Proving the Inequality for Angular Momentum Eigenstates

[neworder1]

Homework Statement

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}.

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi neworder1!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[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}.

 

[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 \leqm\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.

 

[neworder1]

We cannot assume that \psi is the eigenstate of L_{z}. All we now is that it is some eigenstate of L^{2}.

 

[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

 

[neworder1]

That's incorrect, because <L^2> = l(l+1), not l^2.

 

[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?

 

[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.