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Problem: A physical system is in the common eigenstate of $\hat{L^2}$ and $\hat{L_z}$. Calculate the following quantities: $\langle L_x\rangle,\langle L_y\rangle,\langle L_z\rangle,\langle L_x L_y + L_y L_x\rangle$.

Solution: I am not sure on where to start from. Since the physical system is in the common eigenstate of $\hat{L^2}$ and $\hat{L_z}$, I presume that it is described by a wave function $\psi$ that satisfies: $L^2\psi=\hbar^2\ell(\ell+1)\psi$ and $L_z\psi=m\hbar\psi$.

Another thought I made is that $\psi$ could be written as linear combination of $\psi_{\ell m}$, i.e. $\psi=\ldots + C_1 \psi_{\ell 1} + C_0 \psi_{\ell0} + C_{-1} \psi_{\ell,-1} + \ldots$, $C_j$ constants.

Also, by definition it is true that: $\langle \hat{L_x}\rangle = \langle\psi, \hat{L_x}\psi\rangle$ (assuming $\psi$ is normalized).

For $\langle \hat{L_z}\rangle$ specifically, it is easier because: $\langle \psi, \hat{L_z}\psi\rangle = \langle \psi, m\hbar\psi\rangle = m\hbar\langle \psi, \psi\rangle=m\hbar$.

Could you people provide any directions?

EDIT: The missing parts were that $L^2=L_x^2+L_y^2+L_z^2$ and the relationship of creation / annihilation operators $L_{\pm}$ with $L_x,L_y,L_z$.

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  • $\begingroup$ Use \langle \psi \rangle to get $\langle\psi\rangle$ $\endgroup$ – Kyle Kanos Sep 25 '13 at 19:02
  • $\begingroup$ Actually $\langle\hat L_x\rangle=\langle\psi, \hat L_x\psi\rangle$, not the mod squared. $\endgroup$ – Jonas Greitemann Sep 25 '13 at 19:18
  • $\begingroup$ Fixed that too, thanks! I confused it with the formula giving probabilities. $\endgroup$ – stathisk Sep 25 '13 at 19:18
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We can represent the simultaneous eigenstates of $L_z$ and $L^2$ by $|j,m\rangle$. This means

$$L^2|j,m\rangle=\hbar^2j(j+1)|j,m\rangle$$ and

$$L_z|j,m\rangle=\hbar m |j,m\rangle.$$

These states are orthogonal, meaning $$\langle j',m'|j,m \rangle=\delta_{j,j'}\delta_{m,m'}.$$

This immediately yields $$\langle L_z \rangle = \langle j,m | L_z |j,m \rangle =\hbar m.$$ We also can find raising and lowering operators which we denote by

$$L_+=L_x+iL_y, \space\space\space\space L_-=L_x-iL_y.$$ These have the property that

$$L_+|j,m\rangle=\hbar\sqrt{j(j+1)-m(m+1)}|j,m+1\rangle$$ and

$$L_-|j,m\rangle=\hbar\sqrt{j(j+1)-m(m-1)}|j,m-1\rangle.$$

The coefficients out front aren't important for right now -- only that they change the state. So any time you have $$\langle j,m | L_\pm |j,m \rangle$$ you know it must be zero by the orthogonality condition. This holds similarly for any power of $L_\pm$.

To find $\langle L_x \rangle$ we express $L_x$ as a linear combination of $L_+$ and $L_-$: $$\langle L_x \rangle = \langle j,m | L_x |j,m \rangle=\frac{1}{2} \langle j,m | L_+ + L_- |j,m \rangle$$ which must be zero. The case for $L_y$ is identical.

For the last term, we can write

$$L_xL_y=-\frac{i}{4}(L_++L_-)(L_+-L_-)$$ and

$$L_yL_x=-\frac{i}{4}(L_+-L_-)(L_++L_-).$$

Since we are taking the inner product with $| j,m \rangle$ let's ignore the terms $L_\pm^2$ since we know the inner product of those terms will vanish. Then we have

$$\langle L_xL_y \rangle = -\frac{i}{4}\langle L_-L_+-L_+L_- \rangle$$ and $$\langle L_yL_x \rangle = -\frac{i}{4}\langle L_+L_--L_-L_+ \rangle$$ and when we add these together they cancel.

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  • $\begingroup$ Thanks @AlecS for completing the solution. I was able to cross-check my results with yours. $\endgroup$ – stathisk Sep 26 '13 at 1:06
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Recall that $$L^2|l,m\rangle=l(l+1)\hbar|l,m\rangle \\ L_z|l,m\rangle=m\hbar|l,m\rangle \quad\quad\\ L_\pm|l,m\rangle = f\left(l,m\right)|l,m\pm1\rangle$$ where $L^2=L_x^2+L_y^2+L_z^2$, $L_\pm=L_x\pm iL_y$, and $f\left(l,m\right)$ the Clebsch Gordan coefficients. If you close these off with $\langle l',m'|$, you should be able to find the necessary expectation values.

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  • $\begingroup$ Thanks @KyleKanos! You provided enough hints to get me started. (I can't upvote you yet due to my low reputation points.) $\endgroup$ – stathisk Sep 25 '13 at 19:19
  • $\begingroup$ @Zet: no problem, glad I could help! $\endgroup$ – Kyle Kanos Sep 25 '13 at 19:25

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