1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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>

    so:

    |<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

    Avodyne

    User Avatar
    Science Advisor

    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

    turin

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Angular momentum - inequality
  1. Angular momentum (Replies: 4)

  2. Angular momentum (Replies: 5)

  3. Angular Momentum (Replies: 3)

Loading...