# Would I have to find Clecbsh-Gordon coefficients for this question?

I'm new to the Addition of angular momenta. This is a practice question given by my teacher. I don't understand how to find the probability without Clebsch-Gordon coefficients. Is there any other way to find the answer? The question is given below.

Two particles with $$l1=l2=1$$ are in-state $$|l1,l2,l,m>=|1,1,1−1>$$. If measurement is made of L1z in this state, what values may be found and with what probability will these values occur?

Previously I thought it will be tedious to find C_G coefficients and I have to use group theory to solve this but it's easy, I don't have to find all of the C-G coefficients. Thus it can be solved easily using usual methods. The solution is given below.

Given
$$|l_1,l_2,l,m>=|1,1,1,-1>$$.
Now since $$l_1$$ and $$l_2$$ stay constant we can write the ket as just $$|l,m>=|1,-1>$$.
Since $$l_1=l_2=1$$ we have $$m_1,m_2= 1,0,-1$$
We know,
range of $$l$$ is given as, $$|l_1-l_2|{\leq}l{\leq}l_1+l_2;$$ with unit steps.
So, $$l=0,1,2$$.
Now,
$$|l_1,m_1>=|1,1>, |1,0>, |1,-1>$$
$$|l_2,m_2>=|1,1>, |1,0>, |1,-1>$$
And,
$$|l_1,m_1;l_2,m_2>=|l_1,m_1>\otimes|l_2,m_2>$$
$$|l,m>=C_1{|l_1,m_1;l_2,m_2>}+C_2{|l_1,m_1;l_2,m_2>}+C_3{|l_1,m_1;l_2,m_2>}+...$$
$$C_i$$ are the ith Clebsch-Gordan coefficients.
For $$|l,m>=|1,-1>$$ we choose such $$|l_1,m_1;l_2,m_2>$$ for which we satisfy $$m=m_1+m_2$$.
$$\therefore |l,m>=|1,-1>=C_1{|1,0;1,-1>}+C_2{|1,-1;1,0>} \dotsb(1a)$$
Since,$$|l,m>$$ and $$|l_1,m_1;l_2,m_2>$$ are complete orthonormal basis we have,
$$=1$$ and $$=1$$
$$\therefore <1,-1|1-1>={C_1}^2+{C_2}^2 \dotsb(1b)$$
$$\Rightarrow {C_1}^2+{C_2}^2=1 \dotsb(1c)$$
To find $$C_1$$ and $$C_2$$ we consider $$|l,m>=|2,-2>$$ state.
$$|2,-2>=C_0|1,-1;1,-1>$$ and since it's express in orthonormal basis we have $$C_0=1$$ as $$equ^n(1b)$$ and $$equ^n(1c)$$.
$$\therefore |2,-2>=|1,-1;1,-1> \dotsb(2)$$.
Operating $$L_+$$ on $$eqn^n(2)$$
$$L_+|2,-2>=(L_{1+}+L_{2+})|1,-1;1,-1>$$
$$\Rightarrow 2\hbar|2,-1>=\hbar\sqrt{2}{|1,0;1,-1>}+\hbar\sqrt{2}{|1,-1;1,0>}$$
$$\Rightarrow |2,-1>=\frac{1}{\sqrt2}|1,0;1,-1>+\frac{1}{\sqrt2}|1,-1;1,0> \dotsb(3)$$
Taking inner product of $$equ^n(1a)$$ and $$equ^n(3)$$ we get $$\dotsb$$
$$0=\frac{C_1}{\sqrt2}+\frac{C_2}{\sqrt2}$$
$$\Rightarrow C_1=-C_2 \dotsb(4)$$
Equating $$equ^n(1c)$$ and $$equ^n(4)$$ we get $$\dotsb$$
$$C_1=-\frac{1}{\sqrt2}, C_2=\frac{1}{\sqrt2}$$
So, $$|l,m>=|1,-1>=-\frac{1}{\sqrt2}{|1,0;1,-1>}+\frac{1}{\sqrt2}{|1,-1;1,0>}$$.
Thus measuring $$L_{1_z}$$ in this state will give $$m_1= 0$$ or$$-1$$ with probability $$\frac{1}{2}$$ for each $$m_1$$ (Since, $${C_1}^2 = {C_2}^2 = \frac{1}{2}$$).

• Hi Rajiv Das. Welcome to Phys.SE. Note that an answer is supposed to be in an answer segment, not a part of the question segment. Dec 22, 2021 at 6:36

The two possible $$l_Z$$ values are -1,0, and +1. They must add to give the total $$L_z=-1$$, and the only ways of doing this are (-1,0) and (0,-1)
So $$l_z$$ is 0 or -1, with a 50% chance of each.