text: # Homework Help: Angular momentum - inequality
  
  1. Dec 6, 2008
  
  ### neworder1
  
  1. The problem statement, all variables and given/known data
  
  Let $$\psi$$ be an eigenstate of the operator $$L^{2}$$ corresponding to the quantum number $$l$$, i. e. $$L^{2} \psi = \hbar l(l+1) \psi$$. Let $$<A> = <\psi|A|\psi>$$ denote the expectation value of $$A$$ in state $$\psi$$.
  
  Prove that $${|<L_{x}>|}^{2} + {|<L_{y}>|}^{2} + {|<L_{z}>|}^{2}\leq l^{2}$$ and the inequality is strict unless $$\psi$$ happens to be also an eigenstate of the opeator $$L_{\vec{n}}$$ for some axis $$\vec_{n}$$.
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  2. Dec 7, 2008
  
  ### tiny-tim
  
  Hi neworder1!
  
  Show us what you've tried, and where you're stuck, and then we'll know how to help.
  
  3. Dec 7, 2008
  
  ### neworder1
  
  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 $${L_{x}}^{2}$$ and $${L_{y}}^{2}$$ to $${|<L_{z}>|}^{2}$$.
  
  4. Dec 7, 2008
  
  ### martyf
  
  $$\psi$$ is a common eigenfunction of L$$^{}2$$ and L$$_{}z$$. The eigenvalues equation for L$$_{}z$$ is:
  
  L$$_{}z$$ $$\psi$$=m $$\psi$$
  
  with : -l $$\leq$$m$$\leq$$ l
  
  the mean vaue of L$$_{}x$$ and L$$_{}y$$ on an eigenfunction of L$$_{}z$$ ($$\psi$$) is =0. The mean value of L$$_{}z$$ on ($$\psi$$ is =m, and we know that : m^2 <= l^2.
  
  5. Dec 7, 2008
  
  ### neworder1
  
  We cannot assume that $$\psi$$ is the eigenstate of $$L_{z}$$. All we now is that it is some eigenstate of $$L^{2}$$.
  
  6. Dec 7, 2008
  
  ### martyf
  
  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
  
  7. Dec 7, 2008
  
  ### neworder1
  
  That's incorrect, because <L^2> = l(l+1), not l^2.
  
  8. Dec 7, 2008
  
  ### Avodyne
  
  Note that $$\langle\vec L\rangle\!\cdot\!\langle\vec L\rangle$$ 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?
  
  9. Dec 7, 2008
  
  ### turin
  
  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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