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Eigen value of total angular momentum

  1. Nov 8, 2013 #1
    I'm trying to understand why the eigen value of the total angular momentum [itex]L^{2}[/itex] is [itex]\hbar ^2 l(l+1)[/itex]. The proofs I have seen go like this. Using the ladder operators [itex]L_{\pm} = L_x \pm iL_y[/itex] we can see and the [itex]|l, m \rangle[/itex] state with maximum value of m (eigen value of [itex]L_z[/itex] )

    [itex]\langle l,m_{max} | L^2 | l, m_{max} \rangle = \langle l, m_{max} | L_{-}L_+ + L_z^2 + \hbar L^z |l,m_{max}\rangle[/itex]
    = [itex]l^2 \hbar ^2 + \hbar ^2 l = \hbar ^2 l(l + 1)[/itex]

    Because [itex]L_+[/itex] acting on the state with highest [itex]m[/itex] value annihilates it. So far so good, as long as I can prove that [itex]L_+ |l , m_{max} \rangle = 0 [/itex] , I'll be happy.

    Now all proof's I have seen for the eigen values of [itex]L_+[/itex] use the same identity again but now say

    [itex]\langle l, m | L_-L_+| l , m \rangle = \langle l, m | L^2 - L_z^2 - \hbar L_z | l , m \rangle = \hbar ^2 \left( l(l+1) - m(m+1) \right) [/itex]

    and claim that when [itex]l = m[/itex] this is zero. So you are now using the fact that [itex]L^2 |l , m \rangle = \hbar ^2 l(l+1)[/itex] which was what I wanted to prove in the first place. Is there an independent proof that avoids this cycle?
  2. jcsd
  3. Nov 8, 2013 #2


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    There is no circle here. You just have to argue a bit differently. It's clear that there are common eigenvectors of [itex]L^2[/itex] and [itex]L_z[/itex] with
    [tex]L^2 |\alpha,m \rangle=\alpha |\alpha m \rangle, \quad L_z |\alpha,m \rangle=m |\alpha,m \rangle.[/tex]
    I've set [itex]\hbar=1[/itex] to save a bit of typing :-).

    Then you define the ladder operators and prove that there's a maximum value for [itex]L_z[/itex], called [itex]l=m_{\text{max}}[/itex]. Using the ladder operator [itex]L_-[/itex] and the fact that there is also a minimal value [itex]m_{\text{min}}[/itex], you get [itex]m_{\text{min}}=-l[/itex] and that [itex]l \in \mathbb{N}_0/2[/itex].

    Now you use
    [tex]L^2=L_- L_+ +L_z^2+L_z,[/tex]
    which follows from the definition of the ladder operators (no properties of the eigenvectors are used). Then you take the expectation value wrt. [itex]|\alpha,l \rangle[/itex]. You find, because of [itex]L_+|\alpha,l \rangle=0[/itex]
    [tex]\langle \alpha,l | L^2 | \alpha,l \rangle=\alpha=\langle \alpha,l|L_z^2+L_z|\alpha,l\rangle=l^2+l=l(l+1),[/tex]
    where I've only used the eigenvalue equation for [itex]L_z[/itex]. As you see, at the end you get the eigenvalue for [itex]L^2[/itex]
    without assuming this relation before. So there is no circle in the argument!
  4. Nov 8, 2013 #3
    My question is how do you prove that there is a maximum or minimum value for [itex]m[/itex] and also how does this imply that [itex]L_+ | l, m_{max} \rangle = 0[/itex] ? Why not some other state ? And could you explain what is [itex] \mathbb{N}_0/2[/itex]
  5. Nov 8, 2013 #4
    [tex]\langle\alpha,m|J^2|\alpha,m\rangle=\langle\alpha,m|J_x^2+J_y^2+J_z^2| \alpha,m\rangle[/tex]

    but we also have

    [tex]\langle\alpha,m|J_x^2| \alpha,m\rangle=\langle\alpha,m|J_x^\dagger J_x| \alpha,m\rangle[/tex]

    Because Jx is Hermitian. That shows that [itex]\langle\alpha,m|J_x^2| \alpha,m\rangle[/itex] is positive because it is the magnitude (squared) of a vector. Samething is true about [itex]\langle\alpha,m|J_y^2| \alpha,m\rangle[/itex]

    Putting that back into the original equation we have

    [tex]\langle\alpha,m|J^2|\alpha,m\rangle \ge \langle\alpha,m|J_z^2| \alpha,m\rangle[/tex]

    and from that you obtain

    [itex]\alpha \ge m^2.[/itex] Clearly, for fixed alpha, m is limited in scope. Given that the rising operator lifts the eigenvalue m by a unit, eventually that inequality will be violated unless at some point the operator gives a vanishing eigenvector as a result.
  6. Nov 11, 2013 #5
    yes, I see that there must be a upper limit on m for a fixed α. But is it completely rigorous to say that since the equality would be violated [itex]L_+| \alpha , m_{max} \rangle[/itex] is zero? I was reading this proof in Griffith and he mentions that strictly we can only conclude that the function is not normalizable. I am not fully comfortable with the whole argument, is just me or is there some mathematical 'iffyness' here?
  7. Nov 11, 2013 #6
    Also except for the state [itex]| 0 , 0 \rangle[/itex] [itex]\alpha = m[/itex] does not seem to be allowed, why is that?
  8. Nov 11, 2013 #7
    The solutions are [itex]\alpha = l(l+1)[/itex], and [itex]|m| \le l[/itex]. How could they be equal unless they are both zero?
    Last edited: Nov 11, 2013
  9. Nov 12, 2013 #8
    Oh I see. It again follows from the fact that [itex]\alpha = l(l+1)[/itex] which follows from [itex]L_{+} | \alpha, m_{max} \rangle = 0[/itex]
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