# Angular momentum quantum numbers

## Homework Statement

What are the allowed values of the total angular momentum quantum number j for a particle with spin s=3/2 and orbital angular momentum quantum number l=2?

I know that the angular momentum addition theorem states that the allowed values for the total angular momentum quantum number j given two angular momentum j1 and j2 are
j=j1+j2 , j1+j2-1, ...j1-j2
where m=j, j-1,...-j
but I'm not sure how the spin number s relates to the second angular momentum quantum number.

## The Attempt at a Solution

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$\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{S}}$

so j can be $\frac{1}{2},\frac{3}{2},\frac{5}{2}$

So you just consider spin number s to be the second angular momentum quantum number j2 ?

Also, what are the allowed values of m and l for a given value of n ?
Is it that l can be any value from 0 to n-1 and m can be any value from -l to +l ?

thanks

Redbelly98
Staff Emeritus
Homework Helper
So you just consider spin number s to be the second angular momentum quantum number j2 ?
Yes, exactly.

Also, what are the allowed values of m and l for a given value of n ?
Is it that l can be any value from 0 to n-1 and m can be any value from -l to +l ?

thanks
Correct again!

Dear Captainjack2000, I wish I could do some help.

The state of a bounded particle around the nucleus is determined by ( or expressed in ) four quantum numbers:

main quantum number: n

angular magnitude quantum number: L

angular direction quantum number: m

spin quantum number: s

Last edited by a moderator:
Redbelly98
Staff Emeritus
Homework Helper
1. s=3/2, L=2, hence j= {answer omitted}
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Dear Captainjack2000, I wish I could do some help.

The state of a bounded particle around the nucleu is determind by ( or expressed in ) four quantum numbers:

main quantum number: n

angular magnitude quantum number: L

angular direction quantum number: m

spin quantum number: s

As to a determined $n$, $L$ can be $0, 1, 2, ..., n-2, n-1$, and there are $n$ possible values of $L$. Recall that $n >= L+1$.

For each possible value of $L$, there are $2L+1$ possible values of $m$: $-L, -L+1, -L+2, ..., 0, 1, 2, ..., L$. Recall that $L >= |m|$.

For a bounded particle around the nucleu, with spin angular momentum $s$, orbital angular momentum $L$, the allowed values of the total angular momentum quantum number $j$ will be : $j = L+s, L+s-1, L+s-2, ..., |L-s|$.

Attention that the $L$ $s$ and $j$ are always positive, while $m$ can be negative. So the lower boundary of the total angular momentum quantum numbe $j$ is set to be the absolution of $L-s$, because $s$ can be larger than $L$ sometimes.

Now, let's get down to your questions concretely.

1. s=3/2, L=2, hence j={do it yourself}.

2. Does $s$ relates to $m$ ?--- Probaly not.

$n, L, m$ are introduced in Schrodinger's non-relativistic approach, the wave function of the bounded particle depends on $n, L, m$ only. $s$ comes into being naturally only via Dirac's relativistic approach, although in fact $m$ was introduced phenominologically before Dirac's work.

$L$ depends on $n$, $m$ depends on $L$ (or also $n$ !). $n, L, s$ live together in bounded states, yet $s$ is insintric charactars of elementary particles whenever they are bounded or free. And for a certain particle, $s$ is determined and could only has one value. Most particles have $s$ of $\sqrt{3}/2$, with the third component $+-1/2$ (the 3/2 you refered to is the third component of the total s of some special particle). Thus, $s$ depends on none of $n, L, m$. Moreover, no mathematical relations between $s$ and $m$, too.