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Homework Help: Angular momentum - inequality

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Let [tex]\psi[/tex] be an eigenstate of the operator [tex]L^{2}[/tex] corresponding to the quantum number [tex]l[/tex], i. e. [tex]L^{2} \psi = \hbar l(l+1) \psi[/tex]. Let [tex]<A> = <\psi|A|\psi>[/tex] denote the expectation value of [tex]A[/tex] in state [tex]\psi[/tex].

    Prove that [tex]{|<L_{x}>|}^{2} + {|<L_{y}>|}^{2} + {|<L_{z}>|}^{2}\leq l^{2}[/tex] and the inequality is strict unless [tex]\psi[/tex] happens to be also an eigenstate of the opeator [tex]L_{\vec{n}}[/tex] for some axis [tex]\vec_{n}[/tex].

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 7, 2008 #2


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    Hi neworder1! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Dec 7, 2008 #3
    Well, I just don't know how to prove it - there's a similar problem in Griffiths' textbook and the hint is to apply the uncertainty principle fo operators Lx, Ly and Lz, but I don't see how it is supposed to help - the uncertainty principle relates the expectation values of [tex]{L_{x}}^{2}[/tex] and [tex]{L_{y}}^{2}[/tex] to [tex]{|<L_{z}>|}^{2}[/tex].
  5. Dec 7, 2008 #4
    [tex]\psi[/tex] is a common eigenfunction of L[tex]^{}2[/tex] and L[tex]_{}z[/tex]. The eigenvalues equation for L[tex]_{}z[/tex] is:

    L[tex]_{}z[/tex] [tex]\psi[/tex]=m [tex]\psi[/tex]

    with : -l [tex]\leq[/tex]m[tex]\leq[/tex] l

    the mean vaue of L[tex]_{}x[/tex] and L[tex]_{}y[/tex] on an eigenfunction of L[tex]_{}z[/tex] ([tex]\psi[/tex]) is =0. The mean value of L[tex]_{}z[/tex] on ([tex]\psi[/tex] is =m, and we know that : m^2 <= l^2.
  6. Dec 7, 2008 #5
    We cannot assume that [tex]\psi[/tex] is the eigenstate of [tex]L_{z}[/tex]. All we now is that it is some eigenstate of [tex]L^{2}[/tex].
  7. Dec 7, 2008 #6
    In the general case it should be:

    |<L^2>|=|<(L_x ^2 + L_y ^2 + L_z^2)>|= |<L_x^2>|+|<L_y^2>|+|<L_z^2>| = l^2

    The root-mean-square deviation for an osservable is :

    <A^2>-<A>^2>=0 <A>^2 <= <A^2>


    |<L_x>^2+<L_y>^2+<L_z>^2| <= |<L_x^2>|+|<L_y^2>|+|<L_z^2>| = l^2
  8. Dec 7, 2008 #7
    That's incorrect, because <L^2> = l(l+1), not l^2.
  9. Dec 7, 2008 #8


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    Note that [tex]\langle\vec L\rangle\!\cdot\!\langle\vec L\rangle[/tex] is rotationally invariant, so you can choose a convenient coordinate system.

    In general, if there is some direction that is preferred for some reason, it's good to choose that to be the z direction.

    Is there a preferred direction in this case?
  10. Dec 7, 2008 #9


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    Note that if you quantize angular momentum in the z-direction, then an eigenstate of L^2 with eigenvalue l(l+1) is generally a superposition of eigenstates of Lz with eigenvalues m such that |m|<=l. If it is also an eigenvalue of Lz, then it is possible that |m|=l.
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