# How exactly do raising/lowering operators let you know the form of M?

by Silversonic
Tags: form, operators, raising or lowering
 P: 103 There must be some lapse in my understanding of this. I understand that you can have an eigenstate of a system with an angular momentum magnitude value and a value for one component of the angular momentum (z). Using the lowering and raising operators we can create states (or deduce that states exist) with the same angular momentum magnitude value and the value for one component of the angular momentum increased or decreased by one unit. We can keep doing this until the value for the component reaches a maximum or minimum value that is within the bounds of its square not exceeding the square of the magnitude of the whole angular momentum. So we have deduced that there definitely exist states with same L quantum number, but with M values that run from L to -L in steps of one. Using this fact we can deduce L is either a positive half-integer or integer. But how does this let us know that states can ONLY have an M-value that runs in steps of one from L to -L? We only deduced that states with M from L to -L in steps of one exist, but the M values might not be exclusive to those. What if we created some other mythical operator that could act on an eigenstate which created another eigenstate with the same L value, but M increased by a half? For orbital angular momentum the argument is clear. In order for a full rotation to a return a system to itself, the M value (and therefore L) must be limited to integer values. Hence it runs from L to -L in integer steps and in this case, those are the only M-value states that can exist. However L values that are half-integers are mathematically consistent, this is spin and full rotation bringing the same state isn't a necessity. So we know M, in this case, can have values from L to -L in steps of one, but there is nothing telling us that M is exclusive to those values?
P: 3,447
 So we have deduced that there definitely exist states with same L quantum number, but with M values that run from L to -L in steps of one.
This shows that if M is integer, then L is integer. And if M is half-integer, then L is half-integer. Can't have it both ways.
 What if we created some other mythical operator that could act on an eigenstate which created another eigenstate with the same L value, but M increased by a half?
And this would require L to be both integer and half-integer, hence impossible.
P: 103
 Quote by Bill_K This shows that if M is integer, then L is integer. And if M is half-integer, then L is half-integer. Can't have it both ways. And this would require L to be both integer and half-integer, hence impossible.
Ah okay. So the problem occurs when examining the limits of high/low M which determine L. Right?

If half-integer and integer values of M were possible (although it's not exclusive to these combinations). Taking a look at maximum M, it's either integer or half-integer. The stepping up operator shows this maximum value is equal to L, making L integer or half-integer respectively. But if there were a stepping down operator which reduces M by a half, we'd have a state with M-1/2. Using the stepping up operator (which increases M by one unit) on this state, we'd see that the eigenstate created from this would have an M value equal to M+1/2, but this can't be true since M was the maximum value. Hence this state doesn't exist. But this also shows that both M - 1/2 and M have value equal to L. Which is impossible.

P: 708

## How exactly do raising/lowering operators let you know the form of M?

Quote by Bill_K
 Quote by Silversonic So we have deduced that there definitely exist states with same L quantum number, but with M values that run from L to -L in steps of one.
This shows that if M is integer, then L is integer. And if M is half-integer, then L is half-integer. Can't have it both ways.
Bill, this step in your argument seems a bit dodgy. This shows that if M is integer, then L is integer (and likewise for half integer) for the M's we already found. If there were more states, there's no reason to assume these other hypothetical eigenstates, |L,M'>, of LZ would need to satisfy this same property. I think all you've shown is that for the usual |L,M> states we already know about that it's impossible to have two states such that M is integer for one and half integer for the other. I see no reason why you can assume a priori that Silversonic's hypothetical extra states with eigenvalues M' would also need to satisfy M' = -L,...,L in integer steps.
 P: 590 Suppose such hypothetical states of general real z-component of spin (say, m=1/4) existed. You could still act on these states with the raising and lowering operators to increase or decrease m by one; this follows from the commutation relations of the operators. By acting on such a state repeatedly you'd end up with a state of m=j+1/4, which would contradict the requirement that m is less than or equal to j. So if such such states existed in the interval (j-1,j), you'd need them to be annihilated by the raising operator. But then that wouldn't be consistent with the equation $J_{\pm}J_{\mp}=\mathbf{J}^2 -J_{z}^2 \pm \hbar J_z$.