Phy5645/AngularMomentumProblem

We first rewrite the wave vector in Dirac notation:

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

We see that the possible results for a measurement of ${\hat {L}}_{z}$ are $-\hbar ,$ $0,\!$ and $\hbar .$

The probablity for obtaining $-\hbar$ is

$P(-\hbar )=|\langle 1,-1|\psi \rangle |^{2}=\left|{\frac {1}{\sqrt {5}}}\langle 1,-1|1,-1\rangle +{\sqrt {\frac {3}{5}}}\langle 1-1|1,0\rangle +{\frac {1}{\sqrt {5}}}\langle 1,-1|1,1\rangle \right|^{2}={\tfrac {1}{5}}.$

Similarly, the probablites of obtaining $0\!$ and $\hbar$ are

$P(0)=|\langle 1,0|\psi \rangle |^{2}={\tfrac {3}{5}}$

and

$P(\hbar )=|\langle 1,1|\psi \rangle |^{2}={\tfrac {1}{5}}.$

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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