several, or infinitely many.

There are plenty of circumstances where it makes sense to restrict to one irrep. For example, for nuclei the different irreps are often far apart in energy, so you can just restrict to the lowest one. Also, for a particle in 3D space, there are infinitely many irreps (for different values of $L^2$), but you can prove that once you count all of those you have everything, as the resulting basis is complete.

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