Phy5645/AngularMomentumProblem

We may first rewrite the notation for the system as follows;

$|\psi >=1/{\sqrt {5}}|1,-1>+{\sqrt {3/5}}|1,0>+1/{\sqrt {5}}|1,1>$

$L_{z}$ acting on the system produces three values for $l_{z}$;

$l_{z}=-\hbar ,0,\hbar$

The probablity for finding the value $l_{z}=-\hbar$ is;

$P_{1}=|<1,-1|\psi >|^{2}=|1/{\sqrt {5}}<1,-1|1,-1>+{\sqrt {3/5}}<1-1|1,0>+1/{\sqrt {5}}<1,-1|1,1>|^{2}$

$=1/5$

This can easily be verified since;

$<1,-1|1,0>=<1,-1|1,1>=0$ and $<1,-1|1,-1>=1$

The probablites of measuring $l_{z}=\hbar ,0$ are give as follows;

$P_{0}=|<1,0|\psi >|^{2}=|{\sqrt {3/5}}<1,0|1,0>|^{2}=3/5$

$P_{1}=|<1,1|\psi >|^{2}=|{\sqrt {1/5}}<1,1|1,1>|^{2}=1/5$

Now we will calculate the uncertainties $\Delta L_{x}$ and $\Delta L_{y}$ and the product $\Delta L_{x}\Delta L_{y}$

After measuring $l_{z}=-\hbar$ the system will be in the eigenstate $|lm>=|1,-1>$, that is $\psi (\theta ,\phi )=Y_{1},_{-}1(\theta ,|\phi )$. We will first calculate the expectation values of $L_{x},L_{y},L_{x}^{2},L_{y}^{2}$ using $|1,-1>$. Symmetry requires $<1,-1|L_{x}|1,-1>=<1,-1|L_{y}|1,-1>=0$. Using the relation $l-1$ and $m=-1$;

$<L_{x}^{2}>=<L_{y}^{2}>=1/2[<L^{2}>-<L_{z}^{2}>]=\hbar ^{2}/2[l(l+1)-m^{2}]=\hbar ^{2}/2$

$\Delta L_{x}={\sqrt {<L_{x}^{2}>}}=\hbar /{\sqrt {2}}=\Delta L_{y}$

Therefore;

$\Delta L_{x}\Delta L_{y}={\sqrt {<L_{x}^{2}><L_{y}^{2}>}}=\hbar ^{2}/2$

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