# Orbital Angular Momentum Ladder Operator Normalisation Ambiguity

Given $$J_-=L_-+S_-$$, and correspond to total, orbital, and spin, angular momentum lowering operators, respectively.

Where:

$$J_-|j, m_j\rangle=\hbar\sqrt{(j+m_j)(j-m_j+1)}|j,m_j-1\rangle$$,

$$S_-|s\rangle=\hbar|s-1\rangle$$ (where $$s\geq\frac{1}{2})$$ otherwise; $$S_-|-\frac{1}{2}\rangle=\hbar|0\rangle$$

What normalisation is applied to the result of the $$L_-$$ operator?

i.e. If $$b$$ is some value (that may relate directly to $$l$$) such that $$\hbar b$$ corresponds to a normalisation: $$L_-|l\rangle=\hbar b|l-1\rangle$$, what is $$b$$?

Context: in attempting to answer the following question, I have found un-normalised states and have attributed this to improper understanding of the $$L_-$$ operator. Please see question below:

"Hydrogen atoms are prepared such that their electron is in a $$d$$ orbital, carrying total angular momentum $$j=5/2$$, with $$z$$ component $$m_j=1/2$$. Calculate the proportion of elections that will be found on measurement to have their spin pointing upwards, i.e. such that $$m_s=1/2$$. (Hint: write down an expression for the $$m_j=5/2$$ state, and then apply $$J_-$$ in steps.)"

• All angular momentum operators are normalized the same way, i.e. using the exact same rule you wrote down for $J_-$. Jan 15, 2020 at 23:11
• Thanks, would it be possible to get some clarification as to how that would map to the L ladder operator when applied to a state as described by only l, as touched on lower after the "i.e." in my question. Also, would this mean that my findings in my answer below are false? Jan 15, 2020 at 23:14
• Just take your first equation, and everywhere you see a $J$ or $j$, substitute it with $L$ or $\ell$, or $S$ or $s$, to get the analogous equations for orbital or spin angular momentum. There's only one rule here. Jan 15, 2020 at 23:15
• I'm sorry, not quite understanding what would happen in the |l⟩ case, or should I be characterising it differently in order to make said substitutions into the first normalisation formula? Atomic confuses me, would you be willing to provide an example? As a guess from my end, is: $\hbar\sqrt{(l+m_l)(l-m_l+1)}$ the kind of thing you are referring to? Jan 15, 2020 at 23:23

By a separate mark scheme, it appears that $$b=\sqrt{2l}$$ such that:
$$L_-|l\rangle=\hbar\sqrt{2l}|l-1\rangle$$.
$$L_-|3\rangle=\hbar\sqrt{6}|2\rangle$$,
$$L_-|1\rangle=\hbar\sqrt{2}|0\rangle$$.
Note: There is not an explicitly stated $$b$$ value in the notes we have received for this module, or the examination model answers, feel free to dispute my findings.