91影视

The probability of an event occurring. The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

Expand the composite spin $鈭�3,1>$from the individual spins. For spin 2, the expected states are as follows,

$|2,2>,|2,1>,|2,0>,|2,-1>$, and $鈭�2,-2>$.

Write the possible states for spin 1.

localid="1658127583657" $|1,1>,|1,0>,$and$鈭�1,-1>$.

The combinations that have a z projection equal to one are needed, so the expansion can be written as follows,

$|3,1鉄�=伪|2,2>|1,-1>+尾|2,1>|1,0>+纬|2,0>|1,1鉄�$

Determine the three expansion coefficients a ,$尾$ and $纬$ in the Clebsch-Gordon tables. Then the probabilities are $|伪{|}^2,|尾{|}^2$ and ${\left|纬\right|}^2$.

Return to the Clebsch-Gorden table and using the equation.

$|sm鉄�=c_{\begin{array}{c}{m}_1{m}_{2}m\\ m_1+m_2=m\end{array}}^{s_1{s}_{2}s}|s_1{m}_1>|s_2{m}_2>$

Write the outcomes using the above information.

Thus, $2鈩�$ is obtained with a probability equal to 1/15 , or$鈩�$with a probability of 8/15 or with a probability of 6/15 .

Look the table $1脳1/2$ and write the outcome.

$|10鉄�|\frac{1}{2}-\frac{1}{2}鉄�=\sqrt{\frac{2}{3}}\overline{)\frac{3}{4}}-\frac{1}{2}鉄�+\sqrt{\frac{1}{3}}\overline{)\frac{1}{2}}-\frac{1}{2}鉄�$

So the total is 3/2 or 1/2 with $l(l+1){鈩�}^2=\frac{15}{4}{鈩�}^2$and $\frac{3}{4}{鈩�}^2$respectively.

Thus, for $\frac{15}{4}{鈩�}^2$the probability is 2/3 , and for $\frac{3}{4}{鈩�}^2$it is 1/3 .