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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.)"

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    $\begingroup$ All angular momentum operators are normalized the same way, i.e. using the exact same rule you wrote down for $J_-$. $\endgroup$
    – knzhou
    Commented Jan 15, 2020 at 23:11
  • $\begingroup$ 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? $\endgroup$
    – ooertey
    Commented Jan 15, 2020 at 23:14
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    $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Jan 15, 2020 at 23:15
  • $\begingroup$ 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? $\endgroup$
    – ooertey
    Commented Jan 15, 2020 at 23:23

1 Answer 1

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By a separate mark scheme, it appears that $b=\sqrt{2l}$ such that:

$L_-|l\rangle=\hbar\sqrt{2l}|l-1\rangle$.

Source: a similar question from a paper with newly found model answers:

$L_-|3\rangle=\hbar\sqrt{6}|2\rangle$,

$L_-|1\rangle=\hbar\sqrt{2}|0\rangle$.

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.

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