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.