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Quantum Mechanics, Angular momentum, Spin

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    The total angular momentum of a particle with orbital angular momentum l (vector) and spin angular momentum s (vector) is j = l + s (vectors). The eigenvalues of j^2, l^2 and s^2 (vectors) are j(j + 1)ħ^2, l(l + 1)ħ^2 and s(s + 1)ħ^2 respectively. State the possible values of j for l = 1, s = 1/2

    What is the general rule which tells you how many values of j to expect for arbitrary l, s?
    What basic information about the fine structure of hydrogen does all this tell us?

    2. Relevant equations

    j = l + s

    3. The attempt at a solution

    If l = 1 then l (vector) may take values (+ or -) ħ (2)^0.5
    If s = 0.5 then s (vector) may take values (+ or -) ħ (3/4)^0.5

    => j = (+ or -) ħ (2)^0.5 (+ or -) ħ (3/4)^0.5

    How do I know that the directions of these vectors (l and s) will line up, so I can just add their magnitudes? Also, surely the expression I have found won't necessarily satisfy
    j^2 = j(j + 1)ħ^2.

    Any help would be massively appreciated. Thanks
     
  2. jcsd
  3. Apr 7, 2009 #2

    Cyosis

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    From what I can tell there seems to be some confusion about what the vectors and what the quantum numbers are. The orbital angular momentum is given by the vector-operator whose entries are the operators Lx,Ly,Lz => [itex]\boldsymbol{L}=(L_x,L_y,L_z)[/itex]. The same goes for the spin angular momentum just replace L with S and the total angular momentum J=L+S. Therefore L^2 and S^2 are not vectors. Proof: [itex]\boldsymbol{L}^2=\boldsymbol{L} \cdot \boldsymbol{L}=L_x^2+L_y^2+L_z^2 [/itex]. This is the sum of 3 different operators squared not a vector!

    The quantum numbers are j,l and s these are scalars so there is no "lining up" to speak of. This means that given l=1 and s=1/2 l+s=3/2.

    So if j,l and s are scalars then the equation j=l+s for l=1 and s=1/2 yields?

    Another thing that seems to go wrong is the use of the eigenvalue equations.

    Example:
    [tex]\boldsymbol{L}^2 |l,m_l\rangle =l(l+1){\hbar}^2|l,m_l\rangle \Rightarrow \boldsymbol{L}^2 |1,m_l\rangle =2{\hbar}^2|1,m_l\rangle [/tex]

    You then seem to conclude [itex]\boldsymbol{L}^2=2{\hbar}^2 \rightarrow L=\pm \sqrt{2} \hbar[/itex]. This is is not allowed. L^2 is an operator that when acting on an orbital momentum eigenstate yields the same as multiplying said eigenstate by L^2's eigenvalue. This does not mean [itex]L^2=2{\hbar}^2[/itex]

    Try to calculate the eigenvalues of S^2 and J^2 now correctly.

    Since deriving the general rule which tells you which values of j to expect for arbitrary l and s is not that easy and quite cumbersome I have a feeling that you're supposed to know it so I will give you the formula. The values j takes on are [itex]j=|j_1-j_2|,|j_1-j_2|+1,...,j_1+j_2.[/itex]

    Try to find all values of j now.
     
  4. Apr 7, 2009 #3
    OK so if l = 1, s = 1/2, j=l+s = 3/2.

    The eigenvalue of S^2 is therefore s(s+1)ħ^2 = (3/4)ħ^2

    and the eigenvalue of J^2 is therefore j(j+1)ħ^2 = (15/4)ħ^2

    I read on hyperphysics that j = l (+ or-) s, which would give possible values of j of 1/2 and 3/2 for l = 1, s = 1/2, and would imply that there are 2 possible values for j unless s = 0, in which case there is only one possible value of j. Is this true? Thank-you.

    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html
     
  5. Apr 7, 2009 #4

    Cyosis

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    Your eigenvalues and j values are correct. Be careful with the [itex]j=l \pm s[/itex] formula though, although it works in this case. In general [itex]j=j_1 \pm j_2[/itex] is not true and you will have to use the formula I listed in my previous post. For example if j1=2 and j2=2 what would the possible values of j be?
     
  6. Apr 7, 2009 #5
    OK so if j1=2 and j2=2, the formula you give would give the possible values of j as 0, 1, 2, 3 and 4, but in the case j1 = 1, j2= 1/2 this list of values is much shorter, giving just 1/2 and 3/2.

    Thank-you very much for your help.
     
  7. Apr 8, 2009 #6

    Cyosis

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    Yep that's correct!
     
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