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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,
$<l,m|l,m>=1$ and $<l_1,m_1;l_2,m_2|l_1,m_1;l_2,m_2>=1$
$\therefore <1,-1|1-1>={C_1}^2+{C_2}^2 \dotsb(1b)$
$\Rightarrow {C_1}^2+{C_2}^2=1 \dotsb(1c)$
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, ${C_1}^2 = {C_2}^2 = \frac{1}{2}$).

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    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Dec 22, 2021 at 6:36

1 Answer 1

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Gosh, you are making life difficult for yourself. This is just simple arithmetic.

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)

As particles 1 and 2 are identical, the probabilities of these two ways are equal.

So $l_z$ is 0 or -1, with a 50% chance of each.

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  • $\begingroup$ Thanks, professor. $\endgroup$
    – Rajiv Das
    Dec 23, 2021 at 10:57

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