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Angular momentum quantum numbers

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    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.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 20, 2009 #2
    [itex]\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{S}}[/itex]

    so j can be [itex]\frac{1}{2},\frac{3}{2},\frac{5}{2}[/itex]
     
  4. Apr 21, 2009 #3
    So you just consider spin number s to be the second angular momentum quantum number j2 ?
     
  5. Apr 21, 2009 #4
    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
     
  6. Apr 21, 2009 #5

    Redbelly98

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    Yes, exactly.

    Correct again! :smile:
     
  7. Apr 21, 2009 #6
    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: Apr 21, 2009
  8. Apr 21, 2009 #7

    Redbelly98

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    Since we believe students learn more by figuring out the answers themselves, the policy at Physics Forums is to give help in the form of hints but not to give direct answers to the questions.
     
  9. Apr 21, 2009 #8
    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.
     
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